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Although focused exclusively on acid–base equilibriums in aqueous solutions, equilibrium concepts can also be applied to many other kinds of reactions that occur in aqueous solution. In this chapter, we describe the equilibriums involved in the solubility of ionic compounds and the formation of complex ions.
Solubility equilibriums involving ionic compounds are important in fields as diverse as medicine, biology, geology, and industrial chemistry. Carefully controlled precipitation reactions of calcium salts, for example, are used by many organisms to produce structural materials, such as bone and the shells that surround mollusks and bird eggs. In contrast, uncontrolled precipitation reactions of calcium salts are partially or wholly responsible for the formation of scale in coffee makers and boilers, “bathtub rings,” and kidney stones, which can be excruciatingly painful. The principles discussed in this chapter will enable you to understand how these apparently diverse phenomena are related. Solubility equilibriums are also responsible for the formation of caves and their striking features, such as stalactites and stalagmites, through a long process involving the repeated dissolution and precipitation of calcium carbonate. In addition to all of these phenomena, by the end of this chapter you will understand why barium sulfate is ideally suited for x-ray imaging of the digestive tract, and why soluble complexes of gadolinium can be used for imaging soft tissue and blood vessels using magnetic resonance imaging (MRI), even though most simple salts of both metals are toxic to humans.
Scanning electron micrograph of kettle scale. Hard water is a solution that consists largely of calcium and magnesium carbonate in CO2-rich water. When the water is heated, CO2 gas is released, and the carbonate salts precipitate from solution and produce a solid called scale.
We begin our discussion of solubility and complexation equilibriums—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibriums, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:As you will discover in Section 17.4 "Solubility and pH" and in more advanced chemistry courses, basic anions, such as S2−, PO43−, and CO32−, react with water to produce OH− and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
Equation 17.1
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp)The equilibrium constant expression for the dissolution of a sparingly soluble salt that includes the concentration of a pure solid, which is a constant. of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. (For more information on the equilibrium constant expression, see Chapter 15 "Chemical Equilibrium", Section 15.2 "The Equilibrium Constant".) The equilibrium constant expression for the dissolution of calcium phosphate is therefore
Equation 17.2
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43− ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table 17.1 "Solubility Products for Selected Ionic Substances at 25°C"; they show that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equation 17.1, changes in pH can affect the solubility of a compound, as you will discover in Section 17.4 "Solubility and pH".
As with K, the concentration of a pure solid does not appear explicitly in Ksp.
Table 17.1 Solubility Products for Selected Ionic Substances at 25°C
Solid | Color | K sp | Solid | Color | K sp | |
---|---|---|---|---|---|---|
Acetates | Iodides | |||||
Ca(O2CCH3)2·3H2O | white | 4 × 10−3 | Hg2I2* | yellow | 5.2 × 10−29 | |
Bromides | PbI2 | yellow | 9.8 × 10−9 | |||
AgBr | off-white | 5.35 × 10−13 | Oxalates | |||
Hg2Br2* | yellow | 6.40 × 10−23 | Ag2C2O4 | white | 5.40 × 10−12 | |
Carbonates | MgC2O4·2H2O | white | 4.83 × 10−6 | |||
CaCO3 | white | 3.36 × 10−9 | PbC2O4 | white | 4.8 × 10−10 | |
PbCO3 | white | 7.40 × 10−14 | Phosphates | |||
Chlorides | Ag3PO4 | white | 8.89 × 10−17 | |||
AgCl | white | 1.77 × 10−10 | Sr3(PO4)2 | white | 4.0 × 10−28 | |
Hg2Cl2* | white | 1.43 × 10−18 | FePO4·2H2O | pink | 9.91 × 10−16 | |
PbCl2 | white | 1.70 × 10−5 | Sulfates | |||
Chromates | Ag2SO4 | white | 1.20 × 10−5 | |||
CaCrO4 | yellow | 7.1 × 10−4 | BaSO4 | white | 1.08 × 10−10 | |
PbCrO4 | yellow | 2.8 × 10−13 | PbSO4 | white | 2.53 × 10−8 | |
Fluorides | Sulfides | |||||
BaF2 | white | 1.84 × 10−7 | Ag2S | black | 6.3 × 10−50 | |
PbF2 | white | 3.3 × 10−8 | CdS | yellow | 8.0 × 10−27 | |
Hydroxides | PbS | black | 8.0 × 10−28 | |||
Ca(OH)2 | white | 5.02 × 10−6 | ZnS | white | 1.6 × 10−24 | |
Cu(OH)2 | pale blue | 1 × 10−14 | ||||
Mn(OH)2 | light pink | 1.9 × 10−13 | ||||
Cr(OH)3 | gray-green | 6.3 × 10−31 | ||||
Fe(OH)3 | rust red | 2.79 × 10−39 | ||||
*These contain the Hg22+ ion. |
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp, like K, is defined in terms of the molar concentrations of the component ions.
A kidney stone. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O2CCO2)·H2O and Ca3(PO4)2.
© Thinkstock
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: K sp
Strategy:
A Write the balanced dissolution equilibrium and the corresponding solubility product expression.
B Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution:
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2−) are as follows:
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2−] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2− ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise
One crystalline form of calcium carbonate (CaCO3) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer: 4.5 × 10−9
The reaction of weakly basic anions with H2O tends to make the actual solubility of many salts higher than predicted.
A crystal of calcite (CaCO3), illustrating the phenomenon of double refraction. When a transparent crystal of calcite is placed over a page, we see two images of the letters.
Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example 1. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations as we did in Chapter 15 "Chemical Equilibrium", remembering that the concentration of the pure solid is essentially constant.
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
Given: K sp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
A Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
B Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
A The dissolution equilibrium for Ca3(PO4)2 (Equation 17.1) is shown in the following table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43− ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43−] will be +2x. We can insert these values into the table.
Ca3(PO4)2 | [Ca2+] | [PO43−] | |
---|---|---|---|
initial | pure solid | 0 | 0 |
change | — | +3x | +2x |
final | pure solid | 3x | 2x |
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43−] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
Exercise
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
Answer:
The ion product (Q)A quantity that has precisely the same form as the solubility product for the dissolution of a sparingly soluble salt, except that the concentrations used are not necessarily equilibrium concentrations. of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibriums in Chapter 15 "Chemical Equilibrium". Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.
The ion product Q is analogous to the reaction quotient Q for gaseous equilibriums.
As summarized in Figure 17.1 "The Relationship between ", there are three possible conditions for an aqueous solution of an ionic solid:
Figure 17.1 The Relationship between Q and Ksp
If Q is less than Ksp, the solution is unsaturated and more solid will dissolve until the system reaches equilibrium (Q = Ksp). If Q is greater than Ksp, the solution is supersaturated and solid will precipitate until Q = Ksp. If Q = Ksp, the rate of dissolution is equal to the rate of precipitation; the solution is saturated, and no net change in the amount of dissolved solid will occur.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10−10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that NaCl is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
A Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
B Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
C Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution:
A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble. The equation for the precipitation of BaSO4 is as follows:
The solubility product expression is as follows:
B To solve this problem, we must first calculate the ion product—Q = [Ba2+][SO42−]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of BaCl2 solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
Similarly, the concentration of SO42− after mixing is the total number of moles of SO42− in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
We can now calculate Q:
C We now compare Q with the Ksp. If Q > Ksp, then BaSO4 will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that BaSO4 will precipitate when the two solutions are mixed. In fact, BaSO4 will continue to precipitate until the system reaches equilibrium, which occurs when [Ba2+][SO42−] = Ksp = 1.08 × 10−10.
Exercise
The solubility product of calcium fluoride (CaF2) is 3.45 × 10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10−5 M solution of Ca(NO3)2, will CaF2 precipitate?
Answer: yes (Q = 4.7 × 10−11 > Ksp)
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect discussed in Chapter 16 "Aqueous Acid–Base Equilibriums", Section 16.6 "Buffers": adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Châtelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.The exceptions generally involve the formation of complex ions, which is discussed in Section 17.3 "The Formation of Complex Ions".
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains 3 × (1.14 × 10−7 M) = 3.42 × 10−7 M Ca2+ and 2 × (1.14 × 10−7 M) = 2.28 × 10−7 M PO43−, according to the stoichiometry shown in Equation 17.1 (neglecting hydrolysis to form HPO42− as described in Chapter 16 "Aqueous Acid–Base Equilibriums"). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation 17.1 to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43− until Q = Ksp.
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
A Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
B Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution:
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43−] is +2x. We can insert these values into the table.
Ca3(PO4)2 | [Ca2+] | [PO43−] | |
---|---|---|---|
initial | pure solid | 0.20 | 0 |
change | — | +3x | +2x |
final | pure solid | 0.20 + 3x | 2x |
B The Ksp expression is as follows:
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Châtelier’s principle. With one exception, this example is identical to Example 2—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer: 2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Châtelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion.
Write an expression for Ksp for each salt.
Some species are not represented in a solubility product expression. Why?
Describe the differences between Q and Ksp.
How can an ion product be used to determine whether a solution is saturated?
When using Ksp to directly compare the solubilities of compounds, why is it important to compare only the Ksp values of salts that have the same stoichiometry?
Describe the effect of a common ion on the solubility of a salt. Is this effect similar to the common ion effect found in buffers? Explain your answer.
Explain why the presence of MgCl2 decreases the molar solubility of the sparingly soluble salt MgCO3.
For a 1:1 salt, the molar solubility is simply for a 2:1 salt, the molar solubility is Consequently, the magnitudes of Ksp can be correlated with molar solubility only if the salts have the same stoichiometry.
Because of the common ion effect. Adding a soluble Mg2+ salt increases [Mg2+] in solution, and Le Châtelier’s principle predicts that this will shift the solubility equilibrium of MgCO3 to the left, decreasing its solubility.
Predict the molar solubility of each compound using the Ksp values given in Chapter 26 "Appendix B: Solubility-Product Constants (".
Predict the molar solubility of each compound using the Ksp values given.
A student prepared 750 mL of a saturated solution of silver sulfate (Ag2SO4). How many grams of Ag2SO4 does the solution contain? Ksp = 1.20 × 10−5.
Given the Ksp values in Table 17.1 "Solubility Products for Selected Ionic Substances at 25°C" and Chapter 26 "Appendix B: Solubility-Product Constants (", predict the molar concentration of each species in a saturated aqueous solution.
Given the Ksp values in Table 17.1 "Solubility Products for Selected Ionic Substances at 25°C" and Chapter 26 "Appendix B: Solubility-Product Constants (", predict the molar concentration of each species in a saturated aqueous solution.
Silicon dioxide, the most common binary compound of silicon and oxygen, constitutes approximately 60% of Earth’s crust. Under certain conditions, this compound can react with water to form silicic acid, which can be written as either H4SiO4 or Si(OH)4. Write a balanced chemical equation for the dissolution of SiO2 in basic solution. Write an equilibrium constant expression for the reaction.
The Ksp of Mg(OH)2 is 5.61 × 10−12. If you tried to dissolve 24.0 mg of Mg(OH)2 in 250 mL of water and then filtered the solution and dried the remaining solid, what would you predict to be the mass of the undissolved solid? You discover that only 1.0 mg remains undissolved. Explain the difference between your expected value and the actual value.
The Ksp of lithium carbonate is 8.15 × 10−4. If 2.34 g of the salt is stirred with 500 mL of water and any undissolved solid is filtered from the solution and dried, what do you predict to be the mass of the solid? You discover that all of your sample dissolves. Explain the difference between your predicted value and the actual value.
You have calculated that 24.6 mg of BaSO4 will dissolve in 1.0 L of water at 25°C. After adding your calculated amount to 1.0 L of water and stirring for several hours, you notice that the solution contains undissolved solid. After carefully filtering the solution and drying the solid, you find that 22.1 mg did not dissolve. According to your measurements, what is the Ksp of barium sulfate?
In a saturated silver chromate solution, the molar solubility of chromate is 6.54 × 10−5. What is the Ksp?
A saturated lead(II) chloride solution has a chloride concentration of 3.24 × 10−2 mol/L. What is the Ksp?
From the solubility data given, calculate Ksp for each compound.
From the solubility data given, calculate Ksp for each compound.
Given the following solubilities, calculate Ksp for each compound.
Given the following solubilities, calculate Ksp for each compound.
The Ksp of the phosphate fertilizer CaHPO4·2H2O is 2.7 × 10−7 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 3.0 L of water at this temperature?
The Ksp of zinc carbonate monohydrate is 5.5 × 10−11 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 2.0 L of water at this temperature?
Silver nitrate eye drops were formerly administered to newborn infants to guard against eye infections contracted during birth. Although silver nitrate is highly water soluble, silver sulfate has a Ksp of 1.20 × 10−5 at 25°C. If you add 25.0 mL of 0.015 M AgNO3 to 150 mL of 2.8 × 10−3 M Na2SO4, will you get a precipitate? If so, what will its mass be?
Use the data in Chapter 26 "Appendix B: Solubility-Product Constants (" to predict whether precipitation will occur when each pair of solutions is mixed.
What is the maximum volume of 0.048 M Pb(NO3)2 that can be added to 250 mL of 0.10 M NaSCN before precipitation occurs? Ksp = 2.0 × 10−5 for Pb(SCN)2.
Given 300 mL of a solution that is 0.056 M in lithium nitrate, what mass of solid sodium carbonate can be added before precipitation occurs (assuming that the volume of solution does not change after adding the solid)? Ksp = 8.15 × 10−4 for Li2CO3.
Given the information in the following table, calculate the molar solubility of each sparingly soluble salt in 0.95 M MgCl2.
Saturated Solution | K sp |
---|---|
MgCO3·3H2O | 2.4 × 10−6 |
Mg(OH)2 | 5.6 × 10−12 |
Mg3(PO4)2 | 1.04 × 10−24 |
3.37 g
22.4 mg; a secondary reaction occurs, where OH− from the dissociation of the salt reacts with H+ from the dissociation of water. This reaction causes further dissociation of the salt (Le Châtelier’s principle).
1.2 × 10−10
1.70 × 10−5
7.4 × 10−6 M; 2.1 mg
Precipitation will occur in all cases.
8.27 g
The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH. The first two situations are described in this section, the formation of complex ions is discussed in Section 17.3 "The Formation of Complex Ions", and changes in pH are discussed in Section 17.4 "Solubility and pH".
An ion pairA cation and an anion that are in intimate contact in solution rather than separated by solvent. An ion pair can be viewed as a species that is intermediate between the ionic solid and the completely dissociated ions in solution. consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent (Figure 17.2 "Ion-Pair Formation"). The ions in an ion pair are held together by the same attractive electrostatic forces that we discussed in Chapter 8 "Ionic versus Covalent Bonding" for ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently).
Figure 17.2 Ion-Pair Formation
In an ion pair, the cation and the anion are in intimate contact in solution and migrate as a single unit. They are not completely dissociated and individually surrounded by solvent molecules, as are the hydrated ions, which are free to migrate independently.
As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
Equation 17.3
The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions.
Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured Ksp for calcium sulfate is 4.93 × 10−5 at 25°C. The solubility of CaSO4 should be 7.02 × 10−3 M if the only equilibrium involved were as follows:
Equation 17.4
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10−2 M, almost twice the value predicted from its Ksp. The reason for the discrepancy is that the concentration of ion pairs in a saturated CaSO4 solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M2+ and M3+ ions, such as Ca2+ and La3+, and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li+. We therefore expect a saturated solution of CaSO4 to contain a high concentration of ion pairs and its solubility to be greater than predicted from its Ksp.
The formation of ion pairs increases the solubility of a salt.
A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. (For more information about weak organic acids, see Chapter 16 "Aqueous Acid–Base Equilibriums"). Although strong acids (HA) dissociate completely into their constituent ions (H+ and A−) in water, weak acids such as carboxylic acids do not (Ka = 1.5 × 10−5). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH3CO2H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C6H5CO2H), with Ka = 6.25 × 10−5. Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
Equation 17.5
In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H+] = [C6H5CO2−] = 1.4 × 10−3 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10−2 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—C6H5CO2H(aq)—and only about 5% is present as the dissociated ions (Figure 17.3 "Incomplete Dissociation of a Molecular Solute").
Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute.
Figure 17.3 Incomplete Dissociation of a Molecular Solute
In a saturated solution of benzoic acid in water at 25°C, only about 5% of the dissolved benzoic acid molecules are dissociated to form benzoate anions and hydrated protons. The remaining 95% exists in solution in the form of hydrated neutral molecules. (H2O molecules are omitted for clarity.)
Although ion pairs, such as Ca2+·SO42−, and undissociated electrolytes, such as C6H5CO2H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte.
There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes in pH. An ion pair is held together by electrostatic attractive forces between the cation and the anion, whereas incomplete dissociation results from intramolecular forces, such as polar covalent O–H bonds.
Do you expect the actual molar solubility of LaPO4 to be greater than, the same as, or less than the value calculated from its Ksp? Explain your reasoning.
Do you expect the difference between the calculated molar solubility and the actual molar solubility of Ca3(PO4)2 to be greater than or less than the difference in the solubilities of Mg3(PO4)2? Why?
Write chemical equations to describe the interactions in a solution that contains Mg(OH)2, which forms ion pairs, and in one that contains propanoic acid (CH3CH2CO2H), which forms a hydrated neutral molecule.
Draw representations of Ca(IO3)2 in solution
Ferric phosphate has a molar solubility of 5.44 × 10−16 in 1.82 M Na3PO4. Predict its Ksp. The actual Ksp is 1.3 × 10−22. Explain this discrepancy.
9.90 × 10−16; the solubility is much higher than predicted by Ksp due to the formation of ion pairs (and/or phosphate complexes) in the sodium phosphate solution.
In Chapter 4 "Reactions in Aqueous Solution", you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ionAn ionic species that forms between a central metal ion and one or more surrounding ligands because of a Lewis acid–base interaction. The positively charged metal ion acts as the Lewis acid, and the ligand acts as the Lewis base. (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligandsAn ion or a molecule that contains one or more pairs of electrons that can be shared with the central metal in a metal complex., molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion in Figure 16.12 "Effect of a Metal Ion on the Acidity of Water".
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure 17.4 "The Formation of Complex Ions". The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
Figure 17.4 The Formation of Complex Ions
An aqueous solution of CuSO4 consists of hydrated Cu2+ ions in the form of pale blue [Cu(H2O)6]2+ (left). The addition of aqueous ammonia to the solution results in the formation of the intensely blue-violet [Cu(NH3)4(H2O)2]2+ ions, usually written as [Cu(NH3)4]2+ ion (right) because ammonia, a stronger base than H2O, replaces water molecules from the hydrated Cu2+ ion.
The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows:
Equation 17.6
The sum of the stepwise reactions is the overall equation for the formation of the complex ion:The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four NH3 ligands, not six. The reasons for this apparently unusual behavior will be discussed in Chapter 23 "The ".
Equation 17.7
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf)The equilibrium constant for the formation of a complex ion from a hydrated metal ion; that is, for the reaction . The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
Equation 17.8
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table 17.2 "Formation Constants for Selected Complex Ions in Aqueous Solution*".
Table 17.2 Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion | Equilibrium Equation | K f | |
---|---|---|---|
Ammonia Complexes | [Ag(NH3)2]+ | 1.1 × 107 | |
[Cu(NH3)4]2+ | 2.1 × 1013 | ||
[Ni(NH3)6]2+ | 5.5 × 108 | ||
Cyanide Complexes | [Ag(CN)2]− | 1.1 × 1018 | |
[Ni(CN)4]2− | 2.2 × 1031 | ||
[Fe(CN)6]3− | 1 × 1042 | ||
Hydroxide Complexes | [Zn(OH)4]2− | 4.6 × 1017 | |
[Cr(OH)4]− | 8.0 × 1029 | ||
Halide Complexes | [HgCl4]2− | 1.2 × 1015 | |
[CdI4]2− | 2.6 × 105 | ||
[AlF6]3− | 6.9 × 1019 | ||
Other Complexes | [Ag(S2O3)2]3− | 2.9 × 1013 | |
[Fe(C2O4)3]3− | 2.0 × 1020 | ||
*Reported values are overall formation constants. |
Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
If 12.5 g of Cu(NO3)2·6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
A Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation 17.7 to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
B Substitute the final concentrations into the expression for the formation constant (Equation 17.8) to calculate the equilibrium concentration of Cu2+(aq).
Solution:
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation 17.7. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
[Cu2+] | [NH3] | [[Cu(NH3)4]2+] | |
---|---|---|---|
initial | 0.0846 | 1.00 | 0 |
after complete reaction | 0 | 0.66 | 0.0846 |
change | +x | +4x | −x |
final | x | 0.66 + 4x | 0.0846 − x |
B Substituting the final concentrations into the expression for the formation constant (Equation 17.8) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise
The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6].
Answer: 2 × 10−6 M
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography, which was discussed briefly in Chapter 4 "Reactions in Aqueous Solution".
Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
Equation 17.9
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br− ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Châtelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
Equation 17.10
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
Equation 17.11
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2−). Calculate the solubility of AgCl in each situation:
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2−.
Given: Ksp of AgCl, Kf of AgCl2−, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
A Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl− in water.
B Calculate the concentration of Ag+ in the KCl solution.
C Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2− complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
D Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution:
A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl−] = x M. Substituting this value into the solubility product expression,
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl−] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2−] = x and [Cl−] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2− complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise
Calculate the solubility of mercury(II) iodide (HgI2) in each situation:
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer:
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
Equation 17.12
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure 17.5 "An MRI Image of the Heart, Arteries, and Veins"). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Figure 17.5 An MRI Image of the Heart, Arteries, and Veins
When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues.
A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound.
What is the difference between Keq and Kf?
Which would you expect to have the greater tendency to form a complex ion: Mg2+ or Ba2+? Why?
How can a ligand be used to affect the concentration of hydrated metal ions in solution? How is Ksp affected? Explain your answer.
Co(II) forms a complex ion with pyridine (C5H5N). Which is the Lewis acid, and which is the Lewis base? Use Lewis electron structures to justify your answer.
Fe(II) forms the complex ion [Fe(OH)4]2− through equilibrium reactions in which hydroxide replaces water in a stepwise manner. If log K1 = 5.56, log K2 = 4.21, log K3 = −0.10, and log K4 = −1.09, what is Kf? Write the equilibrium equation that corresponds to each stepwise equilibrium constant. Do you expect the [Fe(OH)4]2− complex to be stable? Explain your reasoning.
Zn(II) forms the complex ion [Zn(NH3)4]2+ through equilibrium reactions in which ammonia replaces coordinated water molecules in a stepwise manner. If log K1 = 2.37, log K2 = 2.44, log K3 = 2.50, and log K4 = 2.15, what is the overall Kf? Write the equilibrium equation that corresponds to each stepwise equilibrium constant. Do you expect the [Zn(NH3)4]2+ complex to be stable? Explain your reasoning.
Although thallium has limited commercial applications because it is toxic to humans (10 mg/kg body weight is fatal to children), it is used as a substitute for mercury in industrial switches. The complex ion [TlBr6]3− is highly stable, with log Kf = 31.6. What is the concentration of Tl(III)(aq) in equilibrium with a 1.12 M solution of Na3[TlBr6]?
Thus, Kf = 3.8 × 1033. Because [Fe(OH)4]2− has a very large value of Kf, it should be stable in the presence of excess OH−.
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH.
We begin our discussion by examining the effect of pH on the solubility of a representative salt, M+A−, where A− is the conjugate base of the weak acid HA. When the salt dissolves in water, the following reaction occurs:
Equation 17.13
The anion can also react with water in a hydrolysis reaction:
Equation 17.14
Because of the reaction described in , the predicted solubility of a sparingly soluble salt that has a basic anion such as S2−, PO43−, or CO32− is increased, as described in . If instead a strong acid is added to the solution, the added H+ will react essentially completely with A− to form HA. This reaction decreases [A−], which decreases the magnitude of the ion product (Q = [M+][A−]). According to Le Châtelier’s principle, more MA will dissolve until Q = Ksp. Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH)2 is relatively insoluble in water:
Equation 17.15
When acid is added to a saturated solution that contains excess solid Mg(OH)2, the following reaction occurs, removing OH− from solution:
Equation 17.16
The overall equation for the reaction of Mg(OH)2 with acid is thus
Equation 17.17
As more acid is added to a suspension of Mg(OH)2, the equilibrium shown in is driven to the right, so more Mg(OH)2 dissolves.
Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF2 is a sparingly soluble salt:
Equation 17.18
When strong acid is added to a saturated solution of CaF2, the following reaction occurs:
Equation 17.19
Because the forward reaction decreases the fluoride ion concentration, more CaF2 dissolves to relieve the stress on the system. The net reaction of CaF2 with strong acid is thus
Equation 17.20
CaF2(s) + 2H+(aq) → Ca2+(aq) + 2HF(aq)Example 7 shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt.
Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution.
Lead oxalate (PbC2O4), lead iodide (PbI2), and lead sulfate (PbSO4) are all rather insoluble, with Ksp values of 4.8 × 10−10, 9.8 × 10−9, and 2.53 × 10−8, respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities?
Given: Ksp values for three compounds
Asked for: relative solubilities in acid solution
Strategy:
Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions.
Solution:
The solubility equilibriums for the three salts are as follows:
The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI2 will not greatly affect its solubility; the acid will simply dissociate to form H+(aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO2CCO2H), which is a weak diprotic acid (pKa1 = 1.23 and pKa2 = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions:
These reactions will decrease [C2O42−], causing more lead oxalate to dissolve to relieve the stress on the system.The pKa of HSO4− (1.99) is similar in magnitude to the pKa1 of oxalic acid, so adding a strong acid to a saturated solution of PbSO4 will result in the following reaction:
Because HSO4− has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO4 will be significantly less than for PbC2O4.
Exercise
Which of the following insoluble salts—AgCl, Ag2CO3, Ag3PO4, and/or AgBr—will be substantially more soluble in 1.0 M HNO3 than in pure water?
Answer: Ag2CO3 and Ag3PO4
Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility equilibriums (part (a) in ). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows:
Equation 17.21
Equation 17.22
Equation 17.23
Figure 17.6 The Chemistry of Cave Formation
(a) This cave in Campanet, Mallorca, Spain, and its associated formations are examples of pH-dependent solubility equilibriums. (b) A cave forms when groundwater containing atmospheric CO2, forming an acidic solution, dissolves limestone (CaCO3) in a process that may take tens of thousands of years. As groundwater seeps into a cave, water evaporates from the solution of CaCO3 in CO2-rich water, producing a supersaturated solution and a shift in equilibrium that causes precipitation of the CaCO3. The deposited limestone eventually forms stalactites and stalagmites.
Limestone deposits that form caves consist primarily of CaCO3 from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO3 in CO2-rich water rises toward Earth’s surface or is otherwise heated, CO2 gas is released as the water warms. CaCO3 then precipitates from the solution according to the following equation (part (b) in ):
Equation 17.24
The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated.
When groundwater-containing atmospheric CO2 ( and ) finds its way into microscopic cracks in the limestone deposits, CaCO3 dissolves in the acidic solution in the reverse direction of . The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in to shift to the right. A circular layer of solid CaCO3 is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China ().
Figure 17.7 Solubility Equilibriums in the Formation of Karst Landscapes
Landscapes such as the steep limestone pinnacles of the Stone Forest in Yunnan Province, China, are formed from the same process that produces caves and their associated formations.
One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxidesAn oxide that reacts with water to produce a basic solution or dissolves readily in aqueous acid. and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxidesAn oxide that reacts with water to produce an acidic solution or dissolves in aqueous base. or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. As shown in , there is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:
Equation 17.25
Equation 17.26
Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed!
Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions.
The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete Mn+ cations and O2− anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: Eδ+–Oδ−. The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO3, which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO42−):
Equation 17.27
MoO3(s) + 2OH−(aq) → MoO42−(aq) + H2O(l)As shown in , there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called amphoteric oxidesAn oxide that can dissolve in acid to produce water and dissolve in base to produce a soluble complex. (from the Greek ampho, meaning “both,” as in amphiprotic, which was defined in , ). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in , for example, mixing the amphoteric oxide Cr(OH)3 (also written as Cr2O3·3H2O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH)3 to dissolve to give a bright violet solution of Cr3+(aq), which contains the [Cr(H2O)6]3+ ion, whereas adding strong base gives a green solution of the [Cr(OH)4]− ion. The chemical equations for the reactions are as follows:
Equation 17.28
Equation 17.29
Figure 17.8 Classification of the Oxides of the Main Group Elements According to Their Acidic or Basic Character
There is a gradual transition from basic oxides to acidic oxides from the lower left to the upper right in the periodic table. Oxides of metallic elements are generally basic oxides, which either react with water to form a basic solution or dissolve in aqueous acid. In contrast, oxides of nonmetallic elements are acidic oxides, which either react with water to form an acidic solution or are soluble in aqueous base. Oxides of intermediate character, called amphoteric oxides, are located along a diagonal line between the two extremes. Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex ion. (Radioactive elements are not classified.)
Figure 17.9 Chromium(III) Hydroxide [Cr(OH)3 or Cr2O3·3H2O] Is an Example of an Amphoteric Oxide
All three beakers originally contained a suspension of brownish purple Cr(OH)3(s) (center). When concentrated acid (6 M H2SO4) was added to the beaker on the left, Cr(OH)3 dissolved to produce violet [Cr(H2O)6]3+ ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH)3 to dissolve, producing green [Cr(OH)4]−ions.
Aluminum hydroxide, written as either Al(OH)3 or Al2O3·3H2O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base.
Given: amphoteric compound
Asked for: dissolution reactions in acid and base
Strategy:
Using and as a guide, write the dissolution reactions in acid and base solutions.
Solution:
An acid donates protons to hydroxide to give water and the hydrated metal ion, so aluminum hydroxide, which contains three OH− ions per Al, needs three H+ ions:
Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l)In aqueous solution, Al3+ forms the complex ion [Al(H2O)6]3+.
In basic solution, OH− is added to the compound to produce a soluble and stable poly(hydroxo) complex:
Al(OH)3(s) + OH−(aq) → [Al(OH)4]−(aq)Exercise
Copper(II) hydroxide, written as either Cu(OH)2 or CuO·H2O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base.
Answer:
Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l) Cu(OH)2(s) + 2OH−(aq) → [Cu(OH)4]2−(aq)Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate.
The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations.
Suppose, for example, we have a solution that contains 1.0 mM Zn2+ and 1.0 mM Cd2+ and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilibriums can be written as follows:
Equation 17.30
Equation 17.31
Because the S2− ion is quite basic and reacts extensively with water to give HS− and OH−, the solubility equilibriums are more accurately written as rather than Here we use the simpler form involving S2−, which is justified because we take the reaction of S2− with water into account later in the solution, arriving at the same answer using either equilibrium equation.
The sulfide concentrations needed to cause ZnS and CdS to precipitate are as follows:
Equation 17.32
Equation 17.33
Thus sulfide concentrations between 1.6 × 10−21 M and 8.0 × 10−24 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H2S contains 0.10 M H2S at 20°C. The pKa1 for H2S is 6.97, and pKa2 corresponding to the formation of [S2−] is 12.90. The equations for these reactions are as follows:
Equation 17.34
We can show that the concentration of S2− is 1.3 × 10−13 by comparing Ka1 and Ka2 and recognizing that the contribution to [H+] from the dissociation of HS− is negligible compared with [H+] from the dissociation of H2S. Thus substituting 0.10 M in the equation for Ka1 for the concentration of H2S, which is essentially constant regardless of the pH, gives the following:
Equation 17.35
Substituting this value for [H+] and [HS−] into the equation for Ka2,
Although [S2−] in an H2S solution is very low (1.3 × 10−13 M), bubbling H2S through the solution until it is saturated would precipitate both metal ions because the concentration of S2− would then be much greater than 1.6 × 10−21 M. Thus we must adjust [S2−] to stay within the desired range. The most direct way to do this is to adjust [H+] by adding acid to the H2S solution (recall Le Châtelier's principle), thereby driving the equilibrium in to the left. The overall equation for the dissociation of H2S is as follows:
Equation 17.36
Now we can use the equilibrium constant K for the overall reaction, which is the product of Ka1 and Ka2, and the concentration of H2S in a saturated solution to calculate the H+ concentration needed to produce [S2−] of 1.6 × 10−21 M:
Equation 17.37
Equation 17.38
Thus adding a strong acid such as HCl to make the solution 0.94 M in H+ will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H2S.
A solution contains 0.010 M Ca2+ and 0.010 M La3+. What concentration of HCl is needed to precipitate La2(C2O4)3·9H2O but not Ca(C2O4)·H2O if the concentration of oxalic acid is 1.0 M? Ksp values are 2.32 × 10−9 for Ca(C2O4) and 2.5 × 10−27 for La2(C2O4)3; pKa1 = 1.25 and pKa2 = 3.81 for oxalic acid.
Given: concentrations of cations, Ksp values, and concentration and pKa values for oxalic acid
Asked for: concentration of HCl needed for selective precipitation of La2(C2O4)3
Strategy:
A Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La2(C2O4)3·9H2O.
B Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox2−] needed to precipitate La2(C2O4)3·9H2O into the overall equation for the dissociation of oxalic acid to calculate the required [H+].
Solution:
A Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox2− for oxalate, we write the solubility product expression for calcium oxalate as follows:
The expression for lanthanum oxalate is as follows:
Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between 2.9 × 10−8 M and 2.32 × 10−7 M.
B To prevent Ca2+ from precipitating as calcium oxalate, we must add enough H+ to give a maximum oxalate concentration of 2.32 × 10−7 M. We can calculate the required [H+] by using the overall equation for the dissociation of oxalic acid to oxalate:
Substituting the desired oxalate concentration into the equilibrium constant expression,
Thus adding enough HCl to give [H+] = 6.1 M will cause only La2(C2O4)3·9H2O to precipitate from the solution.
Exercise
A solution contains 0.015 M Fe2+ and 0.015 M Pb2+. What concentration of acid is needed to ensure that Pb2+ precipitates as PbS in a saturated solution of H2S, but Fe2+ does not precipitate as FeS? Ksp values are 6.3 × 10−18 for FeS and 8.0 × 10−28 for PbS.
Answer: 0.018 M H+
The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation.
Which of the following will show the greatest increase in solubility if 1 M HNO3 is used instead of distilled water? Explain your reasoning.
Of the compounds Sn(CH3CO2)2 and SnS, one is soluble in dilute HCl and the other is soluble only in hot, concentrated HCl. Which is which? Provide a reasonable explanation.
Where in the periodic table do you expect to find elements that form basic oxides? Where do you expect to find elements that form acidic oxides?
Because water can autoionize, it reacts with oxides either as a base (as OH−) or as an acid (as H3O+). Do you expect oxides of elements in high oxidation states to be more acidic (reacting with OH−) or more basic (reacting with H3O+) than the corresponding oxides in low oxidation states? Why?
Given solid samples of CrO, Cr2O3, and CrO3, which would you expect to be the most acidic (reacts most readily with OH−)? Which would be the most basic (reacts most readily with H3O+)? Why?
Which of these elements—Be, B, Al, N, Se, In, Tl, Pb—do you expect to form an amphoteric oxide? Why?
A 1.0 L solution contains 1.98 M Al(NO3)3. What are [OH−] and [H+]? What pH is required to precipitate the cation as Al(OH)3? Ksp = 1.3 × 10−33 and Ka = 1.05 × 10−5 for the hydrated Al3+ ion.
A 1.0 L solution contains 2.03 M CoCl2. What is [H+]? What pH is required to precipitate the cation as Co(OH)2? Ksp = 5.92 × 10−15 and Ka = 1.26 × 10−9 for the hydrated Co2+ ion.
Given 100 mL of a solution that contains 0.80 mM Ag+ and 0.80 mM Cu+, can the two metals be separated by selective precipitation as the insoluble bromide salts by adding 10 mL of an 8.0 mM solution of KBr? Ksp values are 6.27 × 10−9 for CuBr and 5.35 × 10−13 for AgBr. What maximum [Br−] will separate the ions?
Given 100 mL of a solution that is 1.5 mM in Tl+, Zn2+, and Ni2+, which ions can be separated from solution by adding 5.0 mL of a 12.0 mM solution of Na2C2O4?
Precipitate | K sp |
---|---|
Tl2C2O4 | 2 × 10−4 |
ZnC2O4·2H2O | 1.38 × 10−9 |
NiC2O4 | 4 × 10−10 |
How many milliliters of 12.0 mM Na2C2O4 should be added to separate Tl+ and Zn2+ from Ni2+?
[H+] = 4.56 × 10−3; [OH−] = 2.19 × 10−12; pH = 2.94
No; both metal ions will precipitate; AgBr will precipitate as Br− is added, and CuBr will begin to precipitate at [Br−] = 8.6 × 10−6 M.
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysisA procedure for determining the identity of metal ions present in a mixture that does not include information about their amounts., a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts).
The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in .
Figure 17.10 Steps in a Typical Qualitative Analysis Scheme for a Solution That Contains Several Metal Ions
Most metal chloride salts are soluble in water; only Ag+, Pb2+, and Hg22+ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M HCl, thereby causing AgCl, PbCl2, and/or Hg2Cl2 to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation.
Next, the acidic solution is saturated with H2S gas. Only those metal ions that form very insoluble sulfides, such as As3+, Bi3+, Cd2+, Cu2+, Hg2+, Sb3+, and Sn2+, precipitate as their sulfide salts under these acidic conditions. All others, such as Fe2+ and Zn2+, remain in solution. Once again, the precipitates are collected by filtration or centrifugation.
Ammonia or NaOH is now added to the solution until it is basic, and then (NH4)2S is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions Co2+, Fe2+, Mn2+, Ni2+, and Zn2+ precipitate as their sulfides, and the trivalent metal ions Al3+ and Cr3+ precipitate as their hydroxides: Al(OH)3 and Cr(OH)3. If the mixture contains Fe3+, sulfide reduces the cation to Fe2+, which precipitates as FeS.
The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When Na2CO3 is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding (NH4)2HPO4 causes the same metal ions to precipitate as insoluble phosphates.
At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+). We now take a second sample from the original solution and add a small amount of NaOH to neutralize the ammonium ion and produce NH3. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. As discussed in , the other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present.
Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing Ag+, Pb2+, and Hg22+, are all quite insoluble in water. Because PbCl2 is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any PbCl2 present. Isolating the solution and adding a small amount of Na2CrO4 solution to it will produce a bright yellow precipitate of PbCrO4 if Pb2+ was in the original sample ().
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any AgCl because Ag+ forms a stable complex with ammonia: [Ag(NH3)2]+. In addition, Hg2Cl2disproportionates in ammonia (2Hg22+ → Hg + Hg2+) to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
Equation 17.39
Hg2Cl2(s) + 2NH3(aq) → Hg(l) + Hg(NH2)Cl(s) + NH4+(aq) + Cl−(aq)Figure 17.11 The Separation of Metal Ions from Group 1 Using Qualitative Analysis
In (a), the cations of group 1 precipitate when HCl(aq) is added to a solution containing a mixture of cations. (b) When a small amount of Na2CrO4 solution is added to a sample containing Pb2+ ions in water, a bright yellow precipitate of PbCrO4 forms. (c) Adding aqueous ammonia to a second portion of the solid sample produces a black solid that is a mixture of finely divided metallic mercury, an insoluble mercury(II) compound [Hg(NH2)Cl], and a stable [Ag(NH3)2]+(aq) complex. (d) The presence of Ag+ is detected by decanting the solution from the precipitated mercury and mercury complex and adding hydrochloric acid to the decanted solution, which causes AgCl to precipitate.
Any silver ion in the solution is then detected by adding HCl, which reverses the reaction and gives a precipitate of white AgCl that slowly darkens when exposed to light:
Equation 17.40
[Ag(NH3)2]+(aq) + 2H+(aq) + Cl−(aq) → AgCl(s) + 2NH4+(aq)Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups.
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
Given a solution that contains a mixture of NaCl, CuCl2, and ZnCl2, propose a method for separating the metal ions.
Problems marked with a ♦ involve multiple concepts.
♦ Gypsum (CaSO4·2H2O) is added to soil to enhance plant growth. It dissolves according to the following equation:
Egyptian blue, which is difficult to prepare, is a synthetic pigment developed about 4500 yr ago. It was the only blue pigment identified in a study of stocks of dry pigments found in color merchants’ shops in Pompeii. The pigment contains 12.9% calcium carbonate (calcite). A major source of CaCO3 is limestone, which also contains MgCO3. Assuming that the masses of CaCO3 and MgCO3 are equal, and that a sample of limestone is dissolved in acidified water to give [Ca2+] = [Mg2+] = 0.010 M in 5.0 L of solution, would selective precipitation be a viable method for purifying enough CaCO3 from limestone to produce 1.0 g of pigment? Why? The Ksp values are 3.36 × 10−9 for CaCO3 and 6.8 × 10−6 for MgCO3.
One method of mining gold is to extract it through the process of cyanidation. Mined ores are milled and treated with aqueous cyanide solution to produce a gold complex ion {[Au(CN)2]−} that is very stable. Given a sample of AuCl, what is the solubility of AuCl in each situation? The Ksp of AuCl = 2.0 × 10−13; log Kf{[Au(CN)2]−} = 38.3.
♦ Almost all barium carbonate is produced synthetically. The compound is used in manufacturing clay tiles and ceramic products as well as in cathode ray tubes and special optical glasses. BaCO3 is synthesized by allowing barium sulfate to react with coal at 1000°C–1200°C in a rotary kiln, followed by treatment of a solution of the product with either CO2 (reaction 1) or Na2CO3 (reaction 2). The reactions are as follows:
BaSO4(s) + 4C(s) → 4BaS(s) + 4CO(g) reaction 1. BaS(s) + CO2(g) + H2O(l) → BaCO3(aq) + H2S(g) reaction 2. BaS(s) + Na2CO3(aq) ⇌ BaCO3(aq) + Na2S(aq)Barium carbonate has a Ksp of 2.58 × 10−9. The pKa for is 6.97, and the pKa for is 12.90. Given this information, answer the following questions:
A person complaining of chronic indigestion continually consumed antacid tablets containing Ca(OH)2 over a two-week period. A blood test at the end of this period showed that the person had become anemic. Explain the reactions that caused this test result.
Although the commercial production of radium has virtually ceased since artificial radionuclides were discovered to have similar properties and lower costs, commercial radium is still isolated using essentially the same procedure developed by Marie Curie, as outlined here. Explain what is happening chemically at each step of the purification process. What is precipitate A? What metal ions are present in solution A? What is precipitate B? What metal ions are present in solution B?
In a qualitative analysis laboratory, a student initially treated his sample of metal ions with 6 M HNO3 instead of 6 M HCl, recognizing his mistake only after the acid-insoluble sulfides had been precipitated. He decided to simply add 6 M HCl to the filtrate from which the sulfides had been removed, but he obtained no precipitate. The student therefore concluded that there were no Ag+, Hg22+, or Pb2+ cations in his original sample. Is this conclusion valid?
Using qualitative analysis, a student decided to treat her sample with (NH4)2S solution directly, skipping the HCl and acidic H2S treatments because she was running out of time. In a sample that contained Ag+, Hg22+, Cd2+, Sb3+, and Zn2+, which metal ions was she most likely to obtain in the resulting precipitate?
No; these cations would precipitate as sulfides.
Chemical reactions obey two fundamental laws. The first of these, the law of conservation of mass, states that matter can be neither created nor destroyed. (For more information on matter, see .) The law of conservation of mass is the basis for all the stoichiometry and equilibrium calculations you have learned thus far in chemistry. The second, the law of conservation of energy, states that energy can be neither created nor destroyed. (For more information on energy, see .) Instead, energy takes various forms that can be converted from one form to another. For example, the energy stored in chemical bonds can be released as heat during a chemical reaction.
In , you also learned about thermochemistry, the study of energy changes that occur during chemical reactions. Our goal in this chapter is to extend the concepts of thermochemistry to an exploration of thermodynamicsThe study of the interrelationships among heat, work, and the energy content of a system at equilibrium. (from the Greek thermo and dynamic, meaning “heat” and “power,” respectively), the study of the interrelationships among heat, work, and the energy content of a system at equilibrium. Thermodynamics tells chemists whether a particular reaction is energetically possible in the direction in which it is written, and it gives the composition of the reaction system at equilibrium. It does not, however, say anything about whether an energetically feasible reaction will actually occur as written, and it tells us nothing about the reaction rate or the pathway by which it will occur. The rate of a reaction and its pathway are described by chemical kinetics. (For more information on reaction rates and kinetics, see .)
The melting of ice is a thermodynamic process. When a cube of ice melts, there is a spontaneous and irreversible transfer of heat from a warm substance, the surrounding air, to a cold substance, the ice cube. The direction of heat flow in this process and the resulting increase in entropy illustrate the second law of thermodynamics.
Chemical thermodynamics provides a bridge between the macroscopic properties of a substance and the individual properties of its constituent molecules and atoms. As you will see, thermodynamics explains why graphite can be converted to diamond; how chemical energy stored in molecules can be used to perform work; and why certain processes, such as iron rusting and organisms aging and dying, proceed spontaneously in only one direction, requiring no net input of energy to occur.
We begin our discussion of thermodynamics by reviewing some important terms introduced in . First, we need to distinguish between a system and its surroundings. A system is that part of the universe in which we are interested, such as a mixture of gases in a glass bulb or a solution of substances in a flask. The surroundings are everything else—the rest of the universe. We can therefore state the following:
Equation 18.1
system + surroundings = universeA closed system, such as the contents of a sealed jar, cannot exchange matter with its surroundings, whereas an open system can; in this case, we can convert a closed system (the jar) to an open system by removing the jar’s lid.
In , we also introduced the concept of a state functionA property of a system whose magnitude depends on only the present state of the system, not its previous history., a property of a system that depends on only the present state of the system, not its history. Thus a change in a state function depends on only the difference between the initial and final states, not the pathway used to go from one to the other. To help understand the concept of a state function, imagine a person hiking up a mountain (). If the person is well trained and fit, he or she may be able to climb almost vertically to the top (path A), whereas another less athletic person may choose a path that winds gradually to the top (path B). If both hikers start from the same point at the base of the mountain and end up at the same point at the top, their net change in altitude will be the same regardless of the path chosen. Hence altitude is a state function. On the other hand, a person may or may not carry a heavy pack and may climb in hot weather or cold. These conditions would influence changes in the hiker’s fatigue level, which depends on the path taken and the conditions experienced. Fatigue, therefore, is not a state function. Thermodynamics is generally concerned with state functions and does not deal with how the change between the initial state and final state occurs.
Figure 18.1 Altitude Is a State Function
When hiking up a mountain, a person may decide to take path A, which is almost vertical, or path B, which gradually winds up to the top. Regardless of the path taken, the net change in altitude going from the initial state (bottom of the climb) to the final state (top of the climb) is the same. Thus altitude is a state function.
The internal energy (E)A state function that is the sum of the kinetic and potential energies of all a system’s components. of a system is the sum of the potential energy and the kinetic energy of all the components; internal energy is a state function. Although a closed system cannot exchange matter with its surroundings, it can exchange energy with its surroundings in two ways: by doing work or by releasing or absorbing heat—the flow of thermal energy. Work and heat are therefore two distinct ways of changing the internal energy of a system. We defined work (w) in as a force F acting through a distance d:
Equation 18.2
w = FdBecause work occurs only when an object, such as a person, or a substance, such as water, moves against an opposing force, work requires that a system and its surroundings be connected. In contrast, the flow of heat, the transfer of energy due to differences in temperature between two objects, represents a thermal connection between a system and its surroundings. Thus doing work causes a physical displacement, whereas the flow of heat causes a temperature change. The units of work and heat must be the same because both processes result in the transfer of energy. In the SI system, those units are joules (J), the same unit used for energy. There is no difference between an energy change brought about by doing work on a system and an equal energy change brought about by heating it.
The connections among work, heat, and energy were first described by Benjamin Thompson (1753–1814), an American-born scientist who was also known as Count Rumford. While supervising the manufacture of cannons, Rumford recognized the relationship between the amount of work required to drill out a cannon and the temperature of the water used to cool it during the drilling process (). At that time, it was generally thought that heat and work were separate and unrelated phenomena. Hence Rumford’s ideas were not widely accepted until many years later, after his findings had been corroborated in other laboratories.
Figure 18.2 The Relationship between Heat and Work
In the 1780s, an American scientist named Benjamin Thompson, also known as Count Rumford, was hired by the Elector of Bavaria to supervise the manufacture of cannons. During the manufacturing process, teams of horses harnessed to a large-toothed wheel supplied the power needed to drill a hole several inches in diameter straight down the center of a solid brass or bronze cylinder, which was cooled by water. Based on his observations, Rumford became convinced that heat and work are equivalent ways of transferring energy.
As we saw in , there are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work, focusing on the work done during changes in the pressure or the volume of a gas. To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions.
Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure Pint and initial volume Vi (). If Pext = Pint, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston (Pext) is less than Pint, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume (Vf) will be greater than Vi. If Pext > Pint, then the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area A, the external pressure exerted by the piston is, by definition, the force per unit area: Pext = F/A. The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height (V = Ah). Rearranging to give F = PextA and defining the distance the piston moves (d) as Δh, we can calculate the magnitude of the work performed by the piston by substituting into :
Equation 18.3
w = Fd = PextAΔhFigure 18.3 PV Work
Using a frictionless piston, if the external pressure is less than Pint (a), the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings. The final volume (Vf) will be greater than Vi. Alternatively, if the external pressure is greater than Pint (b), the gas will be compressed, and the surroundings will perform work on the system.
The change in the volume of the cylinder (ΔV) as the piston moves a distance d is ΔV = AΔh, as shown in . The work performed is thus
Equation 18.4
w = PextΔVThe units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
Figure 18.4 Work Performed with a Change in Volume
The change in the volume (ΔV) of the cylinder housing a piston is ΔV = AΔh as the piston moves. The work performed by the surroundings on the system as the piston moves inward is given by w = PextΔV.
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J. (For more information on the ideal gas law, see .)
Whether work is defined as having a positive sign or a negative sign is a matter of convention. In , we defined heat flow from a system to its surroundings as negative. Using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings.This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, must be written with a negative sign to describe PV work done by the system as negative:
Equation 18.5
w = −PextΔVThe work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings.
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining , in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from (V1, P1) to (V2, P2), which means that work is not a state function.
Internal energy is a state function, whereas work is not.
Figure 18.5 Work Is Not a State Function
In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1). Its pressure is then decreased while the volume remains constant (step 2). Pathway B reverses these steps. Although (V1, P1) and (V2, P2) are identical in both cases, the amount of work done (shaded area) depends on the pathway taken.
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Asked for: work done
Strategy:
A Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio.
B Use to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution:
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
w = −PextΔV = −(40.0 atm)(0.400 L − 0.0400 L) = −14.4 L·atmConverting from liter-atmospheres to joules,
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
Exercise
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?
Answer: −0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
Thermodynamics is the study of the interrelationships among heat, work, and the energy content of a system at equilibrium. The sum of the potential energy and the kinetic energy of all the components of a system is the internal energy (E) of the system, which is a state function. When the pressure or the volume of a gas is changed, any mechanical work done is called PV work. Work done by a system on its surroundings is given a negative value, whereas work done on a system by its surroundings has a positive value.
Thermodynamics focuses on the energetics of the reactants and products and provides information about the composition of the reaction system at equilibrium. What information on reaction systems is not provided by thermodynamics?
Given a system in which a substance can produce either of two possible products, A → B or A → C, which of the following can be predicted using chemical thermodynamics?
In what two ways can a closed system exchange energy with its surroundings? Are these two processes path dependent or path independent?
A microwave oven operates by providing enough energy to rotate water molecules, which produces heat. Can the change in the internal energy of a cup of water heated in a microwave oven be described as a state function? Can the heat produced be described as a state function?
Thermodynamics tells us nothing about the rate at which reactants are converted to products.
heat and work; path dependent
Calculate the work done in joules in each process.
How much work in joules is done when oxygen is compressed from a volume of 22.8 L and an external pressure of 1.20 atm to 12.0 L at a constant temperature? Was work done by the system or the surroundings?
Champagne is bottled at a CO2 pressure of about 5 atm. What is the force on the cork if its cross-sectional area is 2.0 cm2? How much work is done if a 2.0 g cork flies a distance of 8.2 ft straight into the air when the cork is popped? Was work done by the system or the surroundings?
One mole of water is converted to steam at 1.00 atm pressure and 100°C. Assuming ideal behavior, what is the change in volume when the water is converted from a liquid to a gas? If this transformation took place in a cylinder with a piston, how much work could be done by vaporizing the water at 1.00 atm? Is work done by the system or the surroundings?
Acceleration due to gravity on the earth’s surface is 9.8 m/s2. How much work is done by a 175 lb person going over Niagara Falls (approximately 520 ft high) in a barrel that weighs 145 lb?
Recall that force can be expressed as mass times acceleration (F = ma). Acceleration due to gravity on the earth’s surface is 9.8 m/s2.
A gas is allowed to expand from a volume of 2.3 L to a volume of 5.8 L. During the process, 460 J of heat is transferred from the surroundings to the gas.
One mole of an ideal gas is allowed to expand from an initial volume of 0.62 L to a final volume of 1.00 L at constant temperature against a constant external pressure of 1.0 atm. How much work has been done?
−230 kJ
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamicsThe energy of the universe is constant: = + = 0., which states that the energy of the universe is constant. Using , we can express this law mathematically as follows:
Equation 18.6
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the CO2/H2O product mixture is less than that of the isooctane/O2 reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):
Equation 18.7
ΔEsys = q + wAlthough q and w are not state functions on their own, their sum (ΔEsys) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed, as discussed in . Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
Although q and w are not state functions, their sum (ΔEsys) is independent of the path taken and therefore is a state function.
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔE) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
A Determine the sign of q to use in .
B From , calculate w from the values given. Substitute this value into to calculate ΔE.
Solution:
A From , we know that ΔE = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so w must be positive. From ,
Thus
ΔE = q + w = −140 J + 284 J = 144 JIn this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔE) of the gas in joules?
Answer: −216 J
By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
To further understand the relationship between heat flow (q) and the resulting change in internal energy (ΔE), we can look at two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. We will assume that PV work is the only kind of work possible for the system, so we can substitute its definition from into to obtain the following:
Equation 18.8
ΔE = q − PΔVwhere the subscripts have been deleted.
If the reaction occurs in a closed vessel, the volume of the system is fixed, and ΔV is zero. Under these conditions, the heat flow (often given the symbol qv to indicate constant volume) must equal ΔE:
Equation 18.9
No PV work can be done, and the change in the internal energy of the system is equal to the amount of heat transferred from the system to the surroundings or vice versa.
Many chemical reactions are not, however, carried out in sealed containers at constant volume but in open containers at a more or less constant pressure of about 1 atm. The heat flow under these conditions is given the symbol qp to indicate constant pressure. Replacing q in by qp and rearranging to solve for qp,
Equation 18.10
Thus, at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done, as we stated in .
Because conditions of constant pressure are so important in chemistry, a new state function called enthalpy (H)A state function that is the sum of the system’s internal energy and the product of its pressure and volume is defined as H = E + PV. At constant pressure, the change in the enthalpy of a system is as follows:
Equation 18.11
ΔH = ΔE + Δ(PV) = ΔE + PΔVComparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH = qp. This expression is consistent with our definition of enthalpy in , where we stated that enthalpy is the heat absorbed or produced during any process that occurs at constant pressure.
At constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH = qp.
The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔE for the melting of ice at 0.0°C. (For more information on enthalpy, see , .)
Given: enthalpy of fusion for ice, pressure, and molar volumes of ice and water
Asked for: ΔH and ΔE for ice melting at 0.0°C
Strategy:
A Determine the sign of q and set this value equal to ΔH.
B Calculate Δ(PV) from the information given.
C Determine ΔE by substituting the calculated values into .
Solution:
A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = qp = ΔH = 6.01 kJ.
B To find ΔE using , we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so
C Substituting the calculated values of ΔH and PΔV into ,
ΔE = ΔH − PΔV = 6010 J − (−0.0017 J) = 6010 J = 6.01 kJExercise
At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔE for the conversion of C (graphite) to C (diamond) under these conditions.
Answer: ΔH = 1.85 kJ/mol; ΔE = 1.85 kJ/mol
If ΔH for a reaction is known, we can use the change in the enthalpy of the system () to calculate its change in internal energy. When a reaction involves only solids, liquids, liquid solutions, or any combination of these, the volume does not change appreciably (ΔV = 0). Under these conditions, we can simplify to ΔH = ΔE. If gases are involved, however, ΔH and ΔE can differ significantly. We can calculate ΔE from the measured value of ΔH by using the right side of together with the ideal gas law, PV = nRT. Recognizing that Δ(PV) = Δ(nRT), we can rewrite as follows:
Equation 18.12
ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT)At constant temperature, Δ(nRT) = RTΔn, where Δn is the difference between the final and initial numbers of moles of gas. Thus
Equation 18.13
ΔE = ΔH − RTΔnFor reactions that result in a net production of gas, Δn > 0, so ΔE < ΔH. Conversely, endothermic reactions (ΔH > 0) that result in a net consumption of gas have Δn < 0 and ΔE > ΔH. The relationship between ΔH and ΔE for systems involving gases is illustrated in Example 4.
For reactions that result in a net production of gas, ΔE < ΔH. For endothermic reactions that result in a net consumption of gas, ΔE > ΔH.
The combustion of graphite to produce carbon dioxide is described by the equation C (graphite, s) + O2(g) → CO2(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔE for the reaction?
Given: balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant
Asked for: ΔE
Strategy:
A Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.
B Substitute this value and the data given into to obtain ΔE.
Solution:
A In this reaction, 1 mol of gas (CO2) is produced, and 1 mol of gas (O2) is consumed. Thus Δn = 1 − 1 = 0.
B Substituting this calculated value and the given values into ,
To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite.
Exercise
Calculate ΔE for the conversion of oxygen gas to ozone at 298 K: 3O2(g) → 2O3(g). The value of ΔH for the reaction is 285.4 kJ.
Answer: 288 kJ
As the exercise in Example 4 illustrates, the magnitudes of ΔH and ΔE for reactions that involve gases are generally rather similar, even when there is a net production or consumption of gases.
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (E) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔE.
Internal energy change
Enthalpy change
Relationship between Δ H and Δ E for an ideal gas
Describe how a swinging pendulum that slows with time illustrates the first law of thermodynamics.
When air is pumped into a bicycle tire, the air is compressed. Assuming that the volume is constant, express the change in internal energy in terms of q and w.
What is the relationship between enthalpy and internal energy for a reaction that occurs at constant pressure?
An intrepid scientist placed an unknown salt in a small amount of water. All the salt dissolved in the water, and the temperature of the solution dropped several degrees.
For years, chemists and physicists focused on enthalpy changes as a way to measure the spontaneity of a reaction. What arguments would you use to convince them not to use this method?
What is the relationship between enthalpy and internal energy for a reaction that occurs at constant volume?
The enthalpy of combustion (ΔHcomb) is defined thermodynamically as the enthalpy change for complete oxidation. The complete oxidation of hydrocarbons is represented by the following general equation: hydrocarbon + O2(g) → CO2(g) + H2O(g). Enthalpies of combustion from reactions like this one can be measured experimentally with a high degree of precision. It has been found that the less stable the reactant, the more heat is evolved, so the more negative the value of ΔHcomb. In each pair of hydrocarbons, which member do you expect to have the greater (more negative) heat of combustion? Justify your answers.
Using a structural argument, explain why the trans isomer of 2-butene is more stable than the cis isomer. The enthalpies of formation of cis- and trans-2-butene are −7.1 kJ/mol and −11.4 kJ/mol, respectively.
Using structural arguments, explain why cyclopropane has a positive (12.7 kJ/mol), whereas cyclopentane has a negative (−18.4 kJ/mol). (Hint: consider bond angles.)
At constant pressure, ΔH = ΔE + PΔV.
With bond angles of 60°, cyclopropane is highly strained, causing it to be less stable than cyclopentane, which has nearly ideal tetrahedral geometry at each carbon atom.
A block of CO2 weighing 15 g evaporates in a 5.0 L container at 25°C. How much work has been done if the gas is allowed to expand against an external pressure of 0.98 atm under isothermal conditions? The enthalpy of sublimation of CO2 is 25.1 kJ/mol. What is the change in internal energy (kJ/mol) for the sublimation of CO2 under these conditions?
Zinc and HCl react according to the following equation:
Zn(s) + 2HCl(aq) → Zn2+(aq) + 2Cl−(aq) + H2(g)When 3.00 g of zinc metal is added to a dilute HCl solution at 1.00 atm and 25°C, and this reaction is allowed to go to completion at constant pressure, 6.99 kJ of heat must be removed to return the final solution to its original temperature. What are the values of q and w, and what is the change in internal energy?
Acetylene torches, used industrially to cut and weld metals, reach flame temperatures as high as 3000°C. The combustion reaction is as follows:
Calculate the amount of work done against a pressure of 1.0 atm when 4.0 mol of acetylene are allowed to react with 10 mol of O2 at 1.0 atm at 20°C. What is the change in internal energy for the reaction?
When iron dissolves in 1.00 M aqueous HCl, the products are FeCl2(aq) and hydrogen gas. Calculate the work done if 30 g of Fe react with excess hydrochloric acid in a closed vessel at 20°C. How much work is done if the reaction takes place in an open vessel with an external pressure of 1.0 atm?
−350 J; 8.2 kJ
The first law of thermodynamics governs changes in the state function we have called internal energy (E). According to , changes in the internal energy (ΔE) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned in that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔHsoln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in .
Figure 18.6 An Endothermic Reaction
The reaction of barium hydroxide with ammonium thiocyanate is spontaneous but highly endothermic, so water, one product of the reaction, quickly freezes into slush. When water is placed on a block of wood under the flask, the highly endothermic reaction that takes place in the flask freezes water that has been placed under the beaker, so the flask becomes frozen to the wood.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S)The degree of disorder in a thermodynamic system, which is directly proportional to the possible number of microstates., a thermodynamic property of all substances that is proportional to their degree of disorder. In , we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (ΔS > 0) or a decrease in entropy (ΔS < 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: ΔS = Sf − Si.
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
We can illustrate the concepts of microstates and entropy using a deck of playing cards, as shown in . In any new deck, the 52 cards are arranged by four suits, with each suit arranged in descending order. If the cards are shuffled, however, there are approximately 1068 different ways they might be arranged, which corresponds to 1068 different microscopic states. The entropy of an ordered new deck of cards is therefore low, whereas the entropy of a randomly shuffled deck is high. Card games assign a higher value to a hand that has a low degree of disorder. In games such as five-card poker, only 4 of the 2,598,960 different possible hands, or microstates, contain the highly ordered and valued arrangement of cards called a royal flush, almost 1.1 million hands contain one pair, and more than 1.3 million hands are completely disordered and therefore have no value. Because the last two arrangements are far more probable than the first, the value of a poker hand is inversely proportional to its entropy.
Figure 18.7 Illustrating Low- and High-Entropy States with a Deck of Playing Cards
An new, unshuffled deck (top) has only a single arrangement, so there is only one microstate. In contrast, a randomly shuffled deck (bottom) can have any one of approximately 1068 different arrangements, which correspond to 1068 different microstates.
We can see how to calculate these kinds of probabilities for a chemical system by considering the possible arrangements of a sample of four gas molecules in a two-bulb container (). There are five possible arrangements: all four molecules in the left bulb (I); three molecules in the left bulb and one in the right bulb (II); two molecules in each bulb (III); one molecule in the left bulb and three molecules in the right bulb (IV); and four molecules in the right bulb (V). If we assign a different color to each molecule to keep track of it for this discussion (remember, however, that in reality the molecules are indistinguishable from one another), we can see that there are 16 different ways the four molecules can be distributed in the bulbs, each corresponding to a particular microstate. As shown in , arrangement I is associated with a single microstate, as is arrangement V, so each arrangement has a probability of 1/16. Arrangements II and IV each have a probability of 4/16 because each can exist in four microstates. Similarly, six different microstates can occur as arrangement III, making the probability of this arrangement 6/16. Thus the arrangement that we would expect to encounter, with half the gas molecules in each bulb, is the most probable arrangement. The others are not impossible but simply less likely.
Figure 18.8 The Possible Microstates for a Sample of Four Gas Molecules in Two Bulbs of Equal Volume
There are 16 different ways to distribute four gas molecules between the bulbs, with each distribution corresponding to a particular microstate. Arrangements I and V each produce a single microstate with a probability of 1/16. This particular arrangement is so improbable that it is likely not observed. Arrangements II and IV each produce four microstates, with a probability of 4/16. Arrangement III, with half the gas molecules in each bulb, has a probability of 6/16. It is the one encompassing the most microstates, so it is the most probable.
Instead of four molecules of gas, let’s now consider 1 L of an ideal gas at standard temperature and pressure (STP), which contains 2.69 × 1022 molecules (6.022 × 1023 molecules/22.4 L). If we allow the sample of gas to expand into a second 1 L container, the probability of finding all 2.69 × 1022 molecules in one container and none in the other at any given time is extremely small, approximately The probability of such an occurrence is effectively zero. Although nothing prevents the molecules in the gas sample from occupying only one of the two bulbs, that particular arrangement is so improbable that it is never actually observed. The probability of arrangements with essentially equal numbers of molecules in each bulb is quite high, however, because there are many equivalent microstates in which the molecules are distributed equally. Hence a macroscopic sample of a gas occupies all of the space available to it, simply because this is the most probable arrangement.
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know from , , and , a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol·K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol·K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in , the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl− ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔSsoln > 0.
Figure 18.9 The Effect of Solution Formation on Entropy
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Predict which substance in each pair has the higher entropy and justify your answer.
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
Exercise
Predict which substance in each pair has the higher entropy and justify your answer.
Answer:
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible processA process in which every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change., every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible processA process in which the intermediate states between the extremes are not equilibrium states, so change occurs spontaneously in only one direction. is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (Pext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = PextΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev. Whether a process is reversible or irreversible, ΔE = q + w. Because E is a state function, the magnitude of ΔE does not depend on reversibility and is independent of the path taken. So
Equation 18.14
ΔE = qrev + wrev = qirrev + wirrevWork done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev.
In other words, ΔE for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (qrev) to define entropy quantitatively.
Because the quantity of heat transferred (qrev) is directly proportional to the absolute temperature of an object (T) (qrev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ qrev). Combining these relationships for any reversible process,
Equation 18.15
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is wrev = −PΔV, we can express as follows:
Equation 18.16
ΔE = qrev + wrev = TΔS − PΔVThus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the PV work done.
To illustrate the use of and , we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its T constant (). The internal energy of the gas does not change because the temperature of the gas does not change; that is, ΔE = 0 and qrev = −wrev. During expansion, ΔV > 0, so the gas performs work on its surroundings: wrev = −PΔV < 0. According to , this means that qrev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore ΔSsys = +qrev/T, and the entropy change of the surroundings is ΔSsurr = −qrev/T. The corresponding change in entropy of the universe is then as follows:
Equation 18.17
Thus no change in ΔSuniv has occurred.
Figure 18.10 Expansion of Gas at Constant Temperature
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in . The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From , we see that the entropy of fusion of ice can be written as follows:
Equation 18.18
Figure 18.11 Thermograms Showing That Heat Is Absorbed from the Surroundings When Ice Melts at 0°C
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case, ΔSfus = (6.01 kJ/mol)/(273 K) = 22.0 J/(mol·K) = ΔSsys. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so ΔSsurr = qrev/T = −(6.01 kJ/mol)/(273 K) = −22.0 J/(mol·K). Once again, we see that the entropy of the universe does not change:
ΔSuniv = ΔSsys + ΔSsurr = 22.0 J/(mol·K) − 22.0 J/(mol·K) = 0In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamicsThe entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The entropy of the universe increases during a spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔScold = q/Tcold. Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔShot = −q/Thot, where Tcold and Thot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
Equation 18.19
The numerators on the right side of are the same in magnitude but opposite in sign. Whether ΔSuniv is positive or negative depends on the relative magnitudes of the denominators. By definition, Thot > Tcold, so −q/Thot must be less than q/Tcold, and ΔSuniv must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔSuniv is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔSuniv approaches zero will not occur spontaneously as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam ().
Figure 18.12 Spontaneous Transfer of Heat from a Hot Substance to a Cold Substance
When molten lava flows into cold ocean water, so much heat is spontaneously transferred to the water that steam is produced.
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy:
Use to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution:
We know from that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔE are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
Exercise
Elemental sulfur exists in two forms: an orthorhombic form (Sα), which is stable below 95.3°C, and a monoclinic form (Sβ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.
Answer:
A measure of the disorder of a system is its entropy (S), a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.
A Russian space vehicle developed a leak, which resulted in an internal pressure drop from 1 atm to 0.85 atm. Is this an example of a reversible expansion? Has work been done?
Which member of each pair do you expect to have a higher entropy? Why?
Determine whether each process is reversible or irreversible.
Determine whether each process is reversible or irreversible.
Explain why increasing the temperature of a gas increases its entropy. What effect does this have on the internal energy of the gas?
For a series of related compounds, does ΔSvap increase or decrease with an increase in the strength of intermolecular interactions in the liquid state? Why?
Is the change in the enthalpy of reaction or the change in entropy of reaction more sensitive to changes in temperature? Explain your reasoning.
Solid potassium chloride has a highly ordered lattice structure. Do you expect ΔSsoln to be greater or less than zero? Why? What opposing factors must be considered in making your prediction?
Aniline (C6H5NH2) is an oily liquid at 25°C that darkens on exposure to air and light. It is used in dying fabrics and in staining wood black. One gram of aniline dissolves in 28.6 mL of water, but aniline is completely miscible with ethanol. Do you expect ΔSsoln in H2O to be greater than, less than, or equal to ΔSsoln in CH3CH2OH? Why?
No, it is irreversible; no work is done because the external pressure is effectively zero.
Water has a highly ordered, hydrogen-bonded structure that must reorganize to accommodate hydrophobic solutes like aniline. In contrast, we expect that aniline will be able to disperse randomly throughout ethanol, which has a significantly less ordered structure. We therefore predict that ΔSsoln in ethanol will be more positive than ΔSsoln in water.
Liquid nitrogen, which has a boiling point of −195.79°C, is used as a coolant and as a preservative for biological tissues. Is the entropy of nitrogen higher or lower at −200°C than at −190°C? Explain your answer. Liquid nitrogen freezes to a white solid at −210.00°C, with an enthalpy of fusion of 0.71 kJ/mol. What is its entropy of fusion? Is freezing biological tissue in liquid nitrogen an example of a reversible process or an irreversible process?
Using the second law of thermodynamics, explain why heat flows from a hot body to a cold body but not from a cold body to a hot body.
One test of the spontaneity of a reaction is whether the entropy of the universe increases: ΔSuniv > 0. Using an entropic argument, show that the following reaction is spontaneous at 25°C:
4Fe(s) + 3O2(g) → 2Fe2O3(s)Why does the entropy of the universe increase in this reaction even though gaseous molecules, which have a high entropy, are consumed?
Calculate the missing data in the following table.
Compound | ΔHfus (kJ/mol) | ΔSfus [J/(mol·K)] | Melting Point (°C) |
---|---|---|---|
acetic acid | 11.7 | 16.6 | |
CH3CN | 8.2 | 35.9 | |
CH4 | 0.94 | −182.5 | |
CH3OH | 18.2 | −97.7 | |
formic acid | 12.7 | 45.1 |
Based on this table, can you conclude that entropy is related to the nature of functional groups? Explain your reasoning.
Calculate the missing data in the following table.
Compound | ΔHvap (kJ/mol) | ΔSvap [J/(mol·K)] | Boiling Point (°C) |
---|---|---|---|
hexanoic acid | 71.1 | 105.7 | |
hexane | 28.9 | 85.5 | |
formic acid | 60.7 | 100.8 | |
1-hexanol | 44.5 | 157.5 |
The text states that the magnitude of ΔSvap tends to be similar for a wide variety of compounds. Based on the values in the table, do you agree?
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion. Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamicsThe entropy of any perfectly ordered, crystalline substance at absolute zero is zero.: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Figure 18.13 Molecular Motions
Vibrational, rotational, and translational motions of a carbon dioxide molecule are illustrated here. Only a perfectly ordered, crystalline substance at absolute zero would exhibit no molecular motion and have zero entropy. In practice, this is an unattainable ideal.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (Cp) as a function of temperature and then plotting the quantity Cp/T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those we first encountered in .
One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°)The entropy of 1 mol of a substance at a standard temperature of 298 K., which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol·K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol·K)] and 298 K.
As shown in , for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol·K), whereas S° for water vapor is 188.8 J/(mol·K). Likewise, S° is 260.7 J/(mol·K) for gaseous I2 and 116.1 J/(mol·K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in , which is a generalized plot of the entropy of a substance versus temperature.
Table 18.1 Standard Molar Entropy Values of Selected Substances at 25°C
Substance | S° [J/(mol·K)] |
---|---|
Gases | |
He | 126.2 |
H2 | 130.7 |
Ne | 146.3 |
Ar | 154.8 |
Kr | 164.1 |
Xe | 169.7 |
H2O | 188.8 |
N2 | 191.6 |
O2 | 205.2 |
CO2 | 213.8 |
I2 | 260.7 |
Liquids | |
H2O | 70.0 |
CH3OH | 126.8 |
Br2 | 152.2 |
CH3CH2OH | 160.7 |
C6H6 | 173.4 |
CH3COCl | 200.8 |
C6H12 (cyclohexane) | 204.4 |
C8H18 (isooctane) | 329.3 |
Solids | |
C (diamond) | 2.4 |
C (graphite) | 5.7 |
LiF | 35.7 |
SiO2 (quartz) | 41.5 |
Ca | 41.6 |
Na | 51.3 |
MgF2 | 57.2 |
K | 64.7 |
NaCl | 72.1 |
KCl | 82.6 |
I2 | 116.1 |
Figure 18.14 A Generalized Plot of Entropy versus Temperature for a Single Substance
Absolute entropy increases steadily with increasing temperature until the melting point is reached, where it jumps suddenly as the substance undergoes a phase change from a highly ordered solid to a disordered liquid (ΔSfus). The entropy again increases steadily with increasing temperature until the boiling point is reached, where it jumps suddenly as the liquid undergoes a phase change to a highly disordered gas (ΔSvap).
A closer examination of also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol·K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol·K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure.
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example 7 illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
Use the data in to calculate ΔS° for the reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in . Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise
Use the data in to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer: −361.1 J/K
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
We can also calculate a change in entropy using a thermodynamic cycle. As you learned in , the molar heat capacity (Cp) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, Cv is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in is approximately proportional to the heat capacity of the substance.
Recall that the entropy change (ΔS) is related to heat flow (qrev) by ΔS = qrev/T. Because qrev = nCpΔT at constant pressure or nCvΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T1 to T2 is as follows:
As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in .
Equation 18.20
Similarly,
Equation 18.21
Thus we can use a combination of heat capacity measurements ( or ) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved () to calculate the entropy change corresponding to a change in the temperature of a sample.
We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in ): an orthorhombic form with a highly ordered structure (Sα) and a less-ordered monoclinic form (Sβ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from Sα to Sβ can be described by the thermodynamic cycle shown in part (b) in , in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to Sβ (−ΔSfus(β) = ΔS3 in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS3 can be calculated from the overall entropy change (ΔSt) for the Sα–Sβ transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using and tabulated thermodynamic parameters (the heat capacities of Sα and Sβ, ΔHfus(α), and the melting point of Sα.)
Figure 18.15 Two Forms of Elemental Sulfur and a Thermodynamic Cycle Showing the Transition from One to the Other
(a) Orthorhombic sulfur (Sα) has a highly ordered structure in which the S8 rings are stacked in a “crankshaft” arrangement. Monoclinic sulfur (Sβ) is also composed of S8 rings but has a less-ordered structure. (b) At 368.5 K, Sα undergoes a phase transition to Sβ. Although ΔS3 cannot be measured directly, it can be calculated using the values shown in this thermodynamic cycle.
If we know the melting point of Sα (Tm = 115.2°C = 388.4 K) and ΔSt for the overall phase transition [calculated to be 1.09 J/(mol·K) in the exercise in Example 6], we can calculate ΔS3 from the values given in part (b) in where Cp(α) = 22.70 J/mol·K and Cp(β) = 24.77 J/mol·K (subscripts on ΔS refer to steps in the cycle):
Solving for ΔS3 gives a value of −3.24 J/(mol·K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS3 is negative.
The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products.
Temperature dependence of entropy at constant pressure
Temperature dependence of entropy at constant volume
Crystalline MgCl2 has S° = 89.63 J/(mol·K), whereas aqueous MgCl2 has S° = −25.1 J/(mol·K). Is this consistent with the third law of thermodynamics? Explain your answer.
Why is it possible to measure absolute entropies but not absolute enthalpies?
How many microstates are available to a system at absolute zero? How many are available to a substance in its liquid state?
Substance A has a higher heat capacity than substance B. Do you expect the absolute entropy of substance A to be less than, similar to, or greater than that of substance B? Why? As the two substances are heated, for which substance do you predict the entropy to increase more rapidly?
Phase transitions must be considered when calculating entropy changes. Why?
What is the final temperature of water when 5.20 g of ice at 0.0°C are added to 250 mL of water in an insulated thermos at 30.0°C? The value of ΔHfus for water is 6.01 kJ/mol, and the heat capacity of liquid water is 75.3 J/(mol·°C). What is the entropy change for this process?
Calculate the change in both enthalpy and entropy when a 3.0 g block of ice melts at 0.0°C [ΔHfus(H2O) = 6.01 kJ/mol]. For the same block of ice, calculate the entropy change for the system when the ice is warmed from 0.0°C to 25°C. The heat capacity of liquid water over this temperature range is 75.3 J/(mol·°C).
Calculate the entropy change (J/K) when 4.35 g of liquid bromine are heated from 30.0°C to 50.0°C if the molar heat capacity (Cp) of liquid bromine is 75.1 kJ/(mol·K).
Calculate the molar heat capacity (Cp) of titanium tetrachloride if the change in entropy when a 6.00 g sample of TiCl4(l) is heated from 25.0°C to 40.0°C is 0.154 J/K.
When a 1.00 g sample of lead is heated from 298.2 K to just below its melting temperature of 600.5 K, the change in entropy is 0.0891 J/K. Determine the molar heat capacity (Cp) of lead over this temperature range.
Phosphorus oxychloride (POCl3) is a chlorinating agent that is frequently used in organic chemistry to replace oxygen with chlorine. Given ΔSvap = 93.08 J/(mol·K) and ΔHvap = 35.2 kJ/mol, does POCl3 spontaneously convert from a liquid to a gas at 110°C? Does it spontaneously crystallize at 0.0°C if ΔHfus = 34.3 kJ/mol and ΔSfus = 125 J/(mol·K)? Using the information provided, what is the melting point of POCl3?
A useful reagent for the fluorination of alcohols, carboxylic acids, and carbonyl compounds is selenium tetrafluoride (SeF4). One must be careful when using this compound, however, because it is known to attack glass (such as the glass of a reaction vessel).
27.8°C; 0.85 J.
25.0 J/(mol·K)
yes; yes; 274 K
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
The Gibbs free energy (G)A state function that is defined in terms of three other state functions—namely, enthalpy entropy and temperature , often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
Equation 18.22
G = H − TSBecause it is a combination of state functions, G is also a state function.
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Châtelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
The criterion for predicting spontaneity is based on ΔG, the change in G, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as G. At constant temperature and pressure,
Equation 18.23
ΔG = ΔH − TΔSwhere all thermodynamic quantities are those of the system. Recall that at constant pressure, ΔH = q, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce to ΔG = q − qrev. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if ΔG is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite (where qp = ΔH, ) as follows:
Equation 18.24
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, ΔSuniv > 0, so substituting we obtain
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
which is equal to ΔG (). We can therefore see that for a spontaneous process, ΔG < 0.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes (ΔH << 0) that increase the disorder of a system (ΔSsys >> 0) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. (This substance destroyed Texas City, Texas, in 1947; see , .) Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, ΔH = 40,657 J, so the process is highly endothermic. From the definition of ΔS (), we know that for 1 mol of water,
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions. (For more information on standard conditions, see .)
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
At 110°C, ΔG < 0, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate ΔG for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict. (For more information on using a calculator, see Essential Skills 1 in .)
ΔG = 0 only if ΔH = TΔS.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (), setting ΔG = 0, and solving for T,
Thus ΔG = 0 at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, ΔG is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. shows how the ΔH and TΔS terms vary with temperature for the vaporization of water. When the two lines cross, ΔG = 0, and ΔH = TΔS.
Figure 18.16 Temperature Dependence of ΔH and TΔS for the Vaporization of Water
Both ΔH and TΔS are temperature dependent, but the lines have opposite slopes and cross at 373.15 K at 1 atm, where ΔH = TΔS. Because ΔG = ΔH − TΔS, at this temperature ΔG = 0, indicating that the liquid and vapor phases are in equilibrium. The normal boiling point of water is therefore 373.15 K. Above the normal boiling point, the TΔS term is greater than ΔH, making ΔG < 0; hence, liquid water evaporates spontaneously. Below the normal boiling point, the ΔH term is greater than TΔS, making ΔG > 0. Thus liquid water does not evaporate spontaneously, but water vapor spontaneously condenses to liquid.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in . We can rearrange this equation as follows:
Equation 18.25
ΔH = ΔG + TΔSThis equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see .) Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in , the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table 18.2 Approximate Thermodynamic Efficiencies of Various Devices
Device | Energy Conversion | Approximate Efficiency (%) |
---|---|---|
large electrical generator | mechanical → electrical | 99 |
chemical battery | chemical → electrical | 90 |
home furnace | chemical → heat | 65 |
small electric tool | electrical → mechanical | 60 |
space shuttle engine | chemical → mechanical | 50 |
mammalian liver cell | chemical → chemical | 30–50 |
spinach cell | light → chemical | 30 |
internal combustion engine | chemical → mechanical | 25–30 |
fluorescent light | electrical → light | 20 |
solar cell | light → electricity | 10 |
incandescent light bulb | electricity → light | 5 |
yeast cell | chemical → chemical | 2–4 |
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°)The change in free energy when one substance or a set of substances in their standard states is converted to one or more other sustances, also in their standard states: is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using :
Equation 18.26
ΔG° = ΔH° − TΔS°If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are S°(H2O2) = 109.6 J/(mol·K), S°(O2) = 205.2 J/(mol·K), and S°(H2) = 130.7 J/(mol·K). Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
A Calculate ΔS° from the absolute molar entropy values given.
B Use , the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution:
A To calculate ΔG° for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.
B Substituting the appropriate quantities into ,
The negative value of ΔG° indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.
Exercise
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N2H4) = 121.2 J/(mol·K), S°(N2) = 191.6 J/(mol·K), and S°(H2) = 130.7 J/(mol·K). Is the reaction spontaneous as written?
Answer: 149.5 kJ/mol; no
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation The change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states: of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has = 0. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation and the standard entropy of formation using the definition of free energy:
Equation 18.27
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
Equation 18.28
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction. (For more information on chemical kinetics, see .)
The ΔG° of a reaction can be calculated from tabulated values using the “products minus reactants” rule.
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
(isooctane) = −353.2 kJ/mol,
(CO2) = −394.4 kJ/mol, and
(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain , remembering that for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution:
From Example 7, we know that the balanced chemical equation for the reaction is as follows: We are given values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, (O2) is zero. Using the “products minus reactants” rule,
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the of the stable products: water and carbon dioxide.
Exercise
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the data
(benzene) = 124.5 kJ/mol and
(cyclohexane) = 217.3 kJ/mol. Is the reaction spontaneous as written?
Answer: 92.8 kJ; no
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither ΔH nor ΔS vary greatly with temperature.
Calculate (a) ΔG° and (b) ΔG300°C for the reaction assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
S°(N2) = 191.6 J/(mol·K),
S°(H2) = 130.7 J/(mol·K),
S°(NH3) = 192.8 J/(mol·K), and
(NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
A Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that for any element in its standard state is zero.
B Substitute the appropriate values into to obtain ΔG° for the reaction.
C Assuming that ΔH and ΔS are independent of temperature, substitute values into to obtain ΔG for the reaction at 300°C.
Solution:
A To calculate ΔG° for the reaction using , we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of (NH3) is given, and is zero for both N2 and H2:
B Inserting the appropriate values into ,
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into :
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise
Calculate (a) ΔG° and (b) ΔG750°C for the reaction which is important in the formation of urban smog. Assume that ΔH and ΔS do not change between 25.0°C and 750°C and use these data:
S°(NO) = 210.8 J/(mol·K),
S°(O2) = 205.2 J/(mol·K),
S°(NO2) = 240.1 J/(mol·K),
(NO2) = 33.2 kJ/mol, and
(NO) = 91.3 kJ/mol.
Answer
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in , as illustrated in Example 11.
As you saw in Example 10, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 10 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution:
In Example 10, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging to solve for T:
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise
As you found in the exercise in Example 10, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer: 792.6 K
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction.
How does each example illustrate the fact that no process is 100% efficient?
Neither the change in enthalpy nor the change in entropy is, by itself, sufficient to determine whether a reaction will occur spontaneously. Why?
If a system is at equilibrium, what must be the relationship between ΔH and ΔS?
The equilibrium is exothermic in the forward direction. Which has the higher entropy—the products or the reactants? Why? Which is favored at high temperatures?
Is ΔG a state function that describes a system or its surroundings? Do its components—ΔH and ΔS—describe a system or its surroundings?
How can you use ΔG to determine the temperature of a phase transition, such as the boiling point of a liquid or the melting point of a solid?
Occasionally, an inventor claims to have invented a “perpetual motion” machine, which requires no additional input of energy once the machine has been put into motion. Using your knowledge of thermodynamics, how would you respond to such a claim? Justify your arguments.
Must the entropy of the universe increase in a spontaneous process? If not, why is no process 100% efficient?
The reaction of methyl chloride with water produces methanol and hydrogen chloride gas at room temperature, despite the fact that = 7.3 kcal/mol. Using thermodynamic arguments, propose an explanation as to why methanol forms.
In order for the reaction to occur spontaneously, ΔG for the reaction must be less than zero. In this case, ΔS must be positive, and the TΔS term outweighs the positive value of ΔH.
Use the tables in the text to determine whether each reaction is spontaneous under standard conditions. If a reaction is not spontaneous, write the corresponding spontaneous reaction.
Use the tables in the text to determine whether each reaction is spontaneous under standard conditions. If a reaction is not spontaneous, write the corresponding spontaneous reaction.
Nitrogen fixation is the process by which nitrogen in the atmosphere is reduced to NH3 for use by organisms. Several reactions are associated with this process; three are listed in the following table. Which of these are spontaneous at 25°C? If a reaction is not spontaneous, at what temperature does it become spontaneous?
Reaction | (kcal/mol) | [cal/(°·mol)] |
---|---|---|
(a) | 8.0 | −14.4 |
(b) | 21.6 | 2.9 |
(c) | −11.0 | −23.7 |
A student was asked to propose three reactions for the oxidation of carbon or a carbon compound to CO or CO2. The reactions are listed in the following table. Are any of these reactions spontaneous at 25°C? If a reaction does not occur spontaneously at 25°C, at what temperature does it become spontaneous?
Reaction | (kcal/mol) | [cal/(°·mol)] |
---|---|---|
C(s) + H2O(g) → CO(g) + H2(g) | 42 | 32 |
CO(g) + H2O(g) → CO2(g) + H2(g) | −9.8 | −10.1 |
CH4(g) + H2O(g) → CO(g) + 3H2(g) | 49.3 | 51.3 |
Tungsten trioxide (WO3) is a dense yellow powder that, because of its bright color, is used as a pigment in oil paints and water colors (although cadmium yellow is more commonly used in artists’ paints). Tungsten metal can be isolated by the reaction of WO3 with H2 at 1100°C according to the equation WO3(s) + 3H2(g) → W(s) + 3H2O(g). What is the lowest temperature at which the reaction occurs spontaneously? ΔH° = 27.4 kJ/mol and ΔS° = 29.8 J/K.
Sulfur trioxide (SO3) is produced in large quantities in the industrial synthesis of sulfuric acid. Sulfur dioxide is converted to sulfur trioxide by reaction with oxygen gas.
Calculate ΔG° for the general reaction MCO3(s) → MO(s) + CO2(g) at 25°C, where M is Mg or Ba. At what temperature does each of these reactions become spontaneous?
Compound | (kJ/mol) | S° [J/(mol·K)] |
---|---|---|
MCO 3 | ||
Mg | −1111 | 65.85 |
Ba | −1213.0 | 112.1 |
MO | ||
Mg | −601.6 | 27.0 |
Ba | −548.0 | 72.1 |
CO 2 | −393.5 | 213.8 |
The reaction of aqueous solutions of barium nitrate with sodium iodide is described by the following equation:
Ba(NO3)2(aq) + 2NaI(aq) → BaI2(aq) + 2NaNO3(aq)You want to determine the absolute entropy of BaI2, but that information is not listed in your tables. However, you have been able to obtain the following information:
Ba(NO3)2 | NaI | BaI2 | NaNO3 | |
---|---|---|---|---|
(kJ/mol) | −952.36 | −295.31 | −605.4 | −447.5 |
S° [J/(mol·K)] | 302.5 | 170.3 | 205.4 |
You know that ΔG° for the reaction at 25°C is 22.64 kJ/mol. What is ΔH° for this reaction? What is S° for BaI2?
919 K
MgCO3: ΔG° = 63 kJ/mol, spontaneous above 663 K; BaCO3: ΔG° = 220 kJ/mol, spontaneous above 1562 K
We have identified three criteria for whether a given reaction will occur spontaneously: ΔSuniv > 0, ΔGsys < 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. (For more information on the reaction quotient and the equilibrium constant, see .) Recall that if Q < K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q > K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. The relationship between ΔSuniv and ΔGsys was described in . In this section, we explore the relationship between the standard free energy of reaction (ΔG°) and the equilibrium constant (K).
Table 18.3 Criteria for the Spontaneity of a Process as Written
Spontaneous | Equilibrium | Nonspontaneous* |
---|---|---|
ΔSuniv > 0 | ΔSuniv = 0 | ΔSuniv < 0 |
ΔGsys < 0 | ΔGsys = 0 | ΔGsys > 0 |
Q < K | Q = K | Q > K |
*Spontaneous in the reverse direction. |
Because ΔH° and ΔS° determine the magnitude of ΔG° (), and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. As you learned in , ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. Using higher math, the general relationship can be shown as follows:
Equation 18.29
ΔG = VΔP − SΔTIf a reaction is carried out at constant temperature (ΔT = 0), then simplifies to
Equation 18.30
ΔG = VΔPUnder normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the V in by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express ΔG in terms of the initial and final pressures (Pi and Pf, respectively) as in :
Equation 18.31
If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
G − G° = nRTln PThis can be rearranged as follows:
Equation 18.32
G = G° + nRTln PAs you will soon discover, allows us to relate ΔG° and Kp. Any relationship that is true for Kp must also be true for K because Kp and K are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
Equation 18.33
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
Equation 18.34
Substituting for each term into ,
Combining terms gives the following relationship between ΔG and the reaction quotient Q:
Equation 18.35
where ΔG° indicates that all reactants and products are in their standard states. In , you learned that for gases Q = Kp at equilibrium, and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and Kp for gases as follows:
Equation 18.36
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then Kp = 1, and neither reactants nor products are favored: the system is at equilibrium.
For a spontaneous process under standard conditions, Keq and Kp are greater than 1.
In Example 10, we calculated that ΔG° = −32.7 kJ/mol of N2 for the reaction This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: = 2.00 atm, = 7.00 atm, = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
A Using the values given and , calculate Q.
B Substitute the values of ΔG° and Q into to obtain ΔG for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in . Substituting the partial pressures given, we can calculate Q:
B Substituting the values of ΔG° and Q into ,
Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise
Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, = 0.200 atm, and = 1.00 × 10−4 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer: −92.9 kJ/mol of O2; the reaction is spontaneous to the right as written, so products are favored.
Calculate Kp for the reaction of H2 with N2 to give NH3 at 25°C. As calculated in Example 10, ΔG° for this reaction is −32.7 kJ/mol of N2.
Given: balanced chemical equation from Example 10, ΔG°, and temperature
Asked for: K p
Strategy:
Substitute values for ΔG° and T (in kelvins) into to calculate Kp, the equilibrium constant for the formation of ammonia.
Solution:
In Example 10, we used tabulated values of to calculate ΔG° for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging ,
Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in , however, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O2.
Answer: 2.2 × 1012
Although Kp is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of Kp and K in and showed that they are related:
Equation 18.37
Kp = K(RT)Δnwhere Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so Kp = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in can be written in a more general form:
Equation 18.38
ΔG° = −RT ln KOnly when a reaction results in a net production or consumption of gases is it necessary to correct for the difference between Kp and K.Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining and provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
Equation 18.39
ΔG° = ΔH° − TΔS° = −RT ln KNotice that K becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible.
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in , which can be rearranged as follows:
Equation 18.40
Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in agrees with the qualitative predictions made by applying Le Châtelier’s principle, which we discussed in . Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K1 and K2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively. Applying gives the following relationship at each temperature:
Subtracting ln K1 from ln K2,
Equation 18.41
Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K1) allow us to calculate the value of the equilibrium constant at any other temperature (K2), assuming that ΔH° and ΔS° are independent of temperature.
The equilibrium constant for the formation of NH3 from H2 and N2 at 25°C was calculated to be Kp = 5.4 × 105 in Example 13. What is Kp at 500°C? (Use the data from Example 10.)
Given: balanced chemical equation, ΔH°, initial and final T, and Kp at 25°C
Asked for: Kp at 500°C
Strategy:
Convert the initial and final temperatures to kelvins. Then substitute appropriate values into to obtain K2, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from we obtain the following:
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise
In the exercise in Example 13, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the values in the exercise in Example 10 to calculate Kp for this reaction at 1000°C.
Answer: 5.6 × 10−4
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or Kp > 1, and products are favored over reactants. If ΔG° > 0, then K or Kp < 1, and reactants are favored over products. If ΔG° = 0, then K or Kp = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Relationship between standard free-energy change and equilibrium constant
Temperature dependence of equilibrium constant
Calculation of K at second temperature
Do you expect products or reactants to dominate at equilibrium in a reaction for which ΔG° is equal to
The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds?
What happens to the change in free energy of the reaction N2(g) + 3F2(g) → 2NF3(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids?
Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions?
Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO2 and H2, as shown in the equation for which ΔH° = −41.0 kJ/mol and ΔS° = −42.3 J cal/(mol·K) at 25°C and 1 atm.
Methane and water react to form carbon monoxide and hydrogen according to the equation
Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
The gas-phase decomposition of N2O4 to NO2 is an equilibrium reaction with Kp = 4.66 × 10−3. Calculate the standard free-energy change for the equilibrium reaction between N2O4 and NO2.
The standard free-energy change for the dissolution is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C?
Ammonia reacts with water in liquid ammonia solution (am) according to the equation The change in enthalpy for this reaction is 21 kJ/mol, and ΔS° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)?
At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba2+] = [CO32−] = 5.08 × 10−5 M. Determine ΔG° for the dissolution of BaCO3.
Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is for which log K = −1.80. What is ΔG° for this reaction?
The conversion of butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol.
The reaction of CaCO3(s) to produce CaO(s) and CO2(g) has an equilibrium constant at 25°C of 2 × 10−23. Values of are as follows: CaCO3, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO2, −393.5 kJ/mol.
In acidic soils, dissolved Al3+ undergoes a complex formation reaction with SO42− to form [AlSO4+]. The equilibrium constant at 25°C for the reaction is 1585.
13.3 kJ/mol
5.1 × 10−21
10.3 kJ/mol
Because thermodynamics deals with state functions, it can be used to describe the overall properties, behavior, and equilibrium composition of a system. It is not concerned with the particular pathway by which physical or chemical changes occur, however, so it cannot address the rate at which a particular process will occur. Although thermodynamics provides a significant constraint on what can occur during a reaction process, it does not describe the detailed steps of what actually occurs on an atomic or a molecular level.
Thermodynamics focuses on the energetics of the products and the reactants, whereas kinetics focuses on the pathway from reactants to products.
gives the numerical values of the equilibrium constant (K) that correspond to various approximate values of ΔG°. Note that ΔG° ≥ +10 kJ/mol or ΔG° ≤ −10 kJ/mol ensures that an equilibrium lies essentially all the way to the left or to the right, respectively, under standard conditions, corresponding to a reactant-to-product ratio of approximately 10,000:1 (or vice versa). Only if ΔG° is quite small (±10 kJ/mol) are significant amounts of both products and reactants present at equilibrium. Most reactions that we encounter have equilibrium constants substantially greater or less than 1, with the equilibrium strongly favoring either products or reactants. In many cases, we will encounter reactions that are strongly favored by thermodynamics but do not occur at a measurable rate. In contrast, we may encounter reactions that are not thermodynamically favored under standard conditions but nonetheless do occur under certain nonstandard conditions.
Table 18.4 The Relationship between K and ΔG° at 25°C
ΔG° (kJ/mol) | K | Physical Significance |
---|---|---|
500 | 3 × 10−88 | For all intents and purposes, the reaction does not proceed in the forward direction: only reactants are present at equilibrium. |
100 | 3 × 10−18 | |
10 | 2 × 10−2 | Both forward and reverse reactions occur: significant amounts of both products and reactants are present at equilibrium. |
0 | 1 | |
−10 | 60 | |
−100 | 3 × 1017 | For all intents and purposes, the forward reaction proceeds to completion: only products are present at equilibrium. |
−500 | 4 × 1087 |
A typical challenge in industrial processes is a reaction that has a large negative value of ΔG° and hence a large value of K but that is too slow to be practically useful. In such cases, mixing the reactants results in only a physical mixture, not a chemical reaction. An example is the reaction of carbon tetrachloride with water to produce carbon dioxide and hydrochloric acid, for which ΔG° is −232 kJ/mol:
Equation 18.42
The value of K for this reaction is 5 × 1040 at 25°C, yet when CCl4 and water are shaken vigorously at 25°C, nothing happens: the two immiscible liquids form separate layers, with the denser CCl4 on the bottom. In comparison, the analogous reaction of SiCl4 with water to give SiO2 and HCl, which has a similarly large equilibrium constant, occurs almost explosively. Although the two reactions have comparable thermodynamics, they have very different kinetics!
There are also many reactions for which ΔG° << 0 but that do not occur as written because another possible reaction occurs more rapidly. For example, consider the reaction of lead sulfide with hydrogen peroxide. One possible reaction is as follows:
Equation 18.43
for which ΔG° is −886 kJ/mol and K is 10161. Yet when lead sulfide is mixed with hydrogen peroxide, the ensuing vigorous reaction does not produce PbO2 and SO2. Instead, the reaction that actually occurs is as follows:
Equation 18.44
This reaction has a ΔG° value of −1181 kJ/mol, within the same order of magnitude as the reaction in , but it occurs much more rapidly.
Now consider reactions with ΔG° > 0. Thermodynamically, such reactions do not occur spontaneously under standard conditions. Nonetheless, these reactions can be made to occur under nonstandard conditions. An example is the reduction of chromium(III) chloride by hydrogen gas:
Equation 18.45
At 25°C, ΔG° = 35 kJ/mol and Kp = 7 × 10−7. However, at 730°C, ΔG° = −52 kJ/mol and Kp = 5 × 102; at this elevated temperature, the reaction is a convenient way of preparing chromium(II) chloride in the laboratory. Moreover, removing HCl gas from the system drives the reaction to completion, as predicted by Le Châtelier’s principle. Although the reaction is not thermodynamically spontaneous under standard conditions, it becomes spontaneous under nonstandard conditions.
There are also cases in which a compound whose formation appears to be thermodynamically prohibited can be prepared using a different reaction. The reaction for the preparation of chlorine monoxide from its component elements, for example, is as follows:
Equation 18.46
for which is 97.9 kJ/mol. The large positive value of for this reaction indicates that mixtures of chlorine and oxygen do not react to any extent to form Cl2O. Nonetheless, Cl2O is easily prepared using the reaction
Equation 18.47
which has a ΔG° of −22.2 kJ/mol and a Kp of approximately 1 × 104.
Finally, the ΔG° values for some reactions are so positive that the only way to make them proceed in the desired direction is to supply external energy, often in the form of electricity. Consider, for example, the formation of metallic lithium from molten lithium chloride:
Equation 18.48
Even at 1000°C, ΔG is very positive (324 kJ/mol), and there is no obvious way to obtain lithium metal using a different reaction. Hence in the industrial preparation of metallic lithium, electrical energy is used to drive the reaction to the right, as described in .
A reaction that does not occur under standard conditions can be made to occur under nonstandard conditions, such as by driving the reaction to completion using Le Châtelier’s principle or by providing external energy.
Often reactions that are not thermodynamically spontaneous under standard conditions can be made to occur spontaneously if coupled, or connected, in some way to another reaction for which ΔG° << 0. Because the overall value of ΔG° for a series of reactions is the sum of the ΔG° values for the individual reactions, virtually any unfavorable reaction can be made to occur by chemically coupling it to a sufficiently favorable reaction or reactions. In the preparation of chlorine monoxide from mercuric oxide and chlorine (), we have already encountered one example of this phenomenon of coupled reactions, although we did not describe it as such at the time. We can see how the chemical coupling works if we write as the sum of three separate reactions:
Comparing the ΔG° values for the three reactions shows that reaction 3 is so energetically favorable that it more than compensates for the other two energetically unfavorable reactions. Hence the overall reaction is indeed thermodynamically spontaneous as written.
By coupling reactions, a reaction that is thermodynamically nonspontaneous can be made spontaneous.
Bronze Age metallurgists were accomplished practical chemists who unknowingly used coupled reactions to obtain metals from their ores. Realizing that different ores of the same metal required different treatments, they heated copper oxide ore in the presence of charcoal (carbon) to obtain copper metal, whereas they pumped air into the reaction system if the ore was copper sulfide. Assume that a particular copper ore consists of pure cuprous oxide (Cu2O). Using the values given for each, calculate
Given: reactants and products, values for Cu2O and CO, and temperature
Asked for: ΔG° and Kp for the formation of metallic copper from Cu2O in the absence and presence of carbon
Strategy:
A Write the balanced equilibrium equation for each reaction. Using the “products minus reactants” rule, calculate ΔG° for the reaction.
B Substitute appropriate values into to obtain Kp.
Solution:
A The chemical equation for the decomposition of cuprous oxide is as follows:
The substances on the right side of this equation are pure elements in their standard states, so their values are zero. ΔG° for the reaction is therefore
B Rearranging and substituting the appropriate values into ,
This is a very small number, indicating that Cu2O does not spontaneously decompose to a significant extent at room temperature.
A The O2 produced in the decomposition of Cu2O can react with carbon to form CO:
Because ΔG° for this reaction is equal to for CO (−137.2 kJ/mol), it is energetically more feasible to produce metallic copper from cuprous oxide by coupling the two reactions:
B We can find the corresponding value of Kp:
Although this value is still less than 1, indicating that reactants are favored over products at room temperature, it is about 24 orders of magnitude greater than Kp for the production of copper metal in the absence of carbon. Because both ΔH° and ΔS° are positive for this reaction, it becomes thermodynamically feasible at slightly elevated temperatures (greater than about 80°C). At temperatures of a few hundred degrees Celsius, the reaction occurs spontaneously, proceeding smoothly and rapidly to the right as written and producing metallic copper and carbon monoxide from cuprous oxide and carbon.
Exercise
Use the values given to calculate ΔG° and Kp for each reaction.
Answer:
Thermodynamics describes the overall properties, behavior, and equilibrium composition of a system; kinetics describes the rate at which a particular process will occur and the pathway by which it will occur. Whereas thermodynamics tells us what can occur during a reaction process, kinetics tells us what actually occurs on an atomic or a molecular level. A reaction that is not thermodynamically spontaneous under standard conditions can often be made to occur spontaneously by varying the reaction conditions; using a different reaction to obtain the same product; supplying external energy, such as electricity; or coupling the unfavorable reaction to another reaction for which ΔG° << 0.
You are in charge of finding conditions to make the reaction A(l) + B(l) → C(l) + D(g) favorable because it is a critical step in the synthesis of your company’s key product. You have calculated that ΔG° for the reaction is negative, yet the ratio of products to reactants is very small. What have you overlooked in your scheme? What can you do to drive the reaction to increase your product yield?
In a thermodynamic sense, a living cell can be viewed as a low-entropy system that is not in equilibrium with its surroundings and is capable of replicating itself. A constant input of energy is needed to maintain the cell’s highly organized structure, its wide array of precisely folded biomolecules, and its intricate system of thousands of chemical reactions. A cell also needs energy to synthesize complex molecules from simple precursors (e.g., to make proteins from amino acids), create and maintain differences in the concentrations of various substances inside and outside the cell, and do mechanical work (e.g., muscle contraction). In this section, we examine the nature of the energy flow between a cell and its environment as well as some of the chemical strategies cells use to extract energy from their surroundings and store that energy.
One implication of the first and second laws of thermodynamics is that any closed system must eventually reach equilibrium. With no external input, a clock will run down, a battery will lose its charge, and a mixture of an aqueous acid and an aqueous base will achieve a uniform intermediate pH value. In contrast, a cell is an open system that can exchange matter with its surroundings as well as absorb energy from its environment in the form of heat or light. Cells use the energy obtained in these ways to maintain the nonequilibrium state that is essential for life.
Because cells are open systems, they cannot be described using the concepts of classical thermodynamics that we have discussed in this chapter, which have focused on reversible processes occurring in closed chemical systems that can exchange energy—but not matter—with their surroundings. Consequently, a relatively new subdiscipline called nonequilibrium thermodynamics has been developed to quantitatively describe open systems such as living cells.
Because a cell cannot violate the second law of thermodynamics, the only way it can maintain a low-entropy, nonequilibrium state characterized by a high degree of structural organization is to increase the entropy of its surroundings. A cell releases some of the energy that it obtains from its environment as heat that is transferred to its surroundings, thereby resulting in an increase in Ssurr (). As long as ΔSsurr is positive and greater than ΔSsys, the entropy of the universe increases, so the second law of thermodynamics is not violated. Releasing heat to the surroundings is necessary but not sufficient for life: the release of energy must be coupled to processes that increase the degree of order within a cell. For example, a wood fire releases heat to its surroundings, but unless energy from the burning wood is also used to do work, there is no increase in order of any portion of the universe.
Figure 18.17 Life and Entropy
A living cell is in a low-entropy, nonequilibrium state characterized by a high degree of structural organization. To maintain this state, a cell must release some of the energy it obtains from its environment as heat, thereby increasing Ssurr sufficiently that the second law of thermodynamics is not violated. In this example, the cell combines smaller components into larger, more ordered structures; the accompanying release of heat increases the entropy of the surrounding environment so that Suniv > 0.
Any organism in equilibrium with its environment is dead.
Although organisms employ a wide range of specific strategies to obtain the energy they need to live and reproduce, they can generally be divided into two categories: organisms are either phototrophs (from the Greek photos, meaning “light,” and trophos, meaning “feeder”), whose energy source is sunlight, or chemotrophs, whose energy source is chemical compounds, usually obtained by consuming or breaking down other organisms. Phototrophs, such as plants, algae, and photosynthetic bacteria, use the radiant energy of the sun directly, converting water and carbon dioxide to energy-rich organic compounds, whereas chemotrophs, such as animals, fungi, and many nonphotosynthetic bacteria, obtain energy-rich organic compounds from their environment. Regardless of the nature of their energy and carbon sources, all organisms use oxidation–reduction, or redox, reactions to drive the synthesis of complex biomolecules. Organisms that can use only O2 as the oxidant (a group that includes most animals) are aerobic organisms that cannot survive in the absence of O2. Many organisms that use other oxidants (such as SO42−, NO3−, or CO32−) or oxidized organic compounds can live only in the absence of O2, which is a deadly poison for them; such species are called anaerobic organisms.
The fundamental reaction by which all green plants and algae obtain energy from sunlight is photosynthesisThe fundamental reaction by which all green plants and algae obtain energy from sunlight in which is photochemically reduced to a carbon compound such as glucose. Oxygen in water is concurrently oxidized to , the photochemical reduction of CO2 to a carbon compound such as glucose. Concurrently, oxygen in water is oxidized to O2 (recall that hν is energy from light):
Equation 18.49
This reaction is not a spontaneous process as written, so energy from sunlight is used to drive the reaction. Photosynthesis is critical to life on Earth; it produces all the oxygen in our atmosphere.
In many ways, chemotrophs are more diverse than phototrophs because the nature of both the reductant (the nutrient) and the oxidant can vary. The most familiar chemotrophic strategy uses compounds such as glucose as the reductant and molecular oxygen as the oxidant in a process called respirationA process by which chemotrophs obtain energy from their environment; the overall reaction of respiration is the reverse of photosynthesis. Respiration is the combustion of a carbon compound such as glucose to and water.. (For more information on respiration, see .) The overall reaction of respiration is the reverse of photosynthesis:
Equation 18.50
An alternative strategy uses fermentationA process used by some chemotrophs to obtain energy from their environment; a chemical reaction in which both the oxidant and the reductant are organic compounds. reactions, in which an organic compound is simultaneously oxidized and reduced. Common examples are alcohol fermentation, used in making wine, beer, and bread, and lactic acid fermentation, used in making yogurt:
Equation 18.51
Equation 18.52
In these reactions, some of the carbon atoms of glucose are oxidized, while others are reduced. Recall that a reaction in which a single species is both oxidized and reduced is called a disproportionation reaction.
Regardless of the identity of the substances from which an organism obtains energy, the energy must be released in very small increments if it is to be useful to the cell. Otherwise, the temperature of the cell would rise to lethal levels. Cells store part of the energy that is released as ATP (adenosine triphosphate), a compound that is the universal energy currency of all living organisms ().
Figure 18.18 ATP, the Universal Energy Currency of All Cells
The ATP molecule contains a nitrogen-containing base, a sugar, and three phosphate groups, as well as two high-energy phosphoric acid anhydride bonds.
Most organisms use several intermediate species to shuttle electrons between the terminal reductant (such as glucose) and the terminal oxidant (such as O2). In virtually all cases, an intermediate species oxidizes the energy-rich reduced compound, and the now-reduced intermediate then migrates to another site where it is oxidized. The most important of these electron-carrying intermediates is NAD+ (nicotinamide adenine dinucleotide; ), whose reduced form, formally containing H−, is NADH (reduced nicotinamide adenine dinucleotide). The reduction of NAD+ to NADH can be written as follows:
Equation 18.53
NAD+ + H+ + 2e− → NADHFigure 18.19 NAD+ and Its Reduced Form (NADH)
This electron carrier is used by biological systems to transfer electrons from one species to another. The oxidized form (NAD+) is reduced to NADH by energy-rich nutrients such as glucose, and the reduced form (NADH), is then oxidized to NAD+ by O2 during respiration.
Most organisms use NAD+ to oxidize energy-rich nutrients such as glucose to CO2 and water; NADH is then oxidized to NAD+ using an oxidant such as O2. During oxidation, a fraction of the energy obtained from the oxidation of the nutrient is stored as ATP. The phosphoric acid anhydride bonds in ATP can then be hydrolyzed by water, releasing energy and forming ADP (adenosine diphosphate). It is through this sequence of reactions that energy from the oxidation of nutrients is made available to cells. Thus ATP has a central role in metabolism: it is synthesized by the oxidation of nutrients, and its energy is then used by cells to drive synthetic reactions and perform work ().
Figure 18.20 The ATP Cycle
The high-energy phosphoric acid anhydride bond in ATP stores energy released during the oxidation of nutrients. Hydrolysis of the high-energy bond in ATP releases energy, forming adenosine diphosphate (ADP) and phosphate.
Under standard conditions in biochemical reactions, all reactants are present in aqueous concentrations of 1 M at a pressure of 1 atm. For H+, this corresponds to a pH of zero, but very little biochemistry occurs at pH = 0. For biochemical reactions, chemists have therefore defined a new standard state in which the H+ concentration is 1 × 10−7 M (pH 7.0), and all other reactants and products are present in their usual standard-state conditions (1 M or 1 atm). The free-energy change and corresponding equilibrium constant for a reaction under these new standard conditions are denoted by the addition of a prime sign (′) to the conventional symbol: ΔG°′ and K′. If protons do not participate in a biological reaction, then ΔG°′ = ΔG°. Otherwise, the relationship between ΔG°′ and ΔG° is as follows:
Equation 18.54
ΔG°′ = ΔG° + RT ln(10−7)nwhere ΔG°′ and ΔG° are in kilojoules per mole and n is the number of protons produced in the reaction. At 298 K, this simplifies to
Equation 18.55
ΔG°′ = ΔG° − 39.96nThus any reaction that involves the release of protons is thermodynamically more favorable at pH 7 than at pH 0.
The chemical equation that corresponds to the hydrolysis of ATP to ADP and phosphate is as follows:
Equation 18.56
This reaction has a ΔG°′ of −34.54 kJ/mol, but under typical physiological (or biochemical) conditions, the actual value of ΔG′ for the hydrolysis of ATP is about −50 kJ/mol. Organisms use this energy to drive reactions that are energetically uphill, thereby coupling the reactions to the hydrolysis of ATP. One example is found in the biochemical pathway of glycolysis, in which the 6-carbon sugar glucose (C6H12O6) is split into two 3-carbon fragments that are then used as the fuel for the cell. Initially, a phosphate group is added to glucose to form a phosphate ester, glucose-6-phosphate (abbreviated glucose-6-P), in a reaction analogous to that of an alcohol and phosphoric acid:
Equation 18.57
Due to its electrical charge, the phosphate ester is unable to escape from the cell by diffusing back through the membrane surrounding the cell, ensuring that it remains available for further reactions. For the reaction in , ΔG° is 17.8 kJ/mol and K is 7.6 × 10−4, indicating that the equilibrium lies far to the left. To force this reaction to occur as written, it is coupled to a thermodynamically favorable reaction, the hydrolysis of ATP to ADP:
Thus the formation of glucose-6-phosphate is thermodynamically spontaneous if ATP is used as the source of phosphate.
Organisms use energy from the hydrolysis of ATP to drive reactions that are thermodynamically nonspontaneous.
The formation of glucose-6-phosphate is only one of many examples of how cells use ATP to drive an otherwise nonspontaneous biochemical reaction. Under nonstandard physiological conditions, each ATP hydrolyzed actually results in approximately a 108 increase in the magnitude of the equilibrium constant, compared with the equilibrium constant of the reaction in the absence of ATP. Thus a reaction in which two ATP molecules are converted to ADP increases K by about 1016, three ATP molecules by 1024, and so forth. Virtually any energetically unfavorable reaction or sequence of reactions can be made to occur spontaneously by coupling it to the hydrolysis of a sufficiently large number of ATP molecules.
Although all organisms use ATP as the immediate free-energy source in biochemical reactions, ATP is not an efficient form in which to store energy on a long-term basis. If the caloric intake of an average resting adult human were stored as ATP, two-thirds of the body weight would have to consist of ATP. Instead, a typical 70 kg adult human has a total of only about 50 g of both ATP and ADP, and, far from being used for long-term storage, each molecule of ATP is turned over about 860 times per day. The entire ATP supply would be exhausted in less than 2 minutes if it were not continuously regenerated.
How does the body store energy for the eventual production of ATP? Three primary means are as sugars, proteins, and fats. The combustion of sugars and proteins yields about 17 kJ of energy per gram, whereas the combustion of fats yields more than twice as much energy per gram, about 39 kJ/g. Moreover, sugars and proteins are hydrophilic and contain about 2 g of water per gram of fuel, even in very concentrated form. In contrast, fats are hydrophobic and can be stored in essentially anhydrous form. As a result, organisms can store about six times more energy per gram as fats than in any other form. A typical 70 kg adult human has about 170 kJ of energy in the form of glucose circulating in the blood, about 2600 kJ of energy stored in the muscles and liver as glycogen (a polymeric form of glucose), about 100,000 kJ stored in the form of protein (primarily muscle tissue), and almost 500,000 kJ in the form of fats (). Thus fats constitute by far the greatest energy reserve, while accounting for only about 12 kg of the 70 kg body mass. To store this amount of energy in the form of sugars would require a total body mass of about 144 kg, more than half of which would be sugar.
Figure 18.21 Percentages of Forms of Energy Storage in Adult Humans
An average 70 kg adult stores about 6 × 105 kJ of energy in glucose, glycogen, protein, and fats. Fats are the most abundant and most efficient form for storing energy.
Glucose is one form in which the body stores energy.
Given: balanced chemical equation (), values of , conversion efficiency, and ΔG°′ for hydrolysis of ATP
Asked for: ΔG°′ for the combustion reaction and the number of molecules of ATP that can be synthesized
Strategy:
A Using the “products minus reactants” rule, calculate for the respiration reaction.
B Multiply the calculated value of by the efficiency to obtain the number of kilojoules available for ATP synthesis. Then divide this value by ΔG°′ for the hydrolysis of ATP to find the maximum number of ATP molecules that can be synthesized.
Solution:
A Protons are not released or consumed in the reaction, so ΔG°′ = ΔG°. We begin by using the balanced chemical equation in :
C6H12O6 + 6O2 → 6CO2 + 6H2OFrom the given values of (remember that is zero for an element such as O2 in its standard state), we can calculate :
B If we assume that only 50% of the available energy is used, then about 1440 kJ/mol of glucose is available for ATP synthesis. The value of ΔG°′ for the hydrolysis of ATP under biochemical conditions is −34.54 kJ/mol, so in principle an organism could synthesize
Most aerobic organisms actually synthesize about 32 molecules of ATP per molecule of glucose, for an efficiency of about 45%.
Exercise
Some bacteria synthesize methane using the following redox reaction:
CO2(g) + 4H2(g) → CH4(g) + 2H2O(g)Answer:
A living cell is a system that is not in equilibrium with its surroundings; it requires a constant input of energy to maintain its nonequilibrium state. Cells maintain a low-entropy state by increasing the entropy of their surroundings. Aerobic organisms cannot survive in the absence of O2, whereas anaerobic organisms can live only in the absence of O2. Green plants and algae are phototrophs, which extract energy from the environment through a process called photosynthesis, the photochemical reduction of CO2 to a reduced carbon compound. Other species, called chemotrophs, extract energy from chemical compounds. One of the main processes chemotrophs use to obtain energy is respiration, which is the reverse of photosynthesis. Alternatively, some chemotrophs obtain energy by fermentation, in which an organic compound is both the oxidant and the reductant. Intermediates used by organisms to shuttle electrons between the reductant and the oxidant include NAD+ and NADH. Energy from the oxidation of nutrients is made available to cells through the synthesis of ATP, the energy currency of the cell. Its energy is used by cells to synthesize substances through coupled reactions and to perform work. The body stores energy as sugars, protein, or fats before using it to produce ATP.
The tricarboxylic acid (TCA) cycle in aerobic organisms is one of four pathways responsible for the stepwise oxidation of organic intermediates. The final reaction in the TCA cycle has ΔG° = 29.7 kJ/mol, so it should not occur spontaneously. Suggest an explanation for why this reaction proceeds in the forward direction in living cells.
It is coupled to another reaction that is spontaneous, which drives this reaction forward (Le Châtelier’s principle).
Problems marked with a ♦ involve multiple concepts.
Electric utilities have been exploring thermal energy storage as a potentially attractive energy-storage solution for peak use. Thermal energy is extracted as steam and stored in rock, oil, or water for later conversion to electricity via heat exchangers. Which steps involve heat transfer? Which involve work done?
♦ During World War II, German scientists developed the first rocket-powered airplane to be flown in combat, the Messerschmitt 163 Komet. The Komet was powered by the reaction of liquid hydrogen peroxide (H2O2) and hydrazine (N2H4) as follows:
2H2O2(l) + N2H4(l) → N2(g) + 4H2O(g)♦ During the 1950s, pentaborane-9 was tested as a potential rocket fuel. However, the idea was abandoned when it was discovered that B2O3, the product of the reaction of pentaborane-9 with O2, was an abrasive that destroyed rocket nozzles. The reaction is represented by the equation
2B5H9(l) + 12O2(g) → 5B2O3(s) + 9H2O(g)♦ Polar explorers must be particularly careful to keep their clothes from becoming damp because the resulting heat loss could be fatal. If a polar explorer’s clothes absorbed 1.0 kg of water and the clothes dried from the polar wind, what would be the heat loss [ΔHvap(H2O) = 44 kJ/mol]? How much glucose must be consumed to make up for this heat loss to prevent death [ΔHcomb(glucose) = −802 kJ/mol]?
Propane gas is generally preferred to kerosene as a fuel for stoves in the boating industry because kerosene stoves require more maintenance. Propane, however, is much more flammable than kerosene and imposes an added risk because it is denser than air and can collect in the bottom of a boat and ignite. The complete combustion of propane produces CO2 gas and H2O vapor and has a value of ΔH = −2044 kJ/mol at 25°C. What is ΔE?
The propane in Problem 5 can be produced from the hydrogenation of propene according to the following reaction: C3H6(g) + H2(g) → C3H8(g); ΔH = −124 kJ/mol. Given that the reaction has ΔH = −241.8 kJ/mol, what is the
♦ The anaerobic conversion of sucrose, a sweetening agent, to lactic acid, which is associated with sour milk, can be described as follows:
The combustion of sucrose, however, occurs as follows:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)Phosphorus exists as several allotropes, the most common being red, black, and white phosphorus. White phosphorus consists of tetrahedral P4 molecules and melts at 44.15°C; it is converted to red phosphorus by heating at 400°C for several hours. The chemical differences between red and white phosphorus are considerable: white phosphorus burns in air, whereas red phosphorus is stable; white phosphorus is soluble in organic compounds, whereas red phosphorus is not; white phosphorus melts at 44.15°C, whereas red phosphorus melts at 597°C. If the enthalpy of fusion of white phosphorus is 0.659 kJ/mol, what is its ΔS? Black phosphorus is even less reactive than red. Based on this information, which allotrope would you predict to have the highest entropy? the lowest? Why?
♦ Ruby and sapphire have a common mineral name: corundum (Al2O3). Although they are crystalline versions of the same compound, the nature of the imperfections determines the identity of the gem. Outline a method for measuring and comparing the entropy of a ruby with the entropy of a sapphire. How would you expect the entropies to compare with the entropy of a perfect corundum crystal?
♦ Tin has two crystalline forms—α and β—represented in the following equilibrium equation:
The earliest known tin artifacts were discovered in Egyptian tombs of the 18th dynasty (1580–1350 BC), although archaeologists are surprised that so few tin objects exist from earlier eras. It has been suggested that many early tin objects were either oxidized to a mixture of stannous and stannic oxides or transformed to powdery, gray tin. Sketch a thermodynamic cycle similar to part (b) in to show the conversion of liquid tin to gray tin. Then calculate the change in entropy that accompanies the conversion of Sn(l) to α-Sn using the following data: Cp(white) = 26.99, Cp(gray) = 25.77 J/(mol·K), ΔHfus = 7.0 kJ/mol, ΔHβ → α = −2.2 kJ/mol.
The reaction of SO2 with O2 to produce SO3 has great industrial significance because SO3 is converted to H2SO4 by reaction with water. Unfortunately, the reaction is also environmentally important because SO3 from industrial smokestacks is a primary source of acid rain. ΔH for the reaction of SO2 with O2 to form SO3 is −23.49 kJ/mol, and ΔS is −22.66 J/(mol·K). Does this reaction occur spontaneously at 25°C? Does it occur spontaneously at 800°C assuming no change in ΔH and ΔS? Why is this reaction usually carried out at elevated temperatures?
Pollutants from industrial societies pose health risks to individuals from exposure to metals such as lead, mercury, and cadmium. The biological effects of a toxic metal may be reduced by removing it from the system using a chelating agent, which binds to the metal and forms a complex that is eliminated from the system without causing more damage. In designing a suitable chelating agent, one must be careful, however, because some chelating agents form metal complexes that are more toxic than the metal itself. Both methylamine (CH3NH2) and ethylenediamine (H2NCH2CH2NH2, abbreviated en) could, in principle, be used to treat heavy metal poisoning. In the case of cadmium, the reactions are as follows:
Based strictly on thermodynamic arguments, which would you choose to administer to a patient suffering from cadmium toxicity? Why? Assume a body temperature of 37°C.
♦ Explosive reactions often have a large negative enthalpy change and a large positive entropy change, but the reaction must also be kinetically favorable. For example, the following equation represents the reaction between hydrazine, a rocket propellant, and the oxidizer dinitrogen tetroxide:
2N2H4(l) + N2O4(l) → 4H2O(g) + 3N2(g) ΔH° = −249 kJ/mol, ΔS° = 218 J/(mol·K)♦ Cesium, a silvery-white metal used in the manufacture of vacuum tubes, is produced industrially by the reaction of CsCl with CaC2:
2CsCl(l) + CaC2(s) → CaCl2(l) + 2C(s) + 2Cs(g)Compare the free energy produced from this reaction at 25°C and at 1227°C, the temperature at which it is normally run, given these values:
= 32.0 kJ/mol, = 8.0 J/(mol·K); = −0.6 kJ/mol, = 3.6 J/(mol·K).
Dessicants (drying agents) can often be regenerated by heating, although it is generally not economically worthwhile to do so. A dessicant that is commonly regenerated is CaSO4·2H2O:
Regeneration is carried out at 250°C.
The nitrogen triiodide complex with ammonia (NI3·NH3) is a simple explosive that can be synthesized from common household products. When detonated, it produces N2 and I2. It can be painted on surfaces when wet, but it is shock sensitive when dry (even touching it with a feather can cause an explosion). Do you expect ΔG for the explosion reaction to be positive or negative? Why doesn’t NI3·NH3 explode spontaneously?
Adenosine triphosphate (ATP) contains high-energy phosphate bonds that are used in energy metabolism, coupling energy-yielding and energy-requiring processes. Cleaving a phosphate link by hydrolysis (ATP hydrolysis) can be described by the reaction where Pi symbolizes phosphate. Glycerol and ATP react to form glycerol-3-phosphate, ADP, and H+, with an overall K = 6.61 × 105 at 37°C. The reaction of glycerol with phosphate to form glycerol-3-phosphate and water has an equilibrium constant of 2.82 × 10−2. What is the equilibrium constant for ATP hydrolysis? How much free energy is released from the hydrolysis of ATP?
♦ Consider the biological reduction of molecular nitrogen, for which the following is the minimal reaction stoichiometry under optimal conditions (Pi = phosphate):
8H+ + 8e− + N2 + 16ATP → H2 + 2NH3 + 16ADP + 16Pi−2046 kJ/mol
Yes, the reaction is spontaneous at 25°C, but its rate is very slow. The reaction is not spontaneous at 800°C (ΔG = 0.82 kJ/mol), but the reaction rate is much greater.
2.34 × 107; −43.7 kJ/mol
In oxidation–reduction (redox) reactions, electrons are transferred from one species (the reductant) to another (the oxidant). This transfer of electrons provides a means for converting chemical energy to electrical energy or vice versa. The study of the relationship between electricity and chemical reactions is called electrochemistryThe study of the relationship between electricity and chemical reactions., an area of chemistry we introduced in and . In this chapter, we describe electrochemical reactions in more depth and explore some of their applications.
In the first three sections, we review redox reactions; describe how they can be used to generate an electrical potential, or voltage; and discuss factors that affect the magnitude of the potential. We then explore the relationships among the electrical potential, the change in free energy, and the equilibrium constant for a redox reaction, which are all measures of the thermodynamic driving force for a reaction. Finally, we examine two kinds of applications of electrochemical principles: (1) those in which a spontaneous reaction is used to provide electricity and (2) those in which electrical energy is used to drive a thermodynamically nonspontaneous reaction. By the end of this chapter, you will understand why different kinds of batteries are used in cars, flashlights, cameras, and portable computers; how rechargeable batteries operate; and why corrosion occurs and how to slow—if not prevent—it. You will also discover how metal objects can be plated with silver or chromium for protection; how silver polish removes tarnish; and how to calculate the amount of electricity needed to produce aluminum, chlorine, copper, and sodium on an industrial scale.
A view from the top of the Statue of Liberty, showing the green patina coating the statue. The patina is formed by corrosion of the copper skin of the statue, which forms a thin layer of an insoluble compound that contains copper(II), sulfate, and hydroxide ions.
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. As we described in , a redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductantA substance that is capable of donating electrons and in the process is oxidized. is the substance that loses electrons and is oxidized in the process; the oxidantA substance that is capable of accepting electrons and in the process is reduced. is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements. (For more information on valence electrons, see , .)
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactionsReactions that represent either the oxidation half or the reduction half of an oxidation–reduction (redox) reaction., one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
Equation 19.1
Zn(s) + Br2(aq) → Zn2+(aq) + 2Br−(aq)The half-reactions are as follows:
Equation 19.2
reduction half-reaction: Br2(aq) + 2e− → 2Br−(aq)Equation 19.3
oxidation half-reaction: Zn(s) → Zn2+(aq) + 2e−Each half-reaction is written to show what is actually occurring in the system; Zn is the reductant in this reaction (it loses electrons), and Br2 is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
In any redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cellAn apparatus that generates electricity from a spontaneous oxidation–reduction (redox) reaction or, conversely, uses electricity to drive a nonspontaneous redox reaction..
There are two types of electrochemical cells: galvanic cells and electrolytic cells. A galvanic (voltaic) cellAn electrochemical cell that uses the energy released during a spontaneous oxidation–reduction (redox) reaction to generate electricity.Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cellAn electrochemical cell that consumes electrical energy from an external source to drive a nonspontaneous oxidation–reduction (redox) reaction. consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two electrodesA solid metal connected by an electrolyte and an external circuit that provides an electrical connection between systems in an electrochemical cell (galvanic or electrolytic)., which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (). The oxidation half-reaction occurs at one electrode (the anodeOne of two electrodes in an electrochemical cell, it is the site of the oxidation half-reaction.), and the reduction half-reaction occurs at the other (the cathodeOne of two electrodes in an electrochemical cell, it is the site of the reduction half-reaction.). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells. We discuss electrolytic cells in .
Figure 19.1 Electrochemical Cells
A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current. In an electrolytic cell (right), an external source of electrical energy is used to generate a potential difference between the electrodes that forces electrons to flow, driving a nonspontaneous redox reaction; only a single compartment is employed in most applications. In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs.
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
Equation 19.4
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms (). These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
Figure 19.2 The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Single Compartment
When a zinc rod is inserted into a beaker that contains an aqueous solution of copper(II) sulfate, a spontaneous redox reaction occurs: the zinc electrode dissolves to give Zn2+(aq) ions, while Cu2+(aq) ions are simultaneously reduced to metallic copper. The reaction occurs so rapidly that the copper is deposited as very fine particles that appear black, rather than the usual reddish color of copper.
This same reaction can be carried out using the galvanic cell illustrated in part (a) in . To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of Cu2+ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of Zn2+ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridgeA U-shaped tube inserted into both solutions of a galvanic cell that contains a concentrated liquid or gelled electrolyte and completes the circuit between the anode and the cathode., a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are Na+ or K+ and NO3− or SO42−, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to Zn2+ ions at the zinc electrode (the anode), and Cu2+ ions are reduced to Cu metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of Zn2+ ions in the Zn2+ solution increases; simultaneously, the copper strip gains mass, and the concentration of Cu2+ ions in the Cu2+ solution decreases (part (b) in ). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
Figure 19.3 The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Galvanic Cell
(a) A galvanic cell can be constructed by inserting a copper strip into a beaker that contains an aqueous 1 M solution of Cu2+ ions and a zinc strip into a different beaker that contains an aqueous 1 M solution of Zn2+ ions. The two metal strips are connected by a wire that allows electricity to flow, and the beakers are connected by a salt bridge. When the switch is closed to complete the circuit, the zinc electrode (the anode) is spontaneously oxidized to Zn2+ ions in the left compartment, while Cu2+ ions are simultaneously reduced to copper metal at the copper electrode (the cathode). (b) As the reaction progresses, the Zn anode loses mass as it dissolves to give Zn2+(aq) ions, while the Cu cathode gains mass as Cu2+(aq) ions are reduced to copper metal that is deposited on the cathode.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the Zn2+ solution would increase as the zinc metal dissolves, and the total positive charge in the Cu2+ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the Zn2+ solution and a flow of cations into the Cu2+ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A galvanic cell. This galvanic cell illustrates the use of a salt bridge to connect two solutions.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential (Ecell)Related to the energy needed to move a charged particle in an electric field, it is the difference in electrical potential beween two half-reactions. of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (part (a) in ). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example 1.
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
3Sn(s) + 2NO3−(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)For this galvanic cell,
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
A Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
B From the direction of electron flow, assign each electrode as either positive or negative.
Solution:
A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:
reduction: NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l) oxidation: Sn(s) → Sn2+(aq) + 2e−Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
Exercise
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of MnO4− in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of Sn2+ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
2MnO4−(aq) + 5Sn2+(aq) + 16H+(aq) → 2Mn2+(aq) + 5Sn4+(aq) + 8H2O(l)For this galvanic cell,
Answer:
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the Zn/Cu cell shown in part (a) in is written as follows:
Figure 19.4
A cell diagram includes solution concentrations when they are provided.
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
Equation 19.5
Pt(s)∣H2(g)∣HCl(aq)∣AgCl(s)∣Ag(s)This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
Equation 19.6
cathode reaction: AgCl(s) + e− → Ag(s) + Cl−(aq)Equation 19.7
Equation 19.8
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Draw a cell diagram for the galvanic cell described in Example 1. The balanced chemical reaction is as follows:
3Sn(s) + 2NO3−(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution:
The anode is the tin strip, and the cathode is the Pt electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus Sn(s)∣Sn2+(aq). We could include H2SO4(aq) with the contents of the anode compartment, but the sulfate ion (as HSO4−) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (NO) and the Pt electrode. These are written as HNO3(aq)∣NO(g)∣Pt(s), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
Sn(s)∣Sn2+(aq)∥HNO3(aq)∣NO(g)∣Pt(s)The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise
Draw a cell diagram for the following reaction, assuming the concentration of Ag+ and Mg2+ are each 1 M:
Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s)Answer: Mg(s)∣Mg2+(aq, 1 M)∥Ag+(aq, 1 M)∣Ag(s)
Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential.
Is 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) an oxidation–reduction reaction? Why or why not?
If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?
What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?
What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?
One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K+ and Cl− ions do. What would happen if the diffusion rates of the anions and cations differed significantly?
It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?
A large difference in cation/anion diffusion rates would increase resistance in the salt bridge and limit electron flow through the circuit.
Copper(I) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO4 solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of ZnCl2?
Consider the following spontaneous redox reaction: NO3−(aq) + H+(aq) + SO32−(aq) → SO42−(aq) + HNO2(aq).
The reaction Pb(s) + 2VO2+(aq) + 4H+(aq) → Pb2+(aq) + 2V3+(aq) + 2H2O(l) occurs spontaneously.
Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)6]3−, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.
Sulfate is reduced to HS− in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?
Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.
For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.
For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.
Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.
reduction: SO42−(aq) + 9H+(aq) + 8e− → HS−(aq) + 4H2O(l) oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3−(g) + 30H+(aq) + 24e− overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3−(g) + 3H+(aq) + 3HS−(aq)
reduction: 2H+(aq) + 2e− → H2(aq); cathode;
oxidation: Zn(s) → Zn2+(aq) + 2e−; anode;
overall: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(aq)
reduction: AgCl(s) + e− → Ag(s) + Cl−(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e−; anode;
overall: AgCl(s) + H2(g) → 2H+(aq) + Ag(s) + Cl−(aq)
reduction: Fe3+(aq) + e− → Fe2+(aq); cathode;
oxidation: H2(g) → 2H+(aq) + 2e−; anode;
overall: 2Fe3+(aq) + H2(g) → 2H+(aq) + 2Fe2+(aq)
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. (For more information on atomic orbitals, see , .) Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper (). Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work.
Figure 19.5 Potential Energy Difference in the Zn/Cu System
The potential energy of a system consisting of metallic Zn and aqueous Cu2+ ions is greater than the potential energy of a system consisting of metallic Cu and aqueous Zn2+ ions. Much of this potential energy difference is because the valence electrons of metallic Zn are higher in energy than the valence electrons of metallic Cu. Because the Zn(s) + Cu2+(aq) system is higher in energy by 1.10 V than the Cu(s) + Zn2+(aq) system, energy is released when electrons are transferred from Zn to Cu2+ to form Cu and Zn2+.
Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in but instead of copper use a strip of cobalt metal and 1 M Co2+ in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V.
The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potentialThe potential of an electrochemical cell measured under standard conditions (1 M for solutions, 1 atm for gases, and pure solids or pure liquids for other substances) and at a fixed temperature (usually 298 K). (E°cell), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions,Concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for nonideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C.
Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system.
It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured. (This is analogous to measuring absolute enthalpies or free energies. Recall from that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram:
Equation 19.9
This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu2+ in solution at the copper cathode.
All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances. (Standard electrode potentials for various reduction reactions are given in .) The standard cell potential (E°cell) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum:
Equation 19.10
E°cell = E°cathode − E°anodeIn contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to , when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell.
The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials: E°cell = E°cathode − E°anode.
Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE)The electrode chosen as the reference for all other electrodes, which has been assigned a standard potential of 0 V and consists of a Pt wire in contact with an aqueous solution that contains 1 M in equilibrium with gas at a pressure of 1 atm at the Pt-solution interface. is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H+. The [H+] in solution is in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface (). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
Equation 19.11
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure 19.6 The Standard Hydrogen Electrode
The SHE consists of platinum wire that is connected to a Pt surface in contact with an aqueous solution containing 1 M H+ in equilibrium with H2 gas at a pressure of 1 atm. In the molecular view, the Pt surface catalyzes the oxidation of hydrogen molecules to protons or the reduction of protons to hydrogen gas. (Water is omitted for clarity.) The standard potential of the SHE is arbitrarily assigned a value of 0 V.
shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn2+ ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn2+, and H+ ions are reduced to H2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows:
Equation 19.12
Zn(s)∣Zn2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
Equation 19.13
Equation 19.14
Equation 19.15
Figure 19.7 Determining a Standard Electrode Potential Using a Standard Hydrogen Electrode
The voltmeter shows that the standard cell potential of a galvanic cell consisting of a SHE and a Zn/Zn2+ couple is E°cell = 0.76 V. Because the zinc electrode in this cell dissolves spontaneously to form Zn2+(aq) ions while H+(aq) ions are reduced to H2 at the platinum surface, the standard electrode potential of the Zn2+/Zn couple is −0.76 V.
Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potentialThe potential of a half-reaction measured against the SHE under standard conditions. for that half-reaction.In this example, the standard reduction potential for Zn2+(aq) + 2e− → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn2+, often called the Zn/Zn2+redox couple, or the Zn/Zn2+couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E°anode from E°cathode to obtain E°cell: 0 − (−0.76 V) = 0.76 V.
Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.
E° values do not depend on the stoichiometric coefficients for a half-reaction.
To measure the potential of the Cu/Cu2+ couple, we can construct a galvanic cell analogous to the one shown in but containing a Cu/Cu2+ couple in the sample compartment instead of Zn/Zn2+. When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of E°cell indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn2+ couple. Hence the reactions that occur spontaneously, indicated by a positive E°cell, are the reduction of Cu2+ to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H2 is oxidized to H+ at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu2+/Cu couple on the right:
Equation 19.16
Pt(s)∣H2(g, 1 atm)∣H+(aq, 1 M)∥Cu2+(aq, 1 M)∣Cu(s)The half-cell reactions and potentials of the spontaneous reaction are as follows:
Equation 19.17
Equation 19.18
Equation 19.19
Thus the standard electrode potential for the Cu2+/Cu couple is 0.34 V.
In , we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other.
We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas:
Equation 19.20
Al(s) + OH−(aq) → Al(OH)4−(aq) + H2(g)In this reaction, Al(s) is oxidized to Al3+, and H+ in water is reduced to H2 gas, which bubbles through the solution, agitating it and breaking up the clogs.
The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed in , we find the corresponding half-reactions that describe the reduction of H+ ions in water to H2 and the oxidation of Al to Al3+ in basic solution:
Equation 19.21
reduction: 2H2O(l) + 2e− → 2OH−(aq) + H2(g)Equation 19.22
oxidation: Al(s) + 4OH−(aq) → Al(OH)4−(aq) + 3e−The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution.
In , two H+ ions gain one electron each in the reduction; in , the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction () by 3 and the oxidation half-reaction () by 2 to give the same number of electrons in both half-reactions:
Equation 19.23
reduction: 6H2O(l) + 6e− → 6OH−(aq) + 3H2(g)Equation 19.24
oxidation: 2Al(s) + 8OH−(aq) → 2Al(OH)4−(aq) + 6e−Adding the two half-reactions,
Equation 19.25
6H2O(l) + 2Al(s) + 8OH−(aq) → 2Al(OH)4−(aq) + 3H2(g) + 6OH−(aq)Simplifying by canceling substances that appear on both sides of the equation,
Equation 19.26
6H2O(l) + 2Al(s) + 2OH−(aq) → 2Al(OH)4−(aq) + 3H2(g)We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced:
Equation 19.27
2Al + 8O + 14H = 2Al + 8O + 14HThe atoms also balance, so is a balanced chemical equation for the redox reaction depicted in .
The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction.
We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in but instead focus on the atoms whose oxidation states change, as illustrated in the following steps:
Step 1: Write the reduction half-reaction and the oxidation half-reaction.
For the reaction shown in , hydrogen is reduced from H+ in OH− to H2, and aluminum is oxidized from Al0 to Al3+:
Equation 19.28
reduction: OH−(aq) → H2(g)Equation 19.29
oxidation: Al(s) → Al(OH)4−(aq)Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H2O and balance H atoms by adding H+.
Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction:
Equation 19.30
reduction: OH−(aq) → H2(g) + H2O(l)Equation 19.31
oxidation: Al(s) + 4H2O(l) → Al(OH)4−(aq)Balancing H atoms by adding H+, we obtain the following:
Equation 19.32
reduction: OH−(aq) + 3H+(aq) → H2(g) + H2O(l)Equation 19.33
oxidation: Al(s) + 4H2O(l) → Al(OH)4−(aq) + 4H+(aq)We have now balanced the atoms in each half-reaction, but the charges are not balanced.
Step 3: Balance the charges in each half-reaction by adding electrons.
Two electrons are gained in the reduction of H+ ions to H2, and three electrons are lost during the oxidation of Al0 to Al3+:
Equation 19.34
reduction: OH−(aq) + 3H+(aq) + 2e− → H2(g) + H2O(l)Equation 19.35
oxidation: Al(s) + 4H2O(l) → Al(OH)4−(aq) + 4H+(aq) + 3e−Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions.
In this case, we multiply (the reductive half-reaction) by 3 and (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions:
Equation 19.36
reduction: 3OH−(aq) + 9H+(aq) + 6e− → 3H2(g) + 3H2O(l)Equation 19.37
oxidation: 2Al(s) + 8H2O(l) → 2Al(OH)4−(aq) + 8H+(aq) + 6e−Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation.
Adding and, in this case, canceling 8H+, 3H2O, and 6e−,
Equation 19.38
2Al(s) + 5H2O(l) + 3OH−(aq) + H+(aq) → 2Al(OH)4−(aq) + 3H2(g)We have three OH− and one H+ on the left side. Neutralizing the H+ gives us a total of 5H2O + H2O = 6H2O and leaves 2OH− on the left side:
Equation 19.39
2Al(s) + 6H2O(l) + 2OH−(aq) → 2Al(OH)4−(aq) + 3H2(g)Step 6: Check to make sure that all atoms and charges are balanced.
is identical to , obtained using the first method, so the charges and numbers of atoms on each side of the equation balance.
Figure 19.8 The Reaction of Dichromate with Iodide
The reaction of a yellow solution of sodium dichromate with a colorless solution of sodium iodide produces a deep amber solution that contains a green Cr3+(aq) complex and brown I2(aq) ions.
In acidic solution, the redox reaction of dichromate ion (Cr2O72−) and iodide (I−) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green Cr3+(aq) complex and brown I2(aq) ions ():
Cr2O72−(aq) + I−(aq) → Cr3+(aq) + I2(aq)Balance this equation using half-reactions.
Asked for: balanced chemical equation using half-reactions
Strategy:
Follow the steps to balance the redox reaction using the half-reaction method.
Solution:
From the standard electrode potentials listed in , we find the half-reactions corresponding to the overall reaction:
reduction: Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) oxidation: 2I−(aq) → I2(aq) + 2e−Balancing the number of electrons by multiplying the oxidation reaction by 3,
oxidation: 6I−(aq) → 3I2(aq) + 6e−Adding the two half-reactions and canceling electrons,
Cr2O72−(aq) + 14H+(aq) + 6I−(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)We must now check to make sure the charges and atoms on each side of the equation balance:
The charges and atoms balance, so our equation is balanced.
We can also use the alternative procedure, which does not require the half-reactions listed in .
Step 1: Chromium is reduced from Cr6+ in Cr2O72−to Cr3+, and I− ions are oxidized to I2. Dividing the reaction into two half-reactions,
reduction: Cr2O72−(aq) → Cr3+(aq) oxidation: I−(aq) → I2(aq)Step 2: Balancing the atoms other than oxygen and hydrogen,
reduction: Cr2O72−(aq) → 2Cr3+(aq) oxidation: 2I−(aq) → I2(aq)We now balance the O atoms by adding H2O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.
reduction: Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)Next we balance the H atoms by adding H+ to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction.
reduction: Cr2O72−(aq) + 14H+(aq) → + 2Cr3+(aq) + 7H2O(l)Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr+6 to 2Cr+3) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I− to I2) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge:
reduction: Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) oxidation: 2I−(aq) → I2(aq) + 2e−Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3:
oxidation: 6I−(aq) → 3I2(s) + 6e−Step 5: Adding the two half-reactions and canceling substances that appear in both reactions,
Cr2O72−(aq) + 14H+(aq) + 6I−(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance.
Exercise
Copper is commonly found as the mineral covellite (CuS). The first step in extracting the copper is to dissolve the mineral in nitric acid (HNO3), which oxidizes sulfide to sulfate and reduces nitric acid to NO:
CuS(s) + HNO3(aq) → NO(g) + CuSO4(aq)Balance this equation using the half-reaction method.
Answer: 3CuS(s) + 8HNO3(aq) → 8NO(g) + 3CuSO4(aq) + 4H2O(l)
The standard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram:
Equation 19.40
Zn(s)∣Zn2+(aq, 1 M)∥Cu2+(aq, 1 M)∣Cu(s)We know the values of E°anode for the reduction of Zn2+ and E°cathode for the reduction of Cu2+, so we can calculate E°cell:
Equation 19.41
Equation 19.42
Equation 19.43
This is the same value that is observed experimentally. If the value of E°cell is positive, the reaction will occur spontaneously as written. If the value of E°cell is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction.As we shall see in , this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example 4 and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples.
A positive E°cell means that the reaction will occur spontaneously as written. A negative E°cell means that the reaction will proceed spontaneously in the opposite direction.
A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl3, and the other contains a piece of nickel immersed in a 1 M solution of NiCl2. The half-reactions that occur when the compartments are connected are as follows:
cathode: Ni2+(aq) + 2e− → Ni(s) anode: Ga(s) → Ga3+(aq) + 3e−If the potential for the oxidation of Ga to Ga3+ is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni2+?
Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions
Asked for: standard electrode potential of reaction occurring at the cathode
Strategy:
A Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction.
B Use to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions.
Solution:
A We have been given the potential for the oxidation of Ga to Ga3+ under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga3+(aq) + 3e− → Ga(s), E°anode = −0.55 V.
B Using the value given for E°cell and the calculated value of E°anode, we can calculate the standard potential for the reduction of Ni2+ to Ni from :
This is the standard electrode potential for the reaction Ni2+(aq) + 2e− → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni2+ under standard conditions, we must reverse the sign of E°cathode. Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry.
Exercise
A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate [Hg(CH3CO2)2] and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of MgCl2. When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur:
cathode: Hg2+(aq) + 2e− → Hg(l) anode: Mg(s) → Mg2+(aq) + 2e−If the potential for the oxidation of Mg to Mg2+ is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the anode?
Answer: 0.85 V
When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrodeThe electrode of a galvanic cell whose potential is related to the concentration of the substance being measured., whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrodeAn electrode in an galvanic cell whose potential is unaffected by the properties of the solution., must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode.
The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated.
There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrodeA reference electrode that consists of a silver wire coated with a very thin layer of AgCl and dipped into a chloride ion solution with a fixed concentration., which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows:
Equation 19.44
If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE.
A second common reference electrode is the saturated calomel electrode (SCE)A reference electrode that consists of a platinum wire inserted into a moist paste of liquid mercury (calomel; ) and KCl in an interior cell, which is surrounded by an aqueous KCl solution., which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg2Cl2; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in ). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows:
Equation 19.45
Pt(s)∣Hg2Cl2(s)∣KCl(aq, sat)Equation 19.46
Hg2Cl2(s) + 2e− → 2Hg(l) + 2Cl−(aq)Figure 19.9 Three Common Types of Electrodes
(a) The SCE is a reference electrode that consists of a platinum wire inserted into a moist paste of liquid mercury (calomel; Hg2Cl2) and KCl. The interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution. (b) In a glass electrode, an internal Ag/AgCl electrode is immersed in a 1 M HCl solution that is separated from the sample solution by a very thin glass membrane. The potential of the electrode depends on the H+ ion concentration of the sample. (c) The potential of an ion-selective electrode depends on the concentration of only a single ionic species in solution.
At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential.
One of the most common uses of electrochemistry is to measure the H+ ion concentration of a solution. A glass electrodeAn electrode used to measure the ion concentration of a solution and consisting of an internal Ag/AgCl electrode immersed in a 1 M HCl solution that is separated from the solution by a very thin glass membrane. is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in ). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H+] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H+] as follows (recall that pH = −log[H+]:
Equation 19.47
Eglass = E′ + (0.0591 V × log[H+]) = E′ − 0.0591 V × pHThe voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH.
Ion-selective electrodesAn electrode whose potential depends on only the concentration of a particular species in solution. are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in ). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped LaF3 as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in .
Table 19.1 Some Species Whose Aqueous Concentrations Can Be Measured Using Electrochemical Methods
Species | Type of Sample |
---|---|
H+ | laboratory samples, blood, soil, and ground and surface water |
NH3/NH4+ | wastewater and runoff water |
K+ | blood, wine, and soil |
CO2/HCO3− | blood and groundwater |
F− | groundwater, drinking water, and soil |
Br− | grains and plant extracts |
I− | milk and pharmaceuticals |
NO3− | groundwater, drinking water, soil, and fertilizer |
The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E°cell). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E°cell = E°cathode − E°anode). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. If E°cell is positive, the reaction will occur spontaneously under standard conditions. If E°cell is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°.
Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?
List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.
What is the relationship between electron flow and the potential energy of valence electrons? If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?
If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?
Write a cell diagram representing a cell that contains the Ni/Ni2+ couple in one compartment and the SHE in the other compartment. What are the values of E°cathode, E°anode, and E°cell?
Explain why E° values are independent of the stoichiometric coefficients in the corresponding half-reaction.
Identify the oxidants and the reductants in each redox reaction.
Identify the oxidants and the reductants in each redox reaction.
All reference electrodes must conform to certain requirements. List the requirements and explain their significance.
For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?
Ni(s)∣Ni2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)
Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction: H2(g) + Cu2+(aq) → 2H+(aq) + Cu(s).
Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g).
Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.
Balance each reaction and calculate the standard reduction potential for each. Be sure to include the physical state of each product and reactant.
Write a balanced chemical equation for each redox reaction.
Write a balanced chemical equation for each redox reaction.
The standard cell potential for the oxidation of Pb to Pb2+ with the concomitant reduction of Cu+ to Cu is 0.39 V. You know that E° for the Pb2+/Pb couple is −0.13 V. What is E° for the Cu+/Cu couple?
Carbon is used to reduce iron ore to metallic iron. The overall reaction is as follows:
2Fe2O3·xH2O(s) + 3C(s) → 4Fe(l) + 3CO2(g) + 2xH2O(g)Write the two half-reactions for this overall reaction.
Will each reaction occur spontaneously under standard conditions?
Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.
Calculate E°cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?
If you place Zn-coated (galvanized) tacks in a glass and add an aqueous solution of iodine, the brown color of the iodine solution fades to a pale yellow. What has happened? Write the two half-reactions and the overall balanced chemical equation for this reaction. What is E°cell?
Your lab partner wants to recover solid silver from silver chloride by using a 1.0 M solution of HCl and 1 atm H2 under standard conditions. Will this plan work?
Pt(s)∣H2(g, 1 atm) | H+(aq, 1M)∥Cu2+(aq)∣Cu(s)
We can use the procedure described in Section 19.2 "Standard Potentials" to measure the standard potentials for a wide variety of chemical substances, some of which are listed in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C". (Chapter 29 "Appendix E: Standard Reduction Potentials at 25°C" contains a more extensive listing.) These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C". All reactants that lie above the SHE in the table are stronger oxidants than H+, and all those that lie below the SHE are weaker. The strongest oxidant in the table is F2, with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element.
Table 19.2 Standard Potentials for Selected Reduction Half-Reactions at 25°C
Half-Reaction | E° (V) |
---|---|
F2(g) + 2e−→ 2F−(aq) | 2.87 |
H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) | 1.78 |
Ce4+(aq) + e− → Ce3+(aq) | 1.72 |
PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− → PbSO4(s) + 2H2O(l) | 1.69 |
Cl2(g) + 2e− → 2Cl−(aq) | 1.36 |
Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) | 1.23 |
O2(g) + 4H+(aq) + 4e− → 2H2O(l) | 1.23 |
MnO2(s) + 4H+(aq) + 2e− → Mn2+(aq) + 2H2O(l) | 1.22 |
Br2(aq) + 2e− → 2Br−(aq) | 1.09 |
NO3−(aq) + 3H+(aq) + 2e− → HNO2(aq) + H2O(l) | 0.93 |
Ag+(aq) + e− → Ag(s) | 0.80 |
Fe3+(aq) + e− → Fe2+(aq) | 0.77 |
H2SeO3(aq) + 4H+ + 4e− → Se(s) + 3H2O(l) | 0.74 |
O2(g) + 2H+(aq) + 2e− → H2O2(aq) | 0.70 |
MnO4−(aq) + 2H2O(l) + 3e− → MnO2(s) + 4OH−(aq) | 0.60 |
MnO42−(aq) + 2H2O(l) + 2e− → MnO2(s) + 4OH−(aq) | 0.60 |
I2(s) + 2e− → 2I−(aq) | 0.54 |
H2SO3(aq) + 4H+(aq) + 4e− → S(s) + 3H2O(l) | 0.45 |
O2(g) + 2H2O(l) + 4e− → 4OH−(aq) | 0.40 |
Cu2+(aq) + 2e− → Cu(s) | 0.34 |
AgCl(s) + e− → Ag(s) + Cl−(aq) | 0.22 |
Cu2+(aq) + e− → Cu+(aq) | 0.15 |
Sn4+(aq) + 2e− → Sn2+(aq) | 0.15 |
2H+(aq) + 2e− → H2(g) | 0.00 |
Sn2+(aq) + 2e− → Sn(s) | −0.14 |
2SO42−(aq) + 4H+(aq) + 2e− → S2O62−(aq) + 2H2O(l) | −0.22 |
Ni2+(aq) + 2e− → Ni(s) | −0.26 |
PbSO4(s) + 2e− → Pb(s) + SO42−(aq) | −0.36 |
Cd2+(aq) + 2e− → Cd(s) | −0.40 |
Cr3+(aq) + e− → Cr2+(aq) | −0.41 |
Fe2+(aq) + 2e− → Fe(s) | −0.45 |
Ag2S(s) + 2e− → 2Ag(s) + S2−(aq) | −0.69 |
Zn2+(aq) + 2e− → Zn(s) | −0.76 |
Al3+(aq) + 3e− → Al(s) | −1.662 |
Be2+(aq) + 2e− → Be(s) | −1.85 |
Li+(aq) + e− → Li(s) | −3.04 |
Similarly, all species in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" that lie below H2 are stronger reductants than H2, and those that lie above H2 are weaker. The strongest reductant in the table is thus metallic lithium, with a standard electrode potential of −3.04 V. This fact might be surprising because cesium, not lithium, is the least electronegative element. The apparent anomaly can be explained by the fact that electrode potentials are measured in aqueous solution, where intermolecular interactions are important, whereas ionization potentials and electron affinities are measured in the gas phase. Due to its small size, the Li+ ion is stabilized in aqueous solution by strong electrostatic interactions with the negative dipole end of water molecules. These interactions result in a significantly greater ΔHhydration for Li+ compared with Cs+. Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution.
Species in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" that lie below H2 are stronger reductants (more easily oxidized) than H2. Species that lie above H2 are stronger oxidants.
Because the half-reactions shown in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E°cell > 0), as illustrated in Example 5.
The black tarnish that forms on silver objects is primarily Ag2S. The half-reaction for reversing the tarnishing process is as follows:
Given: reduction half-reaction, standard electrode potential, and list of possible reductants
Asked for: reductants for Ag2S, strongest reductant, and potential reducing agent for removing tarnish
Strategy:
A From their positions in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", decide which species can reduce Ag2S. Determine which species is the strongest reductant.
B Use Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to identify a reductant for Ag2S that is a common household product.
Solution:
We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag2S/Ag couple in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" or (2) compare E° for each species with E° for the Ag2S/Ag couple (−0.69 V).
Exercise
Refer to Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to predict
Answer:
Use the data in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to determine whether each reaction is likely to occur spontaneously under standard conditions:
Given: redox reaction and list of standard electrode potentials (Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C")
Asked for: reaction spontaneity
Strategy:
A Identify the half-reactions in each equation. Using Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C", determine the standard potentials for the half-reactions in the appropriate direction.
B Use Equation 19.10 to calculate the standard cell potential for the overall reaction. From this value, determine whether the overall reaction is spontaneous.
Solution:
A Metallic tin is oxidized to Sn2+(aq), and Be2+(aq) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.85 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C".
B Adding the two half-reactions gives the overall reaction:
The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot be used to reduce Be2+ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn2+) by metallic beryllium, which has a positive value of E°cell, will occur spontaneously.
A MnO2 is the oxidant (Mn4+ is reduced to Mn2+), while H2O2 is the reductant (O2− is oxidized to O2). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C".
B The two half-reactions and their corresponding potentials are as follows:
The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO2, and oxygen gas will evolve from the solution.
Exercise
Use the data in Table 19.2 "Standard Potentials for Selected Reduction Half-Reactions at 25°C" to determine whether each reaction is likely to occur spontaneously under standard conditions:
Answer:
Although the sign of E°cell tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics, as described in Section 19.4 "Electrochemical Cells and Thermodynamics".
The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase.
The order of electrode potentials cannot always be predicted by ionization potentials and electron affinities. Why? Do you expect sodium metal to have a higher or a lower electrode potential than predicted from its ionization potential? What is its approximate electrode potential?
Without referring to tabulated data, of Br2/Br−, Ca2+/Ca, O2/OH−, and Al3+/Al, which would you expect to have the least negative electrode potential and which the most negative? Why?
Because of the sulfur-containing amino acids present in egg whites, eating eggs with a silver fork will tarnish the fork. As a chemist, you have all kinds of interesting cleaning products in your cabinet, including a 1 M solution of oxalic acid (H2C2O4). Would you choose this solution to clean the fork that you have tarnished from eating scrambled eggs?
The electrode potential for the reaction Cu2+(aq) + 2e− → Cu(s) is 0.34 V under standard conditions. Is the potential for the oxidation of 0.5 mol of Cu equal to −0.34/2 V? Explain your answer.
No; E° = −0.691 V for Ag2S(s) + 2e− → Ag(s) + S2−(aq), which is too negative for Ag2S to be spontaneously reduced by oxalic acid [E° = 0.49 V for 2CO2(g) + 2H+(aq) + 2e− → H2C2O4(aq)]
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of Co(s) with Ni2+(aq) to form Ni(s) and Co2+(aq) occurs spontaneously, but if we reduce the concentration of Ni2+ by a factor of 100, so that [Ni2+] is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C)The SI unit of measure for the number of electrons that pass a given point in 1 second; it is defined as and relates electron potential (in volts) to energy (in joules): 1 J/1 V = 1 C., an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A)The fundamental SI unit of electric current; it is defined as the flow of 1 C/s past a given point: 1A = 1 C/s.; 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
Equation 19.48
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F)The charge on 1 mol of electrons; it is obtained by multiplying the charge on the electron by Avogadro’s number., named after the English physicist and chemist Michael Faraday (1791–1867):
Equation 19.49
The total charge transferred from the reductant to the oxidant is therefore nF, where n is the number of moles of electrons.
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell (wmax) is equal to the product of the cell potential (Ecell) and the total charge transferred during the reaction (nF):
Equation 19.50
wmax = nFEcellWork is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
As you learned in , the change in free energy (ΔG) is also a measure of the maximum amount of work that can be performed during a chemical process (ΔG = wmax). Consequently, there must be a relationship between the potential of an electrochemical cell and ΔG, the most important thermodynamic quantity discussed in . This relationship is as follows:
Equation 19.51
ΔG = −nFEcellA spontaneous redox reaction is therefore characterized by a negative value of ΔG and a positive value of Ecell, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and E°cell is as follows:
Equation 19.52
ΔG° = −nFE°cellA spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in , calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: ΔG° for the reaction and spontaneity
Strategy:
A From the relevant half-reactions and the corresponding values of E°, write the overall reaction and calculate E°cell using .
B Determine the number of electrons transferred in the overall reaction. Then use to calculate ΔG°. If ΔG° is negative, then the reaction is spontaneous.
Solution:
A As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of E°. From , we can find the reduction and oxidation half-reactions and corresponding E° values:
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of E° is not affected:
B We can now calculate ΔG° using . Because six electrons are transferred in the overall reaction, the value of n is 6:
Thus ΔG° is −81 kJ for the reaction as written, and the reaction is spontaneous.
Exercise
Use the data in to calculate ΔG° for the reduction of ferric ion by iodide:
2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(s)Is the reaction spontaneous?
Answer: −44 kJ/mol I2; yes
Although and list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of Fe3+(aq) to Fe(s) is not listed in the table, but two related reductions are given:
Equation 19.53
Equation 19.54
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because E° is not a state function. However, because ΔG° is a state function, the sum of the ΔG° values for the individual reactions gives us ΔG° for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. To obtain the value of E° for the overall half-reaction, we first must add the values of ΔG° (= −nFE°) for each individual half-reaction to obtain ΔG° for the overall half-reaction:
Equation 19.55
Solving the last expression for ΔG° for the overall half-reaction,
Equation 19.56
ΔG° = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V)Three electrons (n = 3) are transferred in the overall reaction (), so substituting into and solving for E° gives the following:
This value of E° is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of E° for half-reactions cannot be added to give E° for the sum of the half-reactions; only values of ΔG° = −nFE°cell for half-reactions can be added.
We can use the relationship between ΔG° and the equilibrium constant K, defined in , to obtain a relationship between E°cell and K. Recall that for a general reaction of the type aA + bB → cC + dD, the standard free-energy change and the equilibrium constant are related by the following equation:
Equation 19.57
ΔG° = −RT ln KGiven the relationship between the standard free-energy change and the standard cell potential (), we can write
Equation 19.58
−nFE°cell = −RT ln KRearranging this equation,
Equation 19.59
For T = 298 K, can be simplified as follows:
Equation 19.60
Thus E°cell is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of E°cell and vice versa.
Use the data in to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: K
Strategy:
A Write the relevant half-reactions and potentials. From these, obtain the overall reaction and E°cell.
B Determine the number of electrons transferred in the overall reaction. Use to solve for log K and then K.
Solution:
A The relevant half-reactions and potentials from are as follows:
B Two electrons are transferred in the overall reaction, so n = 2. Solving for log K and inserting the values of n and E°,
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise
Use the data in to calculate the equilibrium constant for the reaction of Sn2+(aq) with oxygen to produce Sn4+(aq) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
Answer: 1.2 × 1073
summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between ΔG and the reaction quotient Q developed in .
Figure 19.10 The Relationships among Criteria for Thermodynamic Spontaneity
The three properties of a system that can be used to predict the spontaneity of a redox reaction under standard conditions are K, ΔG°, and E°cell. If we know the value of one of these quantities, then these relationships enable us to calculate the value of the other two. The signs of ΔG° and E°cell and the magnitude of K determine the direction of spontaneous reaction under standard conditions.
Recall from that the actual free-energy change for a reaction under nonstandard conditions, ΔG, is given as follows:
Equation 19.61
ΔG = ΔG° + RT ln QWe also know that ΔG = −nFEcell and ΔG° = −nFE°cell. Substituting these expressions into , we obtain
Equation 19.62
−nFEcell = −nFE°cell + RT ln QDividing both sides of this equation by −nF,
Equation 19.63
is called the Nernst equationAn equation for calculating cell potentials under nonstandard conditions; it can be used to determine the direction of spontaneous reaction for any redox reaction under an conditions: , after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium (ΔG = 0), reduces to because Q = K, and there is no net transfer of electrons (i.e., Ecell = 0).
Substituting the values of the constants into with T = 298 K and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in Q):
Equation 19.64
The Nernst equation can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in that the cell potential changes by 0.0591/n V for each 10-fold change in the value of Q because log 10 = 1.
In the exercise in Example 6, you determined that the following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which you now know means that ΔG° < 0):
Calculate E for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl−] = 0.0030 M, = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution:
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use instead of 19.46. The overall reaction involves the net transfer of two electrons:
2Ce4+(aq) + 2e− → 2Ce3+(aq) 2Cl−(aq) → Cl2(g) + 2e−so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into ,
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture.
Exercise
In the exercise in Example 6, you determined that molecular oxygen will not oxidize MnO2 to permanganate via the reaction
Calculate Ecell for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, = 0.20 atm, [MNO4−] = 1.0 × 10−4 M, and T = 25°C.
Answer: Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell discussed in allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
Equation 19.65
The reaction quotient is therefore Q = [Zn2+]/[Cu2+]. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from :
Equation 19.66
Thus the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with log Q over this range is linear with a slope of −0.0591/n, as illustrated in . As the reaction proceeds still further, Q continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of Q when Ecell = 0 is calculated as follows:
Equation 19.67
Figure 19.11 The Variation of Ecell with Log Q for a Zn/Cu Cell
Initially, log Q < 0, and the voltage of the cell is greater than E°cell. As the reaction progresses, log Q increases, and Ecell decreases. When [Zn2+] = [Cu2+], log Q = 0 and Ecell = E°cell = 1.10 V. As long as the electrical circuit remains intact, the reaction will continue, and log Q will increase until Q = K and the cell voltage reaches zero. At this point, the system will have reached equilibrium.
Recall that at equilibrium, Q = K. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
Equation 19.68
Ag(s)∣Ag+(aq, 0.010 M)∥Ag+(aq, 1.0 M)∣Ag(s)Equation 19.69
cathode: Ag+(aq, 1.0 M) + e− → Ag(s)Equation 19.70
anode: Ag(s) → Ag+(aq, 0.010 M) + e−Equation 19.71
As the reaction progresses, the concentration of Ag+ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the Ag+ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of Ag(s) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cellAn electrochemical cell in which the anode and the cathode compartments are identical except for the concentration of a reactant.. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
A Write the overall reaction that occurs in the cell.
B Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution:
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl− and SO42−) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
Mn2+(aq, 2.0 M) → Mn2+(aq, 5.2 × 10−2 M)B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into :
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water, O2(g) + 4H+(aq) + 4e− → 2H2O(l), what will be the potential when the circuit is closed?
Answer: 0.41 V
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products (Ksp) of sparingly soluble substances. As you learned in , solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in , which is designed to measure the solubility product of silver chloride: Ksp = [Ag+][Cl−]. In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl−] and substituting: [Ag+] = Ksp/[Cl−] = Ksp/1.0 = Ksp. The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
Equation 19.72
Figure 19.12 A Galvanic Cell for Measuring the Solubility Product of AgCl
One compartment contains a silver wire inserted into a 1.0 M solution of Ag+, and the other compartment contains a silver wire inserted into a 1.0 M Cl− solution saturated with AgCl. The potential due to the difference in [Ag+] between the two cells can be used to determine Ksp.
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving for Ksp,
Equation 19.73
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in , which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: K sp
Strategy:
A From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
B Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into and solve for Ksp.
Solution:
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of Pb2+ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)so
Exercise
A concentration cell similar to the one described in Example 11 contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer: 5.7 × 10−17
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example 11, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated in Example 12.
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment (). The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
Zn(s)∣Zn2+(aq, 1.0 M) ∥ H+(aq, ? M)∣H2(g, 1.0 atm)∣Pt(s)What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
A Write the overall cell reaction.
B Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution:
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in ):
B By substituting the given values into the simplified Nernst equation (), we can calculate [H+] under nonstandard conditions:
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows”
Pb(s) ∣Pb2+(aq, ? M)∥H+(aq), 1.0 M∣O2(g, 1.0 atm)∣Pt(s)When the circuit is closed, the cell has a measured potential of 1.62 V. Use and to determine the concentration of Pb2+ in the groundwater.
Answer: 1.2 × 10−9 M
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species.
Charge on a mole of electrons (faraday)
Maximum work from an electrochemical cell
Relationship between Δ G ° and Δ E °
Relationship between Δ G ° and K for a redox reaction
Relationship between Δ E ° and K for a redox reaction at 25°C
Relationship between Δ G ° and Q
Relationship between E cell and Q at 25°C
State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn2+(aq) + 2e−—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.
What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?
In the equation wmax = −nFE°cell, which quantities are extensive properties and which are intensive properties?
For any spontaneous redox reaction, E is positive. Use thermodynamic arguments to explain why this is true.
State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.
Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?
Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?
Blood analyzers, which measure pH, and are frequently used in clinical emergencies. For example, blood is measured with a pH electrode covered with a plastic membrane that is permeable to CO2. Based on your knowledge of how electrodes function, explain how such an electrode might work. Hint: CO2(g) + H2O(l) → HCO3−(aq) + H+(aq).
Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?
Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.
extensive: wmax and n; intensive: E°cell
Gold is highly resistant to corrosion because of its very positive reduction potential.
The chemical equation for the combustion of butane is as follows:
This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?
How many electrons are transferred during the reaction Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg2Cl2 spontaneous? Calculate ΔG° for this reaction.
For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?
Explain why the sum of the potentials for the half-reactions Sn2+(aq) + 2e− → Sn(s) and Sn4+(aq) + 2e− → Sn2+(aq) does not equal the potential for the reaction Sn4+(aq) + 4e− → Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.
For each reaction, calculate E°cell and then determine ΔG°. Indicate whether each reaction is spontaneous.
What is the standard change in free energy for the reaction between Ca2+ and Na(s) to give Ca(s) and Na+? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?
In acidic solution, permanganate (MnO4−) oxidizes Cl− to chlorine gas, and MnO4− is reduced to Mn2+(aq).
Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.
Titrant (mL) | E (mV) |
---|---|
2.00 | 50 |
6.00 | 100 |
9.00 | 255 |
10.00 | 960 |
11.00 | 1325 |
12.00 | 1625 |
14.00 | 1875 |
The standard electrode potential (E°) for the half-reaction Ni2+(aq) + 2e− → Ni(s) is −0.257 V. What pH is needed for this reaction to take place in the presence of 1.00 atm H2(g) as the reductant if [Ni2+] is 1.00 M?
The reduction of Mn(VII) to Mn(s) by H2(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:
Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:
For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?
The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction NAD+ + H+ + 2e− → NADH; E° = −0.32 V.
Given the following biologically relevant half-reactions, will FAD (flavin adenine dinucleotide), a molecule used to transfer electrons whose reduced form is FADH2, be an effective oxidant for the conversion of acetaldehyde to acetate at pH 4.00?
Ideally, any half-reaction with E° > 1.23 V will oxidize water as a result of the half-reaction O2(g) + 4H+(aq) + 4e− → 2H2O(l).
Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction O2(g) + 4H+(aq) + 4e− → 2H2O(l).
Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CN− as a complexing agent for Mn3+/Mn2+?
You have constructed a cell with zinc and lead amalgam electrodes described by the cell diagram Zn(Hg)(s)∣Zn(NO3)2(aq)∥Pb(NO3)2(aq)∣Pb(Hg)(s). If you vary the concentration of Zn(NO3)2 and measure the potential at different concentrations, you obtain the following data:
Zn(NO3)2 (M) | Ecell (V) |
---|---|
0.0005 | 0.7398 |
0.002 | 0.7221 |
0.01 | 0.7014 |
Hydrogen gas reduces Ni2+ according to the following reaction: Ni2+(aq) + H2(g) → Ni(s) + 2H+(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.
The silver–silver bromide electrode has a standard potential of 0.07133 V. What is Ksp of AgBr?
6e−; E°cell = 1.813 V; the reaction is spontaneous; ΔG° = −525 kJ/mol Al.
yes; E° = 0.40 V
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell)A galvanic cell (or series of galvanic cells) that contains all the reactants needed to produce electricity. is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cellA galvanic cell that requires a constant external supply of one or more reactants to generate electricity. is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cellA battery consisting of an electrolyte that is an acidic water-based paste containing graphite, and starch. is actually a “wet cell”: the electrolyte is an acidic water-based paste containing MnO2, NH4Cl, ZnCl2, graphite, and starch (part (a) in ). The half-reactions at the anode and the cathode can be summarized as follows:
Equation 19.74
cathode: 2MnO2(s) + 2NH4+(aq) + 2e− → Mn2O3(s) + 2NH3(aq) + H2O(l)Equation 19.75
anode: Zn(s) → Zn2+(aq) + 2e−The Zn2+ ions formed by the oxidation of Zn(s) at the anode react with NH3 formed at the cathode and Cl− ions present in solution, so the overall cell reaction is as follows:
Equation 19.76
overall: 2MnO2(s) + 2NH4Cl(aq) + Zn(s) → Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the MnO2 that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the Zn anode reacts spontaneously with NH4Cl in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline batteryA battery that consists of a Leclanché cell adapted to operate under alkaline (basic) conditions. is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
Equation 19.77
cathode: 2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq)Equation 19.78
anode: Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e−Equation 19.79
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either HgO or Ag2O as the oxidant rather than MnO2 (part (b) in ). The cathode and overall reactions and cell output for these two types of button batteries are as follows:
Equation 19.80
cathode (Hg): HgO(s) + H2O(l) + 2e− → Hg(l) + 2OH−(aq)Equation 19.81
Equation 19.82
cathode (Ag): Ag2O(s) + H2O(l) + 2e− → 2Ag(s) + 2OH−(aq)Equation 19.83
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as Hg and Ag.
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries.
One of the few commercially successful water-free batteries is the lithium–iodine batteryA battery that consists of an anode of lithium metal and a cathode containing a solid complex of with a layer of solid LiI in between that allows the diffusion of ions.. The anode is lithium metal, and the cathode is a solid complex of I2. Separating them is a layer of solid LiI, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
Equation 19.84
cathode: I2(s) + 2e− → 2I−(LiI)Equation 19.85
anode: 2Li(s) → 2Li+(LiI) + 2e−Equation 19.86
Cardiac pacemaker. An x-ray of a patient showing the location and size of a pacemaker powered by a lithium–iodine battery.
As shown in part (c) in , a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using TiS2, for example, for the cathode.
Figure 19.13 Three Kinds of Primary (Nonrechargeable) Batteries
(a) A Leclanché dry cell is actually a “wet cell,” in which the electrolyte is an acidic water-based paste containing MnO2, NH4Cl, ZnCl2, graphite, and starch. Though inexpensive to manufacture, the cell is not very efficient in producing electrical energy and has a limited shelf life. (b) In a button battery, the anode is a zinc–mercury amalgam, and the cathode can be either HgO (shown here) or Ag2O as the oxidant. Button batteries are reliable and have a high output-to-mass ratio, which allows them to be used in applications such as calculators and watches, where their small size is crucial. (c) A lithium–iodine battery consists of two cells separated by a metallic nickel mesh that collects charge from the anodes. The anode is lithium metal, and the cathode is a solid complex of I2. The electrolyte is a layer of solid LiI that allows Li+ ions to diffuse from the cathode to the anode. Although this type of battery produces only a relatively small current, it is highly reliable and long-lived.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
The nickel–cadmiumA type of battery that consists of a water-based cell with a cadmium anode and a highly oxidized nickel cathode., or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in , the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
Figure 19.14 The Nickel–Cadmium (NiCad) Battery, a Rechargeable Battery
NiCad batteries contain a cadmium anode and a highly oxidized nickel cathode. This design maximizes the surface area of the electrodes and minimizes the distance between them, which gives the battery both a high discharge current and a high capacity.
The electrode reactions during the discharge of a NiCad battery are as follows:
Equation 19.87
cathode: 2NiO(OH)(s) + 2H2O(l) + 2e− → 2Ni(OH)2(s) + 2OH−(aq)Equation 19.88
anode: Cd(s) + 2OH−(aq) → Cd(OH)2(s) + 2e−Equation 19.89
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
NiO(OH)(s) + MH → Ni(OH)2(s) + M(s)The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
The lead–acid batteryA battery consisting of a plate or grid of spongy lead metal, a cathode containing powdered and an electrolyte that is usually an aqueous solution of is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells. As shown in , the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide (PbO2). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M H2SO4). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
Equation 19.90
Equation 19.91
Equation 19.92
Figure 19.15 One Cell of a Lead–Acid Battery
The anodes in each cell of a rechargeable battery are plates or grids of lead containing spongy lead metal, while the cathodes are similar grids containing powdered lead dioxide (PbO2). The electrolyte is an aqueous solution of sulfuric acid. The value of E° for such a cell is about 2 V. Connecting three such cells in series produces a 6 V battery, whereas a typical 12 V car battery contains six cells in series. When treated properly, this type of high-capacity battery can be discharged and recharged many times over.
As the cell is discharged, a powder of PbSO4 forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte.
Source: Photo courtesy of Mitchclanky2008, http://www.flickr.com/photos/25597837@N05/2422765479/.
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and PbSO4 is converted back to metallic lead and PbO2. If the battery is recharged too vigorously, however, electrolysis of water can occur, resulting in the evolution of potentially explosive hydrogen gas. (For more information on electrolysis, see .) The gas bubbles formed in this way can dislodge some of the PbSO4 or PbO2 particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in . The electrode reactions are as follows:
Equation 19.93
cathode: O2(g) + 4H+ + 4e− → 2H2O(g)Equation 19.94
anode: 2H2(g) → 4H+ + 4e−Equation 19.95
Figure 19.16 A Hydrogen Fuel Cell Produces Electrical Energy Directly from a Chemical Reaction
Hydrogen is oxidized to protons at the anode, and the electrons are transferred through an external circuit to the cathode, where oxygen is reduced and combines with H+ to form water. A solid electrolyte allows the protons to diffuse from the anode to the cathode. Although fuel cells are an essentially pollution-free means of obtaining electrical energy, their expense and technological complexity have thus far limited their applications.
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of O2(g) at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of O2.
A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction.
What advantage is there to using an alkaline battery rather than a Leclanché dry cell?
Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?
What type of battery would you use for each application and why?
Why are galvanic cells used as batteries and fuel cells? What is the difference between a battery and a fuel cell? What is the advantage to using highly concentrated or solid reactants in a battery?
This reaction is characteristic of a lead storage battery:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?
[H2SO4] = 3.52 M; E > E°
CorrosionA galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated $100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Corrosion is a galvanic process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide (Fe2O3·xH2O), commonly known as rust, that does not provide a tight protective film (). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil saturated with oxygen will not rust because of the absence of water.
Figure 19.17 Rust, the Result of Corrosion of Metallic Iron
Iron is oxidized to Fe2+(aq) at an anodic site on the surface of the iron, which is often an impurity or a lattice defect. Oxygen is reduced to water at a different site on the surface of the iron, which acts as the cathode. Electrons are transferred from the anode to the cathode through the electrically conductive metal. Water is a solvent for the Fe2+ that is produced initially and acts as a salt bridge. Rust (Fe2O3·xH2O) is formed by the subsequent oxidation of Fe2+ by atmospheric oxygen.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
Equation 19.96
Equation 19.97
Equation 19.98
The Fe2+ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing Fe3+, as represented in the following equation:
Equation 19.99
4Fe2+(aq) + O2(g) + (2 + 4x)H2O → 2Fe2O3·xH2O + 4H+(aq)The sign and magnitude of E° for the corrosion process () indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3− provides a low enough pH to enhance the reaction rate, as does acid rain. (For more information on acid rain, see , .) Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure ().
Figure 19.18 Small Scratches in a Protective Paint Coating Can Lead to the Rapid Corrosion of Iron
Holes in a protective coating allow oxygen to be reduced at the surface with the greater exposure to air (the cathode), while metallic iron is oxidized to Fe2+(aq) at the less exposed site (the anode). Rust is formed when Fe2+(aq) diffuses to a location where it can react with atmospheric oxygen, which is often remote from the anode. The electrochemical interaction between cathodic and anodic sites can cause a large pit to form under a painted surface, eventually resulting in sudden failure with little visible warning that corrosion has occurred.
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process that will be discussed in . With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually made of steel coated with a thin layer of tin. Neither chromium nor tin is intrinsically resistant to corrosion, but both form protective oxide coatings.
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for Sn2+ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in show that Fe is more easily oxidized than Sn. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
Figure 19.19 Galvanic Corrosion
If iron is in contact with a more corrosion-resistant metal such as tin, copper, or lead, the other metal can act as a large cathode that greatly increases the rate of reduction of oxygen. Because the reduction of oxygen is coupled to the oxidation of iron, this can result in a dramatic increase in the rate at which iron is oxidized at the anode. Galvanic corrosion is likely to occur whenever two dissimilar metals are connected directly, allowing electrons to be transferred from one to the other.
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as Zn (E° = −0.76 V for Zn2+ + 2e− → Zn) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
Equation 19.100
cathode: O2(g) + 4e− + 4H+(aq) → 2H2O(l)Equation 19.101
anode: Zn(s) → Zn2+(aq) + 2e−Equation 19.102
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans. In a similar strategy, sacrificial electrodesAn electrode containing a more reactive metal that is attached to a metal object to inhibit that object’s corrosion. using magnesium, for example, are used to protect underground tanks or pipes (). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Figure 19.20 The Use of a Sacrificial Electrode to Protect Against Corrosion
Connecting a magnesium rod to an underground steel pipeline protects the pipeline from corrosion. Because magnesium (E° = −2.37 V) is much more easily oxidized than iron (E° = −0.45 V), the Mg rod acts as the anode in a galvanic cell. The pipeline is therefore forced to act as the cathode at which oxygen is reduced. The soil between the anode and the cathode acts as a salt bridge that completes the electrical circuit and maintains electrical neutrality. As Mg(s) is oxidized to Mg2+ at the anode, anions in the soil, such as nitrate, diffuse toward the anode to neutralize the positive charge. Simultaneously, cations in the soil, such as H+ or NH4+, diffuse toward the cathode, where they replenish the protons that are consumed as oxygen is reduced. A similar strategy uses many miles of somewhat less reactive zinc wire to protect the Alaska oil pipeline.
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
Given: identity of metals
Asked for: corrosion reaction, E°cell, and preventive measures
Strategy:
A Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate E°cell.
B Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution:
A According to , both copper and tin are less active metals than iron (i.e., they have higher positive values of E° than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which O2 is reduced, and the iron screws will act as anodes at which iron dissolves:
Over time, the iron screws will dissolve, and the boat will fall apart.
Exercise
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
Answer:
The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it.
Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?
What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?
Why is it important for automobile manufacturers to apply paint to the metal surface of a car? Why is this process particularly important for vehicles in northern climates, where salt is used on icy roads?
Paint keeps oxygen and water from coming into direct contact with the metal, which prevents corrosion. Paint is more necessary because salt is an electrolyte that increases the conductivity of water and facilitates the flow of electric current between anodic and cathodic sites.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysisAn electrochemical process in which an external voltage is applied to an electrolytic cell to drive a nonspontaneous reaction., occurs: an external voltage is applied to drive a nonspontaneous reaction (). In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a 1 M Cd2+ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (part (a) in ). The overall reaction is as follows:
Equation 19.103
This reaction is thermodynamically spontaneous as written (ΔG° < 0):
Equation 19.104
In this direction, the system is acting as a galvanic cell.
Figure 19.21 An Applied Voltage Can Reverse the Flow of Electrons in a Galvanic Cd/Cu Cell
(a) When compartments that contain a Cd electrode immersed in 1 M Cd2+(aq) and a Cu electrode immersed in 1 M Cu2+(aq) are connected to create a galvanic cell, Cd(s) is spontaneously oxidized to Cd2+(aq) at the anode, and Cu2+(aq) is spontaneously reduced to Cu(s) at the cathode. The potential of the galvanic cell is 0.74 V. (b) Applying an external potential greater than 0.74 V in the reverse direction forces electrons to flow from the Cu electrode [which is now the anode, at which metallic Cu(s) is oxidized to Cu2+(aq)] and into the Cd electrode [which is now the cathode, at which Cd2+(aq) is reduced to Cd(s)]. The anode in an electrolytic cell is positive because electrons are flowing from it, whereas the cathode is negative because electrons are flowing into it.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (part (b) in ). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
Equation 19.105
Equation 19.106
Equation 19.107
Because E°cell < 0, the overall reaction—the reduction of Cd2+ by Cu—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in .
Table 19.3 Comparison of Galvanic and Electrolytic Cells
Property | Galvanic Cell | Electrolytic Cell |
---|---|---|
ΔG | < 0 | > 0 |
E cell | > 0 | < 0 |
Electrode Process | ||
anode | oxidation | oxidation |
cathode | reduction | reduction |
Sign of Electrode | ||
anode | − | + |
cathode | + | − |
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten NaCl, for example, and an electrical potential is applied, Cl− is oxidized at the anode, and Na+ is reduced at the cathode. The overall reaction is as follows:
Equation 19.108
2NaCl(l) → 2Na(l) + Cl2(g)This is the reverse of the formation of NaCl from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of NaCl (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten NaCl in a Downs cell (). In this specialized cell, CaCl2 (melting point = 772°C) is first added to the NaCl to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Figure 19.22 A Downs Cell for the Electrolysis of Molten NaCl
The electrolysis of a molten mixture of NaCl and CaCl2 results in the formation of elemental sodium and chlorine gas. Because sodium is a liquid under these conditions and liquid sodium is less dense than molten sodium chloride, the sodium floats to the top of the melt and is collected in concentric capped iron cylinders surrounding the cathode. Gaseous chlorine collects in the inverted cone over the anode. An iron screen separating the cathode and anode compartments ensures that the molten sodium and gaseous chlorine do not come into contact.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
Equation 19.109
2Al2O3(l) + 3C(s) → 4Al(l) + 3CO2(g)Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
In the Hall–Heroult process, C is oxidized instead of O2− or F− because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
A List all the possible reduction and oxidation products. Based on the electronegativity values shown in , determine which species will be reduced and which species will be oxidized.
B Identify the products that will form at each electrode.
Solution:
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer: Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (). The reactions that occur are as follows:
Equation 19.110
Equation 19.111
Equation 19.112
Figure 19.23 The Electrolysis of Water
Applying an external potential of about 1.7–1.9 V to two inert electrodes immersed in an aqueous solution of an electrolyte such as H2SO4 or Na2SO4 drives the thermodynamically nonspontaneous decomposition of water into H2 at the cathode and O2 at the anode.
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH−] = 1.0 × 10−7. Assuming that = = 1 atm, we can use the standard potentials and to calculate E for the overall reaction:
Equation 19.113
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltageThe voltage that must be applied in electrolysis in addition to the calculated (theoretical) value to overcome factors such as a high activation energy and the formation of bubbles on a surface., represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3−, SO42−, PO43−, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
In a process called electroplatingA process in which a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis., a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in .
Figure 19.24 Electroplating
(a) Electroplating uses an electrolytic cell in which the object to be plated, such as a fork, is immersed in a solution of the metal to be deposited. The object being plated acts as the cathode, on which the desired metal is deposited in a thin layer, while the anode usually consists of the metal that is being deposited (in this case, silver) that maintains the solution concentration as it dissolves. (b) In this commercial electroplating apparatus, a large number of objects can be plated simultaneously by lowering the rack into the Ag+ solution and applying the correct potential.
The half-reactions in electroplating a fork, for example, with silver are as follows:
Equation 19.114
Equation 19.115
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because E°cell = 0 V, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction Ag+(aq) + e− → Ag(s), 1 mol of electrons reduces 1 mol of Ag+ to Ag metal. In contrast, in the reaction Cu2+(aq) + 2e− → Cu(s), 1 mol of electrons reduces only 0.5 mol of Cu2+ to Cu metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,486 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (C) transferred is the product of the current (A) and the time (t, in seconds):
Equation 19.116
C = A × tThe stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of CuSO4 for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
Equation 19.117
The number of moles of electrons transferred to Cu2+ is therefore
Equation 19.118
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
A Calculate the number of moles of metal corresponding to the given mass transferred.
B Write the reaction and determine the number of moles of electrons required for the electroplating process.
C Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution:
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
B The reduction reaction is Ag+(aq) + e− → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer: 5.8 h
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. A Downs cell is used to produce sodium metal from a mixture of salts, and the Hall–Heroult process is used to produce aluminum commercially. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.
Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?
How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?
Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?
Two solutions, one containing Fe(NO3)2·6H2O and the other containing the same molar concentration of Fe(NO3)3·6H2O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer.
The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?
Electrolysis is the most direct way of recovering a metal from its ores. However, the Na+(aq)/Na(s), Mg2+(aq)/Mg(s), and Al3+(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?
What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl2 is electrolyzed using a current of 12.4 A for 1.0 h?
What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO3)2 solution for 1.5 h.
What mass of PbO2 is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery?
Electrolysis of Cr3+(aq) produces Cr2+(aq). If you had 500 mL of a 0.15 M solution of Cr3+(aq), how long would it take to reduce the Cr3+ to Cr2+ using a 0.158 A current?
Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.
5.2 L
Problems marked with a ♦ involve multiple concepts.
The percent efficiency of a fuel cell is defined as ΔG°/ΔH° × 100. If hydrogen gas were distributed for domestic and industrial use from a central electrolysis facility, the gas could be piped to consumers much as methane is piped today. Conventional nuclear power stations have an efficiency of 25%–30%. Use tabulated data to calculate the efficiency of a fuel cell in which the reaction H2(g) + 1/2O2(g) → H2O(g) occurs under standard conditions.
♦ You are about to run an organic reaction and need a strong oxidant. Although you have BrO3− at your disposal, you prefer to use MnO4−. You notice you also have MnO2 in the lab.
It is possible to construct a galvanic cell using amalgams as electrodes, each containing different concentrations of the same metal. One example is the Pb(Hg)(a1)∣PbSO4(soln)∣Pb(Hg)(a2) cell, in which a1 and a2 represent the concentrations of lead in the amalgams. No chemical change occurs; rather, the reaction transfers lead from one amalgam to the other, thus altering the Pb concentration in both amalgams. Write an equation for E for such a cell.
♦ The oldest known metallurgical artifacts are beads made from alloys of copper, produced in Egypt, Mesopotamia, and the Indus Valley around 3000 BC. To determine the copper content of alloys such as brass, a sample is dissolved in nitric acid to obtain Cu2+(aq), and then the pH is adjusted to 7.0. Excess KI is used to reduce the Cu2+ to Cu+ with concomitant oxidation of I− to I2. The iodine that is produced is then titrated with thiosulfate solution to determine the amount of Cu2+ in the original solution. The following reactions are involved in the procedure:
The biological electron transport chain provides for an orderly, stepwise transfer of electrons. Both NADH (reduced nicotinamide adenine dinucleotide) and FADH2 (reduced flavin adenine dinucleotide) are energy-rich molecules that liberate a large amount of energy during oxidation. Free energy released during the transfer of electrons from either of these molecules to oxygen drives the synthesis of ATP (adenosine triphosphate) formed during respiratory metabolism. The reactions are as follows:
The standard potential (E°′) for a biological process is defined at pH = 7.0.
While working at a nuclear reactor site, you have been put in charge of reprocessing spent nuclear fuel elements. Your specific task is to reduce Pu(VI) to elemental Pu without reducing U(VI) to elemental U. You have the following information at your disposal:
Use tabulated data to decide what reductant will accomplish your task in an acidic solution containing 1.0 M concentrations of both UO22+ and PuO22+.
Stainless steels typically contain 11% Cr and are resistant to corrosion because of the formation of an oxide layer that can be approximately described as FeCr2O4, where the iron is Fe(II). The protective layer forms when Cr(II) is oxidized to Cr(III) and Fe is oxidized to Fe(II). Explain how this film prevents the corrosion of Fe to rust, which has the formula Fe2O3.
♦ Ion-selective electrodes are powerful tools for measuring specific concentrations of ions in solution. For example, they are used to measure iodide in milk, copper-ion levels in drinking water, fluoride concentrations in toothpastes, and the silver-ion concentration in photographic emulsions and spent fixing solutions. Describe how ion-selective electrodes work and then propose a design for an ion-selective electrode that can be used for measuring water hardness (Ca2+, Mg2+) in water-conditioning systems.
♦ Enzymes are proteins that catalyze a specific reaction with a high degree of specificity. An example is the hydrolysis of urea by urease:
An enzyme electrode for measuring urea concentrations can be made by coating the surface of a glass electrode with a gel that contains urease.
Gas-sensing electrodes can be constructed using a combination electrode that is surrounded by a gas-permeable membrane. For example, to measure CO2, a pH electrode and a reference electrode are placed in solution on the “inner” side of a CO2-permeable membrane, and the sample solution is placed on the “external” side. As CO2 diffuses through the membrane, the pH of the internal solution changes due to the reaction CO2(g) + H2O(l) → HCO3−(aq) + H+(aq). Thus the pH of the internal solution varies directly with the CO2 pressure in the external sample. Ammonia electrodes operate in the same manner. Describe an electrode that would test for ammonia levels in seawater.
US submarines that are not nuclear powered use a combination of batteries and diesel engines for their power. When submerged, they are battery driven; when on the surface, they are diesel driven. Why are batteries not used when submarines are on the surface?
List some practical considerations in designing a battery to power an electric car.
♦ It is possible to run a digital clock using the power supplied by two potatoes. The clock is connected to two wires: one is attached to a copper plate, and the other is attached to a zinc plate. Each plate is pushed into a different potato; when a wire connects the two potatoes, the clock begins to run as if it were connected to a battery.
♦ The silver–zinc battery has the highest energy density of any rechargeable battery available today. Its use is presently limited to military applications, primarily in portable communications, aerospace, and torpedo-propulsion systems. The disadvantages of these cells are their limited life (they typically last no more than about 2 yr) and their high cost, which restricts their use to situations in which cost is only a minor factor. The generally accepted equations representing this type of battery are as follows:
All metals used in boats and ships are subject to corrosion, particularly when the vessels are operated in salt water, which is a good electrolyte. Based on the data in the following table, where potentials are measured using a glass electrode, explain why
Metal | E versus Ag/AgCl (V) |
---|---|
titanium | 0.02 |
Monel [Ni(Cu)] | −0.06 |
Ni(Al) bronze | −0.16 |
lead | −0.20 |
manganese bronze | −0.29 |
brass | −0.30 |
copper | −0.31 |
tin | −0.31 |
stainless steel | −0.49 |
aluminum | −0.87 |
zinc | −1.00 |
magnesium | −1.60 |
Parents often coat a baby’s first shoes with copper to preserve them for posterity. A conducting powder, such as graphite, is rubbed on the shoe, and then copper is electroplated on the shoe. How much copper is deposited on a shoe if the electrolytic process is run for 60 min at 1.2 A from a 1.0 M solution of CuSO4?
Before 1886, metallic aluminum was so rare that a bar of it was displayed next to the Crown Jewels at the Paris Exposition of 1855. Today, aluminum is obtained commercially from aluminum oxide by the Hall–Heroult process, an electrolytic process that uses molten Al2O3 and cryolite (Na3AlF6). As the operation proceeds, molten Al sinks to the bottom of the cell. The overall reaction is 2Al2O3(l) + 3C(s) → 4Al(l) + 3CO2(g); however, the process is only approximately 90% efficient.
♦ One of the most important electrolytic processes used in industry is the electrolytic reduction of acrylonitrile (CH2CHCN) to adiponitrile [NC(CH2)4CN]. The product is then hydrogenated to hexamethylenediamine [H2N(CH2)6NH2], a key component of one form of nylon. Using this process, Monsanto produces about 200,000 metric tons of adiponitrile annually. The cathode reaction in the electrochemical cell is as follows:
2CH2CHCN + 2H+ + 2e− → NC(CH2)4CNThe cost of electricity makes this an expensive process. Calculate the total number of kilowatt-hours of electricity used by Monsanto each year in this process, assuming a continuous applied potential of 5.0 V and an electrochemical efficiency of 50%. (One kilowatt-hour equals 3.6 × 103 kJ.)
♦ Compact discs (CDs) are manufactured by electroplating. Information is stored on a CD master in a pattern of “pits” (depressions, which correspond to an audio track) and “lands” (the raised areas between depressions). A laser beam cuts the pits into a plastic or glass material. The material is cleaned, sprayed with [Ag(NH3)2]+, and then washed with a formaldehyde solution that reduces the complex and leaves a thin silver coating. Nickel is electrodeposited on the disk and then peeled away to produce a master disk, which is used to stamp copies.
♦ Calculate the total amount of energy consumed in the electrolysis reaction used to make the 16 × 106 metric tons of aluminum produced annually worldwide, assuming a continuous applied potential of 5.0 V and an efficiency of 50%. Express your answer in kilojoules and in kilowatt-hours. (See Problem 19 for the conversion between kilowatt-hours and kilojoules.)
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei.
Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst.
We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe.
The glow caused by intense radiation. The high-energy particles ejected into the surrounding water or air by an intense radioactive source such as this nuclear reactor core produce a ghostly bluish glow.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
As you learned in , each element can be represented by the notation where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleonsThe protons and neutrons that make up the nucleus of an atom., and an atom with a particular number of protons and neutrons is called a nuclideAn atom with a particular number of nucleons.. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
Because the number of neutrons is equal to A − Z, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17.
Any nucleus that is unstable and decays spontaneously is said to be radioactiveAny nucleus that is unstable and decays spontaneously, emitting particles and electromagnetic radiation., emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopesAn isotope that emits radiation.. As you learned in , the rate at which radioactive decay occurs is characteristic of the isotope and is generally reported as a half-life (t1/2), the amount of time required for half of the initial number of nuclei present to decay in a first-order reaction. (For more information on half-life, see , .) An isotope’s half-life can range from fractions of a second to billions of years and, among other applications, can be used to measure the age of ancient objects. Example 1 and its corresponding exercise review the calculations involving radioactive decay rates and half-lives.
Fort Rock Cave in Oregon is the site where archaeologists discovered several Indian sandals, the oldest ever found in Oregon. Analysis of the 14C content of the sagebrush used to make the sandals gave an average decay rate of 5.1 disintegrations per minute (dpm) per gram of carbon. The current 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g C. How long ago was the sagebrush in the sandals cut? The half-life of 14C is 5730 yr.
Given: radioisotope, current 14C/12C ratio, initial decay rate, final decay rate, and half-life
Asked for: age
Strategy:
A Use to calculate N0/N, the ratio of the number of atoms of 14C originally present in the sample to the number of atoms now present.
B Substitute the value for the half-life of 14C into to obtain the rate constant for the reaction.
C Substitute the calculated values for N0/N and the rate constant into to obtain the elapsed time t.
Solution:
We can use the integrated rate law for a first-order nuclear reaction () to calculate the amount of time that has passed since the sagebrush was cut to make the sandals:
A From , we know that A = kN. We can therefore use the initial and final activities (A0 = 15 and A = 5.1) to calculate N0/N:
B Now we can calculate the rate constant k from the half-life of the reaction (5730 yr) using :
Rearranging this equation to solve for k,
C Substituting the calculated values into the equation for t,
Thus the sagebrush in the sandals is about 8900 yr old.
Exercise
While trying to find a suitable way to protect his own burial chamber, the ancient Egyptian pharaoh Sneferu developed the pyramid, a burial structure that protected desert graves from thieves and exposure to wind. Analysis of the 14C content of several items in pyramids built during his reign gave an average decay rate of 8.6 dpm/g C. When were the objects in the chamber created?
Answer: about 4600 yr ago, or about 2600 BC
As discussed in , the nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear forceAn extremely powerful but very short-range attractive force between nucleons that keeps the nucleus of an atom from flying apart (due to electrostatic repulsions between protons)., an extremely powerful but very short-range attractive force between nucleons (). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
Figure 20.1 Competing Interactions within the Atomic Nucleus
Electrostatic repulsions between positively charged protons would normally cause the nuclei of atoms (except H) to fly apart. In stable atomic nuclei, these repulsions are overcome by the strong nuclear force, a short-range but powerful attractive interaction between nucleons. If the attractive interactions due to the strong nuclear force are weaker than the electrostatic repulsions between protons, the nucleus is unstable, and it will eventually decay.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in . The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., and ). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
Figure 20.2 The Relationship between Nuclear Stability and the Neutron-to-Proton Ratio
In this plot of the number of neutrons versus the number of protons, each black point corresponds to a stable nucleus. In this classification, a stable nucleus is arbitrarily defined as one with a half-life longer than 46 billion years (10 times the age of Earth). As the number of protons (the atomic number) increases, the number of neutrons required for a stable nucleus increases even more rapidly. Isotopes shown in red, yellow, green, and blue are progressively less stable and more radioactive; the farther an isotope is from the diagonal band of stable isotopes, the shorter its half-life. The purple dots indicate superheavy nuclei that are predicted to be relatively stable, meaning that they are expected to be radioactive but to have relatively long half-lives. In most cases, these elements have not yet been observed or synthesized.
Data source: National Nuclear Data Center, Brookhaven National Laboratory, Evaluated Nuclear Structure Data File (ENSDF), Chart of Nuclides, http://www.nndc.bnl.gov/chart.
As shown in , more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are with 2 protons and 2 neutrons, and with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Figure 20.3 The Relationship between the Number of Protons and the Number of Neutrons and Nuclear Stability
Most stable nuclei contain even numbers of both neutrons and protons.
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in , the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Classify each nuclide as stable or radioactive.
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
Exercise
Classify each nuclide as stable or radioactive.
Answer:
In addition to the “peninsula of stability,” shows a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elementsAn element with an atomic number near the magic number of 126., with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature.
What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences.
What do chemists mean when they say a substance is radioactive?
What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table?
In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with Z > 83 unstable?
What is the significance of a magic number of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons?
Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers.
Potassium has three common isotopes, 39K, 40K, and 41K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of 40K.
Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is 144Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is 144Sm more stable?
Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not.
Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability.
Write the nuclear symbol for each isotope using notation.
Write the nuclear symbol for each isotope using notation.
Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope.
Which of these nuclides do you expect to be radioactive? Explain your reasoning.
Which of these nuclides do you expect to be radioactive? Explain your reasoning.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiationRadiation that is relatively low in energy. When it collides with an atom in a molecule or ion, most or all of its energy can be absorbed without causing a structural or a chemical change. is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiationRadiation of a high enough energy to transfer some as it passes through matter to one or more atoms with which it collides. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions. is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
Equation 20.23
atom + ionizing radiation → ion+ + e−Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle: 1 MeV/particle = 96 billion J/mol.
Figure 20.11 Radiation Damage
When high-energy particles emitted by radioactive decay interact with matter, they can break bonds or ionize molecules, resulting in changes in physical properties such as ductility or color. The glass electrical insulator on the left has not been exposed to radiation, but the insulator on the right has received intense radiation doses over a long period of time. Radiation damage changed the chemical structure of the glass, causing it to become bright blue.
The effects of ionizing radiation depend on four factors:
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in . Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Figure 20.12 Depth of Penetration of Ionizing Radiation
The depth of penetration of alpha, beta, and gamma radiation varies with the particle. Because α particles interact strongly with matter, they do not penetrate deeply into the human body. In contrast, β particles do not interact as strongly with matter and penetrate more deeply. Gamma rays, which have no charge, are stopped by only very dense materials and can pass right through the human body without being absorbed.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in .
Table 20.3 Some Properties of Ionizing Radiation
Type | Energy Range (MeV) | Penetration Distance in Water* | Penetration Distance in Air* |
---|---|---|---|
α particles | 3–9 | < 0.05 mm | < 10 cm |
β particles | ≤ 3 | < 4 mm | 1 m |
x-rays | <10−2 | < 1 cm | < 3 m |
γ rays | 10−2–101 | < 20 cm | > 3 m |
*Distance at which half of the radiation has been absorbed. |
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
There are many different ways to measure radiation exposure, or the dose. The roentgen (R)A unit that describes the amount of energy absorbed by dry air and measures the radiation exposure or dose., which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose)A unit used to measure the effects of radiation on biological tissues; the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter.; the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
Equation 20.24
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man)A unit that describes the actual amount of tissue damage caused by a given amount of radiation and equal to the number of rads multiplied by the RBE. was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (). One component of background radiation is cosmic rays, high-energy particles and γ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
Figure 20.13 The Radiation Exposure of a Typical Adult in the United States
The average radiation dose from natural sources for an adult in the United States is about 150–600 mrem/yr. Radon accounts for more than half of an adult’s total radiation exposure, whereas background radiation (terrestrial and cosmogenic) and exposure from medical sources account for about 15% each.
Data source: Office of Civilian Radioactive Waste Management
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as 14C:
Equation 20.25
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as 232Th and 238U, as well as radioactive daughter isotopes, such as 226Ra. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the K+ ion. Naturally occurring potassium contains 0.0117% 40K, which decays by emitting both a β particle and a γ ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 40K nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of 238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
Equation 20.26
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of 218Po releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The 218Po isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
A Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
B Determine the number of decays per year for this amount of 40K.
C Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution:
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event:
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
This corresponds to almost half of the normal background radiation most people experience.
Exercise
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer: 5.7 × 103 rad/yr (which is 10 times the fatal dose)
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in . Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table 20.4 The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) | Symptoms/Effects |
---|---|
< 5 | no observable effect |
5–20 | possible chromosomal damage |
20–100 | temporary reduction in white blood cell count |
50–100 | temporary sterility in men (up to a year) |
100–200 | mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded |
> 300 | permanent sterility in women |
> 500 | fatal to 50% within 30 days; destruction of bone marrow and intestine |
> 3000 | fatal within hours |
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in , but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Figure 20.14 Two Possible Relationships between the Number of Genetic Defects and Radiation Exposure
Studies on fruit flies show a linear relationship between the number of genetic defects and the magnitude of the radiation dose and exposure time, which is consistent with a cumulative effect of radiation. In contrast, studies on mice show an S-shaped curve, which suggests that the number of defects is lower when radiation exposure occurs over a longer time. Which of these relationships is more applicable to humans is a matter of considerable debate.
The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure.
Why are many radioactive substances warm to the touch? Why do many radioactive substances glow?
Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation?
Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why?
Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters.
Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why?
List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous?
Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects?
Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases?
Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale.
Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells.
Ten exposures of 10 rem are less likely to cause major damage.
A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is 238U, which has a half-life of 4.46 × 109 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads?
There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 106 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why?
Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy.
The relationship between mass (m) and energy (E) was introduced in and is expressed in the following equation:
Equation 20.27
E = mc2where c is the speed of light (2.998 × 108 m/s), and E and m are expressed in units of joules and kilograms, respectively. Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to , every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? (For more information on the conservation of mass, see .) The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected.This situation is similar to the wave–particle duality discussed in . As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows:
Equation 20.28
Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. As you learned in , when a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as
Equation 20.29
we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy:
Equation 20.30
Because 1 J = 1 (kg·m2)/s2, the change in mass is as follows:
Equation 20.31
This is a mass change of about 3.6 × 10−10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible.
In contrast, for a typical nuclear reaction, such as the radioactive decay of 14C to 14N and an electron (a β particle), there is a much larger change in mass:
Equation 20.32
We can use the experimentally measured masses of subatomic particles and common isotopes given in and to calculate the change in mass directly. The reaction involves the conversion of a neutral 14C atom to a positively charged 14N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral 14N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral 14N atom (14.003074 amu) and the mass of a 14C atom (14.003242 amu):
Equation 20.33
The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of 14C is −0.000168 g = −1.68 × 10−4 g = −1.68 × 10−7 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows:
Equation 20.34
The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of 14C is a relatively low-energy nuclear reaction.
Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 103 eV), megaelectronvolts (1 MeV = 106 eV), and even gigaelectronvolts (1 GeV = 109 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of 14C is thus
Equation 20.35
Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of 238U to 234Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom.
Given: nuclear decay reaction
Asked for: changes in mass and energy
Strategy:
A Use the mass values in and to calculate the change in mass for the decay reaction in atomic mass units.
B Use to calculate the change in energy in joules per mole.
C Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom.
Solution:
A Using particle and isotope masses from and , we can calculate the change in mass as follows:
B Thus the change in mass for 1 mol of 238U is −0.004584 g or −4.584 × 10−6 kg. The change in energy in joules per mole is as follows:
ΔE = (Δm)c2 = (−4.584 × 10−6 kg)(2.998 × 108 m/s)2 = −4.120 × 1011 J/molC The change in energy in electronvolts per atom is as follows:
Exercise
Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium (3H) to 3He and a β particle.
Answer: Δm = −2.0 × 10−5 amu; ΔE = −1.9 × 106 kJ/mol = −19 keV/atom
We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 (1H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium (2H) is 2.014102 amu, although its calculated mass is 2.016490 amu:
Equation 20.36
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defectThe difference between the sum of the masses of the components of an atom (protons, neutrons, and electrons) and the measured atomic mass. of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to , this release of energy must be accompanied by a decrease in the mass of the nucleus.
The amount of energy released when a nucleus forms from its component nucleons is the nuclear binding energyThe amount of energy released when a nucleus forms from its component nucleons, which corresponds to the mass defect of the nucleus. (). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus.
Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components.
Figure 20.15 Nuclear Binding Energy in Deuterium
The mass of a 2H atom is less than the sum of the masses of a proton, a neutron, and an electron by 0.002388 amu; the difference in mass corresponds to the nuclear binding energy. The larger the value of the mass defect, the greater the nuclear binding energy and the more stable the nucleus.
Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in , the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., 4He, 12C, and 16O). As mentioned earlier, these are particularly stable combinations.
Because the maximum binding energy per nucleon is reached at 56Fe, all other nuclei are thermodynamically unstable with regard to the formation of 56Fe. Consequently, heavier nuclei (toward the right in ) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in ) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern.
Figure 20.16 The Curve of Nuclear Binding Energy
This plot of the average binding energy per nucleon as a function of atomic number shows that the binding energy per nucleon increases with increasing atomic number until about Z = 26, levels off, and then decreases. The sharp peaks correspond to light nuclei that have equal numbers of protons and neutrons.
Heavier nuclei spontaneously undergo nuclear reactions that decrease their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that increase their atomic number.
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 56Fe. The experimental mass of the nuclide is given in .
Given: nuclide and mass
Asked for: nuclear binding energy and binding energy per nucleon
Strategy:
A Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of 1H atoms (because its mass is the same as the mass of one electron and one proton).
B Calculate the mass defect by subtracting the experimental mass from the calculated mass.
C Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon.
Solution:
A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 1H atoms, we can calculate the sum of the masses more quickly as follows:
B We subtract to find the mass defect:
C The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon.
Exercise
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 238U.
Answer: 1800 MeV/238U; 7.57 MeV/nucleon
First discussed in , nuclear fissionThe splitting of a heavy nucleus into two lighter ones. is the splitting of a heavy nucleus into two lighter ones. Fission was discovered in 1938 by the German scientists Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with neutrons in an attempt to produce new elements with Z > 92. They observed that lighter elements such as barium (Z = 56) were formed during the reaction, and they realized that such products had to originate from the neutron-induced fission of uranium-235:
Equation 20.37
This hypothesis was confirmed by detecting the krypton-92 fission product. As discussed in , the nucleus usually divides asymmetrically rather than into two equal parts, and the fission of a given nuclide does not give the same products every time.
In a typical nuclear fission reaction, more than one neutron is released by each dividing nucleus. When these neutrons collide with and induce fission in other neighboring nuclei, a self-sustaining series of nuclear fission reactions known as a nuclear chain reactionA self-sustaining series of nuclear fission reactions. can result (). For example, the fission of 235U releases two to three neutrons per fission event. If absorbed by other 235U nuclei, those neutrons induce additional fission events, and the rate of the fission reaction increases geometrically. Each series of events is called a generation. Experimentally, it is found that some minimum mass of a fissile isotope is required to sustain a nuclear chain reaction; if the mass is too low, too many neutrons are able to escape without being captured and inducing a fission reaction. The minimum mass capable of supporting sustained fission is called the critical massThe minimum mass of a fissile isotope capable of supporting sustained fission.. This amount depends on the purity of the material and the shape of the mass, which corresponds to the amount of surface area available from which neutrons can escape, and on the identity of the isotope. If the mass of the fissile isotope is greater than the critical mass, then under the right conditions, the resulting supercritical mass can release energy explosively. The enormous energy released from nuclear chain reactions is responsible for the massive destruction caused by the detonation of nuclear weapons such as fission bombs, but it also forms the basis of the nuclear power industry.
Nuclear fusionThe combining of two light nuclei to produce a heavier, more stable nucleus., in which two light nuclei combine to produce a heavier, more stable nucleus, is the opposite of nuclear fission. As in the nuclear transmutation reactions discussed in , the positive charge on both nuclei results in a large electrostatic energy barrier to fusion. This barrier can be overcome if one or both particles have sufficient kinetic energy to overcome the electrostatic repulsions, allowing the two nuclei to approach close enough for a fusion reaction to occur. The principle is similar to adding heat to increase the rate of a chemical reaction. (For more information on chemical kinetics, see .) As shown in the plot of nuclear binding energy per nucleon versus atomic number in , fusion reactions are most exothermic for the lightest element. For example, in a typical fusion reaction, two deuterium atoms combine to produce helium-3, a process known as deuterium–deuterium fusion (D–D fusion):
Equation 20.38
Figure 20.17 A Nuclear Chain Reaction
The process is initiated by the collision of a single neutron with a 235U nucleus, which undergoes fission, as shown in . Because each neutron released can cause the fission of another 235U nucleus, the rate of a fission reaction accelerates geometrically. Each series of events is a generation.
In another reaction, a deuterium atom and a tritium atom fuse to produce helium-4 (), a process known as deuterium–tritium fusion (D–T fusion):
Equation 20.39
Figure 20.18 Nuclear Fusion
In a nuclear fusion reaction, lighter nuclei combine to produce a heavier nucleus. As shown, fusion of 3H and 2H to give 4He and a neutron releases an enormous amount of energy. In principle, nuclear fusion can produce much more energy than fission, but very high kinetic energy is required to overcome electrostatic repulsions between the positively charged nuclei and initiate the fusion reaction.
Initiating these reactions, however, requires a temperature comparable to that in the interior of the sun (approximately 1.5 × 107 K). Currently, the only method available on Earth to achieve such a temperature is the detonation of a fission bomb. For example, the so-called hydrogen bomb (or H bomb) is actually a deuterium–tritium bomb (a D–T bomb), which uses a nuclear fission reaction to create the very high temperatures needed to initiate fusion of solid lithium deuteride (6LiD), which releases neutrons that then react with 6Li, producing tritium. The deuterium-tritium reaction releases energy explosively. Example 9 and its corresponding exercise demonstrate the enormous amounts of energy produced by nuclear fission and fusion reactions. In fact, fusion reactions are the power sources for all stars, including our sun.
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when the neutron-induced fission of 235U produces 144Cs, 90Rb, and two neutrons:
Given: balanced nuclear reaction
Asked for: energy released in electronvolts per atom and kilojoules per mole
Strategy:
A Following the method used in Example 7, calculate the change in mass that accompanies the reaction. Convert this value to the change in energy in electronvolts per atom.
B Calculate the change in mass per mole of 235U. Then use to calculate the change in energy in kilojoules per mole.
Solution:
A The change in mass that accompanies the reaction is as follows:
The change in energy in electronvolts per atom is as follows:
B The change in mass per mole of is −0.188386 g = −1.88386 × 10−4 kg, so the change in energy in kilojoules per mole is as follows:
Exercise
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when deuterium and tritium fuse to give helium-4 and a neutron:
Answer: ΔE = −17.6 MeV/atom = −1.697 × 109 kJ/mol
Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: ΔE = (Δm)c2. Large changes in energy are usually reported in kiloelectronvolts or megaelectronvolts (thousands or millions of electronvolts). With the exception of 1H, the experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect of the nucleus. The energy corresponding to the mass defect is the nuclear binding energy, the amount of energy released when a nucleus forms from its component particles. In nuclear fission, nuclei split into lighter nuclei with an accompanying release of multiple neutrons and large amounts of energy. The critical mass is the minimum mass required to support a self-sustaining nuclear chain reaction. Nuclear fusion is a process in which two light nuclei combine to produce a heavier nucleus plus a great deal of energy.
How do chemical reactions compare with nuclear reactions with respect to mass changes? Does either type of reaction violate the law of conservation of mass? Explain your answers.
Why is the amount of energy released by a nuclear reaction so much greater than the amount of energy released by a chemical reaction?
Explain why the mass of an atom is less than the sum of the masses of its component particles.
The stability of a nucleus can be described using two values. What are they, and how do they differ from each other?
In the days before true chemistry, ancient scholars (alchemists) attempted to find the philosopher’s stone, a material that would enable them to turn lead into gold. Is the conversion of Pb → Au energetically favorable? Explain why or why not.
Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome.
Imagine that the universe is dying, the stars have burned out, and all the elements have undergone fusion or radioactive decay. What would be the most abundant element in this future universe? Why?
Numerous elements can undergo fission, but only a few can be used as fuels in a reactor. What aspect of nuclear fission allows a nuclear chain reaction to occur?
How are transmutation reactions and fusion reactions related? Describe the main impediment to fusion reactions and suggest one or two ways to surmount this difficulty.
Using the information provided, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole.
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 87Sr if the measured mass of 87Sr is 86.908877 amu.
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 60Ni.
The experimentally determined mass of 53Mn is 52.941290 amu. Find each of the following.
The experimentally determined mass of 29S is 28.996610 amu. Find each of the following.
Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 141Ba, 92Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 90Sr, 143Xe, and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction?
Calculate the amount of energy released by the fusion of 6Li and deuterium to give two helium-4 nuclei. Express your answer in electronvolts per atom and kilojoules per mole.
How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture?
757 MeV/atom, 8.70 MeV/nucleon
−173 MeV/atom; 1.67 × 1010 kJ/mol
ΔE = + 9.0 × 106 kJ/mol beryllium-8; no
D–D fusion: ΔE = −4.03 MeV/tritium nucleus formed = −3.89 × 108 kJ/mol tritium; D–T fusion: ΔE = −17.6 MeV/tritium nucleus = −1.70 × 109 kJ/mol; D–T fusion
The ever-increasing energy needs of modern societies have led scientists and engineers to develop ways of harnessing the energy released by nuclear reactions. To date, all practical applications of nuclear power have been based on nuclear fission reactions. Although nuclear fusion offers many advantages in principle, technical difficulties in achieving the high energies required to initiate nuclear fusion reactions have thus far precluded using fusion for the controlled release of energy. In this section, we describe the various types of nuclear power plants that currently generate electricity from nuclear reactions, along with some possible ways to harness fusion energy in the future. In addition, we discuss some of the applications of nuclear radiation and radioisotopes, which have innumerable uses in medicine, biology, chemistry, and industry.
Pitchblende. Pitchblende, the major uranium ore, consisting mainly of uranium oxide.
When a critical mass of a fissile isotope is achieved, the resulting flux of neutrons can lead to a self-sustaining reaction. A variety of techniques can be used to control the flow of neutrons from such a reaction, which allows nuclear fission reactions to be maintained at safe levels. Many levels of control are required, along with a fail-safe design, because otherwise the chain reaction can accelerate so rapidly that it releases enough heat to melt or vaporize the fuel and the container, a situation that can release enough radiation to contaminate the surrounding area. Uncontrolled nuclear fission reactions are relatively rare, but they have occurred at least 18 times in the past. The most recent event resulted from the damaged Fukushima Dai-ichi plant after the March 11, 2011, earthquake and tsunami that devastated Japan. The plant used fresh water for cooling nuclear fuel rods to maintain controlled, sustainable nuclear fission. When the water supply was disrupted, so much heat was generated that a partial meltdown occurred. Radioactive iodine levels in contaminated seawater from the plant were over 4300 times the regulated safety limit. To put this in perspective, drinking one liter of fresh water with this level of contamination is the equivalent to receiving double the annual dose of radiation that is typical for a person. Dismantling the plant and decontaminating the site is estimated to require 30 years at a cost of approximately $12 billion.
There is compelling evidence that uncontrolled nuclear chain reactions occurred naturally in the early history of our planet, about 1.7 billion years ago in uranium deposits near Oklo in Gabon, West Africa (). The natural abundance of 235U 2 billion years ago was about 3%, compared with 0.72% today; in contrast, the “fossil nuclear reactor” deposits in Gabon now contain only 0.44% 235U. An unusual combination of geologic phenomena in this region apparently resulted in the formation of deposits of essentially pure uranium oxide containing 3% 235U, which coincidentally is identical to the fuel used in many modern nuclear plants. When rainwater or groundwater saturated one of these deposits, the water acted as a natural moderator that decreased the kinetic energy of the neutrons emitted by radioactive decay of 235U, allowing the neutrons to initiate a chain reaction. As a result, the entire deposit “went critical” and became an uncontrolled nuclear chain reaction, which is estimated to have produced about 100 kW of power. It is thought that these natural nuclear reactors operated only intermittently, however, because the heat released would have vaporized the water. Removing the water would have shut down the reactor until the rocks cooled enough to allow water to reenter the deposit, at which point the chain reaction would begin again. This on–off cycle is believed to have been repeated for more than 100,000 years, until so much 235U was consumed that the deposits could no longer support a chain reaction.
Figure 20.19 A “Fossil Nuclear Reactor” in a Uranium Mine Near Oklo in Gabon, West Africa
More than a billion years ago, a number of uranium-rich deposits in West Africa apparently “went critical,” initiating uncontrolled nuclear fission reactions that may have continued intermittently for more than 100,000 years, until the concentration of uranium-235 became too low to support a chain reaction. This photo shows a geologist standing in a mine dug to extract the concentrated uranium ore. Commercial interest waned rapidly after it was recognized that the uranium ore was severely depleted in uranium-235, the isotope of interest.
In addition to the incident in Japan, another recent instance of an uncontrolled nuclear chain reaction occurred on April 25–26, 1986, at the Chernobyl nuclear power plant in the former Union of Soviet Socialist Republics (USSR; now in the Ukraine; ). During testing of the reactor’s turbine generator, a series of mechanical and operational failures caused a chain reaction that quickly went out of control, destroying the reactor core and igniting a fire that destroyed much of the facility and released a large amount of radioactivity. Thirty people were killed immediately, and the high levels of radiation in a 20 mi radius forced nearly 350,000 people to be resettled or evacuated. In addition, the accident caused a disruption to the Soviet economy that is estimated to have cost almost $13 billion. It is somewhat surprising, however, that the long-term health effects on the 600,000 people affected by the accident appear to be much less severe than originally anticipated. Initially, it was predicted that the accident would result in tens of thousands of premature deaths, but an exhaustive study almost 20 yr after the event suggests that 4000 people will die prematurely from radiation exposure due to the accident. Although significant, in fact it represents only about a 3% increase in the cancer rate among the 600,000 people most affected, of whom about a quarter would be expected to eventually die of cancer even if the accident had not occurred.
Figure 20.20 The Chernobyl Nuclear Power Plant
In 1986, mechanical and operational failures during testing at the Chernobyl power plant in the USSR (now in the Ukraine) caused an uncontrolled nuclear chain reaction. The resulting fire destroyed much of the facility and severely damaged the core of the reactor, resulting in the release of large amounts of radiation that was spread over the surrounding area by the prevailing winds. The effects were devastating to the health of the population in the region and to the Soviet economy.
If, on the other hand, the neutron flow in a reactor is carefully regulated so that only enough heat is released to boil water, then the resulting steam can be used to produce electricity. Thus a nuclear reactor is similar in many respects to the conventional power plants discussed in , which burn coal or natural gas to generate electricity; the only difference is the source of the heat that converts water to steam.
We begin our description of nuclear power plants with light-water reactors, which are used extensively to produce electricity in countries such as Japan, Israel, South Korea, Taiwan, and France—countries that lack large reserves of fossil fuels. The essential components of a light-water reactor are depicted in . All existing nuclear power plants have similar components, although different designs use different fuels and operating conditions. Fuel rods containing a fissile isotope in a structurally stabilized form (such as uranium oxide pellets encased in a corrosion-resistant zirconium alloy) are suspended in a cooling bath that transfers the heat generated by the fission reaction to a secondary cooling system. The heat is used to generate steam for the production of electricity. In addition, control rods are used to absorb neutrons and thereby control the rate of the nuclear chain reaction. Control rods are made of a substance that efficiently absorbs neutrons, such as boron, cadmium, or, in nuclear submarines, hafnium. Pulling the control rods out increases the neutron flux, allowing the reactor to generate more heat, whereas inserting the rods completely stops the reaction, a process called “scramming the reactor.”
Figure 20.21 A Light-Water Nuclear Fission Reactor for the Production of Electric Power
The fuel rods are made of a corrosion-resistant alloy that encases the partially enriched uranium fuel; controlled fission of 235U in the fuel produces heat. Water surrounds the fuel rods and moderates the kinetic energy of the neutrons, slowing them to increase the probability that they will induce fission. Control rods that contain elements such as boron, cadmium, or hafnium, which are very effective at absorbing neutrons, are used to control the rate of the fission reaction. A heat exchanger is used to boil water in a secondary cooling system, creating steam to drive the turbine and produce electricity. The large hyperbolic cooling tower, which is the most visible portion of the facility, condenses the steam in the secondary cooling circuit; it is often located at some distance from the actual reactor.
Despite this apparent simplicity, many technical hurdles must be overcome for nuclear power to be an efficient and safe source of energy. Uranium contains only 0.72% uranium-235, which is the only naturally occurring fissile isotope of uranium. Because this abundance is not enough to support a chain reaction, the uranium fuel must be at least partially enriched in 235U, to a concentration of about 3%, for it to be able to sustain a chain reaction. At this level of enrichment, a nuclear explosion is impossible; far higher levels of enrichment (greater than or equal to 90%) are required for military applications such as nuclear weapons or the nuclear reactors in submarines. Enrichment is accomplished by converting uranium oxide to UF6, which is volatile and contains discrete UF6 molecules. Because 235UF6 and 238UF6 have different masses, they have different rates of effusion and diffusion, and they can be separated using a gas diffusion process, as described in . Another difficulty is that neutrons produced by nuclear fission are too energetic to be absorbed by neighboring nuclei, and they escape from the material without inducing fission in nearby 235U nuclei. Consequently, a moderator must be used to slow the neutrons enough to allow them to be captured by other 235U nuclei. High-speed neutrons are scattered by substances such as water or graphite, which decreases their kinetic energy and increases the probability that they will react with another 235U nucleus. The moderator in a light-water reactor is the water that is used as the primary coolant. The system is highly pressurized to about 100 atm to keep the water from boiling at 100°C.
All nuclear reactors require a powerful cooling system to absorb the heat generated in the reactor core and create steam that is used to drive a turbine that generates electricity. In 1979, an accident occurred when the main water pumps used for cooling at the nuclear power plant at Three Mile Island in Pennsylvania stopped running, which prevented the steam generators from removing heat. Eventually, the zirconium casing of the fuel rods ruptured, resulting in a meltdown of about half of the reactor core. Although there was no loss of life and only a small release of radioactivity, the accident produced sweeping changes in nuclear power plant operations. The US Nuclear Regulatory Commission tightened its oversight to improve safety.
The main disadvantage of nuclear fission reactors is that the spent fuel, which contains too little of the fissile isotope for power generation, is much more radioactive than the unused fuel due to the presence of many daughter nuclei with shorter half-lives than 235U. The decay of these daughter isotopes generates so much heat that the spent fuel rods must be stored in water for as long as 5 yr before they can be handled. Even then, the radiation levels are so high that the rods must be stored for many, many more years to allow the daughter isotopes to decay to nonhazardous levels. How to store these spent fuel rods for hundreds of years is a pressing issue that has not yet been successfully resolved. As a result, some people are convinced that nuclear power is not a viable option for providing our future energy needs, although a number of other countries continue to rely on nuclear reactors for a large fraction of their energy.
Deuterium (2H) absorbs neutrons much less effectively than does hydrogen (1H), but it is about twice as effective at slowing neutrons. Consequently, a nuclear reactor that uses D2O instead of H2O as the moderator is so efficient that it can use unenriched uranium as fuel. Using a lower grade of uranium reduces operating costs and eliminates the need for plants that produce enriched uranium. Because of the expense of D2O, however, only countries like Canada, which has abundant supplies of hydroelectric power for generating D2O by electrolysis, have made a major investment in heavy-water reactors. (For more information on electrolysis, see .)
A breeder reactor is a nuclear fission reactor that produces more fissionable fuel than it consumes. This does not violate the first law of thermodynamics because the fuel produced is not the same as the fuel consumed. Under heavy neutron bombardment, the nonfissile 238U isotope is converted to 239Pu, which can undergo fission:
Equation 20.40
The overall reaction is thus the conversion of nonfissile 238U to fissile 239Pu, which can be chemically isolated and used to fuel a new reactor. An analogous series of reactions converts nonfissile 232Th to 233U, which can also be used as a fuel for a nuclear reactor. Typically, about 8–10 yr are required for a breeder reactor to produce twice as much fissile material as it consumes, which is enough to fuel a replacement for the original reactor plus a new reactor. The products of the fission of 239Pu, however, have substantially longer half-lives than the fission products formed in light-water reactors.
Although nuclear fusion reactions, such as those in and , are thermodynamically spontaneous, the positive charge on both nuclei results in a large electrostatic energy barrier to the reaction. (As you learned in , thermodynamic spontaneity is unrelated to the reaction rate.) Extraordinarily high temperatures (about 1.0 × 108°C) are required to overcome electrostatic repulsions and initiate a fusion reaction. Even the most feasible such reaction, deuterium–tritium fusion (D–T fusion; ), requires a temperature of about 4.0 × 107°C. Achieving these temperatures and controlling the materials to be fused are extraordinarily difficult problems, as is extracting the energy released by the fusion reaction, because a commercial fusion reactor would require such high temperatures to be maintained for long periods of time. Several different technologies are currently being explored, including the use of intense magnetic fields to contain ions in the form of a dense, high-energy plasma at a temperature high enough to sustain fusion (part (a) in ). Another concept employs focused laser beams to heat and compress fuel pellets in controlled miniature fusion explosions (part (b) in ).
Figure 20.22 Two Possible Designs for a Nuclear Fusion Reactor
The extraordinarily high temperatures needed to initiate a nuclear fusion reaction would immediately destroy a container made of any known material. (a) One way to avoid contact with the container walls is to use a high-energy plasma as the fuel. Because plasma is essentially a gas composed of ionized particles, it can be confined using a strong magnetic field shaped like a torus (a hollow donut). (b) Another approach to nuclear fusion is inertial confinement, which uses an icosahedral array of powerful lasers to heat and compress a tiny fuel pellet (a mixture of solid LiD and LiT) to induce fusion.
Nuclear reactions such as these are called thermonuclear reactionsA nuclear reaction that requires a great deal of thermal energy to initiate the reaction. because a great deal of thermal energy must be invested to initiate the reaction. The amount of energy released by the reaction, however, is several orders of magnitude greater than the energy needed to initiate it. In principle, a nuclear fusion reaction should thus result in a significant net production of energy. In addition, Earth’s oceans contain an essentially inexhaustible supply of both deuterium and tritium, which suggests that nuclear fusion could provide a limitless supply of energy. Unfortunately, however, the technical requirements for a successful nuclear fusion reaction are so great that net power generation by controlled fusion has yet to be achieved.
Nuclear radiation can damage biological molecules, thereby disrupting normal functions such as cell division (). Because radiation is particularly destructive to rapidly dividing cells such as tumor cells and bacteria, it has been used medically to treat cancer since 1904, when radium-226 was first used to treat a cancerous tumor. Many radioisotopes are now available for medical use, and each has specific advantages for certain applications.
In modern radiation therapy, radiation is often delivered by a source planted inside the body. For example, tiny capsules containing an isotope such as 192Ir, coated with a thin layer of chemically inert platinum, are inserted into the middle of a tumor that cannot be removed surgically. The capsules are removed when the treatment is over. In some cases, physicians take advantage of the body’s own chemistry to deliver a radioisotope to the desired location. For example, the thyroid glands in the lower front of the neck are the only organs in the body that use iodine. Consequently, radioactive iodine is taken up almost exclusively by the thyroid (part (a) in ). Thus when radioactive isotopes of iodine (125I or 131I) are injected into the blood of a patient suffering from thyroid cancer, the thyroid glands filter the radioisotope from the blood and concentrate it in the tissue to be destroyed. In cases where a tumor is surgically inaccessible (e.g., when it is located deep in the brain), an external radiation source such as a 60Co “gun” is used to aim a tightly focused beam of γ rays at it. Unfortunately, radiation therapy damages healthy tissue in addition to the target tumor and results in severe side effects, such as nausea, hair loss, and a weakened immune system. Although radiation therapy is generally not a pleasant experience, in many cases it is the only choice.
Figure 20.23 Medical Imaging and Treatment with Radioisotopes
(a) Radioactive iodine is used both to obtain images of the thyroid and to treat thyroid cancer. Injected iodine-123 or iodine-131 is selectively taken up by the thyroid gland, where it is incorporated into the thyroid hormone: thyroxine. Because iodine-131 emits low-energy β particles that are absorbed by the surrounding tissue, it can be used to destroy malignant tissue in the thyroid. In contrast, iodine-123 emits higher-energy γ rays that penetrate tissues readily, enabling it to image the thyroid gland, as shown here. (b) Some technetium compounds are selectively absorbed by cancerous cells within bones. The yellow spots show that a primary cancer has metastasized (spread) to the patient’s spine (lower center) and ribs (right center).
A second major medical use of radioisotopes is medical imaging, in which a radioisotope is temporarily localized in a particular tissue or organ, where its emissions provide a “map” of the tissue or the organ. Medical imaging uses radioisotopes that cause little or no tissue damage but are easily detected. One of the most important radioisotopes for medical imaging is 99mTc. Depending on the particular chemical form in which it is administered, technetium tends to localize in bones and soft tissues, such as the heart or the kidneys, which are almost invisible in conventional x-rays (part (b) in ). Some properties of other radioisotopes used for medical imaging are listed in .
Table 20.5 Radioisotopes Used in Medical Imaging and Treatment
Isotope | Half-Life | Tissue |
---|---|---|
18F | 110 min | brain |
24Na | 15 h | circulatory system |
32P | 14 days | eyes, liver, and tumors |
59Fe | 45 days | blood and spleen |
60Co | 5.3 yr | external radiotherapy |
99mTc | 6 h | heart, thyroid, liver, kidney, lungs, and skeleton |
125I | 59.4 days | thyroid, prostate, and brain |
131I | 8 days | thyroid |
133Xe | 5 days | lungs |
201Tl | 3 days | heart |
Because γ rays produced by isotopes such as 131I and 99mTc are emitted randomly in all directions, it is impossible to achieve high levels of resolution in images that use such isotopes. However, remarkably detailed three-dimensional images can be obtained using an imaging technique called positron emission tomography (PET). The technique uses radioisotopes that decay by positron emission, and the resulting positron is annihilated when it collides with an electron in the surrounding matter. (For more information on positron emission, see .) In the annihilation process, both particles are converted to energy in the form of two γ rays that are emitted simultaneously and at 180° to each other:
Equation 20.41
With PET, biological molecules that have been “tagged” with a positron-emitting isotope such as 18F or 11C can be used to probe the functions of organs such as the brain.
Another major health-related use of ionizing radiation is the irradiation of food, an effective way to kill bacteria such as Salmonella in chicken and eggs and potentially lethal strains of Escherichia coli in beef. Collectively, such organisms cause almost 3 million cases of food poisoning annually in the United States, resulting in hundreds of deaths. shows how irradiation dramatically extends the storage life of foods such as strawberries. Although US health authorities have given only limited approval of this technique, the growing number of illnesses caused by antibiotic-resistant bacteria is increasing the pressure to expand the scope of food irradiation.
Figure 20.24 The Preservation of Strawberries with Ionizing Radiation
Fruits such as strawberries can be irradiated by high-energy γ rays to kill bacteria and prolong their storage life. The nonirradiated strawberries on the left are completely spoiled after 15 days in storage, but the irradiated strawberries on the right show no visible signs of spoilage under the same conditions.
One of the more unusual effects of radioisotopes is in dentistry. Because dental enamels contain a mineral called feldspar (KAlSi3O8, which is also found in granite rocks), teeth contain a small amount of naturally occurring radioactive 40K. The radiation caused by the decay of 40K results in the emission of light (fluorescence), which gives the highly desired “pearly white” appearance associated with healthy teeth.
In a sign of how important nuclear medicine has become in diagnosing and treating illnesses, the medical community has become alarmed at the global shortage of 99Tc, a radioisotope used in more than 30 million procedures a year worldwide. Two reactors that produce 60% of the world’s radioactive 99Mo, which decays to 99Tc, are operating beyond their intended life expectancies. Moreover, because most of the reactors producing 99Mo use weapons-grade uranium (235U), which decays to 99Mo during fission, governments are working to phase out civilian uses of technology to produce 99Mo because of concerns that the technology can be used to produce nuclear weapons. Engineers are currently focused on how to make key medical isotopes with other alternatives that don’t require fission. One promising option is by removing a neutron from 100Mo, a stable isotope that makes up about 10% of natural molybdenum, transmuting it to 99Mo.
In addition to the medical uses of radioisotopes, radioisotopes have literally hundreds of other uses. For example, smoke detectors contain a tiny amount of 241Am, which ionizes the air in the detector so an electric current can flow through it. Smoke particles reduce the number of ionized particles and decrease the electric current, which triggers an alarm. Another application is the “go-devil” used to detect leaks in long pipelines. It is a packaged radiation detector that is inserted into a pipeline and propelled through the pipe by the flowing liquid. Sources of 60Co are attached to the pipe at regular intervals; as the detector travels along the pipeline, it sends a radio signal each time it passes a source. When a massive leak causes the go-devil to stop, the repair crews know immediately which section of the pipeline is damaged. Finally, radioactive substances are used in gauges that measure and control the thickness of sheets and films. As shown in , thickness gauges rely on the absorption of either β particles (by paper, plastic, and very thin metal foils) or γ rays (for thicker metal sheets); the amount of radiation absorbed can be measured accurately and is directly proportional to the thickness of the material.
Figure 20.25 Using Radiation to Control the Thickness of a Material
Because the amount of radiation absorbed by a material is proportional to its thickness, radiation can be used to control the thickness of plastic film, tin foil, or paper. As shown, a beta emitter is placed on one side of the material being produced and a detector on the other. An increase in the amount of radiation that reaches the detector indicates a decrease in the thickness of the material and vice versa. The output of the detector can thus be used to control the thickness of the material.
In nuclear power plants, nuclear reactions generate electricity. Light-water reactors use enriched uranium as a fuel. They include fuel rods, a moderator, control rods, and a powerful cooling system to absorb the heat generated in the reactor core. Heavy-water reactors use unenriched uranium as a fuel because they use D2O as the moderator, which scatters and slows neutrons much more effectively than H2O. A breeder reactor produces more fissionable fuel than it consumes. A nuclear fusion reactor requires very high temperatures. Fusion reactions are thermonuclear reactions because they require high temperatures for initiation. Radioisotopes are used in both radiation therapy and medical imaging.
In nuclear reactors, two different but interrelated factors must be controlled to prevent a mishap that could cause the release of unwanted radiation. How are these factors controlled?
What are the three principal components of a nuclear reactor? What is the function of each component?
What is meant by the term enrichment with regard to uranium for fission reactors? Why does the fuel in a conventional nuclear reactor have to be “enriched”?
The plot in a recent spy/action movie involved the threat of introducing stolen “weapons-grade” uranium, which consists of 93.3% 235U, into a fission reactor that normally uses a fuel containing about 3% 235U. Explain why this could be catastrophic.
Compare a heavy-water reactor with a light-water reactor. Why are heavy-water reactors less widely used? How do these two reactor designs compare with a breeder reactor?
Conventional light-water fission reactors require enriched fuel. An alternative reactor is the so-called heavy-water reactor. The components of the two different reactors are the same except that instead of using water (H2O), the moderator in a heavy-water reactor is D2O, known as “heavy water.” Because D2O is more efficient than H2O at slowing neutrons, heavy-water reactors do not require fuel enrichment to support fission. Why is D2O more effective at slowing neutrons, and why does this allow unenriched fuels to be used?
Isotopes emit γ rays in random directions. What difficulties do these emissions present for medical imaging? How are these difficulties overcome?
If you needed to measure the thickness of 1.0 mm plastic sheets, what type of radiation would you use? Would the radiation source be the same if you were measuring steel of a similar thickness? What is your rationale? Would you want an isotope with a long or short half-life for this device?
Neutron flow is regulated by using control rods that absorb neutrons, whereas the speed of the neutrons produced by fission is controlled by using a moderator that slows the neutrons enough to allow them to react with nearby fissile nuclei.
It is difficult to pinpoint the exact location of the nucleus that decayed. In contrast, the collision of a positron with an electron causes both particles to be annihilated, and in the process, two gamma rays are emitted in opposite directions, which makes it possible to identify precisely where a positron emitter is located and to create detailed images of tissues.
Palladium-103, which decays via electron capture and emits x-rays with an energy of 3.97 × 10−2 MeV, is often used to treat prostate cancer. Small pellets of the radioactive metal are embedded in the prostate gland. This provides a localized source of radiation to a very small area, even though the tissue absorbs only about 1% of the x-rays. Due to its short half-life, all of the palladium will decay to a stable isotope in less than a year. If a doctor embeds 50 pellets containing 2.50 mg of 103Pd in the prostate gland of a 73.9 kg patient, what is the patient’s radiation exposure over the course of a year?
Several medical treatments use cobalt-60m, which is formed by bombarding cobalt with neutrons to produce a highly radioactive gamma emitter that undergoes 4.23 × 1016 emissions/(s·kg) of pure cobalt-60. The energy of the gamma emission is 5.86 × 10−2 MeV. Write the balanced nuclear equation for the formation of this isotope. Is this a transmutation reaction? If a 55.3 kg patient received a 0.50 s exposure to a 0.30 kg cobalt-60 source, what would the exposure be in rads? Predict the potential side effects of this dose.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed.
In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure 20.26 "The Relative Abundances of the Elements in the Universe and on Earth". The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure 20.26 "The Relative Abundances of the Elements in the Universe and on Earth" illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the trends in nuclear stability discussed in Section 20.1 "The Components of the Nucleus", but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table 20.6 "Relative Abundances of Elements on Earth and in the Known Universe" for some common elements. Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. You may recall from Chapter 1 "Introduction to Chemistry" that the anomalously high iridium content of a 66-million-year-old rock layer was a key finding in the development of the asteroid-impact theory for the extinction of the dinosaurs. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Figure 20.26 The Relative Abundances of the Elements in the Universe and on Earth
In this logarithmic plot, the relative abundances of the elements relative to that of silicon (arbitrarily set equal to 1) in the universe (green bars) and on Earth (purple bars) are shown as a function of atomic number. Elements with even atomic numbers are generally more abundant in the universe than elements with odd atomic numbers. Also, the relative abundances of many elements in the universe are very different from their relative abundances on Earth.
Table 20.6 Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element | Abundance Ratio |
---|---|
H | 0.0020 |
He | 2.4 × 10−8 |
C | 0.36 |
N | 0.02 |
O | 46 |
Ne | 1.9 × 10−6 |
Na | 1200 |
Mg | 48 |
Al | 1600 |
Si | 390 |
S | 0.84 |
K | 5000 |
Ca | 710 |
Ti | 2200 |
Fe | 57 |
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure 20.27 "Nuclear Reactions during the Life Cycle of a Massive Star"). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
Figure 20.27 Nuclear Reactions during the Life Cycle of a Massive Star
At each stage in the lifetime of a star, a different fuel is used for nuclear fusion, resulting in the formation of different elements. Fusion of hydrogen to give helium is the primary fusion reaction in young stars. As the star ages, helium accumulates and begins to “burn,” undergoing fusion to form heavier elements such as carbon and oxygen. As the adolescent star matures, significant amounts of iron and nickel are formed by fusion of the heavier elements formed previously. The heaviest elements are formed only during the final death throes of the star—the formation of a nova or supernova.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
Equation 20.42
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great deal of energy:
Equation 20.43
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
Equation 20.44
Although beryllium-8 has both an even mass number and an even atomic number, the low neutron-to-proton ratio makes it very unstable, decomposing in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
Equation 20.45
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
Equation 20.46
Equation 20.47
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
Equation 20.48
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure 20.27 "Nuclear Reactions during the Life Cycle of a Massive Star"). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure 20.28 "A Supernova"). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
Equation 20.49
Figure 20.28 A Supernova
A view of the remains of Supernova 1987A, located in the Large Magellanic Cloud, showing the circular halo of expanding debris produced by the explosion. Multiple neutron-capture events occur during a supernova explosion, forming both the heaviest elements and many of the less stable nuclides.
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution:
Exercise
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer: 19 neutrons;
By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova.
Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements?
How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed?
Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers.
During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.
A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer.
If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation.
The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons.
Write a balanced nuclear reaction for the formation of each isotope.
At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion.
When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium?
Until the 1940s, uranium glazes were popular on ceramic dishware. One brand, Fiestaware, had bright orange glazes that could contain up to 20% uranium by mass. Although this practice is less common today due to the negative association of radiation, it is still possible to buy Depression-era glassware that is quite radioactive. Aqueous solutions in contact with this “hot” glassware can reach uranium concentrations up to 10 ppm by mass. If 1.0 g of uranium undergoes 1.11 × 107 decays/s, each to an α particle with an energy of 4.03 MeV, what would be your exposure in rem and rad if you drank 1.0 L of water that had been sitting for an extended time in a Fiestaware pitcher? Assume that the water and contaminants are excreted only after 18 h and that you weigh 70.0 kg.
Neutrography is a technique used to take the picture of an object using a beam of neutrons. How does the penetrating power of a neutron compare with alpha, beta, and gamma radiation? Do you expect similar penetration for protons? How would the biological damage of each particle compare with the other types of radiation? (Recall that a neutron’s mass is approximately 2000 times the mass of an electron.)
Spent fuel elements in a nuclear reactor contain radioactive fission products in addition to heavy metals. The conversion of nuclear fuel in a reactor is shown here:
Neglecting the fission products, write balanced nuclear reactions for the conversion of the original fuel to each product.
The first atomic bomb used 235U as a fissile material, but there were immense difficulties in obtaining sufficient quantities of pure 235U. A second fissile element, plutonium, was discovered in 1940, and it rapidly became important as a nuclear fuel. This element was produced by irradiating 238U with neutrons in a nuclear reactor. Complete the series that produced plutonium, all isotopes of which are fissile:
Boron neutron capture therapy is a potential treatment for many diseases. As the name implies, when boron-10, one of the naturally occurring isotopes of boron, is bombarded with neutrons, it absorbs a neutron and emits an α particle. Write a balanced nuclear reaction for this reaction. One advantage of this process is that neutrons cause little damage on their own, but when they are absorbed by boron-10, they can cause localized emission of alpha radiation. Comment on the utility of this treatment and its potential difficulties.
An airline pilot typically flies approximately 80 h per month. If 75% of that time is spent at an altitude of about 30,000 ft, how much radiation is that pilot receiving in one month? over a 30 yr career? Is the pilot receiving toxic doses of radiation?
At a breeder reactor plant, a 72 kg employee accidentally inhaled 2.8 mg of 239Pu dust. The isotope decays by alpha decay and has a half-life of 24,100 yr. The energy of the emitted α particles is 5.2 MeV, and the dust stays in the employee’s body for 18 h.
For many years, the standard source for radiation therapy in the treatment of cancer was radioactive 60Co, which undergoes beta decay to 60Ni and emits two γ rays, each with an energy of 1.2 MeV. Show the sequence of nuclear reactions. If the half-life for beta decay is 5.27 yr, how many 60Co nuclei are present in a typical source undergoing 6000 dps that is used by hospitals? The mass of 60Co is 59.93 amu.
It is possible to use radioactive materials as heat sources to produce electricity. These radioisotope thermoelectric generators (RTGs) have been used in spacecraft and many other applications. Certain Cold War–era Russian-made RTGs used a 5.0 kg strontium-90 source. One mole of strontium-90 releases β particles with an energy of 0.545 MeV and undergoes 2.7 × 1013 decays/s. How many watts of power are available from this RTG (1 watt = 1 J/s)?
Potassium consists of three isotopes (potassium-39, potassium-40, and potassium-41). Potassium-40 is the least abundant, and it is radioactive, decaying to argon-40, a stable, nonradioactive isotope, by the emission of a β particle with a half-life of precisely 1.25 × 109 yr. Thus the ratio of potassium-40 to argon-40 in any potassium-40–containing material can be used to date the sample. In 1952, fragments of an early hominid, Meganthropus, were discovered near Modjokerto in Java. The bone fragments were lying on volcanic rock that was believed to be the same age as the bones. Potassium–argon dating on samples of the volcanic material showed that the argon-40-to-potassium-40 molar ratio was 0.00281:1. How old were the rock fragments? Could the bones also be the same age? Could radiocarbon dating have been used to date the fragments?
6.6 × 10−3 rad; 0.13 rem
130 W
In previous chapters, we used the principles of chemical bonding, thermodynamics, and kinetics to provide a conceptual framework for understanding the chemistry of the elements. Beginning in Chapter 21 "Periodic Trends and the ", we use the periodic table to guide our discussion of the properties and reactions of the elements and the synthesis and uses of some of their commercially important compounds. We begin this chapter with a review of periodic trends as an introduction, and then we describe the chemistry of hydrogen and the other s-block elements. In Chapter 22 "The ", we consider the chemistry of the p-block elements; Chapter 23 "The " presents the transition metals, in which the d-subshell is being filled. In this chapter, you will learn why potassium chloride is used as a substitute for sodium chloride in a low-sodium diet, why cesium is used as a photosensor, why the heating elements in electric ranges are coated with magnesium oxide, and why exposure to a radioactive isotope of strontium is more dangerous for children than for adults.
Flame tests. Heating a compound in a very hot flame results in the formation of its component atoms in electronically excited states. When an excited atom decays to the ground state, it emits light (Chapter 6 "The Structure of Atoms"). Each element emits light at characteristic frequencies. Flame tests are used to identify many elements based on the color of light emitted in the visible region of the electromagnetic spectrum. As shown here, sodium compounds produce an intense yellow light, whereas potassium compounds produce a crimson color.
As we begin our summary of periodic trends, recall from Chapter 7 "The Periodic Table and Periodic Trends" that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries.
The most important periodic trends in atomic properties are summarized in Figure 21.1 "Summary of Periodic Trends in Atomic Properties". Recall from Chapter 7 "The Periodic Table and Periodic Trends" that these trends are based on periodic variations in a single fundamental property, the effective nuclear chargeThe nuclear charge an electron actually experiences because of shielding from other electrons closer to the nucleus. (Zeff), which increases from left to right and from top to bottom in the periodic table (Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom").
The diagonal line in Figure 21.1 "Summary of Periodic Trends in Atomic Properties" separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations.
Figure 21.1 Summary of Periodic Trends in Atomic Properties
Ionization energies, the magnitude of electron affinities, and electronegativities generally increase from left to right and from bottom to top. In contrast, atomic size decreases from left to right and from bottom to top. Consequently, the elements in the upper right of the periodic table are the smallest and most electronegative; the elements in the bottom left are the largest and least electronegative. The semimetals lie along the diagonal line separating the metals from the nonmetals and exhibit intermediate properties.
The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms.
In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group.
Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF3 forms only the four-coordinate, tetrahedral BF4− ion, whereas under the same conditions AlF3 forms the six-coordinate, octahedral AlF63− ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H2O).
Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed.You may recall from Chapter 8 "Ionic versus Covalent Bonding" that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF6 have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results.
One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N2 contains an N≡N bond, but each phosphorus atom in tetrahedral P4 forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements.
Another important trend to note in main group chemistry is the chemical similarity between the lightest element of one group and the element immediately below and to the right of it in the next group, a phenomenon known as the diagonal effect (Figure 21.2 "The Diagonal Effect") There are, for example, significant similarities between the chemistry of Li and Mg, Be and Al, and B and Si. Both BeCl2 and AlCl3 have substantial covalent character, so they are somewhat soluble in nonpolar organic solvents. In contrast, although Mg and Be are in the same group, MgCl2 behaves like a typical ionic halide due to the lower electronegativity and larger size of magnesium.
Figure 21.2 The Diagonal Effect
The properties of the lightest element in a group are often more similar to those of the element below and to the right in the periodic table. For instance, the chemistry of lithium is more similar to that of magnesium in group 2 than it is to the chemistry of sodium, the next member in group 1.
The inert-pair effectThe empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths.
In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.
The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2 electrons, as shown in Table 21.1 "Ionization Energies (" by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.
Table 21.1 Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
Element | Electron Configuration | I1 (kJ/mol) | I1 + I2 + I3 (kJ/mol) | Average M–Cl Bond Energy (kJ/mol) |
---|---|---|---|---|
B | [He] 2s22p1 | 801 | 6828 | 536 |
Al | [Ne] 3s23p1 | 578 | 5139 | 494 |
Ga | [Ar] 3d104s24p1 | 579 | 5521 | 481 |
In | [Kr] 4d105s2p1 | 558 | 5083 | 439 |
Tl | [Xe] 4f145d106s2p1 | 589 | 5439 | 373 |
Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).
Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table 21.1 "Ionization Energies (", the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.
The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.
Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section,
Given: positions of elements in the periodic table
Asked for: classification, oxidation-state stability, and chemical reactivity
Strategy:
From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.
Solution:
Exercise
Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,
Answer:
The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups.
List three physical properties that are important in describing the behavior of the main group elements.
Arrange K, Cs, Sr, Ca, Ba, and Li in order of
Arrange Rb, H, Be, Na, Cs, and Ca in order of
Which periodic trends are affected by Zeff? Based on the positions of the elements in the periodic table, which element would you expect to have the highest Zeff? the lowest Zeff?
Compare the properties of the metals and nonmetals with regard to their electronegativities and preferred oxidation states.
Of Ca, Br, Li, N, Zr, Ar, Sr, and S, which elements have a greater tendency to form positive ions than negative ions?
Arrange As, O, Ca, Sn, Be, and Sb in order of decreasing metallic character.
Give three reasons the chemistry of the second-period elements is generally not representative of their groups as a whole.
Compare the second-period elements and their heavier congeners with regard to
The heavier main group elements tend to form extended sigma-bonded structures rather than multiple bonds to other atoms. Give a reasonable explanation for this tendency.
What is the diagonal effect? How does it explain the similarity in chemistry between, for example, boron and silicon?
Although many of the properties of the second- and third-period elements in a group are quite different, one property is similar. Which one?
Two elements are effective additives to solid rocket propellant: beryllium and one other element that has similar chemistry. Based on the position of beryllium in the periodic table, identify the second element.
Give two reasons for the inert-pair effect. How would this phenomenon explain why Sn2+ is a better reducing agent than Pb2+?
Explain the following trend in electron affinities: Al (−41.8 kJ/mol), Si (−134.1 kJ/mol), P (−72.0 kJ/mol), and S (−200.4 kJ/mol).
Using orbital energy arguments, explain why electron configurations with more than four electron pairs around the central atom are not observed for second-period elements.
Ca > Be > Sn > Sb > As > O
aluminum
The magnitude of electron affinity increases from left to right in a period due to the increase in Zeff; P has a lower electron affinity than expected due to its half-filled 3p shell, which requires the added electron to enter an already occupied 3p orbital.
The following table lists the valences, coordination numbers, and ionic radii for a series of cations. Which would you substitute for K+ in a crystalline lattice? Explain your answer.
Metal | Charge | Coordination Number | Ionic Radius (pm) |
---|---|---|---|
Li | +1 | 4 | 76 |
Na | +1 | 6 | 102 |
K | +1 | 6 | 138 |
Mg | +2 | 6 | 72 |
Ca | +2 | 6 | 100 |
Sr | +2 | 6 | 118 |
Sr2+; it is the ion with the radius closest to that of K+.
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s1 electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in group 17 because the addition of a single electron to a hydrogen atom completes its valence shell.
Although hydrogen has an ns1 electron configuration, its chemistry does not resemble that of the metals of group 1.
Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. (For more information on nuclear fusion, see .) compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is protiumThe most common isotope of hydrogen, consisting of one proton and one electron. (1H or H), followed by deuteriumAn isotope of hydrogen that consists of one proton, one neutron, and one electron. (2H or D), which has an additional neutron. The rarest isotope of hydrogen is tritiumA rare isotope of hydrogen that consists of one proton, two neutrons, and one electron. (3H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to 3He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes.
Table 21.2 The Isotopes of Hydrogen
Protium | Deuterium | Tritium | |
---|---|---|---|
symbol | |||
neutrons | 0 | 1 | 2 |
mass (amu) | 1.00783 | 2.0140 | 3.01605 |
abundance (%) | 99.9885 | 0.0115 | ∼10−17 |
half-life (years) | — | — | 12.32 |
boiling point of X2 (K) | 20.28 | 23.67 | 25 |
melting point/boiling point of X2O (°C) | 0.0/100.0 | 3.8/101.4 | 4.5/? |
The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H2, D2, and T2 differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction.
Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium (2H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis (of Lewis electron structure fame) and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth.
Because the normal boiling point of D2O is 101.4°C (compared to 100.0°C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D+) has twice the mass of a proton (H+), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H2 over D2 or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D2O (1.1044 g/cm3 at 25°C) is greater than that of H2O (0.99978 g/cm3). Heavy water was an important constituent of early nuclear reactors. (For more information on nuclear reactors, see .)
Because deuterons diffuse so much more slowly, D2O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D+ is so slow compared with that of H+ because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions.
The 1s1 electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways ():
Figure 21.3 Three Types of Bonding in Compounds of Hydrogen
Because of its 1s1 electron configuration and the fact that the 1s orbital can accommodate no more than two electrons, hydrogen can (a) bond to other elements by losing an electron to form a proton, which can accept a pair of electrons from a more electronegative atom to form a polar covalent bond; (b) gain an electron from an electropositive metal to form a hydride ion, resulting in an ionic hydride; or (c) share its half-filled 1s orbital with a half-filled orbital on another atom to form a covalent or a polar covalent electron-pair bond.
Hydrogen can also act as a bridge between two atoms. One familiar example is the hydrogen bondAn unusually strong dipole-dipole interaction (intermolecular force) that results when H is bonded to very electronegative elements such as O, N, and F., an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (). An example of this kind of interaction is the hydrogen bonding network found in water (). Hydrogen can also form a three-center bond (or electron-deficient bond)A bond in which a hydride ion bridges two electropositive atoms., in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds. Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp3 orbitals on two boron atoms (). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion in , . Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures.
Figure 21.4 The Hydrogen Bond
The covalent bond between hydrogen and a very electronegative element, such as nitrogen, oxygen, or fluorine, is highly polar. The resulting partial positive charge on H allows it to interact with a lone pair of electrons on another atom to form a hydrogen bond, which is typically a linear arrangement of the three atoms, with the hydrogen atom placed asymmetrically between the two heavier atoms.
Figure 21.5 A Three-Center Bond Uses Two Electrons to Link Three Atoms
In the B–H–B unit shown, a hydride, with a filled 1s orbital, interacts simultaneously with empty sp3 hybrids on the boron atoms of two BH3 units to give three molecular orbitals. The two bonding electrons occupy the lowest-energy (σ) bonding orbital, thereby holding all three atoms together.
Hydrogen can lose its electron to form H+, accept an electron to form H−, share its electron, hydrogen bond, or form a three-center bond.
The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K).
Figure 21.6 The Explosive Properties of Hydrogen
When mixed with air and ignited by a spark, hydrogen gas can explode. The photograph shows the German dirigible Hindenburg on fire at Lakehurst, New Jersey, in 1937, after its hydrogen compartment was accidentally ignited as a consequence of an electrical discharge that caused the dirigible’s flammable skin to catch fire.
The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid:
Equation 21.1
M(s) + 2H+(aq) → H2(g) + M2+(aq)Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base:
Equation 21.2
Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. (For more information on redox reactions like that of Drano, see , .) Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D2O:
Equation 21.3
MH(s) + D2O(l) → HD(g) + M+(aq) + OD−(aq)On an industrial scale, H2 is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H2 known as synthesis gas, or syngas. (For more information on steam reforming, see , .) The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst:
Equation 21.4
Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H2(g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities.
Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework.
The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H+), a hydride ion (H−), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas.
Some periodic tables include hydrogen as a group 1 element, whereas other periodic tables include it as a group 17 element. Refer to the properties of hydrogen to propose an explanation for its placement in each group. In each case, give one property of hydrogen that would exclude it from groups 1 and 17.
If there were a planet where the abundances of D2O and H2O were reversed and life had evolved to adjust to this difference, what would be the effects of consuming large amounts of H2O?
Describe the bonding in a hydrogen bond and the central B–H bond in B2H7−. Why are compounds containing isolated protons unknown?
With which elements does hydrogen form ionic hydrides? covalent hydrides? metallic hydrides? Which of these types of hydrides can behave like acids?
Indicate which elements are likely to form ionic, covalent, or metallic hydrides and explain your reasoning:
Which has the higher ionization energy—H or H−? Why?
The electronegativities of hydrogen, fluorine, and iodine are 2.20, 3.98, and 2.66, respectively. Why, then, is HI a stronger acid than HF?
If H2O were a linear molecule, would the density of ice be less than or greater than that of liquid water? Explain your answer.
In addition to ion–dipole attractions, hydrogen bonding is important in solid crystalline hydrates, such as Na4XeO6·8H2O. Based on this statement, explain why anhydrous Na4XeO6 does not exist.
H has one electron in an s orbital, like the group 1 metals, but it is also one electron short of a filled principal shell, like the group 17 elements. Unlike the alkali metals, hydrogen is not a metal. Unlike the halogens, elemental hydrogen is not a potent oxidant.
Hydrogen bonding with waters of hydration will partially neutralize the negative charge on the terminal oxygen atoms on the XeO64− ion, which stabilizes the solid.
One of the largest uses of methane is to produce syngas, which is a source of hydrogen for converting nitrogen to ammonia. Write a complete equation for formation of syngas from methane and carbon dioxide. Calculate ΔG° for this reaction at 298 K and determine the temperature at which the reaction becomes spontaneous.
Predict the products of each reaction at 25°C and then balance each chemical equation.
Using heavy water (D2O) as the source of deuterium, how could you conveniently prepare
What are the products of reacting NaH with D2O? Do you expect the same products from reacting NaD and H2O? Explain your answer.
A 2.50 g sample of zinc metal reacts with 100.0 mL of 0.150 M HCl. What volume of H2 (in liters) is produced at 23°C and 729 mmHg?
A chemical reaction requires 16.8 L of H2 gas at standard temperature and pressure. How many grams of magnesium metal are needed to produce this amount of hydrogen gas?
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl3), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH3) and hydrogen telluride (H2Te), both of which are highly toxic.
Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line).
Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides:
Equation 21.5
In practice, CaCl2 is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O2 and water because Li reacts with nitrogen gas to form lithium nitride (Li3N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl2. In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg:
Equation 21.6
2RbOH(s) + Mg(s) → 2Rb(l) + Mg(OH)2(s)Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates.
Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al3+ from the ore; basic precipitation to remove Al3+ from the mixture as Al(OH)3; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. illustrates the isolation of liquid lithium from a lithium silicate ore by this process.
Figure 21.7 Isolating Lithium from Spodumene, a Lithium Silicate Ore
The key steps are acid leaching, basic precipitation of aluminum hydroxide, selective precipitation of insoluble lithium carbonate, conversion to lithium chloride, evaporation, and electrolysis. The other alkali metals and the alkaline earth metals are recovered from their ores by similar processes.
Various properties of the group 1 elements are summarized in . In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns2 electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C).
Table 21.3 Selected Properties of the Group 1 Elements
Lithium | Sodium | Potassium | Rubidium | Cesium | Francium | |
---|---|---|---|---|---|---|
atomic symbol | Li | Na | K | Rb | Cs | Fr |
atomic number | 3 | 11 | 19 | 37 | 55 | 87 |
atomic mass | 6.94 | 22.99 | 39.10 | 85.47 | 132.91 | 223 |
valence electron configuration | 2s1 | 3s1 | 4s1 | 5s1 | 6s1 | 7s1 |
melting point/boiling point (°C) | 180.5/1342 | 97.8/883 | 63.5/759 | 39.3/688 | 28.5/671 | 27/— |
density (g/cm3) at 25°C | 0.534 | 0.97 | 0.89 | 1.53 | 1.93 | — |
atomic radius (pm) | 167 | 190 | 243 | 265 | 298 | — |
first ionization energy (kJ/mol) | 520 | 496 | 419 | 403 | 376 | 393 |
most common oxidation state | +1 | +1 | +1 | +1 | +1 | +1 |
ionic radius (pm)* | 76 | 102 | 138 | 152 | 167 | — |
electron affinity (kJ/mol) | −60 | −53 | −48 | −47 | −46 | — |
electronegativity | 1.0 | 0.9 | 0.8 | 0.8 | 0.8 | 0.7 |
standard electrode potential (E°, V) | −3.04 | −2.71 | −2.93 | −2.98 | −3.03 | — |
product of reaction with O2 | Li2O | Na2O2 | KO2 | RbO2 | CsO2 | — |
type of oxide | basic | basic | basic | basic | basic | — |
product of reaction with N2 | Li3N | none | none | none | none | — |
product of reaction with X2 | LiX | NaX | KX | RbX | CsX | — |
product of reaction with H2 | LiH | NaH | KH | RbH | CsH | — |
*The values cited are for four-coordinate ions except for Rb+ and Cs+, whose values are given for the six-coordinate ion. |
The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. (For more information on reduction potentials, see ). Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (). Because Li+ is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li+ more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends.
All alkali metals are electropositive elements with an ns1 valence electron configuration, forming the monocation (M+) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M+ ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds.
All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where X is a halogen:
Equation 21.7
2M(s) + X2(s, l, g) → 2M+X−(s)Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te:
Equation 21.8
2M(s) + Y(s) → M2Y(s)When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na2Sn, where n = 2–6). For example, Na2S3 contains the S32− ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S3−) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine ().
Figure 21.8 The Trisulfide Anion Is Responsible for the Deep Blue Color of Some Gemstones
(a) The rich blue color of lapis lazuli is due to small amounts of the normally unstable S3− anion. (b) The aluminosilicate cages of the minerals (zeolites) that make up the matrix of blue ultramarine stabilize the reactive anion; excess Na+ ions in the structure balance the negative charges on the zeolite framework and the S3− anion.
Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li2O (white), Na2O2 (pale yellow), KO2 (orange), RbO2 (brown), and CsO2 (orange). Only Li2O has the stoichiometry expected for a substance that contains two M+ cations and one O2− ion. In contrast, Na2O2 contains the O22− (peroxide) anion plus two Na+ cations. The other three salts, with stoichiometry MO2, contain the M+ cation and the O2− (superoxide) ion. Because O2− is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li+). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na+ cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO2).
A crystal of spodumene (LiAlSi2O6). This mineral is one of the most important lithium ores.
The chemistry of the alkali metals is largely that of ionic compounds containing the M+ ions.
The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na2O2 is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na2O2 and KO2 are used to purify and regenerate the air by removing the CO2 produced by respiration and replacing it with O2. Both compounds react with CO2 in a redox reaction in which O22− or O2− is simultaneously oxidized and reduced, producing the metal carbonate and O2:
Equation 21.9
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)Equation 21.10
4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g)The presence of water vapor, the other product of respiration, makes KO2 even more effective at removing CO2 because potassium bicarbonate, rather than potassium carbonate, is formed:
Equation 21.11
4KO2(s) + 4CO2(g) + 2H2O(g) → 4KHCO3(s) + 3O2(g)Notice that 4 mol of CO2 are removed in this reaction, rather than 2 mol in .
Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li3N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K+ cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As):
Equation 21.12
12M(s) + Z4(s) → 4M3Z(s)Because of lattice energies, only lithium forms a stable oxide and nitride.
The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K4Si4 whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M2C2 (where M is Li or Na):
Equation 21.13
2M(s) + 2C(s) → M2C2(s)The three-dimensional cage structure of the Si44−ion in the ionic compound K4S4. The Si44− ion is isoelectronic and isostructural with the P4 molecule.
The same compounds can be obtained by reacting the metal with acetylene (C2H2). In this reaction, the metal is again oxidized, and hydrogen is reduced:
Equation 21.14
2M(s) + C2H2(g) → M2C2(s) + H2(g)The acetylide ion (C22−), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq).
The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called graphite intercalation compoundsA compound that forms when heavier alkali metals react with carbon in the form of graphite and insert themselves between the sheets of carbon atoms. (part (a) in ). The stoichiometries of these compounds include MC60 and MC48, which are black/gray; MC36 and MC24, which are blue; and MC8, which is bronze (part (b) in ). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K+C8−.
Figure 21.9 Graphite Intercalation Compounds
Reacting graphite with alkali metals such as K, Rb, and Cs results in partial reduction of the graphite and insertion of layers of alkali metal cations between sets of n layers of carbon atoms. (a) In KC8, layers of K+ ions are inserted between every pair of carbon layers, giving n = 1. (b) The stoichiometry and color of intercalation compounds depend on the number of layers of carbon atoms (n) between each layer of intercalated metal atoms. This schematic diagram illustrates the most common structures that have been observed.
All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M+H−):
Equation 21.15
2M(s) + H2(g) → 2MH(s)All are also capable of reducing water to produce hydrogen gas:
Equation 21.16
Although lithium reacts rather slowly with water, sodium reacts quite vigorously (), and the heavier alkali metals (K, Rb, and Cs) react so vigorously that they invariably explode. This trend, which is not consistent with the relative magnitudes of the reduction potentials of the elements, serves as another example of the complex interplay of different forces and phenomena—in this case, kinetics and thermodynamics. Although the driving force for the reaction is greatest for lithium, the heavier metals have lower melting points. The heat liberated by the reaction causes them to melt, and the larger surface area of the liquid metal in contact with water greatly accelerates the reaction rate.
Figure 21.10 Reacting Sodium with Water
Like most elements in groups 1 and 2, sodium reacts violently with water. The products are the Na+(aq) ion and hydrogen gas, which is potentially explosive when mixed with air.
Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water:
Equation 21.17
2MOH(aq) + H2SO4(aq) → M2SO4(aq) + 2H2O(l)Equation 21.18
MOH(aq) + HNO3(aq) → MNO3(aq) + H2O(l)Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH3CO2Na) by reacting sodium hydroxide and acetic acid:
Equation 21.19
CH3CO2H(aq) + NaOH(s) → CH3CO2Na(aq) + H2O(l)Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH3(CH2)14CO2H] and stearic acid [CH3(CH2)16CO2H]. Lithium salts, such as lithium stearate [CH3(CH2)14CO2Li], are used as additives in motor oils and greases.
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes like those discussed in . Complex formation is most significant for the smallest cation (Li+) and decreases with increasing radius. In aqueous solution, for example, Li+ forms the tetrahedral [Li(H2O)4]+ complex. In contrast, the larger alkali metal cations form octahedral [M(H2O)6]+ complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li+ and Na+ ions (such as Na2SO4) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration.
The tetrahedral [Li(H2O)4]+and octahedral [Rb(H2O)6]+complexes. The Li+ ion is so small that it can accommodate only four water molecules around it, but the larger alkali metal cations tend to bind six water molecules.
Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases.
Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in , crown ethersA cyclic polyether that has four or more oxygen atoms separated by two or three carbon atoms. A central cavity can accommodate a metal ion coordinated to the ring of oxygen atoms. are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li+, whereas 18-crown-6 forms the strongest complexes with K+ (part (a) in ).
CryptandsConsisting of three chains connected by two nitrogen atoms, this compound can completely encapsulate a metal ion of the appropriate size, coordinating to the metal by the lone pairs of electrons on each oxygen and the two nitrogen atoms. are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in ). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO4 in nonpolar organic solvents ().
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2−) rather than hydroxide:
Equation 21.20
Solvated electrons. The presence of solvated electrons (e−, NH3) in solutions of alkali metals in liquid ammonia is indicated by the intense color of the solution and its electrical conductivity.
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in tends to be rather slow. In many cases, the alkali metal amide salt (MNH2) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e−, NH3), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue () and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
Figure 21.11 Alkali Metal–Liquid Ammonia Solutions
Most metals are insoluble in virtually all solvents, but the alkali metals (and the heavier alkaline earth metals) dissolve readily in liquid ammonia to form solvated metal cations and solvated electrons, which give the solution a deep blue color.
In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M+), the neutral metal atom (M), metal dimers (M2), and the metal anion (M−). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M+NH2− and hydrogen gas (). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry.
Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called organometallic compoundsA compound that contains a metal covalently bonded to a carbon atom of an organic species.. The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH3), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi)n, where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in , where each triangular face of the Li4 tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH3 group is using a single pair of electrons in an sp3 hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes discussed in and . Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text.
Figure 21.12 The Tetrameric Structure of Methyllithium
Methyllithium is not an ionic compound; it exists as tetrameric (CH3Li)4 molecules. The structure consists of a tetrahedral arrangement of four lithium atoms, with the carbon atom of a methyl group located above the middle of each triangular face of the tetrahedron. The carbon atoms thus bridge three lithium atoms to form four-center, two-electron bonds.
The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components.
Organosodium and organopotassium compounds are more ionic than organolithium compounds. They contain discrete M+ and R− ions and are insoluble or only sparingly soluble in nonpolar solvents.
Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response.
Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na2CO3, used in the manufacture of glass; K2O, used in porcelain glazes; and Na4SiO4, used in detergents.
Several other alkali metal compounds are also important. For example, Li2CO3 is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder.
For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case.
Given: application and selected alkali metals
Asked for: appropriate metal for each application
Strategy:
Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application.
Solution:
Exercise
Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application.
Answer:
Predict the products of each reaction and then balance each chemical equation.
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
A Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction.
B If a reaction is predicted to occur, balance the chemical equation.
Solution:
A Sodium is a reductant, and oxygen is an oxidant, so a redox reaction is most likely. We expect an electron to be transferred from Na (thus forming Na+) to O2. We now need to determine whether the reduced product is a superoxide (O2−), peroxide (O22−), or oxide (O2−). Under normal reaction conditions, the product of the reaction of an alkali metal with oxygen depends on the identity of the metal. Because of differences in lattice energy, Li produces the oxide (Li2O), the heavier metals (K, Rb, Cs) produce the superoxide (MO2), and Na produces the peroxide (Na2O2).
B The balanced chemical equation is 2Na(s) + O2(g) → Na2O2(s).
A Li2O is an ionic salt that contains the oxide ion (O2−), which is the completely deprotonated form of water and thus is expected to be a strong base. The other reactant, water, is both a weak acid and a weak base, so we can predict that an acid–base reaction will occur.
B The balanced chemical equation is Li2O(s) + H2O(l) → 2LiOH(aq).
A Potassium is a reductant, whereas methanol is both a weak acid and a weak base (similar to water). A weak acid produces H+, which can act as an oxidant by accepting an electron to form This reaction, therefore, is an acid dissociation that is driven to completion by a reduction of the protons as they are released.
B The balanced chemical equation is as follows: .
A One of the reactants is an alkali metal, a potent reductant, and the other is an alkyl halide. Any compound that contains a carbon–halogen bond can, in principle, be reduced, releasing a halide ion and forming an organometallic compound. That outcome seems likely in this case because organolithium compounds are among the most stable organometallic compounds known.
B Two moles of lithium are required to balance the equation: 2Li(s) + CH3Cl(l) → LiCl(s) + CH3Li(soln).
A Lithium nitride and potassium chloride are largely ionic compounds. The nitride ion (N3−) is a very strong base because it is the fully deprotonated form of ammonia, a weak acid. An acid–base reaction requires an acid as well as a base, however, and KCl is not acidic. What about a redox reaction? Both substances contain ions that have closed-shell valence electron configurations. The nitride ion could act as a reductant by donating electrons to an oxidant and forming N2. KCl is not an oxidant, however, and a redox reaction requires an oxidant as well as a reductant.
B We conclude that the two substances will not react with each other.
Exercise
Predict the products of each reaction and balance each chemical equation.
Answer:
The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH2−). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components.
Which of the group 1 elements reacts least readily with oxygen? Which is most likely to form a hydrated, crystalline salt? Explain your answers.
The alkali metals have a significant electron affinity, corresponding to the addition of an electron to give the M− anion. Why, then, do they commonly lose the ns1 electron to form the M+ cation rather than gaining an electron to form M−?
Lithium is a far stronger reductant than sodium; cesium is almost as strong as lithium, which does not agree with the expected periodic trend. What two opposing properties explain this apparent anomaly? Is the same anomaly found among the alkaline earth metals?
Explain why the ionic character of LiCl is less than that of NaCl. Based on periodic trends, would you expect the ionic character of BeCl2 to be greater or less than that of LiCl? Why?
Alkali metals and carbon form intercalation compounds with extremely high electrical conductivity. Is this conductivity through the layers or along the layers? Explain your answer.
Electrolysis is often used to isolate the lighter alkali metals from their molten halides. Why are halides used rather than the oxides or carbonates, which are easier to isolate? With this in mind, what is the purpose of adding calcium chloride to the alkali metal halide?
The only alkali metal that reacts with oxygen to give a compound with the expected stoichiometry is lithium, which gives Li2O. In contrast, sodium reacts with oxygen to give Na2O2, and the heavier alkali metals form superoxides. Explain the difference in the stoichiometries of these products.
Classify aqueous solutions of Li2O, Na2O, and CsO2 as acidic, basic, or amphoteric.
Although methanol is relatively unreactive, it can be converted to a synthetically more useful form by reaction with LiH. Predict the products of reacting methanol with LiH. Describe the visual changes you would expect to see during this reaction.
Lithium reacts with atmospheric nitrogen to form lithium nitride (Li3N). Why do the other alkali metals not form analogous nitrides? Explain why all the alkali metals react with arsenic to form the corresponding arsenides (M3As).
Write a balanced chemical equation to describe each reaction.
What products are formed at the anode and the cathode during electrolysis of
Write the corresponding half-reactions for each reaction.
Sodium metal is prepared by electrolysis of molten NaCl. If 25.0 g of chlorine gas are produced in the electrolysis of the molten salt using 9.6 A (C/s) of current, how many hours were required for the reaction? What mass of sodium was produced?
Sodium peroxide can remove CO2 from the air and replace it with oxygen according to the following unbalanced chemical equation:
Na2O2(s) + CO2(g) → Na2CO3(s) + O2(g)Predict the products of each chemical reaction and then balance each chemical equation.
Predict the products of each reaction.
A 655 mg sample of graphite was allowed to react with potassium metal, and 744 mg of product was isolated. What is the stoichiometry of the product?
Perchloric acid, which is used as a reagent in a number of chemical reactions, is typically neutralized before disposal. When a novice chemist accidentally used K2CO3 to neutralize perchloric acid, a large mass of KClO4 (Ksp = 1.05 × 10−2) precipitated from solution. What mass of potassium ion is present in 1.00 L of a saturated solution of KClO4?
A key step in the isolation of the alkali metals from their ores is selective precipitation. For example, lithium is separated from sodium and potassium by precipitation of Li2CO3 (Ksp = 8.15 × 10−4). If 500.0 mL of a 0.275 M solution of Na2CO3 are added to 500.0 mL of a 0.536 M lithium hydroxide solution, what mass of Li2CO3 will precipitate (assuming no further reactions occur)? What mass of lithium will remain in solution?
5.54 g Li2CO3; 0.82 g Li+
Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium (M2CO3). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides.
Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium:
Equation 21.21
Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure RaCl2. Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses and will not be discussed further.
The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium:
Equation 21.22
CaCl2(l) → Ca(l) + Cl2(g)The group 2 metal chlorides are obtained from a variety of sources. For example, BeCl2 is produced by reacting HCl with beryllia (BeO), which is obtained from the semiprecious stone beryl [Be3Al2(SiO3)6].
A crystal of beryl. Beryl is a gemstone and an important source of beryllium.
Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO3·MgCO3) with an inexpensive iron/silicon alloy at 1150°C. Initially CO2 is released, leaving behind a mixture of CaO and MgO; Mg2+ is then reduced:
Equation 21.23
2CaO·MgO(s) + Fe/Si(s) → 2Mg(l) + Ca2SiO4(s) + Fe(s)An early source of magnesium was an ore called magnesite (MgCO3) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite (SrCO3) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble.
A crystal of strontianite. Both strontianite, one of the most important strontium ores, and strontium are named after the town of Strontian, Scotland, the location of one of the first mines for strontium ores.
Several important properties of the alkaline earth metals are summarized in . Although many of these properties are similar to those of the alkali metals (), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns2 for the alkaline earth metals versus ns1 for the alkali metals).
Table 21.4 Selected Properties of the Group 2 Elements
Beryllium | Magnesium | Calcium | Strontium | Barium | Radium | |
---|---|---|---|---|---|---|
atomic symbol | Be | Mg | Ca | Sr | Ba | Ra |
atomic number | 4 | 12 | 20 | 38 | 56 | 88 |
atomic mass | 9.01 | 24.31 | 40.08 | 87.62 | 137.33 | 226 |
valence electron configuration | 2s2 | 3s2 | 4s2 | 5s2 | 6s2 | 7s2 |
melting point/boiling point (°C) | 1287/2471 | 650/1090 | 842/1484 | 777/1382 | 727/1897 | 700/— |
density (g/cm3) at 25°C | 1.85 | 1.74 | 1.54 | 2.64 | 3.62 | ∼5 |
atomic radius (pm) | 112 | 145 | 194 | 219 | 253 | — |
first ionization energy (kJ/mol) | 900 | 738 | 590 | 549 | 503 | — |
most common oxidation state | +2 | +2 | +2 | +2 | +2 | +2 |
ionic radius (pm)* | 45 | 72 | 100 | 118 | 135 | — |
electron affinity (kJ/mol) | ≥ 0 | ≥ 0 | −2 | −5 | −14 | — |
electronegativity | 1.6 | 1.3 | 1.0 | 1.0 | 0.9 | 0.9 |
standard electrode potential (E°, V) | −1.85 | −2.37 | −2.87 | −2.90 | −2.91 | −2.8 |
product of reaction with O2 | BeO | MgO | CaO | SrO | BaO2 | — |
type of oxide | amphoteric | weakly basic | basic | basic | basic | — |
product of reaction with N2 | none | Mg3N2 | Ca3N2 | Sr3N2 | Ba3N2 | — |
product of reaction with X2 | BeX2 | MgX2 | CaX2 | SrX2 | BaX2 | — |
product of reaction with H2 | none | MgH2 | CaH2 | SrH2 | BaH2 | — |
*The values cited are for six-coordinate ions except for Be2+, for which the value for the four-coordinate ion is given. |
As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns2 valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance.
One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy.
With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M2+ ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent, as discussed in . Some compounds of Mg2+ also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2.
The group 2 elements almost exclusively form ionic compounds containing M2+ ions.
Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds.
All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX2). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M2+ cation and two X− anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl2. These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. (For more information on the VSEPR model, see .) As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons:
Equation 21.24
BeCl2(s) + 2(CH3CH2)2O(l) → BeCl2[O(CH2CH3)2]2(soln)Solid beryllium chloride (BeCl2). The solid has a polymeric, halide-bridged structure.
The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO2) because the larger O22− ion is better able to separate the large Ba2+ ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts:
Equation 21.25
The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Yn2−) are formed.
In the reverse of , the oxides of Ca, Sr, and Ba react with CO2 to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH)2:
Equation 21.26
MO(s) + H2O(l) → M2+(aq) + 2OH−(aq)and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges.
The carbonates of the alkaline earth metals also react with aqueous acid to give CO2 and H2O:
Equation 21.27
MCO3(s) + 2H+(aq) → M2+(aq) + CO2(g) + H2O(l)The reaction in is the basis of antacids that contain MCO3, which is used to neutralize excess stomach acid.
The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N2 to form the nitride (M3N2), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M2+ and N3− ions is apparently sufficient to overcome the chemical inertness of the N2 molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M3Z2.
Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements.
When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC2. The most important alkaline earth carbide is calcium carbide (CaC2), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be2C, which formally contains the C4− ion (although the compound is covalent). Consistent with this formulation, reaction of Be2C with water or aqueous acid produces methane:
Equation 21.28
Be2C(s) + 4H2O(l) → 2Be(OH)2(s) + CH4(g)Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH2 can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH2). The hydrides of the heavier alkaline earth metals are ionic, but both BeH2 and MgH2 have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas:
Equation 21.29
CaH2(s) + 2H2O(l) → Ca(OH)2(s) + 2H2(g)Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size.
The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease.
Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be2+) and decreases rapidly with the increasing radius of the metal ion.
The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals.
The chemistry of Be2+ is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be2+ salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H2O)4]2+ ion. Because of its high charge-to-radius ratio, the Be2+ ion polarizes coordinated water molecules, thereby increasing their acidity:
Equation 21.30
[Be(H2O)4]2+(aq) → [Be(H2O)3(OH)]+(aq) + H+(aq)Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH)4]2−. Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF4]2−. Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases ().
The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg2+ and Ca2+). Thus aqueous solutions of Mg2+ contain the octahedral [Mg(H2O)6]2+ ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in .
Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called Grignard reagents, after Victor Grignard (1871–1935), the French chemist who discovered them. (For more information on the Grignard reagents, see , .) Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines.
Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 105 tn per year). Its low density (1.74 g/cm3 compared with 7.87 g/cm3 for iron and 2.70 g/cm3 for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation:
Equation 21.31
TiCl4(l) + 2Mg(s) → Ti(s) + 2MgCl2(s)Figure 21.13 Magnesium Alloys Are Lightweight and Corrosion Resistant
Because magnesium is about five times lighter than steel and 50% lighter than aluminum, it was long considered the “material of the future,” as shown in this 1950 concept Buick LeSabre sports car made almost entirely of magnesium and aluminum alloys. Modern aluminum alloys usually contain about 5% magnesium to improve their corrosion resistance and mechanical properties.
The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil.
Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl2 is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO3 is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO3 (), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO4 in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks.
For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as Ksp values (), lattice energies (), and band-gap energies (, ).
Given: application and selected alkaline earth metals
Asked for: most appropriate substance for each application
Strategy:
Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use.
Solution:
Exercise
Which of the indicated alkaline earth metals or their compounds is most appropriate for each application?
Answer:
Predict the products of each reaction and then balance each chemical equation.
Given: reactants
Asked for: products and balanced chemical equation
Strategy:
Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation.
Solution:
A Gaseous HCl is an acid, and CaO is a basic oxide that contains the O2− ion. This is therefore an acid–base reaction that produces CaCl2 and H2O.
B The balanced chemical equation is CaO(s) + 2HCl(g) → CaCl2(aq) + H2O(l).
A Magnesium oxide is a basic oxide, so it can either react with water to give a basic solution or dissolve in an acidic solution. Hydroxide ion is also a base. Because we have two bases but no acid, an acid–base reaction is impossible. A redox reaction is not likely because MgO is neither a good oxidant nor a good reductant.
B We conclude that no reaction occurs.
A Because CaH2 contains the hydride ion (H−), it is a good reductant. It is also a strong base because H− ions can react with H+ ions to form H2. Titanium oxide (TiO2) is a metal oxide that contains the metal in its highest oxidation state (+4 for a group 4 metal); it can act as an oxidant by accepting electrons. We therefore predict that a redox reaction will occur, in which H− is oxidized and Ti4+ is reduced. The most probable reduction product is metallic titanium, but what is the oxidation product? Oxygen must appear in the products, and both CaO and H2O are stable compounds. The +1 oxidation state of hydrogen in H2O is a sign that an oxidation has occurred (2H− → 2H+ + 4e−).
B The balanced chemical equation is We could also write the products as Ti(s) + Ca(OH)2(s).
Exercise
Predict the products of each reaction and then balance each chemical equation.
Answer:
Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO2 to form carbonates, which in turn react with acid to produce CO2 and H2O. Except for Be, all the alkaline earth metals react with N2 to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds.
The electronegativities of Li and Sr are nearly identical (0.98 versus 0.95, respectively). Given their positions in the periodic table, how do you account for this?
Arrange Na, Ba, Cs, and Li in order of increasing Zeff.
Do you expect the melting point of NaCl to be greater than, equal to, or less than that of MgCl2? Why?
Which of the group 2 elements would you expect to form primarily ionic rather than covalent organometallic compounds? Explain your reasoning.
Explain why beryllium forms compounds that are best regarded as covalent in nature, whereas the other elements in group 2 generally form ionic compounds.
Why is the trend in the reactions of the alkaline earth metals with nitrogen the reverse of the trend seen for the alkali metals?
Is the bonding in the alkaline earth hydrides primarily ionic or covalent in nature? Explain your answer. Given the type of bonding, do you expect the lighter or heavier alkaline earth metals to be better reducing agents?
Using arguments based on ionic size, charge, and chemical reactivity, explain why beryllium oxide is amphoteric. What element do you expect to be most similar to beryllium in its reactivity? Why?
Explain why the solubility of the carbonates and sulfates of the alkaline earth metals decreases with increasing cation size.
Beryllium oxide is amphoteric, magnesium oxide is weakly basic, and calcium oxide is very basic. Explain how this trend is related to the ionic character of the oxides.
Do you expect the of CaH2 to be greater than, the same as, or less than that of BaH2? Why or why not?
Which of the s-block elements would you select to carry out a chemical reduction on a small scale? Consider cost, reactivity, and stability in making your choice. How would your choice differ if the reduction were carried out on an industrial scale?
Beryllium iodide reacts vigorously with water to produce HI. Write a balanced chemical equation for this reaction and explain why it is violent.
Predict the products of each reaction and then balance each chemical equation.
Predict the products of each reaction and then balance each chemical equation.
Indicate whether each pair of substances will react and, if so, write a balanced chemical equation for the reaction.
Using a thermodynamic cycle and information presented in and , calculate the lattice energy of magnesium nitride (Mg3N2). ( for Mg3N2 is −463 kJ/mol, and ΔH° for N(g) + 3e− → N3− is +1736 kJ.) How does the lattice energy of Mg3N2 compare with that of MgCl2 and MgO? (See for the enthalpy of formation values.)
The solubility products of the carbonate salts of magnesium, calcium, and strontium are 6.82 × 10−6, 3.36 × 10−9, and 5.60 × 10−10, respectively. How many milligrams of each compound would be present in 200.0 mL of a saturated solution of each? How would the solubility depend on the pH of the solution? Why?
The solubility products of BaSO4 and CaSO4 are 1.08 × 10−10 and 4.93 × 10−5, respectively. What accounts for this difference? When 500.0 mL of a solution that contains 1.00 M Ba(NO3)2 and 3.00 M Ca(NO3)2 is mixed with a 2.00 M solution of Na2SO4, a precipitate forms. What is the identity of the precipitate? How much of it will form before the second salt precipitates?
Electrolytic reduction is used to produce magnesium metal from MgCl2. The goal is to produce 200.0 kg of Mg by this method.
A sample consisting of 20.35 g of finely divided calcium metal is allowed to react completely with nitrogen. What is the mass of the product?
What mass of magnesium hydride will react with water to produce 1.51 L of hydrogen gas at standard temperature and pressure?
The Ba2+ ion is larger and has a lower hydration energy than the Ca2+ ion. The precipitate is BaSO4; 117 g of BaSO4.
25.09 g of Ca3N2
The s-block elements play important roles in biological systems. Covalent hydrides, for example, are the building blocks of organic compounds, and other compounds and ions containing s-block elements are found in tissues and cellular fluids. In this section, we describe some ways in which biology depends on the properties of the group 1 and group 2 elements.
There are three major classes of hydrides—covalent, ionic, and metallic—but only covalent hydrides occur in living cells and have any biochemical significance. As you learned in , carbon and hydrogen have similar electronegativities, and the C–H bonds in organic molecules are strong and essentially nonpolar. Little acid–base chemistry is involved in the cleavage or formation of these bonds. In contrast, because hydrogen is less electronegative than oxygen and nitrogen (symbolized by Z), the H–Z bond in the hydrides of these elements is polarized (Hδ+–Zδ−). Consequently, the hydrogen atoms in these H–Z bonds are relatively acidic. Moreover, S–H bonds are relatively weak due to poor s orbital overlap, so they are readily cleaved to give a proton. Hydrides in which H is bonded to O, N, or S atoms are therefore polar, hydrophilic molecules that form hydrogen bonds. They also undergo acid–base reactions by transferring a proton.
Covalent hydrides in which H is bonded to O, N, or S atoms are polar and hydrophilic, form hydrogen bonds, and transfer a proton in their acid-base reactions.
Hydrogen bonds are crucial in biochemistry, in part because they help hold proteins in their biologically active folded structures. Hydrogen bonds also connect the two intertwining strands of DNA (deoxyribonucleic acid), the substance that contains the genetic code for all organisms. (For more information on DNA, see , .) Because hydrogen bonds are easier to break than the covalent bonds that form the individual DNA strands, the two intertwined strands can be separated to give intact single strands, which is essential for the duplication of genetic information.
In addition to the importance of hydrogen bonds in biochemical molecules, the extensive hydrogen-bonding network in water is one of the keys to the existence of life on our planet. Based on its molecular mass, water should be a gas at room temperature (20°C), but the strong intermolecular interactions in liquid water greatly increase its boiling point. Hydrogen bonding also produces the relatively open molecular arrangement found in ice, which causes ice to be less dense than water. Because ice floats on the surface of water, it creates an insulating layer that allows aquatic organisms to survive during cold winter months.
These same strong intermolecular hydrogen-bonding interactions are also responsible for the high heat capacity of water and its high heat of fusion. A great deal of energy must be removed from water for it to freeze. Consequently, as noted in , large bodies of water act as “thermal buffers” that have a stabilizing effect on the climate of adjacent land areas. Perhaps the most striking example of this effect is the fact that humans can live comfortably at very high latitudes. For example, palm trees grow in southern England at the same latitude (51°N) as the southern end of frigid Hudson Bay and northern Newfoundland in North America, areas known more for their moose populations than for their tropical vegetation. Warm water from the Gulf Stream current in the Atlantic Ocean flows clockwise from the tropical climate at the equator past the eastern coast of the United States and then turns toward England, where heat stored in the water is released. The temperate climate of Europe is largely attributable to the thermal properties of water.
Strong intermolecular hydrogen-bonding interactions are responsible for the high heat capacity of water and its high heat of fusion.
The members of group 1 and group 2 that are present in the largest amounts in organisms are sodium, potassium, magnesium, and calcium, all of which form monatomic cations with a charge of +1 (group 1, M+) or +2 (group 2, M2+). Biologically, these elements can be classified as macrominerals ().
For example, calcium is found in the form of relatively insoluble calcium salts that are used as structural materials in many organisms. Hydroxyapatite [Ca5(PO4)3OH] is the major component of bones, calcium carbonate (CaCO3) is the major component of the shells of mollusks and the eggs of birds and reptiles, and calcium oxalate (CaO2CCO2) is found in many plants. Because calcium and strontium have similar sizes and charge-to-radius ratios, small quantities of strontium are always found in bone and other calcium-containing structural materials. Normally this is not a problem because the Sr2+ ions occupy sites that would otherwise be occupied by Ca2+ ions. When trace amounts of radioactive 90Sr are released into the atmosphere from nuclear weapons tests or a nuclear accident, however, the radioactive strontium eventually reaches the ground, where it is taken up by plants that are consumed by dairy cattle. The isotope then becomes concentrated in cow’s milk, along with calcium. Because radioactive strontium coprecipitates with calcium in the hydroxyapatite that surrounds the bone marrow (where white blood cells are produced), children, who typically ingest more cow’s milk than adults, are at substantially increased risk for leukemia, a type of cancer characterized by the overproduction of white blood cells.
The Na+, K+, Mg2+, and Ca2+ ions are important components of intracellular and extracellular fluids. Both Na+ and Ca2+ are found primarily in extracellular fluids, such as blood plasma, whereas K+ and Mg2+ are found primarily in intracellular fluids. Substantial inputs of energy are required to establish and maintain these concentration gradients and prevent the system from reaching equilibrium. Thus energy is needed to transport each ion across the cell membrane toward the side with the higher concentration. The biological machines that are responsible for the selective transport of these metal ions are complex assemblies of proteins called ion pumpsA complex assembly of proteins that selectively transport ions across cell membranes by their high affinity for ions of a certain charge and radius.. Ion pumps recognize and discriminate between metal ions in the same way that crown ethers and cryptands do, with a high affinity for ions of a certain charge and radius.
Defects in the ion pumps or their control mechanisms can result in major health problems. For example, cystic fibrosis, the most common inherited disease in the United States, is caused by a defect in the transport system (in this case, chloride ions). Similarly, in many cases, hypertension, or high blood pressure, is thought to be due to defective Na+ uptake and/or excretion. If too much Na+ is absorbed from the diet (or if too little is excreted), water diffuses from tissues into the blood to dilute the solution, thereby decreasing the osmotic pressure in the circulatory system. The increased volume increases the blood pressure, and ruptured arteries called aneurysms can result, often in the brain. Because high blood pressure causes other medical problems as well, it is one of the most important biomedical disorders in modern society.
For patients who suffer from hypertension, low-sodium diets that use NaCl substitutes, such as KCl, are often prescribed. Although KCl and NaCl give similar flavors to foods, the K+ is not readily taken up by the highly specific Na+-uptake system. This approach to controlling hypertension is controversial, however, because direct correlations between dietary Na+ content and blood pressure are difficult to demonstrate in the general population. More important, recent observations indicate that high blood pressure may correlate more closely with inadequate intake of calcium in the diet than with excessive sodium levels. This finding is important because the typical “low-sodium” diet is also low in good sources of calcium, such as dairy products.
Some of the most important biological functions of the group 1 and group 2 metals are due to small changes in the cellular concentrations of the metal ion. The transmission of nerve impulses, for example, is accompanied by an increased flux of Na+ ions into a nerve cell. Similarly, the binding of various hormones to specific receptors on the surface of a cell leads to a rapid influx of Ca2+ ions; the resulting sudden rise in the intracellular Ca2+ concentration triggers other events, such as muscle contraction, the release of neurotransmitters, enzyme activation, or the secretion of other hormones.
Within cells, K+ and Mg2+ often activate particular enzymes by binding to specific, negatively charged sites in the enzyme structure. Chlorophyll, the green pigment used by all plants to absorb light and drive the process of photosynthesis, contains magnesium. During photosynthesis, CO2 is reduced to form sugars such as glucose. The structure of the central portion of a chlorophyll molecule resembles a crown ether (part (a) in ) with four five-member nitrogen-containing rings linked together to form a large ring that provides a “hole” the proper size to tightly bind Mg2+.
Because the health of cells depends on maintaining the proper levels of cations in intracellular fluids, any change that affects the normal flux of metal ions across cell membranes could well cause an organism to die. Molecules that facilitate the transport of metal ions across membranes are generally called ionophoresA molecule that facilitates the transport of metal ions across membranes. (ion plus phore from the Greek phorein, meaning “to carry”). Many ionophores are potent antibiotics that can kill or inhibit the growth of bacteria. An example is valinomycin, a cyclic molecule with a central cavity lined with oxygen atoms (part (a) in ) that is similar to the cavity of a crown ether (part (a) in ). Like a crown ether, valinomycin is highly selective: its affinity for K+ is about 1000 times greater than that for Na+. By increasing the flux of K+ ions into cells, valinomycin disrupts the normal K+ gradient across a cell membrane, thereby killing the cell (part (b) in ).
Figure 21.14 Valinomycin Is an Antibiotic That Functions Like an Ionophore
(a) This model of the structure of the K+–valinomycin complex, determined by x-ray diffraction, shows how the valinomycin molecule wraps itself around the K+ ion, shielding it from the environment, in a manner reminiscent of a crown ether complex. (For more information on the crown ethers, see , .) (b) Valinomycin kills bacteria by facilitating the transport of K+ ions across the cell membrane, thereby disrupting the normal distribution of ions in the bacterium. At the surface of the membrane, valinomycin binds a K+ ion. Because the hydrophobic exterior of the valinomycin molecule forms a “doughnut” that shields the positive charge of the metal ion, the K+–valinomycin complex is highly soluble in the nonpolar interior of the membrane. After the K+–valinomycin complex diffuses across the membrane to the interior of the cell, the K+ ion is released, and the valinomycin is free to diffuse back to the other side of the membrane to bind another K+ ion. Valinomycin thereby destroys the normal K+ gradient across the membrane, killing the cell.
A common way to study the function of a metal ion in biology is to replace the naturally occurring metal with one whose reactivity can be traced by spectroscopic methods. The substitute metal ion must bind to the same site as the naturally occurring ion, and it must have a similar (or greater) affinity for that site, as indicated by its charge density. Arrange the following ions in order of increasing effectiveness as a replacement for Ca2+, which has an ionic radius of 100 pm (numbers in parentheses are ionic radii): Na+ (102 pm), Eu2+ (117 pm), Sr2+ (118 pm), F− (133 pm), Pb2+ (119 pm), and La3+ (103 pm). Explain your reasoning.
Given: ions and ionic radii
Asked for: suitability as replacement for Ca2+
Strategy:
Use periodic trends to arrange the ions from least effective to most effective as a replacement for Ca2+.
Solution:
The most important properties in determining the affinity of a biological molecule for a metal ion are the size and charge-to-radius ratio of the metal ion. Of the possible Ca2+ replacements listed, the F− ion has the opposite charge, so it should have no affinity for a Ca2+-binding site. Na+ is approximately the right size, but with a +1 charge it will bind much more weakly than Ca2+. Although Eu2+, Sr2+, and Pb2+ are all a little larger than Ca2+, they are probably similar enough in size and charge to bind. Based on its ionic radius, Eu2+ should bind most tightly of the three. La3+ is nearly the same size as Ca2+ and more highly charged. With a higher charge-to-radius ratio and a similar size, La3+ should bind tightly to a Ca2+ site and be the most effective replacement for Ca2+. The order is F− << Na+ << Pb2+ ~ Sr2+ ~ Eu2+ < La3+.
Exercise
The ionic radius of K+ is 138 pm. Arrange the following ions in order of increasing affinity for a K+-binding site in an enzyme (numbers in parentheses are ionic radii): Na+ (102 pm), Rb+ (152 pm), Ba2+ (135 pm), Cl− (181 pm), and Tl+ (150 pm).
Answer: Cl− << Na+ < Tl+ ~ Rb+ < Ba2+
Covalent hydrides in which hydrogen is bonded to oxygen, nitrogen, or sulfur are polar, hydrophilic molecules that form hydrogen bonds and undergo acid–base reactions. Hydrogen-bonding interactions are crucial in stabilizing the structure of proteins and DNA and allow genetic information to be duplicated. The hydrogen-bonding interactions in water and ice also allow life to exist on our planet. The group 1 and group 2 metals present in organisms are macrominerals, which are important components of intracellular and extracellular fluids. Small changes in the cellular concentration of a metal ion can have a significant impact on biological functions. Metal ions are selectively transported across cell membranes by ion pumps, which bind ions based on their charge and radius. Ionophores, many of which are potent antibiotics, facilitate the transport of metal ions across membranes.
Explain the thermochemical properties of water in terms of its intermolecular bonding interactions. How does this affect global climate patterns?
Of the three classes of hydrides, which is (are) biochemically significant? How do you account for this?
Many proteins are degraded and become nonfunctional when heated higher than a certain temperature, even though the individual protein molecules do not undergo a distinct chemical change. Propose an explanation for this observation.
Los Angeles has moderate weather throughout the year, with average temperatures between 57°F and 70°F. In contrast, Palm Springs, which is just 100 miles inland, has average temperatures between 55°F and 95°F. Explain the difference in the average temperature ranges between the two cities.
Although all group 1 ions have the same charge (+1), Na+ and K+ ions are selectively transported across cell membranes. What strategy do organisms employ to discriminate between these two cations?
A 0.156 g sample of a chloride salt of an alkaline earth metal is dissolved in enough water to make 20.5 mL of solution. If this solution has an osmotic pressure of 2.68 atm at 25°C, what is the identity of the alkaline earth metal?
The thermal buffering capacity of water is one of the reasons the human body is capable of withstanding a wide range of temperatures. How much heat (in kilojoules) is required to raise the temperature of a 70.0 kg human from 37.0°C to 38.0°C? Assume that 70% of the mass of the body is water and that body fluids have the same specific heat as water.
During illness, body temperature can increase by more than 2°C. One piece of folklore is that you should “feed a fever.” Using the data in , how many fried chicken drumsticks would a 70.0 kg person need to eat to generate a 2.0°C change in body temperature? Assume the following: there is complete conversion of the caloric content of the chicken to thermal energy, 70% of the mass of the body is solely due to water, and body fluids have the same specific heat as water.
Hydrogen bonding is partly responsible for the high enthalpy of vaporization of water (ΔHvap = 43.99 kJ/mol at 25°C), which contributes to cooling the body during exercise. Assume that a 50.0 kg runner produces 20.0 g of perspiration during a race, and all the perspiration is converted to water vapor at 37.0°C. What amount of heat (in joules) is removed from the runner’s skin if the perspiration consists of only water?
Ba
For each application, which of the indicated substances would you select and why? Base your selections on the properties and reactivities of the alkaline earth metals and their compounds.
Ultrahigh-purity tritium, which is needed in the nuclear weapons industry, is obtained by allowing a mixture of tritium and its nuclear decay product, helium-3, to diffuse through a thin block of palladium metal. Explain why this is an effective method for separating the two substances.
In one technique for harnessing solar energy, the blue-green algae Anabaena cylindrica is used to produce H2 and O2 photosynthetically. The resulting gases are collected, passed through palladium metal, and then recombined in a fuel cell to produce electricity. Draw a diagram showing how this process might work.
Scientists speculate that sodium atoms react with atmospheric ozone to produce a high-energy species, which is then reduced by atomic oxygen. This process is believed to occur as meteors enter Earth’s atmosphere. Write equations for these reactions. Is sodium regenerated in this process?
Propose an effective compound for purifying and regenerating air for breathing in a submarine and justify your choice.
Explain why administering cryptands to a person suffering from iron toxicity could be an effective clinical treatment.
Calcium magnesium carbonate [CaMg(CO3)2], also known as dolomite, is a primary constituent of soils. It is formed when water containing magnesium ions comes in contact with calcium carbonate. Do you expect dolomite to be more or less resistant to acid rain than calcium carbonate? Why?
Few classes of reagents have proved to be as useful for organic syntheses as the Grignard reagents (RMgX), which are produced by reacting an organohalogen compound (RX) with magnesium in an ether solvent. The ease of formation of a Grignard reagent depends on the structure of the organohalogen compound.
The percentage of the population that developed leukemia in the vicinity of the Chernobyl nuclear reactor rose substantially after the nuclear accident in 1986. Why? Based on information in this chapter, what might have been done to reduce the incidence of leukemia in children who lived in the affected region?
Explain how ingesting large amounts of NaCl could induce a heart attack.
The general mechanism by which valinomycin functions is described in the text. If you were asked to develop a new antibiotic that functions like valinomycin, what type of structural features would you want to incorporate into it?
You have been asked to determine the concentrations of both Mg2+ and Ca2+ in a sample of hard water using EDTA (ethylenediaminetetraacetic acid), which forms stable complexes with both metal ions. The procedure requires two titrations. In the first, you adjust the pH of the water to about 10 by adding a solution of KOH. You then add a small amount of an indicator that, by itself, is blue at pH 10 but which forms a red complex with either Mg2+ or Ca2+. Titrating the sample with a solution of EDTA will cause the solution to turn from red to blue when all the metal ions have reacted with the EDTA. The second titration is identical to the first, except that it is carried out at pH 13, and the volume of EDTA solution needed to reach the endpoint is less than in the first titration. What is the difference between what is being measured in these two titrations? Which titration measures the concentration of only a single metal ion? Identify that metal ion and explain how the procedure described allows you to determine the concentrations of both Mg2+ and Ca2+.
The primary component of ordinary glass is SiO2, which is difficult to work with due to its high melting point (1700°C). Adding metal oxides lowers the melting temperature to a workable level (700–900°C) and causes the liquid to flow more easily. Both Na2O and K2O are commonly used for this purpose. Which of these oxides produces a glass that hardens more rapidly and at a higher temperature? Why is there a difference?
Lead poisoning can be treated by administering EDTA, an organic compound that binds to Pb2+ (ETDA is described in Problem 12), thereby removing lead from the system. If you are thinking of administering EDTA as a clinical treatment for Ni2+ poisoning because of its effectiveness in removing Pb2+, what additional factors do you need to consider?
Silicon, aluminum, and oxygen, the most abundant constituents of Earth’s crust, are found in many minerals. For example, two feldspars, which are typical components of granites, are albite and anorthite. Albite has the composition Na[AlSi3O8]; in anorthite, Al3+ replaces one of the Si4+ ions. Suggest a plausible cation in anorthite in place of Na+.
When a plant root is immersed in an aqueous environment, its uptake of metal ions is dictated by equilibrium considerations. Metal ions can be captured by reaction with various types of carrier molecules. The resulting complexes can pass through the cell membrane and dissociate, thereby transporting the metal ions into the cell. The Na+, K+, Mg2+, and Ca2+ ions are all transported using this mechanism. Which of these ions would be captured by
Sodium carbonate has long been used in the glass industry, for making water softeners, and in the wood pulp and paper industries. Recently, sodium carbonate has proved useful for removing SO2 and H2SO4 from the flue gases of coal- and oil-fired power stations. Write balanced chemical equations showing the reactions that occur when flue gases are passed through solid sodium carbonate.
Hard water contains high concentrations of Ca2+ and Mg2+. These ions react with soaps (sodium and potassium salts of naturally occurring carboxylic acids) to form an insoluble scum. Explain how adding Na2CO3 to hard water can soften the water and improve the ability of soaps to remove dirt and grime.
Do you expect MgCO3 or Mg(OH)2 to be more soluble in water? Based on your answer, explain how adding Ca(OH)2 softens water by removing Mg2+. Be sure to show a balanced chemical equation for the reaction.
How does a low-sodium diet help lower blood pressure? Do you think a diet high in KCl would also be effective at lowering blood pressure? Explain your answer.
KO2; the reaction is 4KO2(s) + 4CO2(g) + 2H2O(g) → 4KHCO3(s) + 3O2(g).
Dolomite will be less resistant to acid rain because MgCO3 is more soluble than CaCO3.
Restrict milk consumption and supplement diet with calcium to minimize the uptake of strontium.
The glass containing sodium will have a higher softening or melting temperature because the smaller sodium ions will link the silicate chains more strongly.
Ca2+
Na2CO3(s) + SO2(g) → Na2SO3(s) + CO2(g); Na2CO3(s) + H2SO4(g) → Na2SO4(s) + CO2(g) + H2O(g)