1689. $3 x$ -
3. How do the concentrations of $\mathrm{Ag}^{+}$and $\mathrm{CrO}_{4}{ }^{2-}$ in a saturated solution above 1.0 g of solid $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ change when 100 g of solid $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is added to the system? Explain.
4. How do the concentrations of $\mathrm{Pb}^{2+}$ and $\mathrm{S}^{2-}$ change when $\mathrm{K}_{2} \mathrm{~S}$ is added to a saturated solution of PbS ?
5. What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?
6. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $\mathrm{CoSO}_{3}, \mathrm{CuI}, \mathrm{PbCO}_{3}, \mathrm{PbCl}_{2}, \mathrm{Tl}_{2} \mathrm{~S}, \mathrm{KClO}_{4}$ ?
7. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $\mathrm{AgCl}, \mathrm{BaSO}_{4}, \mathrm{CaF}_{2}, \mathrm{Hg}_{2} \mathrm{I}_{2}, \mathrm{MnCO}_{3}$, and ZnS ?
8. Write the ionic equation for dissolution and the solubility product ( $K_{\mathrm{sp}}$ ) expression for each of the following slightly soluble ionic compounds:
(a) $\mathrm{PbCl}_{2}$
(b) $\mathrm{Ag}_{2} \mathrm{~S}$
(c) $\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
(d) $\mathrm{SrSO}_{4}$
9. Write the ionic equation for the dissolution and the $K_{\mathrm{sp}}$ expression for each of the following slightly soluble ionic compounds:


1689. $3 x$ -
(d) $\mathrm{SrSO}_{4}$
9. Write the ionic equation for the dissolution and the $K_{\mathrm{sp}}$ expression for each of the following slightly soluble ionic compounds:
(a) $\mathrm{LaF}_{3}$
(b) $\mathrm{CaCO}_{3}$
(c) $\mathrm{Ag}_{2} \mathrm{SO}_{4}$
(d) $\mathrm{Pb}(\mathrm{OH})_{2}$
10. The Handbook of Chemistry and Physics (http://openstax.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) $\mathrm{BaSiF}_{6}, 0.026 \mathrm{~g} / 100 \mathrm{~mL}$ (contains $\mathrm{SiF}_{6}{ }^{2-}$ ions)
(b) $\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{4}, 1.5 \times 10^{-2} \mathrm{~g} / 100 \mathrm{~mL}$
(c) $\mathrm{Gd}_{2}\left(\mathrm{SO}_{4}\right)_{3}, 3.98 \mathrm{~g} / 100 \mathrm{~mL}$
(d) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtBr}_{6}, 0.59 \mathrm{~g} / 100 \mathrm{~mL}$ (contains $\mathrm{PtBr}_{6}{ }^{2-}$ ions)
11. The Handbook of Chemistry and Physics (http://openstax.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) $\mathrm{BaSeO}_{4}, 0.0118 \mathrm{~g} / 100 \mathrm{~mL}$


1689. $3 x$ -
(a) $\mathrm{BaSeO}_{4}, 0.0118 \mathrm{~g} / 100 \mathrm{~mL}$
(b) $\mathrm{Ba}\left(\mathrm{BrO}_{3}\right)_{2} \cdot \mathrm{H}_{2} \mathrm{O}, 0.30 \mathrm{~g} / 100 \mathrm{~mL}$
(c) $\mathrm{NH}_{4} \mathrm{MgAsO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}, 0.038 \mathrm{~g} / 100 \mathrm{~mL}$
(d) $\mathrm{La}_{2}\left(\mathrm{MoO}_{4}\right)_{3}, 0.00179 \mathrm{~g} / 100 \mathrm{~mL}$
12. Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: $\mathrm{CaF}_{2}, \mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{PbI}_{2}$, or $\mathrm{Sn}(\mathrm{OH})_{2}$.
13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
(a) $\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}$
(b) $\mathrm{PbI}_{2}$
(c) $\mathrm{Ag}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$, a salt containing the $\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}$ ion
(d) $\mathrm{Hg}_{2} \mathrm{I}_{2}$
14. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
(a) $\mathrm{Ag}_{2} \mathrm{SO}_{4}$
(b) $\mathrm{PbBr}_{2}$
(c) AgI


1689. $3 x$ -
(a) $\mathrm{Ag}_{2} \mathrm{SO}_{4}$
(b) $\mathrm{PbBr}_{2}$
(c) AgI
(d) $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$
15. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
(a) $\mathrm{AgCl}(s)$ in 0.025 M NaCl
(b) $\mathrm{CaF}_{2}(s)$ in 0.00133 MKF
(c) $\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)$ in 0.500 L of a solution containing 19.50 g of $\mathrm{K}_{2} \mathrm{SO}_{4}$
(d) $\mathrm{Zn}(\mathrm{OH})_{2}(s)$ in a solution buffered at a pH of 11.45
16. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
(a) $\mathrm{TlCl}(s)$ in 1.250 M HCl
(b) $\mathrm{PbI}_{2}(s)$ in $0.0355 \mathrm{MCaI}_{2}$
(c) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$ in 0.225 L of a solution containing 0.856 g of $\mathrm{K}_{2} \mathrm{CrO}_{4}$
(d) $\mathrm{Cd}(\mathrm{OH})_{2}(s)$ in a solution buffered at a pH of 10.995
17. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.


1689. $3 x$ -
(a) $\mathrm{TlCl}(s)$ in $0.025 \mathrm{M} \mathrm{TlNO}_{3}$
(b) $\mathrm{BaF}_{2}(s)$ in 0.0313 M KF
(c) $\mathrm{MgC}_{2} \mathrm{O}_{4}$ in 2.250 L of a solution containing 8.156 g of $\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}$
(d) $\mathrm{Ca}(\mathrm{OH})_{2}(s)$ in an unbuffered solution initially with a pH of 12.700
18. Explain why the changes in concentrations of the common ions in Exercise 15.17 can be neglected.
19. Explain why the changes in concentrations of the common ions in Exercise 15.18 cannot be neglected.
20. Calculate the solubility of aluminum hydroxide, $\mathrm{Al}(\mathrm{OH})_{3}$, in a solution buffered at pH 11.00 .
21. Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.
22. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of $\mathrm{BaSO}_{4}$ is possible because of its low solubility. Calculate the molar solubility of $\mathrm{BaSO}_{4}$ and the mass of barium present in 1.00 L of water saturated with $\mathrm{BaSO}_{4}$.


1689. $3 x$ -
23. Public Health Service standards for drinking water set a maximum of $250 \mathrm{mg} / \mathrm{L}\left(2.60 \times 10^{-3} \mathrm{M}\right)$ of $\mathrm{SO}_{4}{ }^{2-}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with $\mathrm{CaSO}_{4}$ ("gyp" water) as a result or passing through soil containing gypsum, $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}$, meet these standards? What is the concentration of $\mathrm{SO}_{4}{ }^{2-}$ in such water?
24. Perform the following calculations:
(a) Calculate $\left[\mathrm{Ag}^{+}\right]$in a saturated aqueous solution of AgBr .
(b) What will $\left[\mathrm{Ag}^{+}\right]$be when enough KBr has been added to make $\left[\mathrm{Br}^{-}\right]=0.050 \mathrm{M}$ ?
(c) What will $\left[\mathrm{Br}^{-}\right]$be when enough $\mathrm{AgNO}_{3}$ has been added to make $\left[\mathrm{Ag}^{+}\right]=0.020 \mathrm{M}$ ?
25. The solubility product of $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ is $2.4 \times 10^{-5}$. What mass of this salt will dissolve in 1.0 L of 0.010 M $\mathrm{SO} 4^{2-}$ ?
26. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).
(a) TlCl
(b) $\mathrm{BaF}_{2}$
(c) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$
(d) $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$
(e) the mineral anglesite, $\mathrm{PbSO}_{4}$


1689. $3 x$ -
(d) $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$
(e) the mineral anglesite, $\mathrm{PbSO}_{4}$
27. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):
(a) AgI
(b) $\mathrm{Ag}_{2} \mathrm{SO}_{4}$
(c) $\mathrm{Mn}(\mathrm{OH})_{2}$
(d) $\mathrm{Sr}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}$
(e) the mineral brucite, $\mathrm{Mg}(\mathrm{OH})_{2}$
28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate $K_{\text {sp }}$ for each of the slightly soluble solids indicated:
(a) $\mathrm{AgBr}:\left[\mathrm{Ag}^{+}\right]=5.7 \times 10^{-7} \mathrm{M},\left[\mathrm{Br}^{-}\right]=5.7 \times 10^{-7} \mathrm{M}$
(b) $\mathrm{CaCO}_{3}:\left[\mathrm{Ca}^{2+}\right]=5.3 \times 10^{-3} \mathrm{M},\left[\mathrm{CO}_{3}{ }^{2-}\right]=9.0 \times 10^{-7} \mathrm{M}$
(c) $\mathrm{PbF}_{2}:\left[\mathrm{Pb}^{2+}\right]=2.1 \times 10^{-3} \mathrm{M},\left[\mathrm{F}^{-}\right]=4.2 \times 10^{-3} \mathrm{M}$
(d) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}:\left[\mathrm{Ag}^{+}\right]=5.3 \times 10^{-5} \mathrm{M}, 3.2 \times 10^{-3} \mathrm{M}$


1689. $3 x$ -
(e) $\mathrm{InF}_{3}:\left[\mathrm{In}^{3+}\right]=2.3 \times 10^{-3} \mathrm{M},\left[\mathrm{F}^{-}\right]=7.0 \times 10^{-3} \mathrm{M}$
29. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate $K_{\text {sp }}$ for each of the slightly soluble solids indicated:
(a) $\mathrm{TlCl}:\left[\mathrm{Tl}^{+}\right]=1.21 \times 10^{-2} \mathrm{M},\left[\mathrm{Cl}^{-}\right]=1.2 \times 10^{-2} \mathrm{M}$
(b) $\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{4}:\left[\mathrm{Ce}^{4+}\right]=1.8 \times 10^{-4} \mathrm{M},\left[\mathrm{IO}_{3}{ }^{-}\right]=2.6 \times 10^{-13} \mathrm{M}$
(c) $\mathrm{Gd}_{2}\left(\mathrm{SO}_{4}\right)_{3}:\left[\mathrm{Gd}^{3+}\right]=0.132 \mathrm{M},\left[\mathrm{SO}_{4}{ }^{2-}\right]=0.198 \mathrm{M}$
(d) $\mathrm{Ag}_{2} \mathrm{SO}_{4}:\left[\mathrm{Ag}^{+}\right]=2.40 \times 10^{-2} \mathrm{M},\left[\mathrm{SO}_{4}{ }^{2-}\right]=2.05 \times 10^{-2} \mathrm{M}$
(e) $\mathrm{BaSO}_{4}:\left[\mathrm{Ba}^{2+}\right]=0.500 \mathrm{M},\left[\mathrm{SO}_{4}{ }^{2-}\right]=4.6 \times 10^{-8} \mathrm{M}$
30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\text {sp }}$ values.)


1689. $3 x$ -
30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\text {sp }}$ values.)
(a) $\mathrm{KClO}_{4}:\left[\mathrm{K}^{+}\right]=0.01 \mathrm{M},\left[\mathrm{ClO}_{4}{ }^{-}\right]=0.01 \mathrm{M}$
(b) $\mathrm{K}_{2} \mathrm{PtCl}_{6}:\left[\mathrm{K}^{+}\right]=0.01 \mathrm{M},\left[\mathrm{PtCl}_{6}{ }^{2-}\right]=0.01 \mathrm{M}$
(c) $\mathrm{PbI}_{2}:\left[\mathrm{Pb}^{2+}\right]=0.003 \mathrm{M},\left[\mathrm{I}^{-}\right]=1.3 \times 10^{-3} \mathrm{M}$
(d) $\mathrm{Ag}_{2} \mathrm{~S}:\left[\mathrm{Ag}^{+}\right]=1 \times 10^{-10} \mathrm{M},\left[\mathrm{S}^{2-}\right]=1 \times 10^{-13} \mathrm{M}$
31. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\mathrm{sp}}$ values.)
(a) $\mathrm{CaCO}_{3}:\left[\mathrm{Ca}^{2+}\right]=0.003 \mathrm{M},\left[\mathrm{CO}_{3}{ }^{2-}\right]=0.003 \mathrm{M}$
(b) $\mathrm{Co}(\mathrm{OH})_{2}:\left[\mathrm{Co}^{2+}\right]=0.01 \mathrm{M},\left[\mathrm{OH}^{-}\right]=1 \times 10^{-7} \mathrm{M}$
(c) $\mathrm{CaHPO}_{4}:\left[\mathrm{Ca}^{2+}\right]=0.01 \mathrm{M},\left[\mathrm{HPO}_{4}{ }^{2-}\right]=2 \times 10^{-6} \mathrm{M}$


1689. $3 x$ -
(d) $\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}:\left[\mathrm{Pb}^{2+}\right]=0.01 \mathrm{M},\left[\mathrm{PO}_{4}{ }^{3-}\right]=1 \times 10^{-13} \mathrm{M}$
32. Calculate the concentration of $\mathrm{Tl}^{+}$when TlCl just begins to precipitate from a solution that is 0.0250 M in $\mathrm{Cl}^{-}$.
33. Calculate the concentration of sulfate ion when $\mathrm{BaSO}_{4}$ just begins to precipitate from a solution that is 0.0758 M in $\mathrm{Ba}^{2+}$.
34. Calculate the concentration of $\mathrm{Sr}^{2+}$ when $\mathrm{SrCrO}_{4}$ starts to precipitate from a solution that is 0.0025 M in $\mathrm{CrO}_{4}{ }^{2-}$.
35. Calculate the concentration of $\mathrm{PO}_{4}{ }^{3-}$ when $\mathrm{Ag}_{3} \mathrm{PO}_{4}$ starts to precipitate from a solution that is 0.0125 M in $\mathrm{Ag}^{+}$.
36. Calculate the concentration of $\mathrm{F}^{-}$required to begin precipitation of $\mathrm{CaF}_{2}$ in a solution that is 0.010 M in $\mathrm{Ca}^{2+}$.
37. Calculate the concentration of $\mathrm{Ag}^{+}$required to begin precipitation of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ in a solution that is $2.50 \times$ $10^{-6} \mathrm{M}$ in $\mathrm{CO}_{3}{ }^{2-}$.
38. What $\left[\mathrm{Ag}^{+}\right]$is required to reduce $\left[\mathrm{CO}_{3}^{2-}\right]$ to $8.2 \times 10^{-4} M$ by precipitation of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ ?


1689. $3 x$ -
39. What $\left[\mathrm{F}^{-}\right]$is required to reduce $\left[\mathrm{Ca}^{2+}\right]$ to $1.0 \times 10^{-4} \mathrm{M}$ by precipitation of $\mathrm{CaF}_{2}$ ?
40. A volume of 0.800 L of a $2 \times 10^{-4}-\mathrm{MBa}\left(\mathrm{NO}_{3}\right)_{2}$ solution is added to 0.200 L of $5 \times 10^{-4} \mathrm{MLi}_{2} \mathrm{SO}_{4}$. Does $\mathrm{BaSO}_{4}$ precipitate? Explain your answer.
41. Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing $\mathrm{NiCO}_{3}$ be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with $\mathrm{NiCO}_{3}\left(K_{\mathrm{sp}}=1.36 \times 10^{-7}\right)$.
(b) If the $\mathrm{NiCO}_{3}$ were a contaminant in a sample of $\mathrm{CoCO}_{3}\left(K_{\text {sp }}=1.0 \times 10^{-12}\right)$, what mass of $\mathrm{CoCO}_{3}$ would have been lost? Keep in mind that both $\mathrm{NiCO}_{3}$ and $\mathrm{CoCO}_{3}$ dissolve in the same solution.
42. Iron concentrations greater than $5.4 \times 10^{-6} M$ in water used for laundry purposes can cause staining. What $\left[\mathrm{OH}^{-}\right]$is required to reduce $\left[\mathrm{Fe}^{2+}\right]$ to this level by precipitation of $\mathrm{Fe}(\mathrm{OH})_{2}$ ?
43. A solution is 0.010 M in both $\mathrm{Cu}^{2+}$ and $\mathrm{Cd}^{2+}$. What percentage of $\mathrm{Cd}^{2+}$ remains in the solution when $99.9 \%$ of the $\mathrm{Cu}^{2+}$ has been precipitated as CuS by adding sulfide?


1689. $3 x$ -
44. A solution is 0.15 M in both $\mathrm{Pb}^{2+}$ and $\mathrm{Ag}^{+}$. If $\mathrm{Cl}^{-}$is added to this solution, what is $\left[\mathrm{Ag}^{+}\right]$when $\mathrm{PbCl}_{2}$ begins to precipitate?
45. What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH . (Hint: Consider the $K_{\mathrm{sp}}$ values given in Appendix J.)
(a) $\mathrm{Hg}_{2}{ }^{2+}$ and $\mathrm{Cu}^{2+}$
(b) $\mathrm{SO}_{4}{ }^{2-}$ and $\mathrm{Cl}^{-}$
(c) $\mathrm{Hg}^{2+}$ and $\mathrm{Co}^{2+}$
(d) $\mathrm{Zn}^{2+}$ and $\mathrm{Sr}^{2+}$
(e) $\mathrm{Ba}^{2+}$ and $\mathrm{Mg}^{2+}$
(f) $\mathrm{CO}_{3}{ }^{2-}$ and $\mathrm{OH}^{-}$
46. A solution contains $1.0 \times 10^{-5} \mathrm{~mol}$ of KBr and 0.10 mol of KCl per liter. $\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgBr or solid AgCl ?
47. A solution contains $1.0 \times 10^{-2} \mathrm{~mol}$ of KI and 0.10 mol of KCl per liter. $\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgI or solid AgCl ?


1689. $3 x$ -
48. The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$. It is necessary to remove all but $1.0 \%$ of the $\mathrm{Ca}^{2+}$ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of $\mathrm{Ca}^{2+}$ per 100 mL of serum, what mass of $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ is required to prevent the coagulation of a 10 mL blood sample that is $55 \%$ serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a $10-\mathrm{mL}$ blood sample is 5.5 mL . Assume that the $\mathrm{K}_{\mathrm{sp}}$ value for $\mathrm{CaC}_{2} \mathrm{O}_{4}$ in serum is the same as in water.)
49. About $50 \%$ of urinary calculi (kidney stones) consist of calcium phosphate, $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$. The normal mid range calcium content excreted in the urine is 0.10 g of $\mathrm{Ca}^{2+}$ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?


1689. $3 x$ -
50. The pH of normal urine is 6.30 , and the total phosphate concentration $\left(\left[\mathrm{PO}_{4}{ }^{3-}\right]+\left[\mathrm{HPO}_{4}{ }^{2-}\right]+\left[\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}\right]\right.$ $\left.+\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]\right)$ is 0.020 M . What is the minimum concentration of $\mathrm{Ca}^{2+}$ necessary to induce kidney stone formation? (See Exercise 15.49 for additional information.)
51. Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:
$\mathrm{Mg}^{2+}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Ca}^{2+}(a q)$
$\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$
$\mathrm{MgCl}_{2}(l) \xrightarrow{\text { electrolysis }} \mathrm{Mg}(s)+\mathrm{Cl}_{2}(g)$
Sea water has a density of $1.026 \mathrm{~g} / \mathrm{cm}^{3}$ and contains 1272 parts per million of magnesium as $\mathrm{Mg}^{2+}(\mathrm{aq})$ by mass. What mass, in kilograms, of $\mathrm{Ca}(\mathrm{OH})_{2}$ is required to precipitate $99.9 \%$ of the magnesium in $1.00 \times$ $10^{3} \mathrm{~L}$ of sea water?


1689. $3 x$ -
52. Hydrogen sulfide is bubbled into a solution that is $0.10 M$ in both $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ and 0.30 M in HCl . After the solution has come to equilibrium it is saturated with $\mathrm{H}_{2} \mathrm{~S}\left(\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.10 \mathrm{M}\right)$. What concentrations of $\mathrm{Pb}^{2+}$ and $\mathrm{Fe}^{2+}$ remain in the solution? For a saturated solution of $\mathrm{H}_{2} \mathrm{~S}$ we can use the equilibrium:


1689. $3 x$ -
$$
\mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q) \quad K=1.0 \times 10^{-26}
$$


1689. $3 x$ -
(Hint: The $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$changes as metal sulfides precipitate.)
53. Perform the following calculations involving concentrations of iodate ions:
(a) The iodate ion concentration of a saturated solution of $\mathrm{La}\left(\mathrm{IO}_{3}\right)_{3}$ was found to be $3.1 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$. Find the $K_{\mathrm{sp}}$.
(b) Find the concentration of iodate ions in a saturated solution of $\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2}\left(K_{\text {sp }}=7.4 \times 10^{-8}\right)$.
54. Calculate the molar solubility of AgBr in $0.035 \mathrm{M} \mathrm{NaBr}\left(K_{\mathrm{sp}}=5 \times 10^{-13}\right)$.
55. How many grams of $\mathrm{Pb}(\mathrm{OH})_{2}$ will dissolve in 500 mL of a $0.050-M \mathrm{PbCl}_{2}$ solution $\left(K_{\text {sp }}=1.2 \times 10^{-15}\right)$ ?
56. Use the simulation (http://openstax.org/l/16solublesalts) from the earlier Link to Learning to complete the following exercise. Using $0.01 \mathrm{~g} \mathrm{CaF}_{2}$, give the $\mathrm{K}_{\mathrm{sp}}$ values found in a $0.2-M$ solution of each of the salts. Discuss why the values change as you change soluble salts.


1689. $3 x$ -
57. How many grams of Milk of Magnesia, $\mathrm{Mg}(\mathrm{OH})_{2}(S)(58.3 \mathrm{~g} / \mathrm{mol})$, would be soluble in 200 mL of water. $K_{\mathrm{sp}}=$ $7.1 \times 10^{-12}$. Include the ionic reaction and the expression for $K_{\text {sp }}$ in your answer. ( $K_{\mathrm{w}}=1 \times 10^{-14}=$ $\left.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\right)$
58. Two hypothetical salts, $\mathrm{LM}_{2}$ and LQ , have the same molar solubility in $\mathrm{H}_{2} \mathrm{O}$. If $K_{\text {sp }}$ for $\mathrm{LM}_{2}$ is $3.20 \times 10^{-5}$, what is the $K_{\mathrm{sp}}$ value for LQ ?
59. The carbonate ion concentration is gradually increased in a solution containing equal concentrations of the divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will precipitate first? Which will precipitate last? Explain.
(a) $\mathrm{MgCO}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O} \quad K_{\text {sp }}=1 \times 10^{-5}$
(b) $\mathrm{CaCO}_{3}$
(c) $\mathrm{SrCO}_{3}$
$K_{\mathrm{sp}}=8.7 \times 10^{-9}$
(d) $\mathrm{BaCO}_{3}$
$K_{\mathrm{sp}}=7 \times 10^{-10}$
(e) $\mathrm{MnCO}_{3}$
$K_{\text {sp }}=1.6 \times 10^{-9}$
$K_{\text {sp }}=8.8 \times 10^{-11}$


1689. $3 x$ -
$K_{\text {sp }}=1.6 \times 10^{-9}$
$K_{\text {sp }}=8.8 \times 10^{-11}$
60. How many grams of $\mathrm{Zn}(\mathrm{CN})_{2}(S)(117.44 \mathrm{~g} / \mathrm{mol})$ would be soluble in 100 mL of $\mathrm{H}_{2} \mathrm{O}$ ? Include the balanced reaction and the expression for $K_{\text {sp }}$ in your answer. The $K_{\text {sp }}$ value for $\mathrm{Zn}(\mathrm{CN})_{2}(s)$ is $3.0 \times 10^{-16}$.


1689. $3 x$ - 1689.1. Lewis Acids and Bases
61. Even though $\mathrm{Ca}(\mathrm{OH})_{2}$ is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_{2}$ ?
62. Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?
63. Explain why the addition of $\mathrm{NH}_{3}$ or $\mathrm{HNO}_{3}$ to a saturated solution of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ in contact with solid $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ increases the solubility of the solid.
64. Calculate the cadmium ion concentration, $\left[\mathrm{Cd}^{2+}\right]$, in a solution prepared by mixing 0.100 L of 0.0100 M $\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}$ with 0.150 L of $0.100 \mathrm{NH}_{3}(\mathrm{aq})$.
65. Explain why addition of $\mathrm{NH}_{3}$ or $\mathrm{HNO}_{3}$ to a saturated solution of $\mathrm{Cu}(\mathrm{OH})_{2}$ in contact with solid $\mathrm{Cu}(\mathrm{OH})_{2}$ increases the solubility of the solid.
66. Sometimes equilibria for complex ions are described in terms of dissociation constants, $K_{d}$. For the complex ion $\mathrm{AlF}_{6}{ }^{3-}$ the dissociation reaction is:
$\mathrm{AlF}_{6}{ }^{3-} \rightleftharpoons \mathrm{Al}^{3+}+6 \mathrm{~F}^{-}$and $K_{\mathrm{d}}=\frac{\left[\mathrm{Al}^{3+}\right]\left[\mathrm{F}^{-}\right]^{6}}{\left[\mathrm{AlF}_{6}{ }^{3-}\right]}=2 \times 10^{-24}$
Calculate the value of the formation constant, $K_{\mathrm{f}}$, for $\mathrm{AlF}_{6}{ }^{3-}$.


1689. $3 x$ - 1689.1. Lewis Acids and Bases
Calculate the value of the formation constant, $K_{\mathrm{f}}$, for $\mathrm{AlF}_{6}{ }^{3-}$.
67. Using the value of the formation constant for the complex ion $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}$, calculate the dissociation constant.
68. Using the dissociation constant, $K_{\mathrm{d}}=7.8 \times 10^{-18}$, calculate the equilibrium concentrations of $\mathrm{Cd}^{2+}$ and $\mathrm{CN}^{-}$in a $0.250-M$ solution of $\mathrm{Cd}(\mathrm{CN})_{4}{ }^{2-}$.
69. Using the dissociation constant, $K_{\mathrm{d}}=3.4 \times 10^{-15}$, calculate the equilibrium concentrations of $\mathrm{Zn}^{2+}$ and $\mathrm{OH}^{-}$in a $0.0465-M$ solution of $\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}$.
70. Using the dissociation constant, $K_{\mathrm{d}}=2.2 \times 10^{-34}$, calculate the equilibrium concentrations of $\mathrm{Co}^{3+}$ and $\mathrm{NH}_{3}$ in a $0.500-M$ solution of $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}{ }^{3+}$.
71. Using the dissociation constant, $K_{\mathrm{d}}=1 \times 10^{-44}$, calculate the equilibrium concentrations of $\mathrm{Fe}^{3+}$ and $\mathrm{CN}^{-}$ in a 0.333 M solution of $\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}$.
72. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve $2.0 \times$ $10^{-2} \mathrm{~mol}$ of silver cyanide, AgCN .
73. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve $3.0 \times 10^{-3} \mathrm{~mol}$ of silver bromide.


1689. $3 x$ - 1689.1. Lewis Acids and Bases
73. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve $3.0 \times 10^{-3} \mathrm{~mol}$ of silver bromide.
74. A roll of $35-\mathrm{mm}$ black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\left(K_{\mathrm{f}}=4.7 \times 10^{13}\right)$ ?
75. We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, $\mathrm{H}^{+}$) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
76. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) $\mathrm{CO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{HCO}_{3}{ }^{-}$
(b) $\mathrm{B}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow \mathrm{B}(\mathrm{OH})_{4}{ }^{-}$
(c) $\mathrm{I}^{-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}_{3}{ }^{-}$


1689. $3 x$ - 1689.1. Lewis Acids and Bases
(c) $\mathrm{I}^{-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}_{3}{ }^{-}$
(d) $\mathrm{AlCl}_{3}+\mathrm{Cl}^{-} \longrightarrow \mathrm{AlCl}_{4}{ }^{-}$(use Al-Cl single bonds)
(e) $\mathrm{O}^{2-}+\mathrm{SO}_{3} \longrightarrow \mathrm{SO}_{4}{ }^{2-}$
77. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
(a) $\mathrm{CS}_{2}+\mathrm{SH}^{-} \longrightarrow \mathrm{HCS}_{3}{ }^{-}$
(b) $\mathrm{BF}_{3}+\mathrm{F}^{-} \longrightarrow \mathrm{BF}_{4}^{-}$
(c) $\mathrm{I}^{-}+\mathrm{SnI}_{2} \longrightarrow \mathrm{SnI}_{3}{ }^{-}$
(d) $\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}{ }^{-}$
(e) $\mathrm{F}^{-}+\mathrm{SO}_{3} \longrightarrow \mathrm{SFO}_{3}{ }^{-}$
78. Using Lewis structures, write balanced equations for the following reactions:
(a) $\mathrm{HCl}(g)+\mathrm{PH}_{3}(g) \longrightarrow$
(b) $\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3}^{-} \longrightarrow$
(c) $\mathrm{CaO}+\mathrm{SO}_{3} \longrightarrow$
(d) $\mathrm{NH}_{4}^{+}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} \longrightarrow$
79. Calculate $\left[\mathrm{HgCl}_{4}{ }^{2-}\right.$ ] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a $0.100-\mathrm{M} \mathrm{HgCl} 2$ solution.


1689. $3 x$ - 1689.1. Lewis Acids and Bases
80. In a titration of cyanide ion, 28.72 mL of $0.0100 \mathrm{MANO}_{3}$ is added before precipitation begins. [The reaction of $\mathrm{Ag}^{+}$with $\mathrm{CN}^{-}$goes to completion, producing the $\mathrm{Ag}(\mathrm{CN})_{2}{ }^{-}$complex.] Precipitation of solid AgCN takes place when excess $\mathrm{Ag}^{+}$is added to the solution, above the amount needed to complete the formation of $\mathrm{Ag}(\mathrm{CN})_{2}{ }^{-}$. How many grams of NaCN were in the original sample?
81. What are the concentrations of $\mathrm{Ag}^{+}, \mathrm{CN}^{-}$, and $\mathrm{Ag}(\mathrm{CN})_{2}{ }^{-}$in a saturated solution of AgCN ?
82. In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point $=19.5^{\circ} \mathrm{C}$ ) is a strong acid. In liquid $\mathrm{HF}, \mathrm{HNO}_{3}$ acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept $\mathrm{F}^{-}$ion (for example, $\mathrm{BF}_{3}$ or $\mathrm{SbF}_{5}$ ). Write balanced chemical equations for the reaction of pure $\mathrm{HNO}_{3}$ with pure HF and of pure HF with $\mathrm{BF}_{3}$.


1689. $3 x$ - 1689.1. Lewis Acids and Bases
83. The simplest amino acid is glycine, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}$. The common feature of amino acids is that they contain the functional groups: an amine group, $-\mathrm{NH}_{2}$, and a carboxylic acid group, $-\mathrm{CO}_{2} \mathrm{H}$. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, and the base strength of the amino group is slightly greater than that of ammonia, $\mathrm{NH}_{3}$.
(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 MHCl and in 1 MKOH .
(b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the $-\mathrm{NH}_{2}$ and $-\mathrm{CO}_{2}{ }^{-}$groups.)
84. Boric acid, $\mathrm{H}_{3} \mathrm{BO}_{3}$, is not a Brønsted-Lowry acid but a Lewis acid.
(a) Write an equation for its reaction with water.
(b) Predict the shape of the anion thus formed.
(c) What is the hybridization on the boron consistent with the shape you have predicted?


1689. $3 x$ - 1689.2. Coupled Equilibria
85. A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
86. Calculate the equilibrium concentration of $\mathrm{Ni}^{2+}$ in a $1.0-M$ solution $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}$.
87. Calculate the equilibrium concentration of $\mathrm{Zn}^{2+}$ in a $0.30-M$ solution of $\mathrm{Zn}(\mathrm{CN})_{4}{ }^{2-}$.
88. Calculate the equilibrium concentration of $\mathrm{Cu}^{2+}$ in a solution initially with $0.050 \mathrm{MCu}^{2+}$ and $1.00 \mathrm{MNH}_{3}$.
89. Calculate the equilibrium concentration of $\mathrm{Zn}^{2+}$ in a solution initially with $0.150 \mathrm{M} \mathrm{Zn}^{2+}$ and $2.50 \mathrm{MCN}^{-}$.
90. Calculate the $\mathrm{Fe}^{3+}$ equilibrium concentration when 0.0888 mole of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is added to a solution with 0.0.00010 $\mathrm{MCN}^{-}$.
91. Calculate the $\mathrm{Co}^{2+}$ equilibrium concentration when 0.010 mole of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}$ is added to a solution with $0.25 \mathrm{M} \mathrm{NH}_{3}$. Assume the volume is 1.00 L .
92. Calculate the molar solubility of $\mathrm{Sn}(\mathrm{OH})_{2}$ in a buffer solution containing equal concentrations of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4}{ }^{+}$.


1689. $3 x$ - 1689.2. Coupled Equilibria
93. Calculate the molar solubility of $\mathrm{Al}(\mathrm{OH})_{3}$ in a buffer solution with $0.100 \mathrm{MNH}_{3}$ and $0.400 \mathrm{MNH}_{4}{ }^{+}$.
94. What is the molar solubility of $\mathrm{CaF}_{2}$ in a $0.100-M$ solution of HF ? $K_{\mathrm{a}}$ for $\mathrm{HF}=6.4 \times 10^{-4}$.
95. What is the molar solubility of $\mathrm{BaSO}_{4}$ in a $0.250-M$ solution of $\mathrm{NaHSO}_{4}$ ? $K_{\mathrm{a}}$ for $\mathrm{HSO}_{4}^{-}=1.2 \times 10^{-2}$.
96. What is the molar solubility of $\mathrm{Tl}(\mathrm{OH})_{3}$ in a $0.10-M$ solution of $\mathrm{NH}_{3}$ ?
97. What is the molar solubility of $\mathrm{Pb}(\mathrm{OH})_{2}$ in a $0.138-M$ solution of $\mathrm{CH}_{3} \mathrm{NH}_{2}$ ?
98. A solution of $0.075 \mathrm{MCoBr}_{2}$ is saturated with $\mathrm{H}_{2} \mathrm{~S}\left(\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.10 \mathrm{M}\right)$. What is the minimum pH at which CoS begins to precipitate?
$\mathrm{CoS}(s) \rightleftharpoons \mathrm{Co}^{2+}(a q)+\mathrm{S}^{2-}(a q) \quad K_{\mathrm{sp}}=2.3 \times 10^{27}$


1689. $3 x$ - 1689.2. Coupled Equilibria
$\mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q) \quad K=8.9 \times 10^{-27}$
99. A $0.125-M$ solution of $\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}$ is saturated with $\mathrm{H}_{2} \mathrm{~S}\left(\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.10 \mathrm{M}\right)$. At what pH does MnS begin to precipitate?
$\operatorname{MnS}(s) \rightleftharpoons \mathrm{Mn}^{2+}(a q)+\mathrm{S}^{2-}(a q) \quad K_{\mathrm{sp}}=2.3 \times 10^{-13}$
$\mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{S}^{2-}(a q) \quad K=1.0 \times 10^{-26}$
100. Both AgCl and AgI dissolve in $\mathrm{NH}_{3}$.
(a) What mass of AgI dissolves in 1.0 L of $1.0 \mathrm{MNH}_{3}$ ?
(b) What mass of AgCl dissolves in 1.0 L of $1.0 \mathrm{M} \mathrm{NH}_{3}$ ?
101. The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Solve the following problem:
$\operatorname{MgF}_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)$


1689. $3 x$ - 1689.2. Coupled Equilibria
$\operatorname{MgF}_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)$
In a saturated solution of $\mathrm{MgF}_{2}$ at $18{ }^{\circ} \mathrm{C}$, the concentration of $\mathrm{Mg}^{2+}$ is $1.21 \times 10^{-3} \mathrm{M}$. The equilibrium is represented by the preceding equation.
(a) Write the expression for the solubility-product constant, $K_{\text {sp }}$, and calculate its value at $18{ }^{\circ} \mathrm{C}$.
(b) Calculate the equilibrium concentration of $\mathrm{Mg}^{2+}$ in 1.000 L of saturated $\mathrm{MgF}_{2}$ solution at $18{ }^{\circ} \mathrm{C}$ to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
(c) Predict whether a precipitate of $\mathrm{MgF}_{2}$ will form when 100.0 mL of a $3.00 \times 10^{-3}-M$ solution of $\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with 200.0 mL of a $2.00 \times 10^{-3}-\mathrm{M}$ solution of NaF at $18^{\circ} \mathrm{C}$. Show the calculations to support your prediction.
(d) At $27{ }^{\circ} \mathrm{C}$ the concentration of $\mathrm{Mg}^{2+}$ in a saturated solution of $\mathrm{MgF}_{2}$ is $1.17 \times 10^{-3} \mathrm{M}$. Is the dissolving of $\mathrm{MgF}_{2}$ in water an endothermic or an exothermic process? Give an explanation to support your conclusion.


1689. $3 x$ - 1689.2. Coupled Equilibria
102. Which of the following compounds, when dissolved in a $0.01-M$ solution of $\mathrm{HClO}_{4}$, has a solubility greater than in pure water: $\mathrm{CuCl}, \mathrm{CaCO}_{3}, \mathrm{MnS}, \mathrm{PbBr}_{2}, \mathrm{CaF}_{2}$ ? Explain your answer.
103. Which of the following compounds, when dissolved in a $0.01-M$ solution of $\mathrm{HClO}_{4}$, has a solubility greater than in pure water: $\mathrm{AgBr}, \mathrm{BaF}_{2}, \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}, \mathrm{ZnS}, \mathrm{PbI}_{2}$ ? Explain your answer.
104. What is the effect on the amount of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ that dissolves and the concentrations of $\mathrm{Mg}^{2+}$ and $\mathrm{OH}^{-}$ when each of the following are added to a mixture of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ and water at equilibrium?
(a) $\mathrm{MgCl}_{2}$
(b) KOH
(c) $\mathrm{HClO}_{4}$
(d) $\mathrm{NaNO}_{3}$
(e) $\mathrm{Mg}(\mathrm{OH})_{2}$
105. What is the effect on the amount of $\mathrm{CaHPO}_{4}$ that dissolves and the concentrations of $\mathrm{Ca}^{2+}$ and $\mathrm{HPO}_{4}{ }^{2-}$ when each of the following are added to a mixture of solid $\mathrm{CaHPO}_{4}$ and water at equilibrium?
(a) $\mathrm{CaCl}_{2}$
(b) HCl
(c) $\mathrm{KClO}_{4}$
(d) NaOH
(e) $\mathrm{CaHPO}_{4}$


1689. $3 x$ - 1689.2. Coupled Equilibria
(a) $\mathrm{CaCl}_{2}$
(b) HCl
(c) $\mathrm{KClO}_{4}$
(d) NaOH
(e) $\mathrm{CaHPO}_{4}$
106. Identify all chemical species present in an aqueous solution of $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ and list these species in decreasing order of their concentrations. (Hint: Remember that the $\mathrm{PO}_{4}{ }^{3-}$ ion is a weak base.)


1689. $3 x$ - 1689.2. Coupled Equilibria
758 15 • Exercises


1690. CHAPTER 16 <br> Electrochemistry -
Figure 16.1 Electric vehicles are powered by batteries, devices that harness the energy of spontaneous redox reactions. (credit: modification of work by Robert Couse-Baker)


1691. CHAPTER OUTLINE - 1691.1. Review of Redox Chemistry
16.2 Galvanic Cells
16.3 Electrode and Cell Potentials
16.4 Potential, Free Energy, and Equilibrium
16.5 Batteries and Fuel Cells
16.6 Corrosion
16.7 Electrolysis

INTRODUCTION Another chapter in this text introduced the chemistry of reduction-oxidation (redox) reactions. This important reaction class is defined by changes in oxidation states for one or more reactant elements, and it includes a subset of reactions involving the transfer of electrons between reactant species. Around the turn of the nineteenth century, chemists began exploring ways these electrons could be transferred indirectly via an external circuit rather than directly via intimate contact of redox reactants. In the two centuries since, the field of electrochemistry has evolved to yield significant insights on the fundamental aspects of redox chemistry as well as a wealth of technologies ranging from industrial-scale metallurgical processes to robust, rechargeable batteries for electric vehicles (Figure 16.1). In this chapter, the essential concepts of electrochemistry will be addressed.


1692. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe defining traits of redox chemistry
- Identify the oxidant and reductant of a redox reaction
- Balance chemical equations for redox reactions using the half-reaction method

Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier text chapter (see chapter on reaction stoichiometry). Readers wishing additional review are referred to the text chapter on reaction stoichiometry.


1693. Oxidation Numbers -
By definition, a redox reaction is one that entails changes in oxidation number (or oxidation state) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description, the oxidation number of an atom in an element is equal to zero. For an atom in a compound, the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic. Consequential to these rules, the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.

Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements' oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl , is comprised of $\mathrm{Na}^{+}$cations and $\mathrm{Cl}^{-}$anions, and so oxidation numbers for sodium and chlorine are, +1 and -1 , respectively. Calcium fluoride, $\mathrm{CaF}_{2}$, is comprised of $\mathrm{Ca}^{2+}$ cations and $\mathrm{F}^{-}$anions, and so oxidation numbers for calcium and fluorine are, +2 and -1 , respectively.

Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent $\mathrm{O}-\mathrm{H}$ bonds. The shared electrons comprising an $\mathrm{O}-\mathrm{H}$ bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral.


1693. Oxidation Numbers -
Imagine that the polarization of shared electrons within the $\mathrm{O}-\mathrm{H}$ bonds of water were $100 \%$ complete-the result would be transfer of electrons from H to O , and water would be an ionic compound comprised of $\mathrm{O}^{2-}$ anions and $\mathrm{H}^{+}$cations. And so, the oxidations numbers for oxygen and hydrogen in water are -2 and +1 , respectively. Applying this same logic to carbon tetrachloride, $\mathrm{CCl}_{4}$, yields oxidation numbers of +4 for carbon and -1 for chlorine. In the nitrate ion, $\mathrm{NO}_{3}^{-}$, the oxidation number for nitrogen is +5 and that for oxygen is -2 , summing to equal the 1 - charge on the molecule:

$$
(1 \mathrm{~N} \text { atom })\left(\frac{+5}{\mathrm{~N} \text { atom }}\right)+(3 \mathrm{O} \text { atoms })\left(\frac{-2}{\mathrm{O} \text { atom }}\right)=+5+-6=-1
$$


1694. Balancing Redox Equations -
The unbalanced equation below describes the decomposition of molten sodium chloride:

$$
\mathrm{NaCl}(l) \longrightarrow \mathrm{Na}(l)+\mathrm{Cl}_{2}(g) \quad \text { unbalanced }
$$

This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes reduction) and that for Cl is increased from -1 to 0 (it undergoes oxidation). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na :

$$
2 \mathrm{NaCl}(l) \longrightarrow 2 \mathrm{Na}(l)+\mathrm{Cl}_{2}(g) \quad \text { balanced }
$$

Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, $\mathrm{H}^{+}(a q)$ and $\mathrm{OH}^{-}(a q)$, as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the half-reaction method is helpful. This approach involves the following steps:

1. Write skeletal equations for the oxidation and reduction half-reactions.
2. Balance each half-reaction for all elements except H and O .
3. Balance each half-reaction for O by adding $\mathrm{H}_{2} \mathrm{O}$.
4. Balance each half-reaction for H by adding $\mathrm{H}^{+}$.
5. Balance each half-reaction for charge by adding electrons.
6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
7. Add the two half-reactions and simplify.
8. If the reaction takes place in a basic medium, add $\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.

The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.


1696. Balancing Equations for Redox Reactions in Acidic Solutions -
Write the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.


1697. Solution -
Following the steps of the half-reaction method:

1. Write skeletal equations for the oxidation and reduction half-reactions.

$$
\begin{array}{ll}
\text { oxidation: } & \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) \\
\text { reduction: } & \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{NO}(g)
\end{array}
$$

2. Balance each half-reaction for all elements except $H$ and $O$.
oxidation: $\quad \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)$
reduction: $\quad \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{NO}(g)$
3. Balance each half-reaction for O by adding $\mathrm{H}_{2} \mathrm{O}$.
oxidation: $\quad \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)$
reduction: $\quad \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{NO}(g)+\mathbf{2 H}_{\mathbf{2}} \mathbf{O}(l)$
4. Balance each half-reaction for $H$ by adding $H^{+}$.

$$
\begin{array}{ll}
\text { oxidation: } & \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) \\
\text { reduction: } & \mathbf{3 H}^{+}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)
\end{array}
$$


1697. Solution -
5. Balance each half-reaction for charge by adding electrons.
oxidation: $\quad \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathbf{e}^{-}$
reduction: $\quad 3 \mathrm{e}^{-}+3 \mathrm{H}^{+}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$
6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
oxidation $(\times 3): \quad 3 \mathrm{Cu}(s) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+62 \mathrm{e}^{-}$
reduction $(\times 2): \quad \quad 63-\mathrm{e}^{-}+63-\mathrm{H}^{+}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow 2 \mathrm{NO}(g)+42 \mathrm{H}_{2} \mathrm{O}(l)$
7. Add the two half-reactions and simplify.
$3 \mathrm{Cu}(s)+\mathbf{6} \mathrm{e}^{-}+6 \mathrm{H}^{+}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+\mathbf{6} \mathrm{e}^{-}+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$
$3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$


1697. Solution -
8. If the reaction takes place in a basic medium, add $\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.


1697. Solution -
This step not necessary since the solution is stipulated to be acidic.
The balanced equation for the reaction in an acidic solution is then

$$
3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)
$$


1698. Check Your Learning -
The reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.

Answer:
$\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$


1700. Balancing Equations for Redox Reactions in Basic Solutions -
Write the balanced equation representing reaction between aqueous permanganate ion, $\mathrm{MnO}_{4}{ }^{-}$, and solid chromium(III) hydroxide, $\mathrm{Cr}(\mathrm{OH})_{3}$, to yield solid manganese(IV) oxide, $\mathrm{MnO}_{2}$, and aqueous chromate ion, $\mathrm{CrO}_{4}{ }^{2-}$ The reaction takes place in a basic solution.


1701. Solution -
Following the steps of the half-reaction method:


1701. Solution -
1. Write skeletal equations for the oxidation and reduction half-reactions.
oxidation: $\quad \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)$
reduction: $\quad \mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)$
2. Balance each half-reaction for all elements except $H$ and $O$.
oxidation: $\quad \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)$
reduction: $\quad \mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)$
3. Balance each half-reaction for O by adding $\mathrm{H}_{2} \mathrm{O}$.
oxidation: $\quad \mathbf{H}_{2} \mathbf{O}(l)+\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)$
reduction: $\quad \mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+2 \mathbf{H}_{\mathbf{2}} \mathbf{O}(l)$
4. Balance each half-reaction for $H$ by adding $H^{+}$.
oxidation: $\quad \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+\mathbf{5 H}^{+}(a q)$
reduction: $\quad 4 \mathbf{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$
5. Balance each half-reaction for charge by adding electrons.


1701. Solution -
5. Balance each half-reaction for charge by adding electrons.
oxidation: $\quad \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+5 \mathrm{H}^{+}(a q)+\mathbf{3 e}^{-}$
reduction: $\quad 3 \mathbf{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$
6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.
This step is not necessary since the number of electrons is already in balance.
7. Add the two half-reactions and simplify.
$\mathbf{H}_{\mathbf{2}} \mathbf{O}(\boldsymbol{(})+\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathbf{3} \mathbf{e}^{-}+\mathbf{- 4 \mathbf { H } ^ { + }}(\mathrm{aq})+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+\mathbf{5}^{-} \mathrm{H}^{+}(a q)$ $+3 \mathrm{e}^{-}+\mathrm{MnO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$
$\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+\mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$


1701. Solution -
8. If the reaction takes place in a basic medium, add $\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.
$\mathbf{O H}^{-}(a q)+\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+\mathrm{H}^{+}(a q)+\mathbf{O H}^{-}(a q)+\mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$
$\mathrm{OH}^{-}(a q)+\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{CrO}_{4}{ }^{2-}(a q)+\mathrm{MnO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$


1702. Check Your Learning -
Aqueous permanganate ion may also be reduced using aqueous bromide ion, $\mathrm{Br}^{-}$, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, $\mathrm{BrO}_{3}{ }^{-}$. Write the balanced equation for this
reaction occurring in a basic medium.
Answer:
$\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q)$


1703. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the function of a galvanic cell and its components
- Use cell notation to symbolize the composition and construction of galvanic cells

As demonstration of spontaneous chemical change, Figure 16.2 shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.

FIGURE 16.2 A copper wire and an aqueous solution of silver nitrate (left) are brought into contact (center) and a spontaneous transfer of electrons occurs, creating blue $\mathrm{Cu}^{2+}(a q)$ and gray $\operatorname{Ag}(s)$ (right).

These observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, $\mathrm{Cu}^{2+}(a q)$, which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:

$$
\begin{array}{ll}
\text { overall reaction: } & \mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s) \\
\text { oxidation half-reaction: } & \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \\
\text {reduction half-reaction: } & 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(s)
\end{array}
$$


1703. LEARNING OBJECTIVES -
Consider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells, and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells).

A galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure 16.3. The cell is comprised of two half-cells, each containing the redox conjugate pair ("couple") of a single reactant. The half-cell shown at the left contains the $\mathrm{Cu}(0) / \mathrm{Cu}(\mathrm{II})$ couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the $\mathrm{Ag}(\mathrm{I}) / \mathrm{Ag}(0)$ couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode. By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell's reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge. The spontaneous reaction in this cell produces $\mathrm{Cu}^{2+}$ cations in the anode half-cell and consumes $\mathrm{Ag}^{+}$ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of $\mathrm{Cu}^{2+}$ in the anode half-cell are balanced by an influx of $\mathrm{NO}_{3}{ }^{-}$from the salt bridge, while a flow of $\mathrm{Na}^{+}$into the cathode half-cell compensates for the decreasing $\mathrm{Ag}^{+}$concentration.


1703. LEARNING OBJECTIVES -
FIGURE 16.3 A galvanic cell based on the spontaneous reaction between copper and silver(I) ions.


1704. Cell Notation -
Abbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics, and they are written following a few guidelines:

- The relevant components of each half-cell are represented by their chemical formulas or element symbols
- All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas
- By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode

A verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure 16.3 consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:

$$
\mathrm{Cu}(s)\left|1 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \| 1 M \mathrm{AgNO}_{3}(a q)\right| \mathrm{Ag}(s)
$$

Consider a different galvanic cell (see Figure 16.4) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:
net cell reaction:
oxidation half-reaction:
reduction half-reaction:

$$
\begin{aligned}
& \mathrm{Mg}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{Fe}^{2+}(a q) \\
& \operatorname{Mg}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{e}^{-} \\
& 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)
\end{aligned}
$$


1704. Cell Notation -
In this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II)
chloride, immersed in which is a platinum cathode. The cell schematic is then written as

$$
\operatorname{Mg}(s)\left|0.1 M \mathrm{MgCl}_{2}(a q) \| 0.2 M \mathrm{FeCl}_{3}(a q), 0.3 M \mathrm{FeCl}_{2}(a q)\right| \operatorname{Pt}(s)
$$

Notice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance ( Pt ) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell's redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes.

FIGURE 16.4 A galvanic cell based on the spontaneous reaction between magnesium and iron(III) ions.


1706. Writing Galvanic Cell Schematics -
A galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a $1 \mathrm{M} \mathrm{CrCl}_{3}$ solution and another in which a copper wire is immersed in $1 \mathrm{M} \mathrm{CuCl}_{2}$. Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode halfreaction, the cathode half-reaction, and the overall cell reaction.


1707. Solution -
Since the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:

$$
\mathrm{Cr}(s)\left|1 M \mathrm{CrCl}_{3}(a q) \| 1 M \mathrm{CuCl}_{2}(a q)\right| \mathrm{Cu}(s)
$$

The half-reactions for this cell are

$$
\begin{array}{ll}
\text { anode (oxidation): } & \mathrm{Cr}(s) \longrightarrow \mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-} \\
\text {cathode (reduction): } & \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)
\end{array}
$$

Multiplying to make the number of electrons lost by Cr and gained by $\mathrm{Cu}^{2+}$ equal yields

$$
\begin{array}{ll}
\text { anode (oxidation): } & 2 \mathrm{Cr}(s) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+6 \mathrm{e}^{-} \\
\text {cathode (reduction): } & 3 \mathrm{Cu}^{2+}(a q)+6 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Cu}(s)
\end{array}
$$

Adding the half-reaction equations and simplifying yields an equation for the cell reaction:

$$
2 \mathrm{Cr}(s)+3 \mathrm{Cu}^{2+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Cu}(s)
$$


1708. Check Your Learning -
Omitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below.

$$
\mathrm{Sn}^{4+}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+\mathrm{Zn}^{2+}(a q)
$$


1709. Answer: -
$\mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q) \| \mathrm{Sn}^{4+}(a q), \mathrm{Sn}^{2+}(a q)\right| \operatorname{Pt}(s)$


1710. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe and relate the definitions of electrode and cell potentials
- Interpret electrode potentials in terms of relative oxidant and reductant strengths
- Calculate cell potentials and predict redox spontaneity using standard electrode potentials

Unlike the spontaneous oxidation of copper by aqueous silver(I) ions described in section 17.2, immersing a copper wire in an aqueous solution of lead(II) ions yields no reaction. The two species, $\mathrm{Ag}^{+}(\mathrm{aq})$ and $\mathrm{Pb}^{2+}(\mathrm{aq})$, thus show a distinct difference in their redox activity towards copper: the silver ion spontaneously oxidized copper, but the lead ion did not. Electrochemical cells permit this relative redox activity to be quantified by an easily measured property, potential. This property is more commonly called voltage when referenced in regard to electrical applications, and it is a measure of energy accompanying the transfer of charge. Potentials are measured in the volt unit, defined as one joule of energy per one coulomb of charge, $\mathrm{V}=\mathrm{J} / \mathrm{C}$.

When measured for purposes of electrochemistry, a potential reflects the driving force for a specific type of charge transfer process, namely, the transfer of electrons between redox reactants. Considering the nature of potential in this context, it is clear that the potential of a single half-cell or a single electrode can't be measured; "transfer" of electrons requires both a donor and recipient, in this case a reductant and an oxidant, respectively. Instead, a half-cell potential may only be assessed relative to that of another half-cell. It is only the difference in potential between two half-cells that may be measured, and these measured potentials are called
cell potentials, E $_{\text {cell }}$, defined as

$$
\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }}
$$


1710. LEARNING OBJECTIVES -
$$
\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }}
$$

where $\mathrm{E}_{\text {cathode }}$ and $\mathrm{E}_{\text {anode }}$ are the potentials of two different half-cells functioning as specified in the subscripts. As for other thermodynamic quantities, the standard cell potential, $\mathbf{E}^{\circ}{ }_{\text {cell }}$, is a cell potential measured when both half-cells are under standard-state conditions (1 M concentrations, 1 bar pressures, 298 K ):

$$
\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}
$$

To simplify the collection and sharing of potential data for half-reactions, the scientific community has designated one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of exactly 0 V . This half-cell is the standard hydrogen electrode (SHE) and it is based on halfreaction below:

$$
2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)
$$

A typical SHE contains an inert platinum electrode immersed in precisely $1 M$ aqueous $\mathrm{H}^{+}$and a stream of bubbling $\mathrm{H}_{2}$ gas at 1 bar pressure, all maintained at a temperature of 298 K (see Figure 16.5).

FIGURE 16.5 A standard hydrogen electrode (SHE).
The assigned potential of the SHE permits the definition of a conveniently measured potential for a single halfcell. The electrode potential ( $\mathbf{E}_{\mathbf{X}}$ ) for a half-cell $X$ is defined as the potential measured for a cell comprised of $X$ acting as cathode and the SHE acting as anode:


1710. LEARNING OBJECTIVES -
$$
\begin{aligned}
& E_{\text {cell }}=E_{X}-E_{\text {SHE }} \\
& E_{S H E}=0 V(\text { defined }) \\
& E_{\text {cell }}=E_{X}
\end{aligned}
$$

When the half-cell $X$ is under standard-state conditions, its potential is the standard electrode potential, $\mathbf{E}^{\circ} \mathbf{x}$. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials.

This approach to measuring electrode potentials is illustrated in Figure 16.6, which depicts a cell comprised of an SHE connected to a copper(II)/copper(0) half-cell under standard-state conditions. A voltmeter in the external circuit allows measurement of the potential difference between the two half-cells. Since the Cu halfcell is designated as the cathode in the definition of cell potential, it is connected to the red (positive) input of the voltmeter, while the designated SHE anode is connected to the black (negative) input. These connections insure that the sign of the measured potential will be consistent with the sign conventions of electrochemistry per the various definitions discussed above. A cell potential of +0.337 V is measured, and so

$$
\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\mathrm{Cu}}^{\circ}=+0.337 \mathrm{~V}
$$

Tabulations of $\mathrm{E}^{\circ}$ values for other half-cells measured in a similar fashion are available as reference literature to permit calculations of cell potentials and the prediction of the spontaneity of redox processes.

FIGURE 16.6 A cell permitting experimental measurement of the standard electrode potential for the half-reaction $\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)$

Table 16.1 provides a listing of standard electrode potentials for a selection of half-reactions in numerical order, and a more extensive alphabetical listing is given in Appendix L.

Selected Standard Reduction Potentials at $25^{\circ} \mathrm{C}$


1710. LEARNING OBJECTIVES -
Selected Standard Reduction Potentials at $25^{\circ} \mathrm{C}$

| Half-Reaction | $E^{\circ}(\mathrm{V})$ |
| :--- | :--- |
| $\mathrm{F}_{2}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~F}^{-}(a q)$ | +2.866 |
| $\mathrm{PbO}_{2}(s)+\mathrm{SO}_{4}{ }^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$ | +1.69 |
| $\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$ | +1.507 |
| $\mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)$ | +1.498 |
| $\mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q)$ | +1.35827 |
| $\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)$ | +1.229 |
| $\mathrm{Pt}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}(s)$ | +1.20 |
| $\mathrm{Br}_{2}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}(a q)$ | +1.0873 |
| $\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)$ | +0.7996 |

TABLE 16.1


1710. LEARNING OBJECTIVES -
| Half-Reaction | $E^{\circ}(\mathrm{V})$ |
| :---: | :---: |
| $\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Hg}(l)$ | +0.7973 |
| $\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)$ | +0.771 |
| $\mathrm{MnO}_{4}{ }^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{OH}^{-}(a q)$ | +0.558 |
| $\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q)$ | +0.5355 |
| $\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q)$ | +0.49 |
| $\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)$ | +0.34 |
| $\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q)$ | +0.26808 |
| $\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)$ | +0.22233 |
| $\mathrm{Sn}^{4+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+}(a q)$ | +0.151 |


1710. LEARNING OBJECTIVES -
| $\mathrm{Sn}^{4+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+}(a q)$ | +0.151 |
| $2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)$ | 0.00 |
| $\mathrm{Pb}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)$ | -0.1262 |
| $\mathrm{Sn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}(s)$ | -0.1375 |
| $\mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s)$ | -0.257 |
| $\mathrm{Co}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(s)$ | -0.28 |
| $\mathrm{PbSO}_{4}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{SO}_{4}{ }^{2-}(a q)$ | -0.3505 |
| $\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)$ | -0.4030 |
| $\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)$ | -0.447 |
| $\mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(s)$ | -0.744 |
| $\mathrm{Mn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}(s)$ | -1.185 |
| $\mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)$ | -1.245 |


1710. LEARNING OBJECTIVES -
| $\mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(s)$ | -0.7618 |
| $\mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s)$ | -1.662 |
| $\mathrm{Mg}^{2}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}(s)$ | -2.372 |
| $\mathrm{Na}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Na}(s)$ | -2.71 |


1710. LEARNING OBJECTIVES -
TABLE 16.1

| Half-Reaction | $E^{0}(\mathrm{~V})$ |
| :--- | :--- |
| $\mathrm{Ca}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}(s)$ | -2.868 |
| $\mathrm{Ba}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ba}(s)$ | -2.912 |
| $\mathrm{~K}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{K}(s)$ | -2.931 |
| $\mathrm{Li}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Li}(s)$ | -3.04 |

TABLE 16.1


1712. Calculating Standard Cell Potentials -
What is the standard potential of the galvanic cell shown in Figure 16.3?


1713. Solution -
The cell in Figure 16.3 is galvanic, the spontaneous cell reaction involving oxidation of its copper anode and reduction of silver(I) ions at its silver cathode:
cell reaction:
anode half-reaction:
cathode half-reaction:

$$
\begin{aligned}
& \mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s) \\
& \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \\
& 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(s)
\end{aligned}
$$

The standard cell potential computed as

$$
\begin{aligned}
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ} \\
& =\mathrm{E}_{\mathrm{Ag}}^{\circ}-\mathrm{E}_{\mathrm{Cu}}^{\circ} \\
& =0.7996 \mathrm{~V}-0.34 \mathrm{~V} \\
& =+0.46 \mathrm{~V}
\end{aligned}
$$


1714. Check Your Learning -
What is the standard cell potential expected if the silver cathode half-cell in Figure 16.3 is replaced with a lead half-cell: $\mathrm{Pb}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)$ ?


1715. Answer: -
-0. 47 V


1716. Intrepreting Electrode and Cell Potentials -
Thinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations in Example 16.4 have just shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness ("strength") with which aqueous $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, $E^{\circ}$. Since by convention electrode potentials are for reduction processes, an increased value of $E^{\circ}$ corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.

Applying this logic to the numerically ordered listing of standard electrode potentials in Table 16.1 shows this listing to be likewise in order of the oxidizing strength of the half-reaction's reactant species, decreasing from strongest oxidant (most positive $E^{\circ}$ ) to weakest oxidant (most negative $E^{\circ}$ ). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, $E^{\circ}{ }_{\text {cell }}$ is positive when $E^{\circ}{ }_{\text {cathode }}>E^{\circ}{ }_{\text {anode }}$, and so any redox reaction in which the oxidant's entry is above the reductant's entry is predicted to be spontaneous.


1716. Intrepreting Electrode and Cell Potentials -
Reconsideration of the two redox reactions in Example 16.4 provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by $\mathrm{Ag}^{+}$is predicted to be spontaneous ( $E^{\circ}{ }_{\text {cathode }}>E^{\circ}$ anode and so $E^{\circ}{ }_{\text {cell }}>0$ ). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by $\mathrm{Pb}^{2+}$ is nonspontaneous ( $E^{\circ}{ }_{\text {cathode }}<E^{\circ}$ anode and so $E^{\circ}{ }_{\text {cell }}<0$ ).

Recalling the chapter on thermodynamics, the spontaneities of the forward and reverse reactions of a reversible process show a reciprocal relationship: if a process is spontaneous in one direction, it is nonspontaneous in the opposite direction. As an indicator of spontaneity for redox reactions, the potential of a cell reaction shows a consequential relationship in its arithmetic sign. The spontaneous oxidation of copper by lead(II) ions is not observed,

$$
\mathrm{Cu}(s)+\mathrm{Pb}^{2+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Pb}(s) E_{\text {forward }}^{\circ}=-0.47 \mathrm{~V} \text { (negative, non-spontaneous) }
$$

and so the reverse reaction, the oxidation of lead by copper(II) ions, is predicted to occur spontaneously:

$$
\mathrm{Pb}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{Cu}(s) E_{\text {forward }}^{\circ}=+0.47 \mathrm{~V} \text { (positive, spontaneous) }
$$


1716. Intrepreting Electrode and Cell Potentials -
Note that reversing the direction of a redox reaction effectively interchanges the identities of the cathode and anode half-reactions, and so the cell potential is calculated from electrode potentials in the reverse subtraction order than that for the forward reaction. In practice, a voltmeter would report a potential of -0.47 V with its red and black inputs connected to the Pb and Cu electrodes, respectively. If the inputs were swapped, the reported voltage would be +0.47 V .


1718. Predicting Redox Spontaneity -
Are aqueous iron(II) ions predicted to spontaneously oxidize elemental chromium under standard state conditions? Assume the half-reactions to be those available in Table 16.1.


1719. Solution -
Referring to the tabulated half-reactions, the redox reaction in question can be represented by the equations below:

$$
\mathrm{Cr}(s)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}(s)
$$

The entry for the putative oxidant, $\mathrm{Fe}^{2+}$, appears above the entry for the reductant, Cr , and so a spontaneous reaction is predicted per the quick approach described above. Supporting this predication by calculating the standard cell potential for this reaction gives

$$
\begin{aligned}
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ} \\
& =\mathrm{E}_{\mathrm{Fe}(\mathrm{II})}^{\circ}-\mathrm{E}^{\circ} \mathrm{Cr} \\
& =-0.447 \mathrm{~V}--0.744 \mathrm{~V}=+0.297 \mathrm{~V}
\end{aligned}
$$

The positive value for the standard cell potential indicates the process is spontaneous under standard state conditions.


1720. Check Your Learning -
Use the data in Table 16.1 to predict the spontaneity of the oxidation of bromide ion by molecular iodine under standard state conditions, supporting the prediction by calculating the standard cell potential for the reaction.

Repeat for the oxidation of iodide ion by molecular bromine.


1721. Answer: -
$$
\begin{array}{ll}
\mathrm{I}_{2}(s)+2 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{I}^{-}(a q)+\mathrm{Br}_{2}(l) & E_{\text {cell }}^{\circ}=-0.5518 \mathrm{~V} \text { (nonspontaneous) } \\
\mathrm{Br}_{2}(s)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(l) & E_{\text {cell }}^{\circ}=+0.5518 \mathrm{~V} \text { (spontaneous) }
\end{array}
$$


1722. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the relations between potential, free energy change, and equilibrium constants
- Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium
- Use the Nernst equation to determine cell potentials under nonstandard conditions

So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties $\Delta \mathrm{G}$ and K .


1723. $\mathrm{E}^{\circ}$ and $\Delta \mathrm{G}^{\circ}$ -
The standard free energy change of a process, $\Delta G^{\circ}$, was defined in a previous chapter as the maximum work that could be performed by a system, $w_{\text {max }}$. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, $w_{\text {elec }}$ :

$$
\Delta G^{\circ}=w_{\max }=w_{\text {elec }}
$$

The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:

$$
\begin{aligned}
\Delta G^{\circ}=w_{\mathrm{elec}} & =-n F E_{\mathrm{cell}}^{\circ} \\
\Delta G^{\circ} & =-n F E_{\mathrm{cell}}^{\circ}
\end{aligned}
$$

where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant, and $E^{\circ}$ cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.


1724. $E^{\circ}$ and $K$ -
Combining a previously derived relation between $\Delta \mathrm{G}^{\circ}$ and K (see the chapter on thermodynamics) and the equation above relating $\Delta \mathrm{G}^{\circ}$ and $E^{\circ}$ cell yields the following:

$$
\begin{aligned}
\Delta G^{\circ} & =-R T \ln K=-n F E_{\text {cell }}^{\circ} \\
E_{\text {cell }}^{\circ} & =\left(\frac{R T}{n F}\right) \ln K
\end{aligned}
$$

This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between $E^{\circ}, \Delta G^{\circ}$ and $K$ is depicted in Figure 16.7, and a table correlating reaction spontaneity to values of these properties is provided in Table 16.2.

FIGURE 16.7 Graphic depicting the relation between three important thermodynamic properties.

| $K$ | $\Delta G^{\circ}$ | $E^{\circ}$ cell |  |
| :--- | :--- | :--- | :--- |
| $>1$ | $<0$ | $>0$ | Reaction is spontaneous under standard conditions
 Products more abundant at equilibrium |
| $<1$ | $>0$ | $<0$ | Reaction is non-spontaneous under standard conditions
 Reactants more abundant at equilibrium |
| $=1$ | $=0$ | $=0$ | Reaction is at equilibrium under standard conditions
 Reactants and products equally abundant |

TABLE 16.2


1726. Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes -
Use data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at $25^{\circ} \mathrm{C}$. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products.

$$
2 \mathrm{Ag}^{+}(a q)+\mathrm{Fe}(s) \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{Fe}^{2+}(a q)
$$


1727. Solution -
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L.

$$
\begin{array}{llrl}
\text { anode (oxidation): } & \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} & E_{\mathrm{Fe} 2+/ \mathrm{Fe}}^{\circ}=-0.447 \mathrm{~V} \\
\text { cathode (reduction): } & 2 \times\left(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\right) & E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.7996 \mathrm{~V} \\
& E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}= & E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=+1.247 \mathrm{~V}
\end{array}
$$

With $n=2$, the equilibrium constant is then

$$
\begin{aligned}
E_{\mathrm{cell}}^{\circ} & =\frac{0.0592 \mathrm{~V}}{n} \log K \\
K & =10^{n \times E_{\mathrm{cell}}^{\circ} / 0.0592 \mathrm{~V}} \\
K & =10^{2 \times 1.247 \mathrm{~V} / 0.0592 \mathrm{~V}} \\
K & =10^{42.128} \\
K & =1.3 \times 10^{42}
\end{aligned}
$$

The standard free energy is then


1727. Solution -
The standard free energy is then

$$
\begin{aligned}
& \Delta G^{\circ}=-n F E_{\mathrm{cell}}^{\circ} \\
& \Delta G^{\circ}=-2 \times 96,485 \frac{\mathrm{C}}{\mathrm{~mol}} \times 1.247 \frac{\mathrm{~J}}{\mathrm{C}}=-240.6 \frac{\mathrm{~kJ}}{\mathrm{~mol}}
\end{aligned}
$$

The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The $K$ value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.


1728. Check Your Learning -
What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?

$$
\mathrm{Sn}(s)+2 \mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Sn}^{2+}(a q)+2 \mathrm{Cu}^{+}(a q)
$$

Answer:
Spontaneous; $n=2 ; E_{\text {cell }}^{\circ}=+0.291 \mathrm{~V} ; \Delta G^{\circ}=-56.2 \frac{\mathrm{~kJ}}{\mathrm{~mol}} ; K=6.8 \times 10^{9}$.


1729. Potentials at Nonstandard Conditions: The Nernst Equation -
Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.

$$
\Delta G=\Delta G^{\circ}+R T \ln Q
$$

Notice the reaction quotient, $Q$, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:

$$
\begin{gathered}
-n F E_{\mathrm{cell}}=-n F E_{\mathrm{cell}}^{\circ}+R T \ln Q \\
E_{\mathrm{cell}}=E_{\mathrm{cell}}^{\circ}-\frac{R T}{n F} \ln Q
\end{gathered}
$$

This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, $n$, the temperature, $T$, and the reaction mixture composition as reflected in $Q$. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants ( $R$ and $F$ ) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:

$$
E_{\mathrm{cell}}=E_{\mathrm{cell}}^{\circ}-\frac{0.0592 \mathrm{~V}}{n} \log Q
$$


1731. Predicting Redox Spontaneity Under Nonstandard Conditions -
Use the Nernst equation to predict the spontaneity of the redox reaction shown below.

$$
\mathrm{Co}(s)+\mathrm{Fe}^{2+}(a q, 1.94 M) \longrightarrow \mathrm{Co}^{2+}(a q, 0.15 M)+\mathrm{Fe}(s)
$$


1732. Solution -
Collecting information from Appendix L and the problem,

$$
\begin{array}{llrl}
\text { Anode (oxidation): } & \mathrm{Co}(s) \longrightarrow \mathrm{Co}^{2+}(a q)+2 \mathrm{e}^{-} & E_{\mathrm{Co}^{2+} / \mathrm{Co}}^{\circ}=-0.28 \mathrm{~V} \\
\text { Cathode (reduction): } & \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.447 \mathrm{~V} \\
& E_{\mathrm{cell}}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}=-0.447 \mathrm{~V}-(-0.28 \mathrm{~V})=-0.17 \mathrm{~V}
\end{array}
$$

Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:

$$
\begin{aligned}
Q & =\frac{\left[\mathrm{Co}^{2+}\right]}{\left[\mathrm{Fe}^{2+}\right]}=\frac{0.15 M}{1.94 M}=0.077 \\
E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592 \mathrm{~V}}{n} \log Q \\
E_{\text {cell }} & =-0.17 \mathrm{~V}-\frac{0.0592 \mathrm{~V}}{2} \log 0.077 \\
E_{\text {cell }} & =-0.17 \mathrm{~V}+0.033 \mathrm{~V}=-0.14 \mathrm{~V}
\end{aligned}
$$

The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.


1733. Check Your Learning -
For the cell schematic below, identify values for $n$ and $Q$, and calculate the cell potential, $E_{\text {cell }}$ -

$$
\mathrm{Al}(s)\left|\mathrm{Al}^{3+}(a q, 0.15 M) \| \mathrm{Cu}^{2+}(a q, 0.025 M)\right| \mathrm{Cu}(s)
$$


1734. Answer: -
$n=6 ; Q=1440 ; E_{\text {cell }}=+1.97 \mathrm{~V}$, spontaneous.

A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same halfreaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.


1736. Concentration Cells -
What is the cell potential of the concentration cell described by

$$
\mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q, 0.10 M) \| \mathrm{Zn}^{2+}(a q, 0.50 M)\right| \mathrm{Zn}(s)
$$


1737. Solution -
From the information given:

| Anode: | $\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q, 0.10 M)+2 \mathrm{e}^{-}$ | $E_{\text {anode }}^{\circ}$ | $=-0.7618 \mathrm{~V}$ |
| :--- | :--- | :--- | :--- |
| Cathode: | $\mathrm{Zn}^{2+}(a q, 0.50 M)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(s)$ | $E_{\text {cathode }}^{\circ}$ | $=-0.7618 \mathrm{~V}$ |
| Overall: | $\mathrm{Zn}^{2+}(a q, 0.50 M) \longrightarrow \mathrm{Zn}^{2+}(a q, 0.10 M)$ | $E_{\text {cell }}^{\circ}$ | $=0.000 \mathrm{~V}$ |

Substituting into the Nernst equation,

$$
E_{\text {cell }}=0.000 \mathrm{~V}-\frac{0.0592 \mathrm{~V}}{2} \log \frac{0.10}{0.50}=+0.021 \mathrm{~V}
$$

The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater ( $E_{\text {cathode }}>E_{\text {anode }}$ ).


1738. Check Your Learning -
The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?


1739. Answer: -
$E_{\text {cell }}=0.000 \mathrm{~V} ;\left[\mathrm{Zn}^{2+}\right]_{\text {cathode }}=\left[\mathrm{Zn}^{2+}\right]_{\text {anode }}=0.30 \mathrm{M}$


1740. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the electrochemistry associated with several common batteries
- Distinguish the operation of a fuel cell from that of a battery

There are many technological products associated with the past two centuries of electrochemistry research, none more immediately obvious than the battery. A battery is a galvanic cell that has been specially designed and constructed in a way that best suits its intended use a source of electrical power for specific applications. Among the first successful batteries was the Daniell cell, which relied on the spontaneous oxidation of zinc by copper(II) ions (Figure 16.8):

$$
\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)
$$

FIGURE 16.8 Illustration of a Daniell cell taken from a 1904 journal publication (left) along with a simplified illustration depicting the electrochemistry of the cell (right). The 1904 design used a porous clay pot to both contain one of the half-cell's content and to serve as a salt bridge to the other half-cell.

Modern batteries exist in a multitude of forms to accommodate various applications, from tiny button batteries that provide the modest power needs of a wristwatch to the very large batteries used to supply backup energy to municipal power grids. Some batteries are designed for single-use applications and cannot be recharged
(primary cells), while others are based on conveniently reversible cell reactions that allow recharging by an external power source (secondary cells). This section will provide a summary of the basic electrochemical aspects of several batteries familiar to most consumers, and will introduce a related electrochemical device called a fuel cell that can offer improved performance in certain applications.


1741. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16batteries) to learn more about batteries.


1742. Single-Use Batteries -
A common primary battery is the dry cell, which uses a zinc can as both container and anode ("-" terminal) and a graphite rod as the cathode ("+" terminal). The Zn can is filled with an electrolyte paste containing manganese(IV) oxide, zinc(II) chloride, ammonium chloride, and water. A graphite rod is immersed in the electrolyte paste to complete the cell. The spontaneous cell reaction involves the oxidation of zinc:

$$
\text { anode reaction: } \mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-}
$$

and the reduction of manganese(IV)
reduction reaction: $2 \mathrm{MnO}_{2}(s)+2 \mathrm{NH}_{4} \mathrm{Cl}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(s)+2 \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}^{-}$
which together yield the cell reaction:
cell reaction: $2 \mathrm{MnO}_{2}(s)+2 \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Mn}_{2} \mathrm{O}_{3}(s)+2 \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}^{-} E_{\text {cell }} \sim 1.5 \mathrm{~V}$
The voltage (cell potential) of a dry cell is approximately 1.5 V . Dry cells are available in various sizes (e.g., D, C, AA, AAA). All sizes of dry cells comprise the same components, and so they exhibit the same voltage, but larger cells contain greater amounts of the redox reactants and therefore are capable of transferring correspondingly greater amounts of charge. Like other galvanic cells, dry cells may be connected in series to yield batteries with greater voltage outputs, if needed.

FIGURE 16.9 A schematic diagram shows a typical dry cell.


1743. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16zinccarbon) to learn more about zinc-carbon batteries.

Alkaline batteries (Figure 16.10) were developed in the 1950s to improve on the performance of the dry cell,
and they were designed around the same redox couples. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are

| $\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q)$
 anode:
 cathode: $2 \mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{OH}^{-}(a q)$ |
| :--- |
| cell: |

An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so they should be removed from devices for longterm storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.

FIGURE 16.10 Alkaline batteries were designed as improved replacements for zinc-carbon (dry cell) batteries.


1744. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16alkaline) to learn more about alkaline batteries.


1745. Rechargeable (Secondary) Batteries -
Nickel-cadmium, or NiCd, batteries (Figure 16.11) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a "jelly-roll" design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are

$$
\left.\begin{array}{l}
\text { anode: } \quad \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \\
\text {cathode: } \mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q)
\end{array}\right]
$$

When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be ruptured or incinerated, and they should be disposed of in accordance with relevant toxic waste guidelines.

FIGURE 16.11 NiCd batteries use a "jelly-roll" design that significantly increases the amount of current the battery can deliver as compared to a similar-sized alkaline battery.


1746. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16NiCdrecharge) for more information about nickel cadmium rechargeable batteries.

Lithium ion batteries (Figure 16.12) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are

$$
\begin{array}{lll}
\text { anode: } & \mathrm{LiCoO}_{2} \rightleftharpoons \mathrm{Li}_{1-x} \mathrm{CoO}_{2}+x \mathrm{Li}^{+}+x \mathrm{e}^{-} \\
\text {cathode: } x \mathrm{Li}^{+}+x \mathrm{e}^{-}+x \mathrm{C}_{6} \rightleftharpoons x \mathrm{LiC}_{6} & \\
\hline \text { cell: } & \mathrm{LiCoO}_{2}+x \mathrm{C}_{6} \rightleftharpoons \mathrm{Li}_{1-x} \mathrm{CoO}_{2}+x \mathrm{LiC}_{6} & E_{\text {cell }} \sim 3.7 \mathrm{~V}
\end{array}
$$

The variable stoichiometry of the cell reaction leads to variation in cell voltages, but for typical conditions, $x$ is usually no more than 0.5 and the cell voltage is approximately 3.7 V . Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.

FIGURE 16.12 In a lithium ion battery, charge flows as the lithium ions are transferred between the anode and cathode.


1747. LINK TO LEARNING -
Visit this site (http://openstax.org/l/16lithiumion) for more information about lithium ion batteries.

The lead acid battery (Figure 16.13) is the type of secondary battery commonly used in automobiles. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are

$$
\begin{aligned}
& \text { anode: } \mathrm{Pb}(s)+\mathrm{HSO}_{4}^{-}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+\mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \\
& \text {cathode: } \mathrm{PbO}_{2}(s)+\mathrm{HSO}_{4}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \\
& \hline \text { cell: }
\end{aligned} \quad \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad E_{\text {cell }} \sim 2 \mathrm{~V} \text { lel }
$$

Each cell produces 2 V , so six cells are connected in series to produce a $12-\mathrm{V}$ car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})$, but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.

FIGURE 16.13 The lead acid battery in your automobile consists of six cells connected in series to give 12 V .


1748. LINK TO LEARNING -
Visit this site (http://openstax.org/1/16leadacid) for more information about lead acid batteries.