Review Questions
1. Describe the basic components of a voltaic cell. Why are the
two half-cells separated from each other?
2. What is an electrode?
3. What is the function of the salt bridge in a voltaic cell?
4. What is the relationship between the emf of a reaction and its spontaneity?
5. Write the Nernst equation and explain all the terms.
6. How does an electrolytic cell differ from a voltaic cell?
Problems
A table of standard electrode potentials is supplied. For more potentials see Table 19.1.
1. Balance the following redox equations.
a. Cr2O72-(aq)
+ Br-(aq)
Æ Cr3+(aq)
+ Br2(g) (acidic soln)
b. MnO4- + I-(aq)
Æ
MnO2(s) + I2(aq)
(basic soln)
2. Given the following standard reduction potentials in acid solution:
E°(V)
Sn4+ + 2e- =
Sn2+ +0.14
Al3+ + 3e- =
Al(s) -1.66
Fe3+ + e-
= Fe2+
+0.77
a. Write the formula of the strongest reducing
agent.
b. Write the formula of the strongest oxidizing
agent.
3. Determine which of the following reactions are spontaneous under standard conditions.
a. Ca2+ + Mg Æ
Ca + Mg2+
b. Pb + Cl2 Æ
Pb2+ + 2Cl-
c. 2NO3- + 8H+
+ 6Cl- Æ 2NO(g) + 4H2O
+ 3Cl2
4. Calculate the standard cell portential for the following reaction
at 25°C:
Zn(s) + Pb2+(aq) Æ
Zn2+(aq) + Pb(s)
5. Calculate Eocell for the following reaction
at 25°C:
2Co2+(aq) + Zn2+(aq)
Æ
2Co3+(aq) + Zn(s)
6. Consider a voltaic cell based on the following cell reaction:
Pu3+(aq) + Cl2(g)
Æ
Pu4+(aq) + Cl-(aq)
Given that the standard cell voltage is 0.35 V, and that the standard
reduction potential of chlorine is 1.36 V:
Cl2 + 2e- Æ
2Cl- Ered° = 1.36 V
What is the value of the following standard
electrode potential?
Pu4+ + e- Æ
Pu3+ Ered°
= ?
7. Determine the standard emf of a cell that uses Ag/Ag+ and Al/Al3+ half-cell reactions. Write the cell reaction that occurs under standard state conditions.
8. Consider a voltaic cell constructed from the following half cells, linked by a KCl salt bridge.
* a Cu electrode in 1.0 M Cu(NO3)2
solution
* a Sn electrode in 1.0 M Sn(NO3)2
solution.
Answer the following questions:
a. Which electrode is the cathode?
b. Write a balanced chemical equation
for the reaction occurring when the cell is discharging.
c. Determine the standard cell potential.
d. Which direction do electrons flow
in the external circuit?
e. Which electrode will K+
of the KCl salt bridge move toward?
f. Calculate the equilibrium constant
for the overall reaction.
g. Calculate the Gibb’s free energy
change for this reaction.
9. Determine the value of the equilibrium constant for the reaction:
2Br-(aq) + I2(s) = Br2(l) + 2I-(aq)
10. A voltaic cell consists of a hydrogen electrode and a Cu/Cu2+ electrode.
Given the cell reaction
Cu2+(aq) + H2(g)
Æ
Cu(s) + 2H+(aq)
a. Determine the cell emf if [Cu2+]
= 0.02 M, [H+] = 1.2
¥
10-4
M, and PH2 = 1.0 atm.
b. Calculate [Cu2+] when [H+]
= 1.0 M, PH2 = 1.0 atm, and Ecell is 0.15 V.
11. A Cu / H2 cell has an Ecell of 0.52 V when
the pressure of hydrogen gas is 1.0 atm and [Cu2+] = 1.0 M.
Determine the pH of the solution in the hydrogen electrode half-cell.
Hint, complete the equation for the cell reaction first:
Cu2+(aq) + H2(g) Æ
12. Which is the half-reaction that occurs at the cathode during the
electrolysis of aqueous lithium bromide?
A. 2Br- Æ
Br2 + 2e-
B. Li+ + e- Æ
Li
C. 2H2O + 2e- Æ
H2 + 2OH-
13. How many coulombs of charge are required to cause reduction of 0.10 mol of Cr3+ to Cr?
14. a. A current of 0.80 A was applied to an electrolytic cell containing molten CdCl2 for 2.5 hours. Decide on the appropriate half-reaction, and calculate the mass of cadmium metal deposited.
A. 3.2 ¥ 10-7 g B. 1.2 ¥ 10-3 g C. 4.2 g D. 8.4 g E. 16.8 g
b. If the cell contained molten iron(III) chloride instead of CdCl2, how many grams of iron could be deposited under the same conditions?
15. How many minutes would be required to electroplate 11.0 grams of
chromium by passing a constant current of 5.8 amperes through a solution
containing CrCl3?
Table 20.1 Standard Reduction Potentials at 25°C
Note equal signs have been used in place of reversible arrows.
Co3+(aq) + e-
= Co2+(aq) Eored
= +1.82 V
Cl2(g) + 2e- =
2Cl-(aq)
Eored = +1.36 V
O2(g) + 4H+(aq) + 4e-
= 2H2O(l) Eored
= +1.23 V
Br2(l) + 2e-
= 2Br-(aq)
Eored = +1.06 V
NO3-(aq) + 4H+
+ 3e- = NO(aq) + 2H2O Eored
= +0.96 V
Ag+(aq) + e-
= Ag(s)
Eored = +0.80 V
I2(s) + 2e-
= 2I-(aq)
Eored = +0.54 V
Cu2+(aq) + 2e-
= Cu(s) Eored
= +0.34 V
2H+(aq) + 2e-
= H2(g)
Eored = 0.00 V
Pb2+(aq) + 2e-
= Pb(s) Eored
= -0.13 V
Sn2+(aq) + 2e-
= Sn(s) Eored
= -0.14 V
Co2+(aq) + 2e-
= Co(s) Eored
= -0.28 V
Zn2+(aq) + 2e-
= Zn(s) Eored
= -0.76 V
2H2O(l) + 2e-
= H2(g) + 2OH- Eored
= -0.83 V
Al3+(aq) + 3e-
= Al(s)
Eored = -1.66 V
Mg2+(aq) + 2e-
= Mg(s) Eored
= -2.37 V
Ca2+(aq) + 2e-
= Ca(s)
Eored = -2.87 V
Faraday constant 9.6500 ¥ 104 C / mol e- 1 J = 1 C·V
DG = DH - TDS DG = DG° + RT ln Q DG° = - RT ln Keq DG = -nFE
Ecell = Eocell - (0.0592/n) log Q Eocell = (0.0592/n) log K
Answers to Study Questions
1. a. 14H+ + Cr2O72-
+ 6Br-(aq) Æ 2Cr3+(aq)
+ 3Br2(l) + 7H2O
b. 2MnO4- + 6I-
+ 4H2O
Æ 2MnO2 +
3I2 + 8OH-
2. a. Al b. Fe3+
3. b. only
4. Eocell = 0.63 V
5. Eocell = -2.58 V
6. E°red = 1.01 V
7. E°red = 2.46 V
8. a. Cu b. Cu2+
+ Sn Æ Cu + Sn2+
c. Eocell = 0.48 V
d. from Sn Æ Cu
e. Cu f. Keq
= 1.6 ¥ 1016
g. -92.6 kJ
9. Keq = 2.7 ¥ 10-18
10. a. 0.52 V
b. [Cu2+] = 3.8 ¥ 10-7
mol/L
11. pH = 3.04 and [H+] = 9.1 ¥10-4
M
12. C. 2H2O + 2e- Æ
H2 + 2OH-
13. 2.9 ¥ 104 C
14. a. C. 4.2 g Cd b. 1.4 g iron
15. 176 min