1. Using standard reduction potentials (Table D in Appendix II), calculate the standard cell potential (EEcell) for each of the
following reactions. Which of these reactions do you expect to occur spontaneously in the forward direction?
(a) Cl2(g) + 2 I – (aq) ® 2 Cl – (aq) + I2(s)
Reduction half-reaction:
Cl2 + 2 e – ® 2 Cl – EEred = 1.36 V
Oxidation half-reaction:
2 I – ® I2 + 2 e – EEox = – 0.54 V
EEcell = EEox + EEred = – 0.54 V + 1.36 V = 0.82 V
This reaction will occur spontaneously in the forward direction since EEcell > 0.
(b) Cu(s) + Ba2+ (aq) ® Cu2+ (aq) + Ba(s)
Reduction half-reaction:
Ba 2+ + 2 e – ® Ba EEred = – 2.90 V
Oxidation half-reaction:
Cu ® Cu 2+ + 2 e – EEox = – 0.34 V
EEcell = EEox + EEred = – 0.34 V + (- 2.90 V) = – 3.24 V
This reaction will not occur spontaneously in the forward direction since EEcell < 0.
2. Using standard reduction potentials (Table D in Appendix II), calculate DGE and K for each of the following reactions
at 25EC.
(a) Fe(s) + Ni2+ (aq) ® Fe2+ (aq) + Ni(s)
Reduction half-reaction:
Ni 2+ + 2 e – ® Ni EEred = – 0.23 V
Oxidation half-reaction:
Fe ® Fe 2+ + 2 e – EEox = 0.45 V
We first need to calculate EEcell.
EEcell = EEox + EEred = 0.45 V + (- 0.23V) = 0.22 V
DGE = – nF EE where n = 2
DGE = – (2)(96485 C/ 1 mol)(0.22 J/C) = – 4.2 x 10 4 J/mol = – 42 kJ/mol
(Recall that 1 V = 1 J/C)
logK = (EEcell)(n / 0.0592) = (0.22 V)(2/0.0592 V) = 7.43
K = 10 7.43 = 2.7 x 10 7
(b) Co(s) + 2 H+ (aq) ® Co2+ (aq) + H2(g)
Reduction half-reaction:
2 H+ + 2 e – ® H2 EEred = 0.00 V (by definition)
Oxidation half-reaction:
Co ® Co 2+ + 2 e – EEox = 0.28 V
We first need to calculate EEcell.
EEcell = EEox + EEred = 0.28 V + 0.00 V = 0.28 V
DGE = – nF EE where n = 2
DGE = – (2)(96485 C/ 1 mol)(0.28 J/C) = – 5.4 x 10 4 J/mol = – 54 kJ/mol
(Recall that 1 V = 1 J/C)
logK = (EEcell)(n / 0.0592) = (0.28 V)(2/0.0592 V) = 9.46
K = 10 9.46 = 2.9 x 10 9
3. A voltaic cell is constructed that uses the following reaction and operates at 25EC.
Zn(s) + Ni2+ (aq) ® Zn2+ (aq) + Ni(s)
(a) Calculate EEcell.
Reduction half-reaction:
Ni 2+ + 2 e – ® Ni EEred = – 0.23 V
Oxidation half-reaction:
Zn ® Zn 2+ + 2 e – EEox = 0.76 V
EEcell = EEox + EEred = 0.76 V + (- 0.23 V) = 0.53 V
(b) Calculate Ecell when [Ni2+] = 3.00 M and [Zn2+] = 0.100 M.
Ecell = EEcell – (0.0592 V / n)(log Q)
n = 2 for this redox reaction
Q = [Zn2+]/[Ni2+] = 0.100M / 3.00M = 0.0333
Ecell = 0.53 V – (0.0592 V / 2){log(0.0333)} = 0.53 V – (- 0.0437 V) = 0.57 V
(c) Calculate Ecell when [Ni2+] = 0.200 M and [Zn2+] = 0.900 M.
Q = [Zn2+]/[Ni2+] = 0.900M / 0.200M = 4.50
Ecell = 0.53 V – (0.0592 V / 2){log(4.50)} = 0.53 V – 0.0193 V = 0.51 V
4. Consider the following voltaic cell
Zn(s) + 2 H2O (l) + 2 OH – (aq) ® [Zn(OH)4] 2 – (aq) + H2 (g)
Half-reaction EEred (V)
2 H2O (l) + 2 e – ® H2 (g) + 2 OH – (aq) – 0.83
[Zn(OH)4] 2 – (aq) + 2 e – ® Zn(s) + 4 OH – (aq) – 1.22
Use the data given above to calculate the pH at 25EC when Ecell = 0.29 V, PH2 = 1.0 bar, and
[[Zn(OH)4] 2 – (aq)] = 0.025 M.
We first need to calculate EEcell. The half-reaction with the less negative reduction potential will be the reduction half-reaction.
EEcell = EEox + EEred = 1.22 V + (- 0.83 V) = 0.39 V
Ecell = EEcell – (0.0592 V / n)(log Q)
EEcell – Ecell = ( 0.0592 V / n)(log Q)
For this cell n = 2.
0.39 V – 0.29 V = (0.0592 V / 2)(log Q)
log Q = (0.10 V)(2)/(0.0592 V) = 3.38
Q = 10 3.38 = 2.4 x 10 3
Q = [[Zn(OH)4] 2 – ] PH2/[OH –] 2 = [(0.025)(1.0)]/ /[OH –] 2 = 2.4 x 10 3
[OH –] 2 = (0.025)(1.0)/(2.4 x 10 3) = 1.0 x 10 – 5
[OH –] = 0.0032 M
pOH = – log [OH –] = 2.49
pH =14.00 – pOH = 14.00 – 2.49 = 11.51
5. A concentration cell consists of two Zn/Zn2+ half-cells. The cell has a potential of 0.067 V at 25EC. What is the ratio of the
Zn2+ concentrations in the two half-cells?
Remember that EEcell = 0 for a concentration cell.
Ecell = EEcell – (0.0592 V / n)(log Q)
EEcell – Ecell = ( 0.0592 V / n)(log Q)
log Q = [Zn 2+]dilute/[Zn 2+]concentrated
For this cell n = 2.
0.00V – 0.067 V = (0.0592 V / 2){log([Zn 2+]dilute/[Zn 2+]concentrated)}
log([Zn 2+]dilute/[Zn 2+]concentrated) = – (0.067 V)(2)/(0.0592 V) = – 2.26
[Zn 2+]dilute/[Zn 2+]concentrated = 10 – 2.26 = 5.5 x 10 – 3
6. Metallic magnesium can be made by the electrolysis of molten MgCl2. (a) What mass of Mg is formed by passing a current
of 5.25 A through molten MgCl2 for 2.50 days? (b) How many minutes are needed to plate out 10.00 g Mg from molten
MgCl2, using 3.50 A of current?
(a) In order to calculate the number of coulombs needed to plate out Mg, time must be expressed in seconds.
time (in sec) = (2.50 days)(24 hr/day)(60 min/hr)(60 sec/min) = 2.16 x 10 5 sec
mass of Mg = (5.25 A)(2.16 x 10 5 sec)(1 mol e –/96485 C)(1mol Mg / 2 mol e –)(24.31 g / mol Mg)
mass of Mg = 143 g
(b) We need to calculate the number of coulombs required to plate out 10.00 g of Mg.
(10.00 g)(1 mol Mg/24.31 g)(2 mol e –/1 mol Mg)(96485 C/1 mol e –) = 7.938 x 10 4 C
Since coulombs = current (in amperes) x time (in sec) we have the following relationship:
time (in sec) = coulombs/amperes
time (in sec) = 7.938 x 104 C/3.50 A = 2.268 x 10 4 sec
We need to express the time in minutes Þ
time (in min) = (2.268 x 10 4 sec)(1 min/60 sec) = 378 min
7. Which of the following metals, if coated onto iron, would prevent the corrosion of iron? Explain your reasoning.
(a) Mg (b) Cr (c) Cu
In order for a metal to be able to protect iron, it must be more easily oxidized than iron. We need a metal that lies below iron in the table of standard reduction potentials (Table 18.1). Both magnesium and chromium meet this criterion.
