Question 21.6: Calculating K and ΔG° from Ecell° Problem Silver occurs in t......
Calculating K and ΔG° from E^\circ _{\text{cell}}
Problem Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution:
Pb(s) + 2Ag^+(aq) ⟶ Pb^{2+}(aq) + 2Ag(s)As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ores. Calculate K and ΔG° at 298.15 K for this reaction.
Plan We divide the spontaneous redox reaction into half-reactions and use values from Appendix D to calculate E^\circ_{\text{cell}}. Then, we substitute this result into Equations 21.7 and 21.8 to find K and into Equation 21.6 to find ΔG°.
E^\circ _{\text{cell}} = \frac{RT}{nF} \ln K (21.7)
E^\circ _{\text{cell}} = \frac{0.0592 V}{n} \log K \text{ or } \log K = \frac{nE^\circ _\text{cell}}{0.0592 V} (\text{at }298.15 K) (21.8)
ΔG^\circ = −nFE^\circ _{\text{cell}} (21.6)
Solution Writing the half-reactions with their E° values:
(1) Ag^+(aq) + e^− ⟶ Ag(s) E^\circ = 0.80 V
(2) Pb^{2+}(aq) + 2e^− ⟶ Pb(s) E^\circ = −0.13 V
Calculating E^\circ_{\text{cell}}: We double half-reaction (1) to obtain a gain of 2 mol of electrons, reverse (2) since this is the oxidation half-reaction, add the half-reactions, and subtract E^\circ_{\text{lead}} from E^\circ_{\text{silver}}:
2Ag^+(aq) + 2e^− ⟶ 2Ag(s) E^\circ = 0.80 V
Pb(s) ⟶ Pb^{2+}(aq) + 2e^− E^\circ = −0.13 V
Calculating K with Equations 21.7 and 21.8: The adjusted half-reactions show that 2 mol of e^− are transferred per mole of reaction as written, so n = 2. Then, performing the substitutions for R and F that we just discussed, changing to common logarithms, and running the cell at 25°C (298.15 K), we have
E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K = \frac{0.0592 V}{n} \log K = \frac{0.0592 V}{2} \log K = 0.93 V
So, \log K = \frac{0.93 V × 2}{0.0592 V} = 31.42\text{ and }K = 10^{31.42} = 2.6×10^{31}
Calculating ΔG° (Equation 21.6):
ΔG^\circ = −nFE^\circ_{\text{cell}} = −\frac{2\text{ mol} e^−}{\text{mol rxn}} × \frac{96.5 kJ}{V·\text{mol }e^−} × 0.93 V = −1.8×10^2 \text{kJ/mol rxn}Check The three variables are consistent with the reaction being spontaneous at standard-state conditions: E^\circ_{\text{cell}} > 0, ΔG° < 0, and K > 1. Be sure to round and check the order of magnitude: to find ΔG°, for instance, ΔG° ≈ −2 × 100 × 1 = −200, so the overall math seems right. Another check would be to obtain ΔG° directly from its relation with K:
ΔG^\circ = −RT \ln K = −8.314\text{ J/mol rxn}·K × 298.15 K × \ln (2.6×10^{31})
= −1.8×10^5\text{ J/mol rxn} = −1.8×10^2\text{ kJ/mol rxn}