Question 11.2: Calculating ΔG° for a Cell Reaction Using the data in Table ...
Calculating ΔG° for a Cell Reaction
Using the data in Table 11.1, calculate ΔG° for the reaction
Cu^{2+} (aq) + Fe (s) → Cu (s) + Fe^{2+} (aq)
Is this reaction spontaneous?
Table 11.1
Standard Reduction Potentials at 25°C (298 K) for Many Common Half-reactions
| Half-reaction | \xi° ( V ) | Half-reaction | \xi° ( V ) |
| F_{2} + 2 e^{-} → 2 F^{-} | 2.87 | O_{2} + 2 H_{2}O + 4 e^{-} → 4 OH^{-} | 0.40 |
| Ag^{2+} + e^{-} → Ag^{+} | 1.99 | Cu^{2+} + 2 e^{-} → Cu | 0.34 |
| Co^{3+} + e^{-} → Co^{2+} | 1.82 | Hg_{2}Cl_{2} + 2 e^{-} → 2 Hg + 2 Cl^{-} | 0.27 |
| H_{2}O_{2} + 2 H^{+} + 2 e^{-} → 2 H_{2}O | 1.78 | AgCl + e^{-} → Ag + Cl^{-} | 0.22 |
| Ce^{4+} + e^{-} → Ce^{3+} | 1.70 | SO_{4}^{2-} + 4 H^{+} + 2 e^{-} → H_{2}SO_{3} + H_{2}O | 0.20 |
| PbO_{2} + 4 H^{+} + SO_{4}^{2-} + 2 e^{-} → Pb SO_{4} + 2 H_{2}O | 1.69 | Cu^{2+} + e^{-} → Cu^{+} | 0.16 |
| MnO_{4}^{-} + 4 H^{+} + 3 e^{-} → Mno_{2} + 2 H_{2}O | 1.68 | 2 H^{+} + 2 e^{-} → H _{2} | 0.00 |
| IO_{4}^{-} + 2 H^{+} + 2 e^{-} → IO_{3}^{-} + H_{2}O | 1.60 | Fe^{3+} + 3 e^{-} → Fe | – 0.036 |
| MnO_{4}^{-} + 8 H^{+} + 5 e^{-} → Mn^{2+} + 4 H_{2}O | 1.51 | Pb^{2+} + 2 e^{-} → Pb | – 0.13 |
| Au^{3+} + 3 e^{-} → Au | 1.50 | Sn^{2+} + 2 e^{-} → Sn | – 0.14 |
| PbO_{2} + 4 H^{+} + 2 e^{-} → Pb^{2+} + 2 H_{2}O | 1.46 | Ni^{2+} + 2 e^{-} → Ni | – 0.23 |
| Cl_{2} + 2 e^{-} → 2Cl ^{-} | 1.36 | Pb SO_{4} + 2 e^{-} → Pb + SO_{4} ^{2-} | – 0.35 |
| Cr_{2}O_{7} ^{2-} + 14 H^{+} + 6 e^{-} →2 Cr ^{3+}+ 7 H_{2}O | 1.33 | Cd ^{2+} + 2 e^{-} → Cd | – 0.40 |
| O_{2} + 4 H^{+} + 4 e^{-} → 2 H_{2}O | 1.23 | Fe^{2+} + 2 e^{-} → Fe | – 0.44 |
| MnO_{2} + 4 H^{+} + 2 e^{-} → Mn ^{2+} + 2 H_{2}O | 1.21 | Cr ^{3+} + e^{-} → Cr^{2+} | – 0.50 |
| IO_{3}^{-} + 6 H^{+} + 5 e^{-} → \frac{1}{2}I _{2} + 3 H_{2}O | 1.20 | Cr ^{3+} + 3 e^{-} → Cr | – 0.73 |
| Br _{2} + 2 e^{-} → 2 Br^{-} | 1.09 | Zn^{2+} + 2 e^{-} → Zn | – 0.76 |
| VO_{2}^{+} + 2 H^{+} + e^{-} → VO^{2+} + H_{2}O | 1.00 | 2 H_{2}O + 2 e^{-} → H_{2} + 2 OH^{-} | – 0.83 |
| AuCl_{4}^{-} + 3 e^{-} → Au + 4 Cl^{-} | 0.99 | Mn^{2+} + 2 e^{-} → Mn | – 1.18 |
| NO_{3}^{-} + 4 H^{+} + 3 e^{-} → NO + 2 H_{2}O | 0.96 | Al^{3+} + 3 e^{-} → Al | – 1.66 |
| ClO_{2} + e^{-} → ClO_{2}^{-} | 0.954 | H _{2} + 2 e^{-} → 2 H ^{-} | – 2.23 |
| 2 Hg ^{2+} + 2 e^{-} → Hg_{2} ^{2+} | 0.91 | Mg^{2+} + 2 e^{-} → Mg | – 2.37 |
| Ag^{+} + e^{-} →Ag | 0.80 | La^{3+} + 3 e^{-} → La | – 2.37 |
| Hg_{2} ^{2+} + 2 e^{-} → 2 Hg | 0.80 | Na^{+} + e^{-} → Na | – 2.71 |
| Fe^{3+} + e^{-} → Fe^{2+} | 0.77 | Ca^{2+} + 2 e^{-} → Ca | – 2.76 |
| O_{2} + 2 H^{+} + 2 e^{-} → H_{2} O_{2} | 0.68 | Ba^{2+} + 2 e^{-} → Ba | – 2.90 |
| MnO_{4}^{-} + e^{-} → MnO_{4}^{2-} | 0.56 | K^{+} + e^{-} → K | – 2.92 |
| I_{2} + 2 e^{-} → 2 I^{-} | 0.54 | Li^{+} + e^{-} → Li | – 3.05 |
| Cu ^{+} + e^{-} → Cu | 0.52 |
What are we trying to solve?
We want to know if the reaction is spontaneous. Which leads to the question: What do we mean by spontaneous?
The reaction is spontaneous if it proceeds without outside intervention, and we can tell this by looking at the signs of both \xi °_{cell} (positive for a spontaneous process) and ΔG° (negative for a spontaneous process).
The half-reactions are
Cu^{2+} + 2 e ^{-} → Cu \xi ° (cathode) = 0.34 V
\underline{ Fe → Fe ^ {2+} + 2 e ^{-} – \xi °(anode) = 0.44 V}
Cu^{2+} + Fe → Fe ^ {2+} + Cu \xi °_{cell} = 0.78 V
We can calculate ΔG° from the equation
ΔG° = -nF \xi °
Since two electrons are transferred in the reaction, 2 moles of electrons are required per mole of reactants and products. Thus n = 2 moles of e^{-}, F = 96,485 C/mol e^{-}, and \xi ° = 0.78 V = 0.78 J/C. Therefore,
ΔG° = (- 2 mol e^{-}) (96,485 \frac{C}{ mol e^{-}} )(0.78 \frac{J}{ C}) = – 1.5 × 10 ^{5} J
The process is spontaneous, as indicated both by the negative sign of ΔG° and by the positive sign of \xi ° _{cell} .
This reaction is used industrially to deposit copper metal from solutions containing dissolved copper ores.