U9.8 · Cell Potential and Free EnergyPractice · 电池电势与自由能
Lessons covered:U9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)U9.8 Cell Potential and Free Energy
- 1Numeric
Calculate E°cell for 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s). E°: Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V; Al³⁺(aq) + 3e⁻ → Al(s), E° = −1.66 V.
VExplanationCu²⁺ is reduced (cathode) and Al is oxidized (anode). E°cell = E°(cathode) − E°(anode) = 0.34 − (−1.66) = 2.00 V. The equation multiplies Al by 2 and Cu by 3, but E° is an intensive quantity and isn’t multiplied.
- 2Numeric
For 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), E°cell = 2.00 V. Calculate ΔG° in kJ/mol. F = 96485 C/mol.
kJ/molExplanationTwo Al each lose 3 electrons and three Cu²⁺ each gain 2, so n = 6. ΔG° = −nFE° = −6 × 96485 × 2.00 = −1157820 J/mol ≈ −1158 kJ/mol. Using n = 2 or 3 are common mistakes.
- 3Multiple choice
Based on these E° values, which is the strongest oxidizing agent? Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V; Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V; Fe²⁺(aq) + 2e⁻ → Fe(s), E° = −0.45 V; Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V.
ExplanationAn oxidizing agent is itself reduced. The higher the reduction potential, the more readily the half-reaction runs to the right (reduction), so the species on the left is the stronger oxidizer. Ag⁺ has the highest E° (+0.80 V). Conversely, Zn is the strongest reducing agent among these metals.
- 4Multiple choice
Each of these metals is placed in 1 M hydrochloric acid. Which one does NOT dissolve (releases no H₂)? E°: 2H⁺(aq) + 2e⁻ → H₂(g), E° = 0.00 V; Ni²⁺(aq) + 2e⁻ → Ni(s), E° = −0.26 V; Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V; Fe²⁺(aq) + 2e⁻ → Fe(s), E° = −0.45 V; Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V.
ExplanationFor M + 2H⁺ → M²⁺ + H₂, E°cell = 0.00 − E°(M). Ni, Zn, and Fe have negative E° values, so E°cell > 0 and they dissolve; Ag has E° = +0.80 V, giving E°cell = −0.80 V < 0, so it can’t reduce H⁺ to H₂.
- 5Numeric
For Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s), E°cell = 0.79 V. Find log K for this reaction at 25 °C. (At 25 °C, 2.303RT/F = 0.0591 V.)
ExplanationCombining ΔG° = −nFE° with ΔG° = −RT ln K gives log K = nE° ÷ (2.303RT/F) = 2 × 0.79 ÷ 0.0591 = 26.7, so K ≈ 1026.7. Less than 1 V of E° still gives an enormous K, which is why iron displaces Cu²⁺ from solution almost completely.
Review: U9.8 Cell Potential and Free EnergyU9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)
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