Cu(s) + Sn (0.001M) → Cu (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the half-reactions and determine E°cell** Oxidation: Sn(s) → Sn²⁺ + 2e⁻ (E° = –0.14 V) Reduction: Cu²⁺ + 2e⁻ → Cu(s) (E° = 0.34 V) E°cell = E°cathode – E°anode = 0.34 – (–0.14) = 0.48 V **Step 2: Calculate ΔG° using the Nernst equation concept** For the reaction: ΔG° = –nFE°cell where n = 2 (moles of electrons), F = 96500 C/mol ΔG° = –2 × 96500 × 0.48 = –92,640 J/mol = –92.64 kJ/mol **Step 3: Apply the Nernst equation to find actual ΔG** ΔG = ΔG° + RT ln(Q) At 298 K: RT = 8.314 × 298 = 2,481.6 J/mol Q = [Cu²⁺]/[Sn²⁺] = 0.01/0.001 = 10 **Step 4: Calculate ln(Q) and ΔG** ln(10) = 2.303 ΔG = –92,640 + 2,481.6 × 2.303 ΔG = –92,640 + 5,713 = –86,927 J/mol ≈ –86.93 kJ/mol **Step 5: Express in required form** ΔG = –86.93 kJ/mol = –0.8693 × 10² kJ/mol ≈ –983 × 10⁻¹ kJ/mol Therefore, the answer is **983**.