For the given reactions Sn + 2e → Sn Sn + 4e → Sn the electrode potentials are; = –0.140 V and = 0.0 — Electrochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the half-reactions and given potentials** - Sn²⁺ + 2e⁻ → Sn, E° = –0.140 V - Sn⁴⁺ + 4e⁻ → Sn, E° = 0.010 V We need to find E°(Sn⁴⁺/Sn²⁺). **Step 2: Set up the relationship using Gibbs free energy** For each half-reaction: ΔG° = –nFE° For reaction 1: ΔG₁° = –2F(–0.140) = 0.280F For reaction 2: ΔG₂° = –4F(0.010) = –0.040F **Step 3: Determine the target reaction** Sn⁴⁺ + 2e⁻ → Sn²⁺ This is obtained by: Reaction 2 – Reaction 1 ΔG° = ΔG₂° – ΔG₁° = –0.040F – 0.280F = –0.320F **Step 4: Calculate the electrode potential** For the target reaction (n = 2 electrons): –0.320F = –2F × E°(Sn⁴⁺/Sn²⁺) E°(Sn⁴⁺/Sn²⁺) = 0.320/2 = 0.160 V **Step 5: Express in required format** 0.160 V = 16 × 10⁻² V = 16 × 10⁻² V The magnitude is **16** (in units of × 10⁻² V) Therefore, the answer is **16**.