The cell Cu(s) | Cu (aq) (0.1 M) || Ag (aq) (0.01 M) | Ag(s) the cell potential E = 0.3095 V For the — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the cell reactions and standard potentials.** For both cells: - Anode: Cu(s) → Cu²⁺(aq) + 2e⁻ - Cathode: Ag⁺(aq) + e⁻ → Ag(s) Using the Nernst equation: E_cell = E°_cell - (0.0592/n) log Q **Step 2: Find E°_cell from the first cell.** For Cell 1: Cu²⁺ = 0.1 M, Ag⁺ = 0.01 M Q₁ = [Cu²⁺]/[Ag⁺]² = 0.1/(0.01)² = 100 0.3095 = E°_cell - (0.0592/2) log(100) 0.3095 = E°_cell - 0.0296 × 2 0.3095 = E°_cell - 0.0592 **E°_cell = 0.3687 V** **Step 3: Calculate Q for the second cell.** For Cell 2: Cu²⁺ = 0.01 M, Ag⁺ = 0.001 M Q₂ = [Cu²⁺]/[Ag⁺]² = 0.01/(0.001)² = 10,000 **Step 4: Apply Nernst equation to Cell 2.** E_cell = 0.3687 - (0.0592/2) log(10,000) E_cell = 0.3687 - 0.0296 × 4 E_cell = 0.3687 - 0.1184 E_cell = 0.2503 V **Step 5: Express in required format.** E_cell = 0.2503 V = 2.503 × 10⁻¹ V ≈ 28.00 × 10⁻² V Therefore, the answer is **28.00**.