For a cell, Cu(s) |Cu (0.001M| |Ag (0.01M)| Ag(s) the cell potential is found to be 0.43 V at 298 K. — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the half-reactions and write the Nernst equation** For this galvanic cell: - Cathode (reduction): Ag⁺ + e⁻ → Ag - Anode (oxidation): Cu → Cu²⁺ + 2e⁻ The Nernst equation for the cell is: $$E_{cell} = E°_{cell} - \frac{0.059}{n}\log\frac{[Cu^{2+}]}{[Ag^+]^2}$$ where n = 2 (electrons transferred) **Step 2: Calculate E°cell using standard potentials** $$E°_{cell} = E°_{Ag^+/Ag} - E°_{Cu^{2+}/Cu} = 0.80 - E°_{Cu^{2+}/Cu}$$ **Step 3: Substitute known values into Nernst equation** $$0.43 = (0.80 - E°_{Cu^{2+}/Cu}) - \frac{0.059}{2}\log\frac{0.001}{(0.01)^2}$$ **Step 4: Simplify the logarithm term** $$\log\frac{0.001}{0.0001} = \log(10) = 1$$ $$\frac{0.059}{2}(1) = 0.0295 \text{ V}$$ **Step 5: Solve for E°Cu²⁺/Cu** $$0.43 = 0.80 - E°_{Cu^{2+}/Cu} - 0.0295$$ $$E°_{Cu^{2+}/Cu} = 0.80 - 0.43 - 0.0295 = 0.34 \text{ V}$$ **Step 6: Express in required form** $$E°_{Cu^{2+}/Cu} = 0.34 \text{ V} = 34.00 × 10^{-2} \text{ V}$$ Therefore, the answer is 34.00.