The cell potential for the given cell at 298 K is 0.31 V. The pH of the acidic solution is found to — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the cell reaction and half-reactions** This is a Cu/H⁺ cell. The half-reactions are: - Oxidation: H₂ → 2H⁺ + 2e⁻ (E° = 0 V) - Reduction: Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V) **Step 2: Calculate standard cell potential** E°cell = E°cathode - E°anode = 0.34 - 0 = 0.34 V **Step 3: Apply the Nernst equation** $$E_{cell} = E°_{cell} - \frac{0.059}{n} \log Q$$ Where n = 2 electrons transferred, and: $$Q = \frac{[H^+]^2}{[Cu^{2+}]}$$ **Step 4: Calculate Q using given concentrations** - pH = 3, so [H⁺] = 10⁻³ M - [Cu²⁺] = 10⁻ˣ M $$Q = \frac{(10^{-3})^2}{10^{-x}} = \frac{10^{-6}}{10^{-x}} = 10^{x-6}$$ **Step 5: Substitute into Nernst equation** $$0.31 = 0.34 - \frac{0.059}{2} \log(10^{x-6})$$ $$0.31 = 0.34 - 0.0295(x - 6)$$ $$-0.03 = -0.0295(x - 6)$$ $$1.017 ≈ x - 6$$ $$x ≈ 7$$ Therefore, the answer is 7.00.