[Single-digit Integer] A volume of 500 ml of 0.1 M - solution is electrolysed for 5 min at a current — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Calculate charge passed.} \[ Q = I \times t = 0.161 \times (5 \times 60) = 0.161 \times 300 = 48.3\ \text{C} \] \[ n_e = \frac{Q}{F} = \frac{48.3}{96500} = 5.0 \times 10^{-4}\ \text{mol} \] \textbf{Step 2: Cathode reaction — copper deposition.} \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \] \[ n(\text{Cu deposited}) = \frac{n_e}{2} = 2.5 \times 10^{-4}\ \text{mol} \] \textbf{Step 3: Anode reaction — oxygen evolution.} \[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \] \[ n(\text{H}^+) = n_e = 5.0 \times 10^{-4}\ \text{mol} \quad \text{(4H}^+\text{ per 4e}^-\text{)} \] \textbf{Step 4: Calculate [H⁺] and pH.} \[ [\text{H}^+] = \frac{5.0 \times 10^{-4}}{0.500} = 1.0 \times 10^{-3}\ \text{M} \] \[ \text{pH} = -\log(10^{-3}) = \boxed{3} \]