The pH of 100 L of concentrated KCl solution after the electrolysis for 10 s using 9.65 A at 298 K i — Electrochemistry Chemistry Question

💡 Solution & Explanation

Charge $Q = 9.65 \text{ A} \times 10 \text{ s} = 96.5$ C. Faradays = $96.5 / 96500 = 0.001$ F. At the cathode, water is reduced: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. 1 Faraday produces 1 mole of $OH^-$, so 0.001 F produces 0.001 mol $OH^-$. Concentration $[OH^-] = 0.001 \text{ mol} / 100 \text{ L} = 10^{-5}$ M. pOH = 5, thus pH = 14 - 5 = 9. Therefore, correct answer is 9.