Electrolysis of 600 mL aqueous — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Initial pH of solution = 7 & [H+] = 10–7 Final pH of solution = 12  pOH = 2 so [OH–] = 10–2 At At cathode 2H2O(g) + 2e–   H2(g) + 2OH– 6 × 10–3 mole 10–2 × 0.6 mole charge (g) = it = i × 5 × 60 = [6 ×10–3] × 96500 I = 1.93 A Ans. = 2