Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2 as per the reaction: Mn^2+(aq) + 2H2O - MnO2(s) + 2H^+(aq) + H2(g). Passing a current of 19.3 A f

Question

Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2 as per the reaction: Mn^2+(aq) + 2H2O -> MnO2(s) + 2H^+(aq) + H2(g). Passing a current of 19.3 A for 2 h gives only 52.2 g of MnO2. The current efficiency is (Mn = 55)

Prashant Kotian · Accepted Answer

Correct answer: B. The oxidation is Mn^2+ -> Mn^4+ + 2e^–, meaning n = 2. Total charge Q = 19.3 A × 2 × 3600 s = 138960 C. Faradays = 138960 / 96500 = 1.44 F. Theoretical moles of MnO2 = 1.44 / 2 = 0.72 mol. Molar mass of MnO2 = 55 + 32 = 87 g/mol. Theoretical mass = 0.72 × 87 = 62.64 g. Current efficiency = (Actual mass / Theoretical mass) × 100 = (52.2 / 62.64) × 100 = 83.33%. Therefore, correct answer is B.

💡 Solution & Explanation

Practice 22,000+ questions like this