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Practice question

MCQ

Question

A dry cell delivers 0.1 A for 19300 s. What mass of MnO₂ (molar mass 87 g/mol) is reduced at the cathode? (F = 96500 C/mol)

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Explanation

Charge = 0.1 × 19300 = 1930 C . Cathode: MnO₂ + H⁺ + e⁻ → MnO(OH) , 1 mol MnO₂ requires 1F. Faradays = (1930/96500) = 0.02 F , Moles = 0.02 mol , Mass = 0.02 × 87 = 1.74 g .

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