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Practice question

MCQ

Question

In a dry cell, 0.435 g of MnO₂ (molar mass 87 g/mol) is reduced to MnO(OH). How many coulombs are consumed? (F = 96500 C/mol)

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Explanation

MnO₂ + H⁺ + e⁻ → MnO(OH) . Moles = (0.435/87) = 0.005 mol , Charge = 0.005 × 96500 = 482.5 C .

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