Practice question
Question
In a dry cell, the cathode reaction produces MnO(OH). If 0.87 g of MnO₂ (atomic mass Mn = 55, O = 16) is reduced, how many coulombs are consumed? (F = 96500 C/mol)
Explanation
MnO₂ + H⁺ + e⁻ → MnO(OH) . Molar mass = 55 + 32 = 87 g/mol . Moles = (0.87/87) = 0.01 mol , Charge = 0.01 × 96500 = 965 C .
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