During electrolysis of aqueous NiSO₄ with Ni electrodes, 0.59 g of Ni (atomic mass 59 g/mol) is deposited at the cathode

Practice question

Question

During electrolysis of aqueous NiSO₄ with Ni electrodes, 0.59 g of Ni (atomic mass 59 g/mol) is deposited at the cathode. If the same charge oxidizes water at the anode, what volume of O₂ (STP) is produced? (F = 96500 C/mol)

Options

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0.224 L 0.056 L 0.448 L 0.112 L Correct

Explanation

Cathode: Ni²⁺ + 2e⁻ → Ni . Moles = (0.59/59) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ . 1 mol O₂ (22.4 L) requires 4F. Faradays = (1930/96500) = 0.02 F , Moles O₂ = (0.02/4) = 0.005 mol , Volume = 0.005 × 22.4 = 0.112 L .

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