A current of 0.5 A is passed through a CuSO₄ solution for 9650 s, and then the same current is passed through an AlCl₃ s

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Question

A current of 0.5 A is passed through a CuSO₄ solution for 9650 s, and then the same current is passed through an AlCl₃ solution for 6433.33 s. What is the total mass of metal deposited? (Atomic masses: Cu = 63.5 g/mol, Al = 27 g/mol, F = 96500 C/mol)

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1.8875 g Correct 1.5875 g 2.1875 g 1.2875 g

Explanation

Charge = 0.5 × 9650 = 4825 C . Cu: Cu²⁺ + 2e⁻ → Cu , Mass = (4825/96500) × (63.5/2) = 0.05 × 31.75 = 1.5875 g . Charge = 0.5 × 6433.33 = 3216.665 C . Al: Al³⁺ + 3e⁻ → Al , Mass = (3216.665/96500) × (27/3) = 0.03333 × 9 = 0.3 g . Total mass = 1.5875 + 0.3 = 1.8875 g .

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