Practice question
Question
A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂(SO₄)₃ with the same current? (Atomic masses: Cu = 63.5 g/mol, Al = 27 g/mol, F = 96500 C/mol)
Explanation
Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .
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