Practice question
Question
A current of 0.2 A deposits 0.127 g of Cu from CuSO₄ in 9650 s. What is the current required to deposit 0.27 g of Al from AlCl₃ in the same time? (Atomic masses: Cu = 63.5 g/mol, Al = 27 g/mol, F = 96500 C/mol)
Explanation
Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .
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