Practice question
Question
How many coulombs are required to deposit 0.27 g of aluminum from Al(NO₃)₃ solution? (Atomic mass of Al = 27 g/mol, F = 96500 C/mol)
Explanation
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C .
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