Practice question
Question
What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A? (Atomic mass of Cr = 52 g/mol, F = 96500 C/mol)
Explanation
Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .
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