How many grams of copper are deposited by passing 9650 C of charge through CuSO4 solution? (M(Cu) = 63.5, F = 96500 C/mol)

3.18 g (n = q/(zF) = 9650/(2×96500)) Cu^{2+} + 2e⁻ → Cu. Moles e⁻ = 9650/96500 = 0.100. Moles Cu = 0.0500. Mass = 0.0500 × 63.5 = 3.175 g.

3.18 g (n = q/(zF) = 9650/(2×96500))

Electrolysis and Faraday's Law

AP Chemistry· difficulty 4/5

_4

Explanation

^{2+}

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