Q.7 Calculate the time required to deposit 2·4 g of Cu, when 2·03 A of current passed through CuSO4 solution.
(At. mass of Cu = 63.5 g.mol–1)
Best answer
Answer: t ≈ 3593 s (about 59.9 min)
Cathode reaction: Cu\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Cu. Deposition of 1 mol Cu (63.5 g) needs 2 F = 2 × 96500 C.
Charge required: \(Q = \dfrac{W \times n \times F}{M} = \dfrac{2.4 \times 2 \times 96500}{63.5} = \dfrac{463200}{63.5} = 7294.5\) C
\(t = \dfrac{Q}{I} = \dfrac{7294.5}{2.03} = 3593\) s \(\approx\) 59.9 min
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