Q10 of 26 Page 1

Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO4.

(Molar mass of Cu = 63.5 g mol-1, 1F = 96500 C mol-1).


When a Cu electrode is placed in a solution of CuSO4 and current is passed through it, copper from the solution is deposited on the electrode.

Cu2+ + 2e-→ Cu


63.5 g of copper is deposited by 2 × 1F = 2 × 96500 C = 193000C


∴ 1.27 g of copper will be deposited by



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