CuSO 4 solution is electrolysed for 15 minutes to deposit 0 . 4725 g of copper at the cathode. The current in amperes required is ( Faraday = 96 , 500 Cmol - 1 , atomic weight of copper = 63 )

Step-by-step solution

The reaction at cathode takes place as below: Cu 2 +   +   2 e -   →   Cu s For   63   g   of   Cu   =   2 F   electricity   required So ,     for   0 . 475   g ,   electricity   required = 2 F 63 × 0 . 4725 = 2 × 96500 × 0 . 4725 63 = 1447 . 5   C According to Faradays's law: Q = It → I = Q t I = 1447 . 5 15 × 60 = 1 . 608   C