In the electrolysis of cupric sulphate, the reaction that occurs at cathode is Cu2++2e−→Cu Thus 2F or 2×96500C of electricity is required to deposit =1mol of Cu=63.5g of Cu It means that to deposit 63.5g of Cu, the amount of electricity required =2×96500C So, to deposit 0.634g of Cu, the amount of electricity required =