VITEEE 2021 Paper

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
=

2×96500

63.5

×0.634≈1930C