Question

A current of 20 ampere is passed through copper sulphate solution for 1 hour how much copper will be deposited

Answer 1

No of copper ions =Charge on one ionCharge

Number of Cu ions liberated, n=e×valence electronI×t

n=1.6×10−19×21×10

n=3.125×1019

∴n≈3.1×1019.

Hence, option C is correct.