Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: (At. mass of Cu = 63.5 amu)

Cu2+ + 2e- → Cu. From the above-given half-reaction, two moles of electrons react with one mole of Cu(II) to deposit one mole of Cu. This corresponds to 63.5 g (molecular weight) of copper. Two Faraday's of electricity corresponds to 2 moles of electrons. Hence, they will deposit 63.5 g of copper.