Two Faraday of electricity is passed through a solution of ${\mathrm{CuSO}}_4$ . The mass of copper deposited at the cathode is (Given: atomic mass of Cu= 63.5 amu)

The deposition reaction is ${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)$

Two faraday of electricity amounts to 2mol of electrons. Hence, from the above reaction, 1mol of Cu will be deposited. Its mass will be its molar mass, that is 63.5 g.