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

Given, Q - 2F, Atomic mass of Cu - 63.5 u
Valency of the metal  Z = 2
We have, ${\mathrm{CuSO}}_4⟶{\mathrm{Cu}}^{2+}+{\mathrm{SO}}_4^2$
              $\underset{1\mathrm{mol}}{{\mathrm{Cu}}^{2+}}+\underset{2\mathrm{mol}}{2{\mathrm{e}}^{-}}⟶\underset{1\mathrm{mol}=63.5\mathrm{g}}{\mathrm{Cu}}$
Alternatively,  $\mathrm{w}=\mathrm{zQ}=\frac{\mathrm{E}}{\mathrm{F}}\cdot 2\mathrm{F}=2\mathrm{E}=\frac{2\times 63.5}{2}=63.5\mathrm{g}$