When a quantity of electricity is passed through CuSO₄ solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H₂ liberated at STP will be [given : at. wt. of Cu = 64]

According to Faraday's law of electrolysis;

$$\begin{gathered} \frac{{\mathrm{W}}_{\mathrm{Cu}}}{{\mathrm{E}}_{\mathrm{Cu}}}=\frac{{\mathrm{W}}_{{\mathrm{H}}_2}}{{\mathrm{E}}_{{\mathrm{H}}_2}} \\ \frac{0.16}{{\displaystyle \raisebox{1ex}{$63.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{{\mathrm{W}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{H}}_2}/2} \\ \mathrm{on} \mathrm{solving} \\ \frac{{\mathrm{W}}_{{\mathrm{H}}_2}}{{\mathrm{M}}_{{\mathrm{H}}_2}}= \mathrm{moles} \mathrm{of} {\mathrm{H}}_2=2.515 \mathrm{X}\hspace{0.17em}{10}^{-3}\mathrm{mol}. \end{gathered}$$

Volume of H₂ at STP = moles X 22400 mL 

                                  = 2.515 X 10^-3 X 22400 

Volume of H₂ liberated at STP  = 56 mL