In the electrolysis of a CuSO 4 solution, how many grams of Cu are plated out on the cathode, in the time…

The mole of oxygen = $\frac{\mathrm{Given} \mathrm{volume}}{\mathrm{Molar} \mathrm{volume}}=\frac{5.6}{22.4}=0.25 \mathrm{mol}$

Equivalent of oxygen = $\mathrm{n}-\mathrm{factor}\times \mathrm{mole}$

$\mathrm{n}-\mathrm{factor}=4$

Equivalent of oxygen=$4\times 0.25=1$

So, Equivalent of oxygen = Equivalent of copper

Equivalent mass of copper= $\frac{\mathrm{Molar} \mathrm{mass}}{\mathrm{n}-\mathrm{factor}}=\frac{63.5}{2}$

Equivalent of coper = $\frac{\mathrm{mass} \mathrm{of} \mathrm{copper}}{\mathrm{Equivalent} \mathrm{mass}}=\frac{\mathrm{mass} \mathrm{of} \mathrm{copper}}{{\displaystyle \raisebox{1ex}{$63.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Equivalent of oxygen = Equivalent of copper

$$\begin{gathered} 1=\frac{\mathrm{mass} \mathrm{of} \mathrm{copper}}{{\displaystyle \raisebox{1ex}{$63.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \mathrm{mass} \mathrm{of} \mathrm{copper}=\raisebox{1ex}{$63.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=31.75 \mathrm{g} \end{gathered}$$

Hence, $31.75 \mathrm{g}$ of copper are plated out on the cathode.