An acidic solution of Cu²⁺ salt containing 0.4 g of Cu²⁺ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis.

$\text{ aq. }{\mathrm{CuSO}}_4⟶\mathrm{Cu}+1/2{\mathrm{O}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4$

Number of equivalents of Cu = $\frac{0.4\times 2}{63.5}$= equivalents of O₂ liberated

volume of O₂ liberated at STP = 5.6 x number of equivalents of oxygen liberated

= $\frac{5.6\times 0.4\times 2\times 1000}{63.5}=70.55 \mathrm{ml}$(1eq of O₂ = 5.6 lit)
$\text{ But aq. }{\mathrm{CuSO}}_4\frac{\text{ prolonged }}{\text{ electrolysis }}{\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2$
$2\times 96500\text{ coul }\to 22.4\text{ ltrs of }{\mathrm{H}}_2\mathrm{\&}11.2\text{ ltrs of }{\mathrm{O}}_2$
$\begin{array}{l}(1.2\times 7\times 60)\mathrm{coul}\to {\mathrm{V}}_{{\mathrm{H}}_2}=?\mathrm{\&}{\mathrm{V}}_{{\mathrm{O}}_2\left(2\right)}=?\\ {\mathrm{V}}_{{\mathrm{H}}_2}=58.495\mathrm{ml};{\mathrm{V}}_{{\mathrm{O}}_2\left(2\right)}=29.248\mathrm{ml}\\ {\mathrm{V}}_{{\mathrm{O}}_2}\left(1\right)+\left(2\right)=99.799\mathrm{ml}\\ {\mathrm{V}}_{{\mathrm{H}}_2}=58.5\mathrm{ml},{\mathrm{V}}_{{\mathrm{O}}_2}=99.8\mathrm{ml}\end{array}$