A current of$0.75 \mathrm{A}$ is passed through an acidic ${\mathrm{CuSO}}_4$ solution for 5 min. The volume of oxygen at STP liberated at Pt anode is

Quantity of electricity passed , $Q=It=(0.75\times 5\times 60)\mathrm{C}$

Amount o electricity passed $n=\frac{Q}{F}=\frac{0.75\times 5\times 60}{96500}\mathrm{mol}$

The reaction at anode is $2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+4\mathrm{e}$

Thus,$4 \mathrm{mol}$ of electrons causes the release of $1 \mathrm{mol}$ of ${\mathrm{O}}_2 (= 22.414 \mathrm{Lat} \mathrm{STP}).$ Hence, for the amount n of electricity, volume of ${\mathrm{O}}_2$ released will be

$V=\frac{1}{4}\left(\frac{0.75\times 5\times 60}{96500}\mathrm{mol}\right)\left(22.414 \mathrm{L} {\mathrm{mol}}^{-1}\right)=0.01306 \mathrm{L}=13.06 \mathrm{mL}$