Reduction potential of Cu²⁺ is higher than H⁺ hence Cu²⁺ will be reduced at cathode more easily.

Reaction at the cathode:

${\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$

Reaction at the Anode:

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

According to Faraday's first law

$\mathrm{W}=\mathrm{Zit}$

$$\begin{gathered} \mathrm{W}=\frac{\mathrm{E}}{96500}\mathrm{it} \\ \mathrm{W}=\frac{\mathrm{Atomic} \mathrm{weight} \mathrm{of} \mathrm{Cu}}{\mathrm{n}\times 96500}\mathrm{it} \\ =\frac{63.54}{2\times 96500}\times 38.6\times 100 \\ =1.27\mathrm{g} \end{gathered}$$

When 63.54 grams of copper deposited 11.2 litres of oxygen gas is released at anode.

Hence, for 1.27 grams of copper 0.224 Litres of oxygen gas is released.