During the preparation of ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$ (per disulphuric acid) by the electrolysis of 75% aqueous Sulphuric acid solution, ${\mathrm{O}}_2$ gas is also released at anode as by-product. If 15.12 L of ${\mathrm{H}}_2$ gas is released at cathode and 5.04 L $O_{\mathit{2}}$ gas is released at anode at STP, the weight ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$ produced in gram is 

$$\begin{gathered} 2{\mathrm{H}}_2{\mathrm{SO}}_4=4{\mathrm{H}}^{+1}+2{\mathrm{SO}}_4^{-2} \\ \mathrm{Upon} \mathrm{electrolysis} \mathrm{by} \mathrm{the} \mathrm{application} \mathrm{of} \mathrm{voltage} \mathrm{of} \mathrm{about} 2.1\mathrm{V}, \\ \mathrm{At} \mathrm{Anode} \left(\mathrm{Oxidation}\right): 2{\mathrm{SO}}_4^{-2}\to {\mathrm{S}}_2{\mathrm{O}}_8^{-2}+2{\mathrm{e}}^{-2}; \\ \mathrm{Also}, \mathrm{water} \mathrm{undergoes} \mathrm{Oxidation} \mathrm{as} \mathrm{the} \mathrm{required} \mathrm{potential} \mathrm{is} \mathrm{only} 1.23\mathrm{V} \\ 2{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{H}}^{+1}+{\mathrm{O}}_2 +4{\mathrm{e}}^{-1} \\ \mathrm{Altogether}, 6{\mathrm{e}}^{-1} \mathrm{are} \mathrm{released} \mathrm{during} \mathrm{the} \mathrm{process} \mathrm{of} \mathrm{Oxidation}. \\ \mathrm{At} \mathrm{cathode} \left(\mathrm{Reduction}\right): 6{\mathrm{H}}^{+1}+6{\mathrm{e}}^{-1}\to 3{\mathrm{H}}_2 \\ \mathrm{The} \mathrm{over} \mathrm{all} \mathrm{redox} \mathrm{reaction} \mathrm{is}: 2{\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}\stackrel{6{\mathrm{e}}^{-1}}{\to }{\mathrm{O}}_2+ 3{\mathrm{H}}_2+{\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8 \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{the} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2, {\mathrm{O}}_2 \mathrm{and} {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8 \mathrm{will} \mathrm{be} \mathrm{in} 3:1:1 \mathrm{ratio}. \\ \mathrm{Number} \mathrm{of} \mathrm{mole} \mathrm{of} {\mathrm{H}}_2=\frac{\mathrm{volume} \mathrm{of} {\mathrm{H}}_{2 }\mathrm{at} \mathrm{STP} 15.12\mathrm{L}}{\mathrm{Gram} \mathrm{molar} \mathrm{volume} 22.4\mathrm{L}}=0.675 \mathrm{mole} \\ \mathrm{Number} \mathrm{of} \mathrm{mole} \mathrm{of} {\mathrm{O}}_2=\frac{\mathrm{volume} \mathrm{of} {\mathrm{O}}_2 \mathrm{at} \mathrm{STP} 5.04\mathrm{L}}{\mathrm{Gram} \mathrm{molar} \mathrm{volume} 22.4\mathrm{L}}=0.225 \mathrm{mole} \\ \mathrm{No}. \mathrm{of} \mathrm{mole} \mathrm{of} {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8 \mathrm{formed}=0.225 \mathrm{mole}, \mathrm{same} \mathrm{as} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{mole} \mathrm{of} {\mathrm{O}}_2 \\ \mathrm{Gram} \mathrm{molecular} \mathrm{weight} \mathrm{of} {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8=194 \mathrm{g} {\mathrm{mole}}^{-1} \\ \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8 \mathrm{formed}=194 \mathrm{x} 0.225=43.65\mathrm{g} \end{gathered}$$