Chemistry - Electrochemistry Question with Solution | TestHub

Let moles of ${\mathrm{O}}_2$ liberated $=\mathrm{x}$ mol moles of ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$ formed $=3\times \mathrm{mol}$

Cathode : $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

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

$2S{O}_4^{2-}\to S_2{O}_8^{2-}+2e^{-}$

$\left(1\mathrm{mol}{\mathrm{O}}_2=4\mathrm{F},x\mathrm{mol}{\mathrm{O}}_2=4\mathrm{xF},1\mathrm{mol}{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}=2\mathrm{F},3\times \mathrm{mol}{\mathrm{S}}_2{\mathrm{O}}_8{}^{2-}=6\mathrm{xF}\right)$

Faraday at cathode $2\mathrm{F}=1\mathrm{mol}{\mathrm{H}}_2$

Total Faraday at anode $=4\times F+6\times F=10\times F$

Total Faraday at cathode $=$ Total Faraday at anode 10x F

At cathode $=\frac{1}{2\mathrm{F}}\times 10\times \mathrm{F}=5\times$ mole of ${\mathrm{H}}_2$

$\frac{\text{ moles of }{\mathrm{H}}_2}{\text{ moles of }{\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8}=\frac{5\mathrm{x}}{3\mathrm{x}}=\frac{5}{3}$

Moles of ${\mathrm{H}}_2=3\times \frac{5}{3}=5\mathrm{mol}$