JEE Main 2018 — Electrochemistry Question with Solution

Reduction at cathode: $2{\mathrm{e}}^{-}+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$ (valence factor) ${\mathrm{H}}_2=2$ At NTP $22400\mathrm{mL}$ of ${\mathrm{H}}_2=1$ mole of ${\mathrm{H}}_{2}112\mathrm{mL}$ of ${\mathrm{H}}_2=\frac{1}{22400}\times 112=0.005\text{ mole of }{\mathrm{H}}_2$ Moles of ${\mathrm{H}}_2$ produced $\begin{aligned} &=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times \text { mole ratio } \\ &0.005=\frac{\mathrm{I} \times 965}{96500} \times \frac{1 \text { mole of } \mathrm{H}_2}{2 \text { mole of }e^{-}} \\ &\mathrm{I}=1.0 \mathrm{~A} \end{aligned}$