Chemistry - Electrochemistry Question with Solution | TestHub

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 $\begin{array}{rl}& =\frac{\mathrm{I}\times \mathrm{t}}{96500}\times \text{ mole ratio }\\ & 0.005=\frac{\mathrm{I}\times 965}{96500}\times \frac{1\text{ mole of }{\mathrm{H}}_2}{2\text{ mole of }{e}^{-}}\\ & \mathrm{I}=1.0\mathrm{A}\end{array}$