Reduction at cathode: 2e−+2H2O→H2+2OH− (valence factor) H2=2 At NTP 22400 mL of H2=1 mole of H2 112 mL of H2=224001×112=0.005 mole of H2 Moles of H2 produced $\begin{aligned}
&=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times \text { mole ratio } \
&0.005=\frac{\mathrm{I} \times 965}{96500} \times \frac{1 \text { mole of } \mathrm{H}_2}{2 \text { mole of }e^{-}} \
&\mathrm{I}=1.0 \mathrm{~A}
\end{aligned}$