Explanation:
Reduction at cathode: \(2 H^{+}(a q)+2 e^{-} \rightarrow H_{2}(g)\)
At \(N.T.P, 22.4\, L\) (or \(22400 mL\) ) of \(H _{2}=1\, mole\) of \(H _{2}\)
\(112 \,mL\) of \(H _{2}=\frac{112}{22400} \times 1=0.005\, mole\) of \(H_{2}\)
Moles of \(H_{2}\) produced \(=\frac{I(A) \times t(s)}{96500\left(C / m o l e^{-}\right)} \times\) mole ratio
\(0.005 \,mol =\frac{I(A) \times 965\, s}{96500\left(C / m o l e^{-}\right)} \times \frac{1\, mol \,H_{2}}{2 \,mol e^{-}}\)
\(I=1 \,A\)