The electrolysis of acidified water is represented by:
\(4 OH ^{-} \rightarrow 2 H _{2} O + O _{2}+4 e ^{-}\)
1 mole of involves \(1 F\) of electricity.
Thus 1 mole of \(O _{2}\) is produced by \(4 F\). i.e. \(22.4 dm ^{3}\) of ox-
ygen requires \(4 F .1 dm ^{3}\) will require
\((4 / 22.4) F =0.1786 F\)
\begin{aligned}
&0.1786 F =(0.1786 \times 96500) C\\
&=17232.14 C\\
&Q=I \times t, t=\frac{Q}{I}=\left(\frac{17232.14}{4.5}\right) s\\
&=3829.37 s =63.82 min
\end{aligned}