Explanation

(d)
$H_{2}O\ \longrightarrow \ H_{2} + \ \frac{1}{2}\ O_{2}.$
${2H}^{+} + \ {2e}^{-}\ \longrightarrow \ H_{2}$
∴ 1 mole of $H_{2}$ is liberated from 2 moles of electrons.
2 moles electrons = 2 F charge
$= 2\ \times \ 96500\ C\ charge$
Given , $i = 4\ A,\ t\ = 30\ \min$
∴ $2\ \times \ 96500$ C charge liberates
= 22400mL $H_{2}$ at NTP
∴ $4\ \times \ 30\ \times \ 60\ C$ charge liberates
$= \ \frac{22400}{2\ \times 96500}\ \times 4\ \times \ 30\ \times \ 60\ mL\ H_{2}$
= 835.6mL
= 0.836 L of $H_{2}$