Explanation

(c)
Electrolysis of water takes place as follows
$H_{2}O\ \ \ \ \ \rightleftharpoons \ {\ \ \ \ \ \ \ \ \ H}^{+}\ \ \ \ \ \ + \ \ \ \ \ \ {OH}^{-}$
Cathode anode
At anod
${OH}^{-}\ \ \ \overset{oxidation\ \ \ \ }{\rightarrow}\ OH + \ e^{-}$
$4OH\ \longrightarrow \ {2H}_{2}O\ + \ O_{2}$
At cathode
${2H}^{+} + \ {2e}^{-}\ \overset{Reduction\ \ \ \ }{\rightarrow}\ H_{2}$
Given, time, $t\ = \ 1930s$
Number of moles of hydrogen collected
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{1120\ \times \ 10^{- 3}}{22.4}\ moles$
= 0.05 moles
∵ 1 mole of hydrogen is deposited by = 2 moles of electrons
∵ 0.05 moles of hydrogen will be deposited by
$= \ 2\ \times \ 0.05$
= 0.10 mole of electrons
Charge$\mathcal{,\ Q\ = \ }nF\$
$= \ 0.1\ \times \ 96500$
Charge,$\mathcal{\ Q\ = \ }it$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1\ \times \ 96500\ = \ i\ \times \ 1930\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \ \frac{0.1\ \times 96500}{1930}$
= 5.0 A