Electrolysis of water takes place as follows.
$H _{2} O \rightleftharpoons \underset{\text { cathode }}{ H ^{+}}+\underset{\text { anode }}{ OH ^{-}}$
At anode
$OH ^{-} \xrightarrow{\text { Oxidation }} OH + e ^{-}$ $4 OH \longrightarrow 2 H _{2} O + O _{2}$
At cathode
$2 H ^{+}+2 e ^{-} \xrightarrow{\text { Reduction }}H _{2}$
Given, time, $t=1930\, s$
Number of moles of hydrogen collected
$=\frac{1120 \times 10^{-3}}{22.4}$ moles
$=0.05$ moles
$\because 1$ mole of hydrogen is deposited by $=2$ moles of
electrons
$\therefore 0.05$ moles of hydrogen will be deposited by
$=2 \times 0.05$
$=0.10$ mole of electrons
Charge, $Q = nF$
$=0.1 \times 96500$
Charge, $Q=$ it
$0.1 \times 96500=i \times 1930$
$i=\frac{0.1 \times 96500}{1930}$
$i=5.0\, A$