Chemistry - Electrochemistry Question with Solution | TestHub

${\text{22400 mL H}}_{\text{2}}$at STP$\text{=}\text{1}\text{mol}{\text{H}}_{\text{2}}$
So$\text{450}\text{ml}{\text{H}}_{\text{2}}$at STP
$\text{=}\frac{\text{450}\text{mL}}{\text{22400}\text{mL}}\text{×1}\text{mol}{\text{H}}_{\text{2}}$
$=\mathrm{0.02\; mol}$
Electrode reaction
$2{\mathrm{H}}^{+}+\underset{2\mathrm{F}}{2{\mathrm{e}}^{-}}\to \underset{1\mathrm{m}\mathrm{o}\mathrm{l}}{{\mathrm{H}}_2}$
The liberation of$\text{1}\text{mol}{\text{H}}_{\text{2}}$requires$=2\times 96500\text{C}$
So the liberation of$\text{0}\text{.02}\text{mol}{\text{H}}_{\text{2}}$requires$=2\times 96500\times 0.02\text{C}$
$=3860$C
Time required to produce$\text{0}\text{.02}\text{mol}{\text{H}}_{\text{2}}\text{,}\text{t}\text{=}\frac{\text{Q}}{\text{i}}$
$\text{=}\frac{\text{3860}}{\text{2A}}\text{=}\text{1930}\text{s}\simeq \text{0}\text{.503}\text{hour}$
$\approx 0.5\text{hour}$