Chemistry - Electrochemistry Question with Solution | TestHub

22400 mL${\mathrm{H}}_2$at STP$=1\text{mol}{\text{H}}_2$
So 450 ml${\mathrm{H}}_2$at STP
$=\frac{450\text{mL}}{22400\text{mL}}\times 1\text{mol}{\text{H}}_2$
$=0.02$mol
Electrode reaction
$2{\mathrm{H}}^{+}+\underset{2\text{F}}{2{\mathrm{e}}^{-}}\to \underset{1\text{mol}}{{\mathrm{H}}_2}$
The liberation of 1 mol${\mathrm{H}}_2$requires$=2\times 96500\text{C}$
So the liberation of 0.02 mol${\mathrm{H}}_2$requires$=2\times 96500\times 0.02\text{C}$
$=3860$C
Time required to produce 0.02 mol${\mathrm{H}}_2$,$\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{i}}$
$=\frac{3860}{2\text{A}}=1930\text{ sec}$