Chemistry - Electrochemistry Question with Solution | TestHub

22,400 mL of hydrogen at STP (or NTP) = 2g

So 112 mL of hydrogen at STP$=\frac{2\mathrm{g}\times 112\mathrm{m}\mathrm{L}}{\mathrm{22,400}\mathrm{m}\mathrm{L}}=10^{-2}\mathrm{g}$

$2{\mathrm{H}}^{+}+\underset{2\mathrm{F}}{2{\mathrm{e}}^{-}}\to \underset{1\text{ mol}}{{\mathrm{H}}_2}$

$=2\times 96500\mathrm{ }\mathrm{C}=2\mathrm{g}$

2 g hydrogen is deposited by$=2\times 96500\mathrm{C}$

So$10^{-2}$g hydrogen will be deposited by

$=\frac{2\times 96500\times 10^{-2}\mathrm{g}}{2\mathrm{g}}=965\mathrm{ }\mathrm{C}$

Charge$\left(\mathrm{Q}\right)=\mathrm{i}\mathrm{t}$

So$\mathrm{i}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{965\text{ C}}{965\text{ s}}=1\text{ C}{\text{s}}^{-1}=1\mathrm{ }\mathrm{A}$