Chemistry - Electrochemistry Question with Solution | TestHub

As per the first law of electrolysis, The mass deposited or volume liberated at the electrode is directly proportional to the charge passing between the electrode.

$\mathrm{m}=\mathrm{ZQ}$

Where $\mathrm{Z}=\frac{\mathrm{E}}{\mathrm{F}}$ (electrovalent equivalent)

$\mathrm{Q}=\mathrm{i}\times \mathrm{t}$.

mass of ${\mathrm{O}}_2$ $= \frac{\mathrm{E}\times \mathrm{i}\times \mathrm{t}}{\mathrm{F}}$ (Where $\mathrm{F}=96500 \mathrm{C}$)

$=\frac{8\times 5\times 193}{96500} \mathrm{g}$

$=0.08 \mathrm{g}$

$\because$  At $S\mathrm{TP}$, volume of $32 \mathrm{g} {\mathrm{O}}_2 = 22400 \mathrm{mL}$

$\therefore$ The volume of $0.08 \mathrm{g} {\mathrm{O}}_2$ $=\frac{22400\times 0\text{.}08}{32}$

$=56 \mathrm{ml}$