JEE Main 2020 — Electrochemistry Question with Solution

${\left(n_{Ag}\right)}_{deposit}=\frac{108}{108}=\mathrm{1 }mole$
${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Ag}$
$1\mathrm{F}$ charge is required to deposit $1 \mathrm{mole}$ of $\mathrm{Ag}$
$H_{2}O$ $⟶\frac{1}{2}{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$
$2F$ charge is required to deposit $\frac{1}{2}mole$
$1F$ charge will deposit $\frac{1}{4}mole$ of $O_2$  gas.
$V_{O_2}=\frac{\mathrm{n}RT}{P}$
$=\frac{1}{4}\times \frac{0.0821\times 273}{1}=\frac{1}{4}\times 22.4$
$V_{O_2}=\mathrm{5.6 }L$

Faraday's second law of electrolysis states that, when the same quantity of electricity is passed through several electrolytes, the mass of the substances deposited are proportional to their respective equivalent weight.