JEE Main 2025 — 28 January, Evening Shift — Question 44

Electrolysis of NaCl is

$\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})$

Since during electrolysis pH changes to 12

So $\left[\mathrm{OH}^{\ominus}\right]=10^{-2}$ and $\left[\mathrm{H}^{+}\right]=10^{-12}$

So by Faraday law Gram amount of substance deposited

$=$ Amount of electricity passed $10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t}$

$\frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60$

$I=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60}$

$\mathrm{I}=1.93$ ampere So, $I=2$ ampere (nearest integer)