JEE Main 2025 — Electrochemistry Question with Solution

Electrolysis of NaCl is

$$\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}(\mathrm{aq})\to \mathrm{NaOH}(\mathrm{aq})+\frac{1}{2}{\mathrm{Cl}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})$$

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

$$\begin{array}{rl}& 10^{-2}\times \frac{600}{1000}\times 96500=\mathrm{I}\times \mathrm{t}\\ & \frac{10^{-2}\times 600}{1000}\times 96500=\mathrm{I}\times 5\times 60\\ & \mathrm{I}=\frac{10^{-2}\times 600\times 96500}{1000\times 5\times 60}\\ & \mathrm{I}=1.93\text{ ampere }\end{array}$$

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