Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ____. (Nearest integer).

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Aicharya · Accepted Answer

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
$\begin{aligned} & 10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t} \\ & \frac{10^{-2} \times 600}{1000} \times 96500=\mathrm{I} \times 5 \times 60 \\ & \mathrm{I}=\frac{10^{-2} \times 600 \times 96500}{1000 \times 5 \times 60} \\ & \mathrm{I}=1.93 \text { ampere } \\ & \text { So }, I=2 \text { ampere (nearest integer) }\end{aligned}$

Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current…

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