Question

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).

Answer 1

Answer is : 2

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 \(\left[\mathrm{OH}^{\ominus}\right]=10^{-2}\ \text{and}\ \left[\mathrm{H}^{+}\right]=10^{-12}\) 

So by Faraday law

Gram amount of substance deposited = Amount of electricity passed

\(10^{-2} \times \frac{600}{1000} \times 96500=\mathrm{I} \times \mathrm{t}\) 

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

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

\(\mathrm{I}=1.93\ \text{ampere}\) 

So, \(I=2\) ampere (nearest integer)