Question

$0.05M$ aqueous solution of $NaCl$ is electrolyzed. If a current of strength $0.5A$ is used for $193$ sec, then the final concentration of $N{a}^{+}$ ions in the electrolyte will be:

(volume of solution will be constant)

• A. $0.05M$
• B. $0.049M$
• C. $0.051M$
• D. $0.04M$

Answer 1

The correct option is A $0.05M$

Total charge used $Q=0.5\times 193$.

$\therefore$ Moles of electron used $=\frac{0.5\times 193}{96500}=0.001$.

But $N{a}^{+}$ is not reduced in aqueous solution as the reduction potential of $H^{+}$ is higher.

$\therefore H^{+}$ is reduced.

Hence, the concentration of $N{a}^{+}$ does not change.

Hence, the correct option is A.