Question

On electrolysis of very dilute aqueous solution of NaCl using platinum electrodes :

• A. H$_2$ gas is evolved at anode.
• B. Na is produced at cathode.
• C. O$_2$ gas is evolved at anode.
• D. H$_2$ gas is evolved at cathode.

Hint:

In aqueous solutions of alkali metal salts, alkali metal ions are generally not discharged. Water is reduced instead, producing hydrogen gas at the cathode.

Answer 1

The Correct Option is D

Step 1: Species present in very dilute NaCl solution.
The solution contains $Na^+$, $Cl^-$, and $H_2O$. In very dilute solution, $Cl^-$ concentration is negligibly low.
Step 2: Cathode reaction.
Reduction of $Na^+$ requires very high energy. Water is preferentially reduced: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. So $H_2$ is evolved at the cathode.
Step 3: Anode reaction.
At low $Cl^-$ concentration, water is preferentially oxidized: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. So $O_2$ evolves at the anode, not $Cl_2$.
\[ \boxed{H_2 \text{ gas is evolved at cathode}} \]