Q25 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
During the electrolysis of aqueous NaCl the product formed at cathode is?
- A$\text{Cl}_{2(g)}$
- B$\text{O}_{2(g)}$
- C$\text{Na}_{(s)}$
- Correct answerD$\text{H}_{2(g)}$
Solution
In the electrolysis of aqueous $\text{NaCl}$, the ions present in the solution are $\text{Na}^{+}$, $\text{Cl}^{-}$, $\text{H}^{+}$, and $\text{OH}^{-}$.
At the cathode, reduction occurs. The competing reactions are:
$\text{Na}^{+} + e^{-} \rightarrow \text{Na}_{(s)}$ ($E^{\circ} = -2.71 \text{ V}$)
$2\text{H}_{2}\text{O}_{(l)} + 2e^{-} \rightarrow \text{H}_{2(g)} + 2\text{OH}^{-}_{(aq)}$ ($E^{\circ} = -0.83 \text{ V}$)
Since the standard reduction potential of water is greater than that of sodium ions, water is preferentially reduced at the cathode.
Thus, hydrogen gas ($\text{H}_{2(g)}$) is liberated at the cathode.
Answer: $\text{H}_{2(g)}$
At the cathode, reduction occurs. The competing reactions are:
$\text{Na}^{+} + e^{-} \rightarrow \text{Na}_{(s)}$ ($E^{\circ} = -2.71 \text{ V}$)
$2\text{H}_{2}\text{O}_{(l)} + 2e^{-} \rightarrow \text{H}_{2(g)} + 2\text{OH}^{-}_{(aq)}$ ($E^{\circ} = -0.83 \text{ V}$)
Since the standard reduction potential of water is greater than that of sodium ions, water is preferentially reduced at the cathode.
Thus, hydrogen gas ($\text{H}_{2(g)}$) is liberated at the cathode.
Answer: $\text{H}_{2(g)}$
Hence, the correct answer is option: 'D'.