Question

During the electrolysis of aqueous NaCl the product formed at cathode is?

• A. \(\text{Cl}_{2(g)}\)
• B. \(\text{O}_{2(g)}\)
• C. \(\text{Na}_{(s)}\)
• D. \(\text{H}_{2(g)}\)

Hint:

Water is reduced more easily than Na+, so hydrogen gas forms at the cathode.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In electrolysis of an aqueous salt, water competes with the ions. At the cathode, the species that is reduced most easily (higher reduction potential) wins.

Step 2: Possible Cathode Reactions:
\(\text{Na}^+ + e^- \rightarrow \text{Na}\) has \(E^{\circ} = -2.71\) V.
\(2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-\) has \(E^{\circ} = -0.83\) V.
The water reaction has the less negative potential, so it occurs.

Step 3: Product at the Cathode:
Hydrogen gas bubbles out of the cathode, and the solution near it becomes basic because of \(\text{OH}^-\) (this is the basis of the chlor-alkali process).

Step 4: Check the Other Options:
\(\text{Cl}_2\) forms at the anode, not the cathode. \(\text{O}_2\) is an anode product of water oxidation, and it is not formed at the cathode. Sodium metal is not deposited because \(\text{Na}^+\) is harder to reduce than water. So (D) is correct.

Final Answer:
Hydrogen gas is evolved at the cathode, option (D). \[ \boxed{\text{(D) } \text{H}_{2(g)}} \]