In electrolysis of `NaCl` when `Pt` electrode is taken `H_(2)` is liberated at cathode while `Hg` cathode it forms sodium amalgam because

Options

A. Hg is more inert than Pt
B. more voltage is required to reduce `H^(+)` at Hg than at Pt
C. Na is dissovled in Hg while it does not dissolved in Pt
D. concentration of `H^(+)` ion is larger when Pt electrode is taken

Explanation

To solve the question regarding the electrolysis of NaCl with different electrodes (Pt and Hg), we will analyze the behavior of these electrodes during the electrolysis process.

### Step-by-Step Solution:

1. **Understanding Electrolysis of NaCl**:
- When NaCl is dissolved in water and subjected to electrolysis, it dissociates into Na⁺ and Cl⁻ ions. Water also dissociates into H⁺ and OH⁻ ions.
- The overall reaction can be summarized as:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \]

2. **Identifying the Electrodes**:
- The cathode is the electrode where reduction occurs (gaining electrons).
- The anode is where oxidation occurs (losing electrons).
- In this case, we are comparing a platinum (Pt) electrode and a mercury (Hg) electrode.

3. **Behavior at the Cathode**:
- **With Pt Electrode**:
- At the Pt cathode, both H⁺ and Na⁺ ions can be reduced. However, hydrogen ions (H⁺) are preferentially reduced to form hydrogen gas (H₂) because the reduction potential for H⁺ to H₂ is more favorable than that for Na⁺ to Na.
- Thus, H₂ is liberated at the Pt cathode.

- **With Hg Electrode**:
- At the Hg cathode, the reduction of H⁺ to H₂ requires a higher voltage compared to the reduction of Na⁺ to form sodium amalgam (Na-Hg).
- Therefore, Na⁺ ions are preferentially reduced at the Hg cathode, forming sodium amalgam (Na-Hg) instead of liberating hydrogen gas.
- The reaction can be represented as:
\[ \text{Na}^+ + \text{Hg} \rightarrow \text{Na-Hg (sodium amalgam)} \]

4. **Conclusion**:
- The reason why hydrogen is liberated at the Pt cathode while sodium amalgam is formed at the Hg cathode is due to the different reduction potentials and the solubility of sodium in mercury.
- The correct answer to the question is that more voltage is required to reduce H⁺ at Hg than at Pt, making option B the correct choice.