Electrolysis of a concentrated aquoous solution of NaCl botween two Pt electrodes gives at cathode and anodo, respectively.

Options

A. `Na and Cl_(2(g))`
B. `H_(2(g)) and Cl_(2(g))`
C. `H_(2(g)) and O_(2(g))`
D. `O_(2(g)) and H_(2(g))`

Answer

To solve the question regarding the products of electrolysis of a concentrated aqueous solution of NaCl between two platinum electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolysis Setup**:
- We have a concentrated aqueous solution of NaCl, which means it contains Na⁺, Cl⁻, H⁺, and OH⁻ ions in solution.
- The electrodes used are platinum (Pt), which are inert and do not participate in the reaction.

2. **Determine the Reactions at the Electrodes**:
- The anode is positively charged and will attract negatively charged ions (anions).
- The cathode is negatively charged and will attract positively charged ions (cations).

3. **Identify the Cations and Anions**:
- In the solution, the cations are Na⁺ and H⁺, while the anions are Cl⁻ and OH⁻.

4. **Reduction at the Cathode**:
- At the cathode, we need to determine which cation will be reduced.
- The reduction potentials are:
- \( \text{Na}^+ + e^- \rightarrow \text{Na} \) (E° = -2.7 V)
- \( \text{2H}^+ + 2e^- \rightarrow \text{H}_2 \) (E° = 0 V)
- Since hydrogen ions (H⁺) have a higher reduction potential than sodium ions (Na⁺), H⁺ will be reduced to form hydrogen gas (H₂).

5. **Oxidation at the Anode**:
- At the anode, we need to determine which anion will be oxidized.
- The oxidation potentials are:
- \( \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \) (E° = +1.36 V)
- \( \text{4OH}^- \rightarrow \text{O}_2 + 2H_2O + 4e^- \) (E° = +0.40 V)
- In a concentrated solution of NaCl, the concentration of Cl⁻ is higher than that of OH⁻, so Cl⁻ will be oxidized to form chlorine gas (Cl₂).

6. **Write the Overall Reactions**:
- At the cathode: \( 2H^+ + 2e^- \rightarrow H_2(g) \)
- At the anode: \( 2Cl^- \rightarrow Cl_2(g) + 2e^- \)

7. **Conclusion**:
- The products of the electrolysis of a concentrated aqueous solution of NaCl are hydrogen gas (H₂) at the cathode and chlorine gas (Cl₂) at the anode.

### Final Answer:
- The products at the cathode and anode, respectively, are: **H₂ (hydrogen gas) at the cathode and Cl₂ (chlorine gas) at the anode.**