Electrolysis of a concentrated aqueous solution of NaCl between two Pt electrodes gives at cathode and anode, 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 electrolysis of a concentrated aqueous solution of NaCl between two platinum electrodes, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolytic Process**:
- In the electrolysis of concentrated NaCl solution, we have sodium ions (Na⁺) and chloride ions (Cl⁻) in the solution. Water (H₂O) is also present.

2. **Determine the Reactions at the Electrodes**:
- At the cathode (negative electrode), reduction occurs. The possible species that can be reduced are:
- Sodium ions (Na⁺)
- Water (H₂O)

3. **Compare Reduction Potentials**:
- The standard reduction potential (E°) for Na⁺ is approximately -2.71 V.
- The standard reduction potential (E°) for water (to form hydrogen gas) is approximately -0.83 V.
- Since -0.83 V (water) is greater than -2.71 V (sodium), water will be reduced instead of sodium ions.

4. **Write the Cathode Reaction**:
- The reduction of water at the cathode can be represented as:
\[ 2H₂O + 2e⁻ \rightarrow H₂ + 2OH⁻ \]
- Thus, hydrogen gas (H₂) is produced at the cathode.

5. **Identify the Anode Reaction**:
- At the anode (positive electrode), oxidation occurs. The possible species that can be oxidized are:
- Chloride ions (Cl⁻)
- Water (H₂O)

6. **Compare Oxidation Potentials**:
- The oxidation potential for Cl⁻ (to form chlorine gas) is approximately -1.36 V.
- The oxidation potential for water (to form oxygen gas) is approximately -1.23 V.
- Since -1.36 V (chlorine) is less than -1.23 V (water), chloride ions will be oxidized instead of water.

7. **Write the Anode Reaction**:
- The oxidation of chloride ions at the anode can be represented as:
\[ 2Cl⁻ \rightarrow Cl₂ + 2e⁻ \]
- Thus, chlorine gas (Cl₂) is produced at the anode.

### Final Summary of Reactions:
- **At the Cathode**:
\[ 2H₂O + 2e⁻ \rightarrow H₂ + 2OH⁻ \]
- **At the Anode**:
\[ 2Cl⁻ \rightarrow Cl₂ + 2e⁻ \]