In Kolbe's electrolysis sodium propanoate gives `:`

Options

A. `CH_(3)-CH_(3)`
B. `CH_(2)=CH_(2)`
C. `CH_(3)-CH_(2)-CH_(2)-CH_(3)`
D. A mixture of 1,2 and 3

Answer

To solve the question regarding the products formed during Kolbe's electrolysis of sodium propanoate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Sodium Propanoate**: Sodium propanoate is the sodium salt of propanoic acid (C2H5COOH). Its formula is CH3CH2COONa.

2. **Dissociation in Aqueous Solution**: In an aqueous solution, sodium propanoate dissociates into sodium ions (Na+) and propanoate ions (CH3CH2COO-).

3. **Electrolysis Setup**: During electrolysis, the anode is where oxidation occurs, and the cathode is where reduction occurs. The propanoate ions will migrate to the anode.

4. **Oxidation at the Anode**: At the anode, the propanoate ion (CH3CH2COO-) will lose electrons and undergo oxidation. This will lead to the formation of a radical species.

5. **Formation of Ethyl Radical**: The propanoate ion can break down to form an ethyl radical (C2H5•) and carbon dioxide (CO2) as a byproduct. The reaction can be summarized as:
\[ CH3CH2COO^- \rightarrow C2H5• + CO2 \]

6. **Dimerization of Ethyl Radical**: The ethyl radicals can undergo dimerization to form butane (C4H10):
\[ 2 C2H5• \rightarrow C4H10 \]

7. **Disproportionation of Ethyl Radical**: The ethyl radicals can also undergo disproportionation. One ethyl radical can abstract a hydrogen atom from another, leading to the formation of ethane (C2H6) and another radical:
\[ C2H5• + C2H5• \rightarrow C2H6 + C2H4 \]

8. **Reduction at the Cathode**: At the cathode, water is reduced to form hydrogen gas (H2) and hydroxide ions (OH-):
\[ 2 H2O + 2 e^- \rightarrow H2 + 2 OH^- \]

9. **Summary of Products**: The products formed during Kolbe's electrolysis of sodium propanoate are:
- Ethane (C2H6)
- Butane (C4H10)
- Carbon dioxide (CO2)
- Hydrogen gas (H2)

10. **Final Answer**: The correct answer is a mixture of ethane, butane, and carbon dioxide.