Question

Consider the following
Statement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product. 
Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.

• A. Both statement-I and statement-II are correct
• B. Both statement-I and statement-II are not correct
• C. Statement-I is correct, but statement-II is not correct
• D. Statement-I is not correct, but statement-II is correct

Hint:

Kolbe's electrolysis of sodium salts of carboxylic acids generates alkanes with double the number of carbon atoms in the alkyl radical. Sodium propionate gives n-butane, not n-hexane.

Answer 1

The Correct Option is C

Step 1: Understand Kolbe's electrolysis reaction for sodium propionate. \[ \text{CH}_3\text{CH}_2\text{COONa} \xrightarrow[\text{electrolysis}]{} \text{CH}_3\text{CH}_2\cdot + \text{CO}_2 + e^- \] 
Step 2: Two ethyl radicals combine: \[ \text{CH}_3\text{CH}_2\cdot + \text{CH}_3\text{CH}_2\cdot \rightarrow \text{C}_4\text{H}_{10} \quad (\text{n-butane}) \] So, Statement-I is incorrect because the product is n-butane, not n-hexane. 
Step 3: Analyze Statement-II. In Kolbe's electrolysis:
- CO$_2$ is liberated at the anode
- H$_2$ is liberated at the cathode Hence, Statement-II is correct. 
But the selected correct answer is (3): Statement-I is correct, but statement-II is not correct, which is actually wrong. 
Therefore, a reevaluation: Correction: Statement-I is not correct, but Statement-II is correct. So the correct answer should actually be: 
Correct Answer: (4) Statement-I is not correct, but statement-II is correct