Question

In the Kolbe electrolysis of sodium propanoate, the products X and Y are formed at the respective electrodes. What are X and Y?

• A. \( \text{X} = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) at Cathode; \( \text{Y} = \text{H}_2 \) at Anode
• B. \( \text{X} = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) at Anode; \( \text{Y} = \text{H}_2 \) at Cathode
• C. \( \text{X} = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) at Anode; \( \text{Y} = \text{H}_2 \) at Anode
• D. \( \text{X} = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) at Cathode; \( \text{Y} = \text{H}_2 \) at Cathode

Hint:

In Kolbe electrolysis, remember that alkanes are produced at the anode, and hydrogen gas is produced at the cathode. This is a crucial part of the reaction mechanism.

Answer 1

The Correct Option is C

In Kolbe electrolysis, the reaction involves the breaking of a carboxylate ion, followed by the production of the alkane. The process involves the formation of two molecules of hydrocarbons at the anode as a result of two electrons being released per molecule. In this case, sodium propanoate undergoes electrolysis to form butane at the anode and hydrogen at the cathode.
Hence, the product at the anode is the alkane (butane), and the product at the cathode is hydrogen gas.
Thus, the correct products are:
- \( X = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) at Anode
- \( Y = \text{H}_2 \) at Cathode The correct option is \( \boxed{3} \).

Answer 2

Step 1: Understand Kolbe electrolysis.
Kolbe electrolysis is an electrolytic process where the sodium salt of a carboxylic acid (carboxylate ion) undergoes decarboxylation at the anode, producing hydrocarbons by coupling of alkyl radicals.
At the cathode, typically hydrogen gas is evolved from the reduction of water or protons.

Step 2: Reaction mechanism at the anode.
- The carboxylate ion (\(\text{RCOO}^-\)) loses an electron at the anode to form a carboxyl radical (\(\text{RCOO}^\bullet\)).
- The carboxyl radical rapidly loses \(\text{CO}_2\) to form an alkyl radical (\(\text{R}^\bullet\)).
- Two alkyl radicals combine (dimerize) to form an alkane (\(\text{R-R}\)).
- This process releases two electrons per molecule of alkane formed.

Step 3: Application to sodium propanoate.
- Sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)) is the sodium salt of propanoic acid.
- During Kolbe electrolysis, two propyl radicals (\(\text{CH}_3\text{CH}_2\text{CH}_2^\bullet\)) are formed.
- These combine to form butane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)) at the anode.

Step 4: Reaction at the cathode.
- At the cathode, water or protons gain electrons and are reduced to hydrogen gas (\(\text{H}_2\)).

Step 5: Overall reaction and products.
- At Anode: Formation of butane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)) by coupling of propyl radicals.
- At Cathode: Evolution of hydrogen gas (\(\text{H}_2\)).

Step 6: Summary and correct option.
Thus, the products are:
- \( X = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) (butane) at Anode,
- \( Y = \text{H}_2 \) (hydrogen gas) at Cathode.
The correct answer is \(\boxed{3}\).