Question

In Kolbe's electrolysis of sodium propanoate, products formed at anode and cathode are respectively:

• A. \( C_2H_6, H_2 \)
• B. \( C_3H_8, H_2 \)
• C. \( C_4H_{10}, H_2 \)
• D. \( H_2, C_4H_{10} \)

Hint:

In Kolbe's electrolysis, the key product at the anode is an alkane formed by decarboxylation of carboxylate ions. The cathode always releases hydrogen gas in such reactions.

Answer 1

The Correct Option is C

Step 1: Understanding Kolbe's Electrolysis of Sodium Propanoate Kolbe's electrolysis is a type of electrolysis that involves the decarboxylation of carboxylate salts, such as sodium propanoate. In this process, at the anode, two carboxylate ions combine to form an alkane (a hydrocarbon) and carbon dioxide. At the cathode, hydrogen gas is liberated due to the reduction of protons (H+). The reaction at the anode is: \[ 2 \, RCOONa \rightarrow R-R + CO_2 \, (alkane) \] In this case, the carboxylate ion is derived from sodium propanoate (C2H5COONa), and two of these combine to form butane (C4H10) at the anode. Hydrogen gas (H2) is released at the cathode.
Step 2: Identifying Products at the Anode and Cathode Given that sodium propanoate (C2H5COONa) is used, the products formed at the anode and cathode will be: - At the anode: The two propanoate ions combine to form butane (C4H10).
- At the cathode: Hydrogen gas (H2) is released. Thus, the correct products are butane (C4H10) at the anode and hydrogen (H2) at the cathode.
Step 3: Conclusion The correct products are \( C_4H_{10} \) and \( H_2 \), which corresponds to option (3).