Number of alkanes obtained on electrolysis of a mixture of CH₃ COONa and C₂ H₅ COONa is ________.

Solution

Chemistry ma’am · 3 steps

Kolbe electrolysis makes radicals, and any two radicals can pair up, including a mixed pair.

1. Kolbe electrolysis. At the anode each carboxylate ion loses an electron and then CO₂, leaving an alkyl radical. Two radicals then join to give an alkane.

   RCOO⁻ → RCOO• + e⁻, RCOO• → R• + CO₂

2. Radicals formed. CH₃COO⁻ gives CH₃•, and C₂H₅COO⁻ gives C₂H₅•. Both radicals are present at the anode together.

   Radicals that join	Alkane
   CH₃• + CH₃•	CH₃-CH₃, ethane
   CH₃• + C₂H₅•	CH₃-C₂H₅, propane
   C₂H₅• + C₂H₅•	C₂H₅-C₂H₅, butane

3. So. Ethane, propane and butane form, so the number of alkanes is 3.

The common slip: 2. Pairing each radical only with itself gives ethane and butane, which is 2. That misses the cross product, propane.

The idea to keep. For Kolbe electrolysis of a mixture of two salts, count every pairing of the radicals: two like pairs and one cross pair. Two different radicals always give 3 alkanes.

Answer from the NTA final answer key. Solution by PrepWiser.