Question

In Kolbe’s electrolysis reaction:
This question has multiple correct options
a) \[{{H}_{2}},\,C{{O}_{2}}\] are liberated at cathode
b) Hydrocarbon, \[C{{O}_{2}}\] are liberated at anode
c) pH of the solution increases
d) aq. solution of sodium acetate forms ethane

Answer 1

Hint: In Kolbe’s reaction there is an electrochemical oxidative decarboxylation of carboxylic acid salts that leads to radicals, which dimerize. It is best applied to the synthesis of symmetrical dimers, in some cases can be used with a mixture of two carboxylic acids to furnish unsymmetrical dimers.

Complete step by step answer:
1. Suppose there is an electrolytic solution of sodium acetate (\[C{{H}_{3}}COONa\]) as it is a salt it will break into cation and anion on electrolysis and will form acetate ion (\[C{{H}_{3}}COO-\]) which will move towards the anode.
2. Now this sodium acetate ion will release an electron, forming a free radical which is unstable.
3. This free radical will try to stabilize itself, So, it takes an electron from the \[C{{H}_{3}}-C\] bond, therefore, forming \[C{{O}_{2}}\] which will be liberated at anode.
4. There will be formation of \[C{{H}_{3}}\] free radical, now let's take 2 molecules of sodium acetate ion from the beginning then two free radicals of \[C{{H}_{3}}\] will be formed.
5. Both of the free radicals will dimerize to form ethane, thereby a hydrocarbon is formed which will be liberated at the anode.
6. Simultaneously the \[N{{a}^{+}}\] ion will react with the \[{{H}_{2}}O\] forming, thus in this reaction NaOH is formed as a product so as reaction proceeds solution becomes alkaline and pH of solution increases.
The chemical reaction involved is:
\[2C{{H}_{3}}COONa\,+\,2{{H}_{2}}O\,\to C{{H}_{3}}C{{H}_{3}}\,+\,NaOH\,+C{{O}_{2}}\,+\,{{H}_{2}}\,\]

At Anode
\[\begin{align}
  & 2C{{H}_{3}}CO{{O}^{\bullet }}\to 2CH_{3}^{\bullet }+2C{{O}_{2}}(g) \\
 & 2CH_{3}^{\bullet }\to {{C}_{2}}{{H}_{6}}(g) \\
\end{align}\]
At Cathode
\[2N{{a}^{+}}+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}\]

Therefore, from the above statements we can conclude that the correct options are (b) and (c).

Note: There’s a difference between Kolbe’s electrolysis and Kolbe’s reaction:
When phenol is treated with sodium hydroxide, phenoxide ion is generated. This phenoxide ion generated is more reactive than phenol towards electrophilic aromatic substitution reaction. Hence it undergoes electrophilic substitution reaction with carbon dioxide which is a weak electrophile and Salicylic acid is formed as the major product. This reaction is popularly known as Kolbe’s reaction.