Question

The compounds formed at the anode in the electrolysis of an aqueous solution of potassium acetate ,are __________________respectively.
A . ${C_2}{H_6}$ and $C{O_2}$
B . ${C_2}{H_4}$ and $C{O_2}$
C . $C{H_4}$ and ${H_2}$
D . $C{H_4}$ and $C{O_2}$

Answer 1

Hint : We know that potassium acetate is the potassium salt of the acetic acid which can be prepared with the help of strong base $KOH$. We have studied in electro-chemistry that ethane is formed at the anode in the electrolysis of an aqueous solution of potassium acetate. This process is an example of Kolbe electrolysis. Lets know about Kolbe’s electrolysis reaction.

Complete step by step solution:
The Kolbe reaction or Kolbe electrolysis is named after Hermann Kolbe. Kolbe's electrolysis involves the formation of symmetric hydrocarbons with the help of radicals generated from carboxylic acid at anode. To furnish unsymmetrical hydrocarbons this reaction can be performed with two carboxylic acids.
Thus the compound formed at the anode in the electrolysis of an aqueous solution of potassium acetate are ${C_2}{H_6}$ and $C{O_2}$.Let's take a look at the reaction of Kolbe’s electrolysis;
$2C{H_3}COOK + 2{H_2}O \to {C_2}{H_6} + 2C{O_2} + {H_2} + 2KOH$

At cathode - $2{K^ + } + 2{e^ - } \to 2K$
                                 $2{H_2}O + {e^ - } \to O{H^ - } + {H^ \bullet } \\$
                                 $2{H^ \bullet } \to {H_2} \uparrow \\$


At anode - $2C{H_3}CO{O^ - } \to 2C{H_3}CO{O^ \bullet } + 2{e^ - } \to {C_2}{H_6} + C{O_2}$

Hence Option A is the correct answer to this problem, that is ethane and carbon dioxide is formed at anode .Ethane is an organic compound which is a symmetric hydrocarbon.

Note : We have approached this problem with the help of Kolbe’s electrolysis reaction. We have learned that In Kolbe’s electrolysis the electrochemical oxidation of carboxylic acid and salt acid (Here in the problem it is potassium acetate) that leads to radicals, which dimerize. That means the mechanism of this reaction involves two stages. In the first stage electrochemical decarboxylation gives free radical intermediate and in the second stage it combines to form a covalent bond. Here $C{H_3}CO{O^ \bullet }$ is the radical which gives symmetric ethane and carbon dioxide.