Question

Sodium acetate was electrolysed by Kolbe's method to form two gases $A$ and $B$ at anode. $C$ and $D$ are formed, when $B$ is heated with regulated supply of $O_2$, or air in the presence of ${\left(C{H}_{3}COO\right)}_{2}Mn.C$ reacts with $NaOH$ to form a salt. $A$ and $D$ are respectively,

• A. $\mathrm{CO}{}_2,\mathrm{CH}{}_3\mathrm{COOH}$
• B. $\mathrm{CO}{}_2,\mathrm{H}{}_2\mathrm{O}$
• C. $\mathrm{C}{}_2\mathrm{H}{}_6,\mathrm{H}{}_2\mathrm{O}$
• D. $\mathrm{C}{}_2\mathrm{H}{}_2,\mathrm{H}{}_2\mathrm{O}{}_2$

Answer 1

Answer: (C)

Kolbe electrolysis of $C{H}_{3}CO{O}^{-}N{a}^{+}$.

At anode,



$C{H}_3^{•}+C{H}_3^{•}⟶\underset{(A)}{C{H}_3}⟶C{H}_3$

$2C{H}_3-C{H}_3+7O_2\stackrel{{\left(C{H}_{3}COO\right)}_{2}Mn}{⟶}4C{O}_2+\underset{(D)}{6H_{2}O}$

$C{O}_2+O_2⟶\underset{(C)}{\mathrm{CO}}+O_3$

$CO+NaOH\underset{\text{ (Salt })}{\rightleftharpoons }HCOONa$

Hence, $A$ is $C_2{H}_6$ and $D$ is $H_{2}O.$