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 $\left( CH _{3} COO \right)_{2} Mn . C$ reacts with $NaOH$ to form a salt. $A$ and $D$ are respectively,

• A. $\ce{CO_2, CH_3COOH}$
• B. $\ce{CO_2, H_2O}$
• C. $\ce{C_2H_6,H_2O}$
• D. $\ce{C_2H_2,H_2O_2}$

Answer 1

Answer: (C)

Kolbe electrolysis of $CH _{3} COO ^{-} Na ^{+}$.

At anode,



$CH _{3}^{•}+ CH _{3}^{•} \longrightarrow \underset{(A)}{C H _{3}} \longrightarrow CH _{3}$

$2 CH _{3}- CH _{3}+7 O _{2} \stackrel{\left( CH _{3} COO \right)_{2} Mn }{\longrightarrow } 4 CO _{2}+\underset{(D)}{6 H _{2} O}$

$CO _{2}+ O _{2} \longrightarrow \underset{(C)}{ CO }+ O _{3}$

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

Hence, $A$ is $C _{2} H _{6}$ and $D$ is $H _{2} O .$