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 \(\mathrm{O}_2\) or air in the presence of \(\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Mn}\). \(\mathrm{C}\) reacts with \(\mathrm{NaOH}\) to form a salt. \(A\) and \(D\) are respectively,

• A

  \(\mathrm{CO}_2, \mathrm{CH}_3 \mathrm{COOH}\)
• B

  \(\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}\)
• Correct answer
  C

  \(\mathrm{C}_2 \mathrm{H}_6, \mathrm{H}_2 \mathrm{O}\)
• D

  \(\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}_2\)

Solution

Kolbe electrolysis of \(\mathrm{CH}_3 \mathrm{COO}^{-} \mathrm{Na}^{+}\). At anode,
\(\begin{aligned} & \mathrm{CH}_3^*+\mathrm{CH}_3^* \longrightarrow \mathrm{CH}_3 \underset{(\mathrm{A})}{-\mathrm{CH}_3} \\ & 2 \mathrm{CH}_3-\mathrm{CH}_3+7 \mathrm{O}_2 \stackrel{\left(\mathrm{CH}_3 \mathrm{COO}_2 \mathrm{Mn}\right.}{\longrightarrow} 4 \mathrm{CO}_2+\underset{\mathrm{(D)}}{6 \mathrm{H}_2 \mathrm{O}} \end{aligned}\) \(\mathrm{CO}_2+\mathrm{O}_2 \longrightarrow \underset{\mathrm{(C)}}{\mathrm{CO}}+\mathrm{O}_3\) \(\mathrm{CO}+\mathrm{NaOH} \rightleftharpoons \underset{\text{(Salt)}}{\mathrm{HCOONa}}\) Hence, \(A\) is \(\mathrm{C}_2 \mathrm{H}_6\) and \(D\) is \(\mathrm{H}_2 \mathrm{O}\).

Hence, the correct answer is option: 'C'.