STD 11 - 9. Hydricarbon — NEET STD 11 Science — Question

$\begin{array}{*{20}{l}}
  {CH} \\ 
  {|\,|\,|} \\ 
  {\underbrace {CH}_{Anode}} 
\end{array} + 2\,C{O_2} + 2\,KOH + \underbrace {{H_2}}_{{\text{Cathode}}}$

$Ca{C_2} + 2\,{H_2}O \to Ca{\left( {OH} \right)_2} + CH \equiv CH$

$\,\mathop {\mathop {C{H_2}}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{Br\,\,\,\,\,}  - \mathop {C{H_2}}\limits_{\mathop {|\,\,\,\,\,\,\,\,\,}\limits_{Br\,\,\,\,} }  + 2\,KOH \to CH \equiv CH + 2\,KBr + 2\,{H_2}O$