A) Ethane
B) Ethyne
C) Ethene
D) Ethane-1, 2-diol
Correct Answer: C
Solution :
\[\begin{array}{*{35}{l}} \underset{|}{\mathop{C}}\,{{H}_{2}}COOK \\ C{{H}_{2}}COOK \\ \end{array}+2{{H}_{2}}O\xrightarrow{\text{Electrolysis}}\]\[\underset{\text{Anode}}{\mathop{\underset{\,C{{H}_{2}}\,\,\,}{\mathop{\underset{||\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,}}\,\,+2C{{O}_{2}}+2KOH\,\,\,\,+\underset{\begin{matrix} {} \\ \text{Cathode} \\ \end{matrix}}{\mathop{{{H}_{2}}}}\,\]
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