Question

The gas liberated by the electrolysis of dipotassium succinate solution is/are :

• A. Ethene
• B. Ethyne
• C. $C{O}_2$
• D. $H_2$

Answer 1

Answer: (A), (C), (D)

$H_2$

During the electrolysis of potassium succinate, the anion produced is succinate anion. It gives one ethene molecule and $2C{O}_2$ gas. The reaction is as follows:

$(C{H}_{2}CO{O}^{-}{)}_2\to C{H}_2=C{H}_2+2C{O}_2(g)+2e^{-}$

At cathode, water molecule get reduce to produce $H_2$ gas as shown below:
$2H_{2}O+2e^{-}\to 2O{H}^{-}+H_2\left(g\right)$

$O{H}^{-}$ generated here further reacts with $K^{+}$ ion present near cathode to form KOH.
$2K^{+}+2O{H}^{-}\to 2KOH$

So, the gas generated during electrolysis of dipotassium succinate solution is ethene, $C{O}_2$ and $H_2$