Question

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

• A. ethene
• B. ethane
• C. ethyne
• D. propene

Answer 1

Answer: (A)

ethene

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 electrolysis of potassium succinate produces $C{O}_2$ as well as $H_2$ gas.

Option A is the correct answer.