During the electrolysis of potassium succinate, the anion produced is succinate anion.
It gives one ethene molecule and $2 CO _{2}$ gas.
The reaction is as follows:
$\left( CH _{2} COO ^{-}\right)_{2} \rightarrow CH _{2}= CH _{2}+2 CO _{2}( g )+2 e ^{-}$
At cathode, water molecule get reduce to produce $H _{2}$ gas as shown below:
$2 H _{2} O +2 e ^{-} \rightarrow 2 OH ^{-}+ H _{2}( g )$
$OH ^{-}$generated here further reacts with $K ^{+}$ion present near cathode to form $KOH$.
$2 K ^{+}+2 OH ^{-} \rightarrow 2 KOH$
So, the electrolysis of potassium succinate produces $CO _{2}$ as well as $H _{2}$ gas.