Electrolysis of a concentrated aqueous solution of potassium fumarate (potassium (E)-but-2-enedioate) gives, at the anode, besides $\mathrm{CO}{}_2$, the hydrocarbon

Kolbe electrolysis removes both carboxylate groups as $\mathrm{CO}{}_2$ and joins the radicals left behind. For ${}_{}^{}{}_{}^{-}\mathrm{OOC}-\mathrm{CH}=\mathrm{CH}-\mathrm{COO}{}^{-}$ the two carbons that remain form $\mathrm{HC}\equiv \mathrm{CH}$. Potassium succinate would give ethene, and potassium acetate would give ethane.