In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of ${CO}_2$ is

Reduction half reaction is

${MnO}_4{}^{-}+8H^{+}+5e^{-}⟶{Mn}^{2+}+4H_{2}O$

Oxidation state of manganese in reactant is $+7$ an in product is $+2$ net change is $+5$

Oxidation half reaction is

$C_2{O}_4{}^{2-}⟶{CO}_2+5e^{-}$

Multiply reduction half reaction with 2 and oxidation half reaction with 5 to get equal number of electrons, we get

$2{MnO}_4^{-}+16H^{+}+10e^{-}+5C_2{O}_4^{2-}⟶2{Mn}^{2+}+10{CO}_2+8H_{2}O$

There are total ten electrons involved in the reaction and produces 10 molecules of carbon dioxide.

One electron produces one molecule of carbon dioxide.