In the reaction $\ce{C2O4^2- → CO2}$ the $n$-factor of $\ce{C2O4^2-}$ is:
So I think the answer to this question is $2$ because in $\ce{C2O4^2-}$ the oxidation number of carbon is $+3$ and in $\ce{CO2}$ the oxidation number of carbon is $+4$. Now according to the formula of calculating $n$-factor, i.e.:
$$|(\text{no. of atoms})\times(\text{change in oxidation no.}) + (\text{do the same for other atoms in the compound})|$$
So we get $n$-factor of $\ce{C2O4^2-} = |2\times(-1) + 4\times(0)| = 2$.
I guess I have used the right formula and got the right answer but still I think that my answer is wrong. Can somebody please clarify that is my answer wrong or right and if wrong why?