Finding the n-factor of C₂O₄²⁻


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> In the reaction $\ce{C2O4^2- → CO2}$ the $n$-factor of $\ce{C2O4^2-}$ is:
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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?




By using formula:


$\ce{\overset{(+3)}{C}\_2O4^2- ->[oxidation]~\overset{(+4)}{C}O2}$


$n$-factor $=$ change in oxidation number per $\ce{C}$ atom $\times$ total number of $\ce{C}$ atoms in $\ce{C2O4^2-}$


$=1 \times 2=2$


By using logic (as Poutnik suggested in comment):


$\ce{C2O4^2- ->[oxidation]~2CO2 + 2e-}$


$n-\text{factor} =2$


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