# Finding the n-factor of C₂O₄²⁻

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?

• You may be using a cannon against sparrows. What happens if you cut away the net charge and cut the rest in 2 pieces ? How many electrons is there, aside of the product. ? Apr 18, 2019 at 12:18
• @poutnik Hi thanks for responding to my question but I am not able to understand what you are trying to say. Apr 18, 2019 at 16:16
• I mean, why to count or even use a formula, when you must see it. $\ce{C2O4^2- -> 2 CO2 + 2e-}$ Apr 18, 2019 at 16:22
• @Poutnik Oh... I had to use that formula because that was the one taught to me. But overall my answer is correct, right? Apr 18, 2019 at 16:26
• Well there are basically 2 methods to know, how many apples were taken from the box. 1/ Count apples at the start and at the end, and then calculate the difference. 2/ Count just the taken apples. 1/ is the summing the differences of oxidation numbers. 2/ is counting the exchanged electrons. BTW, I have not studed chemistry in English, so I may have a gap, but I have not heart about the term "n-factor". Apr 18, 2019 at 17:22

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):

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

$$n-\text{factor} =2$$