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?

  • 1
    $\begingroup$ 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. ? $\endgroup$ – Poutnik Apr 18 at 12:18
  • $\begingroup$ @poutnik Hi thanks for responding to my question but I am not able to understand what you are trying to say. $\endgroup$ – gucci Apr 18 at 16:16
  • 2
    $\begingroup$ I mean, why to count or even use a formula, when you must see it. $\ce{C2O4^2- -> 2 CO2 + 2e-}$ $\endgroup$ – Poutnik Apr 18 at 16:22
  • $\begingroup$ @Poutnik Oh... I had to use that formula because that was the one taught to me. But overall my answer is correct, right? $\endgroup$ – gucci Apr 18 at 16:26
  • $\begingroup$ 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". $\endgroup$ – Poutnik Apr 18 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.