Should I write formula for oxalate or oxalic acid while balancing a redox reaction using the ion-electron method?

**Problem:**


Balance the following reaction using the [ion-electron method](https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Introduction_to_Inorganic_Chemistry_(Wikibook)/04%3A_Redox_Stability_and_Redox_Reactions/4.02%3A_Balancing_Redox_Reactions#:%7E:text=systematic%20method.%20The-,ion%2Delectron%20method,-allows%20one%20to):


$$\ce{H2C2O4 + KMnO4 + H2SO4 -> CO2 + MnSO4 + K2SO4 + H2O}$$


**My book's solution:**


Oxidation half-reaction:


$$\ce{C2O4^2- -> 2CO2 +2e^-}\tag{1}$$


$$...$$


**My question:**


1. Should oxalate ion $(\ce{C2O4^2-})$ be written in $(1)$? I'm asking this because oxalic acid $(\ce{H2C2O4})$ is a weak acid and doesn't dissociate/ionize much into $\ce{H+}$ and $\ce{C2O4^2-}$ ions. So, should I write $\ce{H2C2O4}$ instead of writing $\ce{C2O4^2-}$? To elucidate $$\ce{H2C2O4->2CO2 + 2H+ +2e^-}\tag{2}$$ Should I write $(2)$ instead of $(1)$?




---


[Related](https://chemistry.stackexchange.com/questions/165057/why-did-my-book-write-s2-if-hydrogen-sulfide-isnt-ionic)




One should report the species which is the most important one in the given system. As oxalic acid has two $\mathrm{p}K\_\mathrm{a}$ values ($1.23$, and $4.19$), it means that you should report the molecular formula $\ce{H2C2O4}$ if the pH is lower than $1.23$. You should write the ionic formula $\ce{HC2O4^{-}}$ is the pH is between $1.23$ and $4.19$. It is improbable that you should work at a pH greater than $4.19$, because the reaction requires $\ce{H2SO4}$ to proceed, and this acid is not supposed to be present at such a low concentration (about $\pu{10^{-5} M}$).


Please note that, whatever your choice, the final equation would be the same, as your equation has no ions.


 4