Balance the following reaction using the ion-electron method:

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

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 $(1)$?



1 Answer 1


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.


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