# 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:

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

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

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 $$\pu{10^{-5} M}$$).