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:
- 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 $(1)$?