# Oxidation - Reduction Ionic Half Reaction

Trying to figure out how to balance the following molecular equation:

$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$

I have compiled all of their oxidation states

$$\ce{KMnO4 + Na2C2O4 + H2SO4 -> MnSO4 + CO2 + Na2SO4 + H2O}$$

$$\ce{(+1)(+7)(-2 * 4) (+1 * 2)(+3 * 2)(-2 * 4) -> (+2)(-6)(-2 * 4) + (+4)(-2 * 2) + (+1 * 2)(+6)(-2 * 4) + (+1 * 2)(-2)}$$

Now I know that theoretically Mn changes from +7 to +2, so it gains 5 electrons. I also know Carbon in this situation goes from a state of +3 to +4, so it loses one electron.

However, after this I am drawing blanks on how to setup my ionic half reactions with these polyatomic ions, and I am suspicious that I may have set this up incorrectly. Any help at all would be appreciated, I can not find much to explain this type of equation with this particular molecular formula.

The idea of Redox reactions make sense when talking about individual atoms or more clearly depicted changes, but I am having trouble figuring out what is even changing into what in this example.

• Welcome to ChemSE. Be invited to familiarize yourself with formatting this site allows (especially when dealing with math, phys, chem), briefly outlined here: chemistry.meta.stackexchange.com/questions/86/… Secondly, both reaction equations are not yet balanced (two atoms of carbon on the left, but only one on the right; potassium vanishes entirely). Hence let me suggest you start formulating the balanced (atom- and charge-wise) half reactions, e.g. $\ce{MnO4^-}$ is reduced to $\ce{Mn^{2+}}$. Then put the equations together. – Buttonwood Jun 21 '17 at 0:28
• @Buttonwood I noticed the equation seemed a bit off with the Potassium vanishing and the lack of atomic balance, but I thought I couldn't derive the half reactions without the equation first being balanced. – Christopher Keenan Jun 21 '17 at 0:39   