# How can I balance the following equation atomically and electrically?

$$\ce{C2O4^2- + MnO2 -> Mn^2+ + CO2}$$

I think that the half reactions are

$$\ce{C2O4^2- -> CO2}$$ $$\ce{MnO2 -> Mn^2+}$$

I am supposed to balance these by adding water, $\ce{H+}$ atoms and by adding $\ce{e-}$’s, but I’m just not sure on the method to do this as we’ve covered it extremely quickly.

First, I found the oxidation numbers for the overall equation, and I think that $\ce{C2O4^2-}$ is the reducing agent because $\ce{C}$ is losing charge from +3 to +4, I just don’t know how to use that to balance this. Any help would be appreciated.

Deal with the two half equations separately and then combine them.

Starting with the oxalate equation: $$\ce{C2O4^{2-} -> CO2}$$

Balance the atoms: $$\ce{C2O4^{2-} -> 2CO2}$$

Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -> CO2 + 2e-}$$

Now for the manganese dioxide reduction: $$\ce{MnO2 -> Mn^{2+}}$$

We can balance the half equations by adding water, hydrogen ions or electrons. Since water is the only one which contains oxygen we should add this first: $$\ce{MnO2 -> Mn^{2+} + 2H2O}$$

Now add hydrogen ions to balance the hydrogens: $$\ce{MnO2 +4H+ -> Mn^{2+} + 2H2O}$$

Finally add electrons the balance the charge: $$\ce{MnO2 +4H+ + 2e- -> Mn^{2+} + 2H2O}$$

Now combine the equations to cancel all the electrons. In this case there are two on both sides so we don't need to multiply any of the equations by anything.

$$\ce{MnO2 +4H+ + C2O4^{2-} -> Mn^{2+} + 2CO2 + 2H2O}$$

It might be a good idea to balance your half reactions first and then sum up:

\begin{align*} \ce{C2O4^{2-} &-> 2CO2 +2e-\\ MnO2 + 4H+ + 2e- &-> Mn^2+ + 2H2O}\\ \hline \ce{MnO2 + 4H+ + C2O4^{2-} &-> Mn^2+ + 2CO2 + 2H2O}\\ \end{align*}