# 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.

## 2 Answers

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*}