# Oxidation of Methanol - how to write the half reaction

I am trying to write the half reaction equations for:

$$\ce{CH3OH + O2 -> HCOOH + H2O}$$

The primary goal for me here is to find out which reactant is oxidising without using prior knowledge of alcohol reactions.

So I wrote

$$\ce{CH3OH -> HCOOH}$$

and balanced the non hydrogens and non oxygens. I then added $$\ce{H2O}$$ and $$\ce{O}$$ to the sides appropriately I calculated the charge of the alcohol by adding up the individual ionic charges of each atom in the molecule Each side ended up with the equal charge of -2. However I had no need to add any electrons into the equation to balance it out.

So how can I figure out whether it is oxidising if it doesn't indicate electron transfer? Is it some fundamental mistake in my method? Also, do alcohols have 0 net charge when dissolved in water?

This problem is not as easy as it looks, because there are two sorts of oxygen atoms. Some change their oxidation numbers, and some not. The first half-reaction is an oxidation of the reducing agent $$\ce{CH3OH}$$ which produces $$\ce{HCOOH}$$ on the right-hand side. The second half-reaction is a reduction of the oxidizing agent $$\ce{O2}$$ which produces $$\ce{H2O}$$ on the right-hand side. The two half-equations are : $$\ce{CH3OH + H2O -> HCOOH + 4 H+ + 4 e-}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$If you make the sum of these two half-equations, the $$4$$ e- and the $$4$$ $$\ce{H+}$$ disappear, and the final equation is : $$\ce{CH3OH + O2 -> HCOOH + H2O}$$