Oxidation of Methanol - how to write the half reaction

Question

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?

Answer 1

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