# How do you write the balanced half equation for the oxidation of methanal (formaldehyde), CH_2O to CO_2?

Sep 4, 2016

${H}_{2} C = O + {H}_{2} O \rightarrow C {O}_{2} + 4 {H}^{+} + 4 {e}^{-}$ $i .$

#### Explanation:

Carbon is oxidized from ${C}^{0}$ to ${C}^{I V}$.

Where do the electrons go? To oxygen:

${O}_{2} + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$ $i i .$

We add (i) and (ii) to get:

${H}_{2} C = O + {H}_{2} O + {O}_{2} + \cancel{4 {e}^{-}} \rightarrow 2 {O}^{2 -} + C {O}_{2} + 4 {H}^{+} + \cancel{4 {e}^{-}}$

Now on the right side we have $2 {O}^{2 -} + 4 {H}^{+} \equiv 2 {H}_{2} O$

Thus:

${H}_{2} C = O + \cancel{{H}_{2} O} + {O}_{2} \rightarrow C {O}_{2} + \cancel{2} {H}_{2} O$

To give (finally!):

${H}_{2} C = O + {O}_{2} \rightarrow C {O}_{2} + {H}_{2} O$

As the balanced redox equation.

Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons