Wirte the half equation for the reduction of acetic acid.

I Know that a colourless odourless gas is formed in the process but I can't seem to find an appropriate product to accompany it. Pleasae help.

I think I would look for carbon dioxide.

So would the half equation be

CH3COOH -> CO2 + CH4

and if so, where is the oxidation/reduction - electron transfer?

I don't know about the products; however, for what you have written the redox is like this.

In CH3COOH:
O is -2 each for total -4.
H is +1 each for total +4.
Therefore each C = 0
In CO2, O is -2 and C is +4. Carbon has gone from 0 to +4 which is a loss of electrons which makes it oxidized.
In CH4, C is -4; C has gone from 0 to -4 which is a gain of electrons which makes it reduced.

To determine the half equation for the reduction of acetic acid, we need to know the chemical formula of acetic acid, which is CH3COOH. Reducing agents commonly used for the reduction of carboxylic acids include lithium aluminum hydride (LiAlH4) or sodium borohydride (NaBH4).

If we consider the reduction of acetic acid using LiAlH4, here is the half equation:

CH3COOH + 4H+ + 4e- -> CH3CH2OH + 2H2O

In this equation, acetic acid (CH3COOH) is reduced to ethanol (CH3CH2OH) while consuming four hydrogen ions (H+), and four electrons (e-). The hydrogen ions are often provided by an acid, such as sulfuric acid (H2SO4) or hydrochloric acid (HCl).

The product accompanying the reduction of acetic acid is ethanol (CH3CH2OH). It is a colorless, odourless liquid. The reaction generates two molecules of water (H2O) as a by-product.

Please note that if you are referring to a different reduction process or conditions, the reaction and product might be different.