Oxidation in organic chem is slightly different to oxidation in inorganic....in organic chem, oxidation may be considered as the gain of oxygen by a compound / loss of hydrogen, while reduction is gain of hydrogen/loss of oxygen.
Hence in this example the ethanol is oxidised to form acetic acid (gain of oxygen by ethanol), while the water can be said to reduced as it loses oxygen.
If you want something more formal, then you can look at electronegativities (x) in organic chem to assign oxidation numbers, since covalent bonds are considered to be bonds where electrons are shared - ie while there is electron density on the carbons (x=2.1) connected to oxygen (x = 3.5), it can nonetheless be treated as if the carbon atom gave up the electron pair to oxygen.
So, in assigning oxidation states:
for a C-O bond, we assign C the oxidation state of +1, O of -1
for a C-H bond, we assign C the oxidation state of -1, H of +1, as H has x=2.1, C has x=2.5
I'm not entirely sure how to assign the oxidation number for the CH3-C bond, but we'll assume the carbon joined to the OH functional group is more electronegative (either way works)...so for this carbon, it changes from oxidation state of -1-1+1-1=-2, to +2(for the double bond) +1 -1 =2, which accounts for your 4e- out the other end.
Obviously, this is simply a half reaction, and so you'd need an oxidising reagent to cause this reaction to occur - dichromates, chromic oxide, potassium permanganate are all useful here.
BTW, we don't need any of this
