The 3CO has gained oxygen, so it undergoes oxidation. The CH3CH2OH loses hydrogen, therefore undergoes oxidation. The CH3CHO acquired from hydrogen then undergoes reduction. Finally, there is electron transfer, which I will focus on because this was what we were taught. The two topics some time ago were just examples of other oxidation-reduction reactions. In electron transfer, remember the same mnemonic “OIL RIG” because in electron transfer, oxidation is the loss of electrons, while, of course, reduction is a gain of electrons.[2] Example: [6]Cu2+ + Mg Cu + Mg2+B. Solving Oxidation Part of the Red-Ox Reaction is the Half-Reaction. It can be reduction or oxidation. Now we can say that the half-reaction has occurred when a change in the oxidation state occurs. Oxidation state/number is the number assigned to an element in a chemical combination that corresponds to the number of electrons lost, gained, if the number is negative, by an atom of that element in the compound. [7]So now, let's find out how to get the oxidation number of an element. Example: [8]Na2SO4How do we find S?