# Can oxidation and reduction be described in terms of transfer of oxygen in the following way?

After reading and thinking about oxidation and reduction a bit, I tried to come up with a way to explain it to myself… Could someone verify if this is technically correct? Consider 2 elements A and B where A is more electronegative than B. (Also O represents Oxygen) If in a reaction $$\ce{AO + B -> BO + A}$$ So assuming that O is more electronegative than both A and B In AO,

• Oxidation Number of O = -2
• Oxidation Number of A = +2

So we could say that O is “negatively charged” (or partially negative if it’s a covalent bond) while A is “positively charged” ( or partially positive if it’s covalent) and hence this is why they are attracted to each other (This part I’m not sure). Now given that A is more electronegative than B, when B comes near AO, A attracts electrons from B and oxidises B.

• Oxidation Number of A gets reduced to 0 (A is reduced)
• Oxidation Number of B increases to +2 (B is oxidised)

Now we could say that A is now “neutral” and B is “positively charged” while O remains “negatively charged”. Hence O is attracted to B forming BO. Thus $$\ce{AO + B -> BO + A}$$

From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation. Is this how redox reactions work or am I exaggerating everything?

• You may overthink it. Why not to stay at oxidation/reduction as change of oxidation numbers, what is not limited to oxygen presence? Dec 22, 2021 at 11:22

Your explanation reminds me of the historical definition of the words "oxidation" and "reduction". In the early 19th century, oxidation was the reaction an element with oxygen. The only thing that the chemist could measure with precision was the weight, the mass, before and after the reaction. When iron or copper gets oxidized, it produces oxides which are heavier than the element. Now when one of these oxides reacts with hydrogen, it produces the original metal, but its weight is reduced. So the chemist said that this oxide has been reduced in metal by hydrogen. The oxide has lost some mass to be transformed into the metal. Its mass is reduced. So the element is also "reduced".

• With the Latin root ducere (leading), one may understand reduction not only as action to diminish/to scale something down, yet equally to guide back/to return something into its earlier (elemental) state. Like education rooting in educere, to lead outwards. Dec 22, 2021 at 12:42
• I initially thought of reduction occurring when oxidation number got reduced to a lower value :) Dec 23, 2021 at 1:49
• @Unknown. You are right. But this is the present definition of reduction. In the time of Berzelius and Dalton, oxidation numbers were not known. Dec 23, 2021 at 9:33

An easier approach would be to think about redox-reactions as an exchange of electrons, rather than of oxygen. This is more general and equally allows oxygen to be oxidized (e.g., fluorine is even more electronegative than oxygen, as in oxygen fluorides).

Retain that an oxidation is the removal of electrons, and reduction is the addition of electrons. The «who gets the electrons» depends on the partners reacting with each other. In an allegory, think about an ox whose horns pierce and remove material from a tree (oxydation). And think of reduction as bringing electrons back (based on the Latin root reducere) to heal the wound.

You can find examples where your thoughts fit what is actually happening. For example, a reaction of magnesium oxide with barium (I don't know if this reaction would actually happen):

$$\ce{MgO(s) + Ba(s) -> Mg(s) + BaO(s)}$$

Here, it is correct to speak of the negative and positive charges because we are looking at ionic compounds. You could show the ions explicitly:

$$\ce{Mg^2+ + O^2- + Ba -> Mg + O^2- + Ba^2+}$$

On the other hand, you could have a redox reaction involving molecular compounds:

$$\ce{H2O + C -> H2 + CO}$$

Here it does not make sense to talk about charges of atoms in the compounds (partial charges maybe, but they are also not necessarily correlated with oxidation states). The easiest way to figure out oxidation states in this case is to define the oxidation state of oxygen as -2 (except when oxygen is bound to itself or to more electronegative atoms), and use that to assign +1 to hydrogen in water and +2 to carbon in carbon monoxide.

From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation?

No, unless this is the only change and the binding partner of the oxygen is less electronegative, as in your examples. Here is a counterexample:

$$\ce{F2 + O2 -> O2F2}$$

When you add water across a carbon-carbon double bond, this is usually not called oxidation either, even though one carbon gains a bond with oxygen and is oxidized. The reason it is not called oxidation is that at the same time, the other carbon gains a bond with hydrogen and is reduced.

Is this how redox reactions work or am I exaggerating everything?

This is how a subset of redox reactions work. In many redox reactions, there is no oxygen at all, e.g.:

$$\ce{Fe^2+ + Cu^2+ -> Fe^3+ + Cu+}$$

For a general modern definition of a redox reaction, we look at the transfer of electrons (as opposed to transfer of protons in acid/base reactions).

• Yes I do know and agree that electron transfer is an easier way to understand redox. However for exam purposes I am supposed to learn redox in terms of gain/loss in electron, oxygen and hydrogen. So given that I am talking only when gain in oxygen does cause oxidation, then is what I said applicable? I was specifically concerned with whether when B is oxidised when it gains an oxygen, it’s not the oxygen which gains the electrons but rather it’s A? And for reactions where a gain in oxygen does cause oxidation, is what I said above applicable or are some details too frivolous ? Thanks Dec 23, 2021 at 1:45
• For covalent substances, especially organic compounds, counting bonds with more electronegative atoms (oxygen) vs less electronegative (hydrogen) works. In counting, you have to count double bonds twice, though. Artificially, you are breaking all the bonds, giving both electrons in a bonded pair to the more electronegative atom.
– Karsten
Dec 23, 2021 at 3:36