# What happens if an element has two different oxidation numbers on the same side of reaction?

I am taking General Chemistry 1 and I am studying redox reactions. I was looking at the following combustion reaction: $$\ce{4CH3 + \underset{O: 0}{7 O2} \to \underset{O: -2}{4CO2} + \underset{O: -2}{6H2O}}$$ Under the reaction, I am labeling how the oxidation numbers for each oxygen atom. I noticed how the oxidation state on the left hand side of the reaction for O changes from 0 to -2 on the right hand side, even though there are now two sets of oxygen atoms in two different molecules.

What would happen if—disregarding the fact that O always has an oxidation of -2—the first molecule on the right hand side happened to have an O atom with an oxidation number of -2, but the second molecule happened to have an O with an oxidation number of -1? Is a situation like this possible in any reaction, with an element different than O? If so, how could the general change in oxidation number be described from the left hand side to the right hand side of the reaction?

EDIT: I found a reaction that details what I'm confused about: $$\ce{2 \underset{N: +5;O: -2}{N2O5}(g) \to 4 \underset{N: +4; O: -2}{NO2} (g) + \underset{O: 0}{O2} (g)}$$ Here, it is obvious that nitrogen is reduced from an oxidation number of $$+5$$ to an oxidation number of $$+4$$, but how would you describe how the oxidation number of oxygen changes? On one molecule, its oxidation number stays the same, but on another, the oxidation number has gone up—has O been oxidized, or has it not changed? (I have now written this edit as a separate question).

First things first: your equation presumably is looking at the combustion of methane, and that should be $$\ce{CH4}$$, not $$\ce{CH3}$$. With that in mind, you might want to re-balance your equation.

Next, O doesn't always have an oxidation state of −2: peroxides, for example, exist, where oxygen is in an oxidation state of −1. That's actually useful to know, because I'll use it as an example later.

Lastly, to answer the actual question: sure, there are plenty of cases where you can't just describe the changes as one oxidation state to another oxidation state. I guess, you're asking specifically about something like:

$$\ce{A}^{(x)} + \ce{A}^{(y)} + \ce{O} \longrightarrow \ce{A}^{(z)}$$

where $$\ce{A}$$ is some element, $$\ce{O}$$ is some oxidising agent, parenthesised superscripts denote oxidation state, and $$x < y < z$$. (This isn't standard notation, so if someone wants to fix it please be my guest.) Off the top of my head, I can't think of anything that has this pattern;* but it's fairly common to have a reaction where $$x < z < y$$, i.e. the oxidation states get 'averaged' out. In this case you don't need another oxidising agent, and this has a special name: a comproportionation reaction, and essentially features the element $$\ce{A}$$ being both the oxidising and reducing agent at the same time. The Wikipedia page provides several examples.

The reverse reaction,

$$\ce{A}^{(z)} \longrightarrow \ce{A}^{(x)} + \ce{A}^{(y)}$$

with $$x < z < y$$ is a disproportionation reaction, the simplest being the decomposition of hydrogen peroxide:

$$\ce{2H2O2 -> 2H2O + O2}.$$

* The reason is probably because if you had a reaction like that, you could just split it into the sum of two redox reactions, one where you have the oxidation state change $$x \to z$$, and one with $$y \to z$$. In both cases this is a simple oxidation of the element $$\ce{A}$$.

• I copied this equation from one of my chemistry lectures on redox reactions, which was covering combustion. Maybe I copied the molecule wrong, but why is it "wrong?" if I have a $\ce{CH3}$ there? Feb 2, 2023 at 4:27
• $\ce{CH3}$ does not exist as a usable substance. It is an intermediate entity that does not exist longer than $1$ microsecond. You cannot fill a bottle with $\ce{CH3}$. It will immediately dimerize and become ethane $\ce{C2H6}$. Feb 2, 2023 at 9:13
• How does it become ethane? Feb 2, 2023 at 21:34

First it is a moot point if CH3 is a molecule or not. Think of it as the empirical formula of ethane C2H6. In the compounds CO2 and H2O oxygen has the oxidation number of -2; you seem to have understood that.

The decomposition of N2O5 requires that nitrogen gain an electron; this electron comes from oxygen the oxygen eventually is oxidized to O2 molecules, oxidation number zero [it is elemental oxygen]. The combined oxygen in NO2 is still minus 2 oxidation number, it has yet to react. NO2 can be decomposed further to N2 and O2 via a series of reactions. Write out the final equation.

The O2 molecule can react in steps to give intermediate negative oxidation numbers such as the superoxide ion, O2-, oxidation number -1/2 and the peroxide ion O2[minus 2], ox. no. minus one, but these are special compounds that are reactive.

The answer is part of the oxygen has been oxidized, as you can see from your oxidation state labels. We often see part of an element change oxidation state and the rest nit change or even change differently.

Such situations are handled by identifying whole molecules or ions as oxidizing or reducing agents. Here we may regard all of $$\ce{N2O5}$$ as an oxidizing agent, which when it takes electrons loses a virtual oxide ion:

$$\ce{N2O5 +2e^- -> 2NO2 + O^{2-}}$$

This ion is then the effective reducing agent, giving off the electrons that the nitrogen atoms in the $$\ce{N2O5}$$ capture.

$$\ce{2O^{2-} -> O2 +4e^-}$$

Thus the oxygen that is displaced from bonding with nitrogen is what gets oxidized; the oxygen that remains bound to nitrogen is redox-inert.

Sometimes acids can exist in dehydrated form.

SO3 is dehydrated H2SO4. N2O5 is dehydrated 2 HNO3.

$$\ce{N2O5 + H2O -> 2 HNO3}$$

Now we see 3 Oxygens at -2, Nitrogen at +5, and Hydrogen at +1. What this is saying is that Oxygen has a greater affinity for electrons than Nitrogen or Hydrogen.

HNO3 takes 1 electron away from Silver:

$$\ce{2 (H +1 N +5 O3 -6) + Ag 0 -> Ag +1(N +5 O3 -6) + N +4 O2 -4 + H2 +2 O-2}$$

and gives it to one of the NO3- moieties, because NO3- has a higher affinity for electrons than Silver.

We can write this reaction "short hand" as:

$$\ce{2 HNO3 + Ag -> AgNO3 + NO2 + H2O}$$.

How would you describe the way the oxidation number of Oxygen changes?

$$\ce{2 N2O5 <-> 4 NO2 + O2}$$

It makes perfect sense. 2 Oxygens went from -2 to 0. 4 Nitrogens went from +5 to +4. That can only mean that, in spite of convention, N+5 (as NO3-) has a greater affinity for electrons than Oxygen.

It then becomes important to see where the equilibrium of the stated reaction lies. Studying redox potential charts can be helpful in further understanding electron affinity beyond individual atoms.