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Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule. In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule. Does this mean an actual electron was transferred given that oxidation numbers are fictitious? I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

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    $\begingroup$ Yes, electron transfer can happen. Of course, in some cases, the molecule fragments, or there is a transfer of a fragment, which is considered to be "carrying" the electron. $\endgroup$ – Shoubhik R Maiti Apr 5 at 17:05
  • $\begingroup$ Usually a change in oxidation number is an electron transfer. But this transfer is not easy to discover. For example, the air oxidation of carbon to produce $\ce{CO2}$ is due to $$\ce{C +O2->CO2}$$ The electron transfer is not obvious in this case. It is usually forgotten. It may even be ignored, as nobody has ever taken the initiative of explaining the combustion of carbon by the following half-equations $$\ce{C + 2 H2O -> CO2 + 4 H+ + 4 e^-}$$ $$\ce{O2 + 4 H+ + 4 e^- -> 2 H2O}$$ whose sum is $\ce{C + O2->CO2}$ $\endgroup$ – Maurice Apr 6 at 12:55
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Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule.

Sort of. In calculating oxidation numbers we pretend that covalent bonds are ionic. Perhaps it is valid to say we are pretending the entire molecule is an ionic compound.

In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule.

More accurate, we find the that oxidation states of atoms change going from reactants to products. Charge conservation, coupled with the way we assign oxidation states, guarantees that the sum of oxidation states is constant. So if one atom shows an increase in oxidation state, at least one other has to show a decrease in oxidation state.

Does this mean an actual electron was transferred given that oxidation numbers are fictitious?

It is not clear what "an actual electron" means. Electrons are indistinguishable, you can't paint one red and follow its path during a reaction. If the electron transfer (according to oxidation numbers) is intramolecular, it becomes a bit of a philosophical question whether there was an electron transfer. For example, is there an electron transfer during keto-enol tautomerism?

I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

It is not clear what "physical electron jumps" are.

For some reactions, it is clear that you have an electron transfer because the only thing that distinguishes reactants from products (pairwise) is the number of electrons:

$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$

If this reaction is run as two half-cells, this is even more clear-cut.

For other reactions, the changes are more subtle. If a iodine radical combines with a methyl radical (as part of a chain termination reaction), it is formally a redox process:

$$\ce{I. + .CH3 -> I-CH3}$$

However, the electronegativities of iodine and carbon are quite similar, so it is not intuitive that an electron transfer happened. As Maurice points out in the comments for a similar case, you would have trouble writing this in terms of half-reactions.

In the reaction of methyl lithium with a carbonyl, the oxidation numbers change. You might not see it as a electron transfer and would rather think of it as methyl transfer. However, because carbon is more electronegative than lithium, the oxidation numbers of the carbonyl carbon and the methyl anion change when a bond is formed between them.

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Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?

If I answer this like a certain politician: it depends upon what the meaning of the phrase "actual electron transfer" is. So it is mostly a question about how we talk about redox reactions rather than a question about the underlying chemistry.

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The sum of oxidation numbers is equal to the difference between the total proton count and total electron count.

So yes, it means the real electron transfer, as oxidation means the increase of the sum of oxidation numbers and reduction the opposite.

It also means that while the summary oxidation number is the real thing, oxidation numbers of particular atoms are the arbitrary convention.

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Yes, the electron transfer does take place...oxidation is the loss of electrons and reduction is the gain of electrons. In a redox reaction, the oxidizing agent gains electrons and the reducing agent loses electrons, i.e, actual electron transfer takes place.

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