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I realise this is a rather pedantic question, but I'm wondering which application of terminology is more correct here.

I know that, in the complete combustion of methane, methane is the reducing agent. But I also know the oxidation state of hydrogen doesn't change in this reaction.

If we were asked to identify what is oxidised in this reaction, could we simply say that methane is oxidised? Or would we have to be more specific and say that 'carbon atoms in methane' are oxidised? Would either be fine?

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    $\begingroup$ As for me, either would be fine. $\endgroup$
    – Ivan Neretin
    Commented Jan 25, 2022 at 22:30
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    $\begingroup$ Since the four hydrogen atoms combine with oxygen, I would not say only the carbon is oxidized. $\endgroup$
    – DrMoishe Pippik
    Commented Jan 25, 2022 at 23:47
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    $\begingroup$ @DrMoishePippik Well, formally, H in CH4 already IS oxidized. By partial oxidation $\ce{CH4 + O2 -> C + 2 H2O}$, only carbon is oxidized. Similarly, by replacing oxygen by fluorine, hydrogen is not oxidized either. // But I do agree that factually, considering homolytic cleavage and radical mechanism, rather both C and H are oxidized by full combustion. $\endgroup$
    – Poutnik
    Commented Jan 26, 2022 at 6:43

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In the $19$th century, oxidation numbers were not known. The oxidation was the reaction of a molecule or an atom with $\ce{O2}$. Every chemist would say that, when burning, methane is oxidized into $\ce{CO2 + H2O}$. When studying this reaction, nobody was ready to consider that only one atom could be oxidized. Oxidation was the reaction of the whole molecule. There are still people today using the word "oxidation" with this definition.

Of course today the notion of oxidation has been extended. An oxidation is not necessarily a reaction of $\ce{O2}$. And oxidation numbers have been introduced. I will not develop this well-known question here. Of course when burning methane, the Carbon atom is oxidized.

But I would like to add here a personal observation. In my country, there are two sorts of high school teachers. Most are teaching that in the $\ce{C-H}$ bond, the $\ce{H}$ atom is at the oxidation number $+1$. But some are teaching that the difference in electronegativity between carbon and hydrogen is so small that the $\ce{C-H}$ bond may be considered as non-polar, so that in hydrocarbons, both $\ce{C}$ and $\ce{H}$ atoms are at the oxidation number zero. In a hydrocarbon, all atoms are at the oxidation number zero. For these teachers, both $\ce{C}$ and $\ce{H}$ atoms are oxidized in the methane combustion.

I would be interested to hear what you think of this strange situation.

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Oxidation numbers can be very arbitrary and are a tool, not a be all and end all, describing a redox reaction. The C-H bond, in alkanes especially, is particularly nonpolar compared to the resultant O-H and C=O bonds in H2O and CO2. So if you examine the equation and the products the conclusion [to me at least] is that both are oxidized. [I would prefer an ox. no. of zero for both.] For a more polar molecule such as the oxidation of aqueous HCl to Cl2 and water by permanganate the question doesn't arise; H is +1 all along.

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