# In a combustion reaction, why is carbon in methane oxidised even though it starts and finishes with 8 electrons?

• In $\ce{CH4}$, carbon has 8 electrons due to 4 covalent bonds.
• In $\ce{CO2}$, carbon has 8 electrons due to 2 double bonds.

In both cases, carbon has 8 electrons. So why is carbon oxidised? Doesn't oxidation occur when electrons are lost?

The crux of my question is how to treat covalent bonds in redox reactions.

I know we can say carbon has gained oxygen so is oxidised, but I am after an answer in terms of the electron definition of reduction and oxidation.

• Oxygen is more electronegative than hydrogen and carbon , so it attracts the electron cloud more. – user14857 Oct 30 '16 at 20:06
• So then is every single reaction a redox reaction because one of the elements in a molecule (unless its a diatomic molecule with the same atoms) will be more electronegative than the other and will have the electrons slightly towards it? – K-Feldspar Oct 30 '16 at 20:09
• Actually you need to check the oxidation numbers for identifying redox reactions. In methane carbon has -4 oxidation number but in carbon dioxide it is +4. Increase in oxidation number means oxidation. – user14857 Oct 30 '16 at 20:14

In methane, carbon is more electronegative than hydrogen ($2.5 > 2.1$), so when setting up oxidation states, we treat methane as if it were composed of $\ce{C^4-}$ and $\ce{4 H+}$ — leading to the oxidation states of $\mathrm{-IV}$ on carbon and $\mathrm{+I}$ on hydrogen.
Likewise, oxygen is more electronegative than carbon ($2.5 < 3.5$), so in carbon dioxide we assume the electrons to be completely transferred over to oxygen; as if it were made up of $\ce{C^4+}$ and $\ce{2 O^2-}$ leading to the oxidation states of $\mathrm{+IV}$ and $\mathrm{-II}$, respectively.