Consider a combustion reaction like the complete combustion of ethane. Here is what I have for the complete combustion as well as the oxidation and reduction:

$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$

$$ \begin{align} \ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\ \ce{\frac{1}{2} O2 + 10 H+ + 10 e- &-> 5 H2O}\tag{red} \end{align} $$

  • Why is it that when writing the reduction half equation for the reduction of oxygen that we don't consider oxygen being reduced into carbon dioxide? i.e. why isn't the reduction equation something of the form $\ce{O2 -> H2O + CO2}~?$

  • Also the oxidation reaction suggests that water is required initially for the reaction to proceed, is this true?


I realise my reduction half equation is not balanced and so I now have this equation:

$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$

$$ \begin{align} \ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\ \ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red} \end{align} $$

which is clearly not correct as the reduction and oxidation half equations do not add together to give the combustion reaction and the reduction equation does not have the $\ce{CO2}$ product as well. I guess my question now stands as how might I write (or begin to write) the half equations for a combustion reaction?

Edit 2

I realise I wrote my equations wrong above $$ \begin{align} \ce{4 H2O + C2H6 &-> 2 CO2 + 14 H+ + 14 e-}\tag{ox}\\ \ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red} \end{align} $$

But my question still remains. Why do we not consider the reduction of $\ce{O2}$ into $\ce{CO2}$ as well?

  • 3
    $\begingroup$ Don't use this redox approach for balancing organic reactions. It will make miserable. The approach is not only wrong, but it is confusing you for unnecessary reasons. Use simple algebra or do trial guesses for practice. If a teacher is using this way, please tell him/her that this is not a good way. $\endgroup$
    – AChem
    Apr 19, 2019 at 3:46
  • 3
    $\begingroup$ I'm not sure what you mean by simple algebra or trial guesses. I don't quite understand what I have done wrong. Could you please elaborate $\endgroup$
    – sab hoque
    Apr 19, 2019 at 5:19
  • 1
    $\begingroup$ Consider the fact that O(0) as in O2 is being reduced to O(-2) as in H2O and CO2 in the reduction half reaction. $\endgroup$
    – glucose
    Apr 19, 2019 at 5:40
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    $\begingroup$ I will have to strongly back M. Farooq's comment. It is really not advisable to use redox approach in organic reactions. $\endgroup$ Apr 19, 2019 at 17:13
  • 2
    $\begingroup$ Note that near all redox reactions can be expressed as redox half reactions with related standard redox potentials, which are equivalent to the Gibbs energy of the half reaction, oxidizing the hydrogen. But many of them have only formal significance in electrochemistry, as such reactions due slow kinetics may not be ever observable and inert electrodes like Pt may not measure such a potential. $\endgroup$
    – Poutnik
    Apr 19, 2019 at 17:37

1 Answer 1


Your redox equations look awkward (and wrong, too) because you are trying apply the half-reactions method as if it were happening in aquatic media to the combustion reaction, which effectively occurs in the gas phase. Of course, there should be no requirements for water to be present. As it's been noted in the comments, ionic half-reactions method is not the most elegant way of find coefficients in this case.

If you really want to use it, I'd suggest to write the elements undergoing redox reactions separately from the corresponding compounds (and yes, we do consider oxygen being reduced into carbon dioxide too):

$$ \begin{align} \ce{2 \overset{-3}{C} &→ 2 \overset{+4}{C} + 14 e-} &|\cdot 2 \tag{ox} \\ \ce{\overset{0}{O}_2 + 4 e- &→ 2 \overset{-2}{O}} &|\cdot 7 \tag{red} \\ \hline \ce{4 \overset{-3}{C} + 7 \overset{0}{O}_2 &→ 4 \overset{+4}{C} + 14 \overset{-2}{O}} \end{align} $$

resulting in a balanced equation:

$$\ce{2 C2H6 + 7 O2 → 4 CO2 + 6 H2O}$$

Note that in general there is an average oxidation number of carbon involved, and it might as well be fractional.

  • 2
    $\begingroup$ This is a very interesting approach that I have not seen before. It seems to suggest that both reactants atomize and undergo their respective half reactions or are redox half equations generally not an indicator of reaction mechanisms? $\endgroup$
    – sab hoque
    Apr 20, 2019 at 1:39
  • 2
    $\begingroup$ @sabhoque No, half-reactions method is a simplistic construct helping to illustrate electron transfer during the redox process (and to balance the reaction along the way) and has little to do with the mechanism. Whereas one would include oxo- and hydroxocomplexes in "typical" red+ox half-reactions (e.g. $\ce{SO4^2-}$ or $\ce{[Cr(OH)6]^3-}$) occurring in aqueous phase as the units that do participate in the process, the same cannot be done in the case of combustion as there are tons of radicals generated, and it's overall easier to track oxidation numbers of the elements instead. $\endgroup$
    – andselisk
    Apr 20, 2019 at 1:52

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