Consider the reaction of a mixture of $\ce{C2H4}$ and $\ce{CH4}$ with $\ce{O2}$. Using the hit-and-trial method, this is the balanced chemical reaction:

$\ce{C2H4 + CH4 + 5 O2 -> 3 CO2 + 4 H2O}$

The above reaction can be easily balanced using hit-and-trial or the conventional arbitrary coefficient method. However, I attempted to balance it using the ion-electron method, but it seems like it cannot be balanced using that method.

My attempt:

Ox-half: $\ce{2 C2H4 + 2 CH4 -> 6 CO2 + 40 e}$

Red-half: $\ce{10 O2 + 40 e -> 5 CO2 + 10 H2O}$

However, when we add these, we end up with this, which is obviously incorrect:

$\ce{2 C2H4 + 2 CH4 + 10 O2 + 40 e -> 11 CO2 + 10 H2O + 40 e}$

Why does this happen?

  • $\begingroup$ For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations. $\endgroup$
    – Buttonwood
    Sep 15 at 19:01
  • 1
    $\begingroup$ Though the reactions are now easier to read, please revise the reactions. In a balanced reaction equation, the number of atoms on the left to the arrow equates the number of atoms on the right of arrow; the same applies to charges (see especially the half reaction labeled as oxidation). $\endgroup$
    – Buttonwood
    Sep 15 at 19:04
  • $\begingroup$ Well, technically this equation would be a sum of two equations. $\endgroup$ Sep 15 at 19:46

1 Answer 1


Your equations are wrong for many reasons.

First, $\ce{CH4}$ and $\ce{C2H4}$ do not react together, even if you would like them to react. Both react with $\ce{O2}$ and both produce $\ce{CO2}$ and $\ce{H2O}$.

Second, $\ce{CH4}$ and $\ce{C2H4}$ react independently with $\ce{O2}$. These reactions occur independently. So it is nonsense to introduce both $\ce{CH4}$ and $\ce{C2H4}$ together in the same equation. The amounts of $\ce{CH4}$ and of $\ce{C2H4}$ reacting with $\ce{O2}$ may be different and are not related to one another. Both oxidation equations are : $$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$ $$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O \tag{2}}$$ Pedagocially speaking, these equations can be decomposed into half-equations. For the oxidation of $\ce{CH4}$ in $(1)$, one gets : $$\ce{CH4 + 2 H2O -> CO2 + 8 H+ + 8 e-} \tag{3}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O} \tag{4}$$ Doubling the second equation $(4)$ and adding it to $(3)$ produces $(1)$ $$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$ For the oxidation of $\ce{C2H4}$ as in $(2)$, the half-equations are $$\ce{C2H4 + 4 H2O -> 2 CO2 + 12 H+ + 12 e-} \tag{5}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O} \tag{6}$$ so that multiplying the second equation $(6)$ by $3$ and adding it to the previous one $(5)$ gives (2) :

$$\ce{C2H4 + 3 O2 -> 2 CO2 + 2H2O} \tag{2}$$

  • $\begingroup$ I apologize for any inconvenience, and I appreciate your patience. I will be deleting this question as I believe it may not contribute to the understanding of concepts for others. Thank you for the clarification! $\endgroup$ Sep 15 at 19:34
  • $\begingroup$ One last question, why didn't you consider CO2 in product in reduction halves where oxygen gets reduced in both reactions? $\endgroup$ Sep 15 at 19:40
  • 1
    $\begingroup$ Since you're trying to be exact, I'll point out that you're giving the "overall" reactions. Oxidation of the hydrocarbons doesn't occur in one step. It involves numerous intermediate species. $\endgroup$
    – MaxW
    Sep 16 at 3:51
  • $\begingroup$ @MaxW Thank you for your observation. You are right. The given equations should be decomposed into a lot of successive equations. For example, half-equations (3) and (5) are the result of a lot of intermediate steps. Half-equations like (3) and (5) have only pedagogical values. $\endgroup$
    – Maurice
    Sep 16 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.