Can all equations be balanced using ion-electron method?

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?

• 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. Commented Sep 15, 2023 at 19:01
• 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). Commented Sep 15, 2023 at 19:04
• Well, technically this equation would be a sum of two equations. Commented Sep 15, 2023 at 19:46

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) :

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

• 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! Commented Sep 15, 2023 at 19:34
• One last question, why didn't you consider CO2 in product in reduction halves where oxygen gets reduced in both reactions? Commented Sep 15, 2023 at 19:40
• 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.
– MaxW
Commented Sep 16, 2023 at 3:51
• @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. Commented Sep 16, 2023 at 9:39