Can all equations be balanced using ion-electron method?

Question

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?

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) :

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