$\ce{C2H4 + O2 → CO2 + H2O}$

I want to balance the above reaction using the ion-electron method.

My attempt:

Ox-half: $\ce{C2^(^-^2^)H4 + 4H2O → 2C^(^+^4^)O2 + 12H^+ +12e^-}$

Red-half: $\ce{2O2^(^0^) + 4H^+ + 8e^-→ CO2^(^-^2^) + 2H2O^(^-^2^)}$

After making the electrons gained and lost equal and adding them, and finally summing up, the carbon atoms and the oxygen atoms get unbalanced. How is it supposed to be balanced?

  • $\begingroup$ Your red-half itself is not balanced. All in all, this method is hardly ever used in organic chemistry, though still technically applicable. $\endgroup$ Sep 16, 2023 at 7:24
  • $\begingroup$ @IvanNeretin How would we balance it then? $\endgroup$ Sep 16, 2023 at 7:57
  • $\begingroup$ The way kids do. How did you balance reactions before you ever heard of the ion-electron method? $\endgroup$ Sep 16, 2023 at 8:41
  • 3
    $\begingroup$ Does this answer your question? Can all equations be balanced using ion-electron method? $\endgroup$
    – Mithoron
    Sep 16, 2023 at 13:23
  • 2
    $\begingroup$ Asking again about the same thing isn't a good idea... $\endgroup$
    – Mithoron
    Sep 16, 2023 at 13:24

1 Answer 1


The reduction half equation starts from $\ce{O2}$ and goes to $\ce{H2O}$, with the help of $\ce{H+}$. So it is something like $$\ce{O2 + .. H+ + .. e- -> 2 H2O}$$As $\ce{2 H2O}$ contains $4$ $ \ce{H}$ atoms, these atoms must be provided at left by $4$ $\ce{H+}$ ions, and, as a consequence by $4$ electrons. This makes the final reduction half-equation $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$ There is no carbon atom here.


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