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Reaction:

$\ce{C2H4 + O2 → CO2 + H2O}$

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?

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

1 Answer 1

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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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