[Evaldas]

Hi, how do I balance this
KMnO₄ + C₃H₈O₃  K₂CO₃ + Mn₂O₃ + CO₂ + H₂O
equation using the electron balance method?
First, please tell me the oxidation states of atoms in glycerol, because C^-4, O^-2, H⁺ do not work...

[Borek]

Not sure what you mean by electron balance method. Of two methods I know and use one uses oxidation numbers (and apparently that's what you are aiming at), the other is the half reaction method.

Assigning oxidation numbers to carbon in organic compounds can be tricky. 8*1+3*(-2)=2 - for the glycerol molecule to be neutral carbons have to be -2/3 on average. Yes, it doesn't make sense, no, there are no better values. You can try to assign separate ON to different carbons, but it is not worth the effort, as the same reaction can be balanced much faster and much easier with half reactions.

[Evaldas]

Ok, so:
C^-2/3-14/3e^- C⁺⁴ |x3
Mn⁺⁷+e^- 14Mn⁺⁶ |x14
3C^-2/3-14e^-+14Mn⁺⁷+14e^- 3C⁺⁴+14Mn⁺⁶
Then:
14KMnO₄+?C₃H₈O₃ 7K₂CO₃+7Mn₂O₃+?CO₂+?H₂O
The three next to C is useless, how do I balance the carbon atoms?

[Borek]

What a mess.

C^-2/3-14/3e^- C⁺⁴ |x3

You don't have separate C^-2/3 on the LHS, if anything, it is (C^-2/3)₃.

Mn⁺⁷+e^- 14Mn⁺⁶ |x14

14² = 196. Why Mn⁶⁺? It should be Mn³⁺, or even better 2Mn³⁺ as that's how they occur in the sesquioxide.

Once these things are sorted out you should have a correct KMnO₄/C₃H₈O₃ ratio.

[Evaldas]

Well, to be perfectly honest you can't write (C^-2/3)₃ in a half reaction, it's supposed to be 3C^-2/3. Talk about a mess.
Sorry, when I was rewriting the reaction on paper, I wrote down MnO₃ instead of Mn₂O₃
And I didn't mean to write 14 next to Mn, so I would get 14x14=196.

3C^-2/3-14e^- 3C⁺⁴ | x4
Mn⁺⁷+4e^- Mn⁺³ | x14
12C^-2/3-56e^-+14Mn⁺⁷+56e^- 12C⁺⁴ + 14Mn⁺³
14KMnO₄ + 4C₃H₈O₃  7K₂CO₃ + 7Mn₂O₃ + 5CO₂ + 16H₂O