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What is the purpose of calculating an average oxidation state for a certain element in a compound?

Hence the “average oxidation state” of carbon in acetic acid is 0.

And

the oxidation state value for carbon atoms in acetic acid is, however, computed as zero. This latter value represents an average of the values for the two carbon atoms derived by the rigorous assignment rules.

What does this tell me? Why not just stick with individual oxidation states? Is there something the overall oxidation state is supposed to tell me about acetic acid?

The only reason I can think of is that perhaps this average oxidation state of 0 for the carbon atoms tells me that acetic acid has some ionic character. Hence the oxidation state of 0 for carbon (as opposed to +4 for carbon in its elemental form).

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Ionic half-reactions method is represented here:

enter image description here

enter image description here

Then, as the same number of electrons is transfered in both processes, you just sum these equations to get the balanced reaction. If the number of transfered electrons wasn't the same, then you need to find the greatest common denominator and multiply each equation with a number so the number of transfered electrons gets equal to the greatest common denominator. Then you sum the equations.

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It helps you when you want to balance a red-ox reaction by the mean of the oxidation number change. For example, if you want to balance the reaction of thiosulfate with iodine to tetrathionate and iodide, you realize that the average oxidation number of sulfur changed from +2 to +2.5. The oxidation half-reaction looks like this: enter image description here

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    $\begingroup$ still, it gives some strange ideas without definitive simplification. So the practice is questionable. $\endgroup$ – permeakra Aug 10 '14 at 16:33
  • $\begingroup$ Oxidation number per se is an invented model to simplify some problems and explain some phenomena. The usage of the oxidation number change method or the ionic half-reactions method (which I prefer more) is up to the person. $\endgroup$ – Marko Aug 10 '14 at 16:41
  • $\begingroup$ What's the ionic half-reactions method? $\endgroup$ – Dissenter Aug 11 '14 at 2:55

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