Specifying which compound is oxidized/reduced in a redox reaction [closed]








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I wrote the oxidation numbers in a redox reaction:


$\rm{2 \overset{+7}{Mn} \overset{-2}{O\_4^{-}}~+~5\overset{+4}{S} \overset{-2}{O\_2}~+~6\overset{+1}{H}\_2 \overset{-2}{O}~~\rightarrow~~ 5\overset{+6}{S}\overset{-2}{O\_4^{2-}}~+~2\overset{+2}{Mn^{2+}}~+~4\overset{+1}{H}\_3 \overset{-2}{O^{+}}}$


It is clear that, S is oxidized and Mn is reduced.


But the text says that:
$\rm{SO\_2}$ is oxidized and $\rm{MnO\_4^{-}}$ is reduced.


It is easy to see that S suffers oxidation. But writing "$\rm{SO\_2}$" as a whole seems to take our concentration away from S.


Is there any rule by which we can write the "compound as a whole" which is oxidized/reduced?




Both concepts can be accepted. Some chemists state that Sulfur atom is oxidized and the Manganese atom is reduced. Some other chemists may state that $\ce{SO2}$ is oxidized, and permanganate ion is reduced. These two concepts are not contradictory : both can be accepted and considered as valid.


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If you can separate the redox reaction into a reduction and an oxidation half reaction, you can say that the entire molecular entity undergoes reduction or oxidation.



> 
> $\rm{2 \overset{+7}{Mn} \overset{-2}{O\_4^{-}}~+~5\overset{+4}{S} \overset{-2}{O\_2}~+~6\overset{+1}{H}\_2 \overset{-2}{O}~~\rightarrow~~ 5\overset{+6}{S}\overset{-2}{O\_4^{2-}}~+~2\overset{+2}{Mn^{2+}}~+~4\overset{+1}{H}\_3 \overset{-2}{O^{+}}}$
> 
> 
> 


can be obtained from the two half-reactions


$$\ce{MnO4- + 5 e- + 8H3O+ -> Mn^2+ + 12 H2O}$$


and


$$\ce{SO2 + 6 H2O -> SO4^2- + 4H3O+ + 2e-}$$


We know that hydronium and water are acid/base pairs, so they don't show changes in oxidation state. By elimination, the electrons are coming from the sulfur dioxide and are going to the permanganate.


It is not always trivial to separate a redox reaction into oxidation and reduction half reaction, for example if one part of a molecule gets oxidized and another gets reduced. In those cases, it is best to assign oxidation states (atom by atom) and specify the atom that gets reduced or oxidized.


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Chemical reactions are reactions of MOLECULES. A useful definition being the identity of a chemical species in the system being studied. Oxidation numbers are an artifact invented to simplify certain types of reactions. The oxidation number of an atom is dependent on the molecular structure it is in. In the case of SO2 or MnO4- the molecule or molecular ion is reduced or oxidized and our current understanding of atomic and molecular structure allows the simplification that most of the change involves the sulfur or manganese atoms. This is manifested in assigning constant oxidation numbers to oxygen[-2] and hydrogen[+1] atoms in most compounds.


This idea carries over to Organic chemistry where the reactions of certain functional groups are to a large extent independent of the bulk of the molecule. As one studies and practices chemistry One learns that there are nuances and variations.


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Thank you for the comments and replies.
Based on the comments and replies, I thought about five simple steps which may be applicable to each species in any redox reaction.


Applying the steps to $\rm{SO\_2}$:


1. The partners are S and O.
2. During the reaction, S loses electrons.
3. During the reaction, the other partner O neither loses nor gains electrons.
4. So the net effect is that, $\rm{SO\_2}$ loses electrons.
5. Consequently, we write: $\rm{SO\_2}$ is oxidized.


Applying the steps to $\rm{MnO\_4^{-}}$:


1. The partners are Mn and O.
2. During the reaction, Mn gains electrons.
3. During the reaction, the other partner O neither loses nor gains electrons.
4. So the net effect is that, $\rm{MnO\_4^{-}}$ gains electrons.
5. Consequently, we write: $\rm{MnO\_4^{-}}$ is reduced.


Are these five steps valid? Can we apply these steps to any redox reaction?


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