Redox Reaction when a compound is both oxidized and reduced

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Homework Statement
Is OCl- both reduced and oxidized in the reaction between sodium hypochlorite and hydrogen peroxide?
Relevant Equations
Redox reactions
For the reaction between hydrogen peroxide and sodium hypochlorite, is the OCl- both reduced and oxidized?

H2O2 + NaOCl -> NaCl + H2O + O2
Oxidation states:
+1 -1 +1 -2 +1 +1 -1 +1 -2 0
From the oxidation states, it seems that Cl is reduced from +1 to -1, and O (of NaOCl) is oxidized from -2 to 0.

On the contrary, the half reactions given by the solution are:
2 e- + 2H+ + OCl- -> Cl- + H2O
H2O2 -> O2 + 2H+ + 2 e-
In this case, the OCl- is reduced only.

What is the real case? Thanks.
 
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So do you ask about what happens to Cl, O, or OCl-? Net effect on the whole is a sum of effects on the individual atoms.

(There are many reasons why this way of approaching this case is wrong, but they will probably confuse you even more).
 
I think my question was stated clearly enough: OCl-. Thanks!
 
Oxidation state is a property of a single atom! Doesn't really make sense to ask whether ##\mathrm{ClO^-}## is oxidised or reduced.
 
Compounds (in this case, e.g. NaOCl) can be reduced or oxidized in a redox reaction - I didn't ask about oxidation states. Please correct me if I'm wrong.
 
I don't think so, compounds can be oxidising or reducing agents, but you only talk about individual atoms being be oxidised or reduced.

The oxidation number is the resulting charge on the atom if all bonds in the structure are (for accounting purposes) split heterolytically with both electrons in each pair assigned to the more electronegative atom in each case. So an oxidation number is a property of a single atom in the structure. [But of course by construction, the total charge on a molecule equals the total of the oxidation numbers].
 
etotheipi said:
I don't think so, compounds can be oxidising or reducing agents, but you only talk about individual atoms being be oxidised or reduced.

This is tricky. You can say when MnO42- gets oxidized to MnO4- it is Mn that gets oxidized, but that's an oversimplification stemming from the accounting based on oxidation number, and not following the orbital reality.
 
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Borek said:
This is tricky. You can say when MnO42- gets oxidized to MnO4- it is Mn that gets oxidized, but that's an oversimplification stemming from the accounting based on oxidation number, and not following the orbital reality.

Yeah, as far as I'm aware the oxidation number formalism is just that - a formalism - and very often won't accurately depict the distribution of charge around the molecule.
 
OK. So the answer is?

BTW. The way I see it, this statement:

"...and O (of NaOCl) is oxidized from -2 to 0."

... is clearly wrong. What is the oxidation state of O in H2O? It's not 0, is it?
 
You are right! Of course it is possible to have the following two reactions
##\mathrm{OCl^- \rightarrow Cl^- + \frac{1}{2} O_2}## and
##\mathrm{H_2O_2 \rightarrow \frac{1}{2} O_2 + H_2O}##,
which somehow autocatalyse each other.
A possibility to distinguish the two possibilities is to use isotopically marked oxygen in e.g. H2O2.