# Identifying the oxygen atoms that undergo reduction

Identify the oxygen atoms that undergo reduction in the reaction below:

$$\ce{O2 + Cr^3+ -> H2O2 + Cr2O7^2-}$$

I know that the oxygen atoms in $$\ce{O2}$$ have an oxidation state of $$0$$.

In the product $$\ce{H2O2}$$, the oxygen atoms have an oxidation state of $$-1$$; and in $$\ce{Cr2O7^2-}$$, oxygen atoms have an oxidation state of $$-2$$.

Based on how the problem was solved on the answer key it was assumed that all the oxygen atoms from $$\ce{O2}$$ are reduced to peroxide atoms.

How do we know that there are no oxygen atoms that are oxidized to become chromate oxygens with a $$-2$$ oxidation state?

I understand that there are other $$\ce{H2O}$$ oxygen atoms available to provide the chromate oxygens that would change the oxidation state.

• It would be nice if we started with a reaction that at least listed all of the component participating. Even nicer would be if that reaction were balanced.
– Zhe
Jan 24 '17 at 15:28

The oxygens in the dichromate ion on the right have -2 oxidation number, the chromium being +6. http://ch302.cm.utexas.edu/worksheets/basic-redox-wkst-KEY.pdf https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-the-dichromate-ion-cr2o7-2

As you seem to have deduced, the peroxide ions are -1 so all oxygens, as shown, have been reduced.

• This does not answer the question. How do you know that the chromate oxygens do not come from dioxygen?
– DHMO
Jan 25 '17 at 8:42