Identifying the oxygen atoms that undergo reduction

Question

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.

Answer 1

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.