Balancing redox equations with oxygens in multiple oxidation states
This is a rather unusual case of what is discussed in answers like this one, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents.

Here, the whole-compound redox-active material is $\ce{CrO5}$, and as in peroxide disproportionations generally this is both an oxidizing agent and a reducing agent.  We thereby render

$\ce{CrO5 -&gt; Cr^{3+} + (5/2) O2 + 3 e^-}$

$\ce{CrO5 + 10H^+ + 7 e^- -&gt; Cr^{3+} + 5 H2O}$

We then apply the usual method of multiplying the first reaction by $7$ and the second one by $3$ to balance the electrons leading to

$\ce{10 CrO5 + 30 H^+ -&gt; 10 Cr^{3+} + (35/2) O2 + 15 H2O}$

and reducing to "lowest (whole number) terms"

$\ce{4 CrO5 + 12 H^+ -&gt; 4 Cr^{3+} + 7 O2 + 6 H2O}$

All that remains now is to add the spectator ions if desired, and we're done.