Balancing redox equations with oxygens in multiple oxidation states

Question

How do you balance this using the ion-electron method;

$\ce{CrO5 + H2SO4 -> Cr2(SO4)3 + H2O + O2}$

The oxygen has multiple oxidation states in $\ce{CrO5}$, and none of the sites I looked this up on dealt with that.

Answer 1

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

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

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

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

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

and reducing to "lowest (whole number) terms"

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

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

Answer 2

Without taking oxidation numbers into account, I have first balanced the Cr atoms, then the H atoms, and finally the Oxygen atoms. And I have found :$$\ce{4CrO_5 + 12 H^+ -> 4 Cr^{3+} + 7 O_2 + 6 H_2O}$$ With the sulfate ions it gives : $$\ce{6 H_2SO_4 + 4 CrO_5 -> 2 Cr_2(SO_4)_3 + 7 O_2 + 6 H_2O}$$

Answer 3

I found a 2019 source work on Chromium peroxide. To quote the key preparation equation:

Chromium peroxide (Cr(O2)2.H2O or briefly CrO5) is an extremely potent oxidant. This compound is a product of the following reaction:

$\ce{(NH4)2Cr2O7 + 4 H2O2 + 2 H+ -> 2 Cr(O2)2.H2O + 4 H2O + ammonium salt}$ (1)

Here are associated corresponding comments from Wikipedia:

Chromate or dichromate reacts with hydrogen peroxide and an acid to give chromium peroxide and water.

$\ce{CrO4^{2−} + 2 H2O2 + 2 H+ → CrO5 + 3 H2O}$

After a few seconds, the chromium(VI) peroxide decomposes to turn green as chromium(III) compounds are formed.[1] To avoid this decomposition, it is possible to stabilize chromium(VI) oxide peroxide in a water-immiscible organic solvent such as diethyl ether, butan-1-ol or amyl acetate by adding a layer of the organic solvent above the chromate/dichromate solution and shaking during the addition of hydrogen peroxide. In this way, the chromium(VI) peroxide (unstable in the aqueous phase in which newly formed) is dissolved in the immiscible organic solvent. In this condition it can be observed over a much longer period.

$\ce{2 CrO5 + 7 H2O2 + 6 H+ → 2 Cr^{3+} + 10 H2O + 7 O2}$

So, in my opinion, due to instability, the reaction mechanics commences with chromium peroxide decomposition to $\ce{Cr^{3+}}$ and then the action of acid as follows:

$\ce{4 Cr(O2)2.H2O -> 2 Cr2O3 + 5 O2 + 4 H2O}$

$\ce{2 Cr^{3+} + 3 SO4^{2-} ⟶ Cr2(SO4)3}$

[EDIT] Some more comments per Wikipedia source above:

This compound contains one oxo ligand and two peroxo ligands, making a total of five oxygen atoms per chromium atom [...] Therefore, CrO5 is a good oxidizing agent (even better than chromates and dichromates due to the presence of two reactive peroxo ligands) but due to its low stability it is never used in organic syntheses.

So, one source claims one oxo ligand ($\ce{O^{2-}}$, but perhaps not reactive as the other 2019 source ignores it) and both agree on two reactive peroxo ligands, $\ce{O_2}{^{2−}}$. Based on the last statement on stability, and in particular, per both sources, stable in polar organic solvents (and not aqueous H2SO4), I suspect, the only valid path is likely multi-step, and further doubt that the reaction would ever be conducted in practice.