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.
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$\begingroup$ In my opinion, this question itself is questionable in regard to its possible advancement of academics (accept, to perhaps, illustrate the limitation of current scientific knowledge and the state of academic education). $\endgroup$ – AJKOER Mar 25 '20 at 14:54
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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.
answered Mar 24 '20 at 12:42
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$\begingroup$ You just have to separate out the species that reacted and deal with them separately as usual after that, is that right? $\endgroup$ – Harry Holmes Mar 24 '20 at 13:23
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1$\begingroup$ Yes. Here the "separation" is part of the CrO5 got oxidized to make oxygen and the rest got reduced. $\endgroup$ – Oscar Lanzi Mar 24 '20 at 13:24
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$\begingroup$ But doesn't it go from +6 to +3 in both the cases? $\endgroup$ – Harry Holmes Mar 25 '20 at 1:42
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1$\begingroup$ The oxygen behaves differently, going up to 0 in one component but all of it down to -2 in the other. We need to look at both, so the CrO5, not just the Cr, is being oxidized or reduced. $\endgroup$ – Oscar Lanzi Mar 25 '20 at 9:48
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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 : $$\ce{6 H_2SO_4 + 4 CrO_5 -> 2 Cr_2(SO_4)_3 + 7 O_2 + 6 H_2O}$$
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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.
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$\begingroup$ @ Ajkoer. How can you say that Cr(O2)2.H2O is ... briefly CrO5 ? Two H atoms have disappeared. $\endgroup$ – Maurice Mar 24 '20 at 12:57
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$\begingroup$ I do not, that is the quote from the source paper, which implies, politely, that CrO5 is misleading shorthand, for the actual salt. Note the author use of this depiction of the salt in the key formation reaction which is likely more complex (radical intermediates, I suspect). $\endgroup$ – AJKOER Mar 24 '20 at 13:03
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1$\begingroup$ That being so, $\ce{CrO5}$ does exist. As a neat compound it is too unstable to be of any use, but when stabilized by complexation with a suitable electron donor such as pyridine is is an effective oxidizing agent in organic synthesis. $\endgroup$ – Oscar Lanzi Mar 24 '20 at 13:11
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$\begingroup$ Oscar: Do note per the Wikipedia comment on its stabilization with organics cites also "shaking during the addition of hydrogen peroxide", so perhaps Cr(O2)2⋅H2O2 shorthand for CrO5 seems speculatively more accurate. Also, the Wiki prep equation disagrees from the actual path, which apparently requires more H2O2. $\endgroup$ – AJKOER Mar 24 '20 at 13:34
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1$\begingroup$ "The structure of the pyridyl complex has been determined crystallographically.[3]" Looks like a CrO5 complex to me. $\endgroup$ – Oscar Lanzi Mar 24 '20 at 17:20
