fusso - 5-11-2019 at 18:54
MnO4 react with H2O2 to form O2
MnO4- + 5 H2O2 + 6 H+ = 2 Mn2+ + 5 O2 + 8 H2O
or
2 MnO4- + H2O2 = 2 OH- + 2 MnO2 + 2 O2
It is said that H2O2 is oxidised to O2 and MnO4- is reduced to Mn2+/MnO2.
This infers that the O in O2 comes from H2O2 and not MnO4-.
How do they know that the O2 isn't from MnO4-?
Why can't the O2 come from both species, or only from MnO4-?
What is the mechanism that proves that the O2 is from H2O2 and not MnO4-?
j_sum1 - 5-11-2019 at 22:09
These things can be tracked using radioisotopic marking. I am not sure if experiments have been done for this particular reaction.
They are also modelled by theory and by computer models. This rules out a number of seemingly-plausible reaction mechanisms and leaves us with the
ones that are actually possible.
woelen - 12-11-2019 at 07:05
This indeed can be experimentally verified, using special isotopes of oxygen.
One option is to use H2O2, which has 18O instead of 16O, while all other chemicals (the water, the acid and the KMnO4) have
normal oxygen. Then, the oxygen, which escapes in the reaction is tested for its 18O contents. If this is (nearly) 100% 18O,
then one knows that the oxygen comes from the H2O2.
Whether this particular reaction is tested in this way I do not know, but I know of another interesting reaction, which is tested in this way. That
reaction is the formation of the tetraperoxo complex of chromium(V) from chromate, H2O2 and hydroxide. In this test, H2O2, enriched in 18O
was used and both the resulting oxygen and the potassium tetraperoxochromate(V), formed in the reaction, had the same amount of enrichment.
This reaction can be written as
2 CrO4(2-) + 2 OH(-) + 9 H2O2 --> 2 CrO8(3-) + 10 H2O + O2
I made this compound myself, it is quite easy to make in a pure state.
But real understanding comes when this reaction is reformulated with the peroxo-group as ligand. Write hydrogen peroxide as H2Lig and the peroxo
ligand as the ion Lig(2-), and oxygen, formed by oxidation of Lig(2-), as Lig. Then the reaction can be reformulated as
2 CrO4(2-) + 2 OH(-) + 9 H2Lig --> 2 CrLig4(3-) + 10 H2O + Lig
Now it is clear that of each 9 ligands, 8 are attached to chromium and one of the ligands is used to reduce two hexavalent chromium cores to
pentavalent chromium cores.
[Edited on 12-11-19 by woelen]