woelen
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Riddle: adding reductor and acid to dichromate, order of addition makes big difference
I want to share the result of a simple experiment with you, with a remarkable outcome (at least for me). I would like comments and suggestions on the
solution of this riddle.
http://woelen.homescience.net/science/chem/riddles/dichromat...
Simply changing the order in which reductor and acid are added to a dichromate solution can make a big difference.
[Edited on 11-6-08 by woelen]
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[Edited on 7.1.14 by bfesser]
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pantone159
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A kind of stupid question... I assume that 'hydroxyl ammonium sulfate' and 'hydroxylamine sulfate' are the same thing? I have heard the second name
more than the first.
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woelen
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Yes, both are the same. I'll change the webpage and make mention of the other name as well. Thanks for that remark.
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Klute
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Maybe the Cr III formed some kind of complex with hydroxylamine? What was the pH of the solution before adding the acid?
Maybe as the redox reaction consumed acid, the pH got high enough to liberate hydroxylamine freebase, whcih formed a somewhat stable complex with the
Cr3+, more stable than the hydroxylamine salt: the complex doesn't get destroyed when adding acid?
Looks like you've discovered a pretty unusual reaction indeed! It would be interezsting to see if there is any mention of this in the litterature..
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Jor
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Maybe a very inert complex of Cr(VI) is formed. if this is true, maybe no CrO5 is formed, and thus no blue color. If that is true, it could be
possible it is not reduced in acidic environment, because the complex is too inert. Probably this complex would not form when acid is first added,
because the NH3OH+ would react immediatly, before it could even form the complex. Just a guess. What happens if you add some NaOH-solution to the left
test-tube, instead of adding H2SO4 ?
What happens if you try to reduce a yellow potassium chromate (alkaline) solution with NH3OH+ (wich would turn in NH2OH in alkaline environment) ?
Because if there is actually a oxidation of NH3OH+ , when no acid is added, the solution must have become alkaline. And because no CrO5 is detected,
the reaction must have gone to completion. So if K2CrO4 does not react with hydroxylamine, then the reaction products of the reduction must somehow
catalyse further reduction of Cr(VI).
It is possible I'm talking rubbish, I have not done any uni yet, just high school, so not yet familiar yet with the real 'rules' of ligand-chemistry.
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kmno4
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Possibly, this brown-yellow "thing" is some nitrosyl complex(es) of Cr(III). Hydroxylamine is known to generate such complexes (it depends on pH) and
is using for making such complexes on preparative scale. Unfortunately, propably the most important reference articles are unavailable in electronic
form (some Russian journal). Few less interesting articles, can be found on SD (for example: Inorganica Chimica Acta, Volume 51,
1981, Pages 123-130)
[Edited on 13-6-2008 by kmno4]
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Nicodem
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In organic chemistry the addition order, especially in oxidations, is very important since reversing the addition order often gives very different
products compositions. This is particularly true for fast reactions.
Woelen, the IUPAC chemical nomenclature does not allow names based on substitution to be separated by a blank so "hydroxyl ammonium" should be either
hydroxylammonium (more common) or hydroxyammonium (correct but not used). So "hydroxyl ammonium sulfate" is correctly "hydroxylammonium sulfate"
though "hydroxylamine sulfate" is for some reason more commonly used name.
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woelen
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Nicodem, thanks for your comment. I removed the blank, according to IUPAC nomenclature.
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kmno4
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| Quote: | Originally posted by kmno4
Possibly, this brown-yellow "thing" is some nitrosyl complex(es) of Cr(III).
(...) |
I have just found (by chance) an article from ACS titled "Electron Spin Relaxation in Chromium-Nitrosyl Complexes"
(doi : 10.1021/ic981063+)
, where "A new method to prepare Cr(NO)(H2O)5
2+ from dichromate and NH2OH is reported."
It seems that this "thing" is really nitrosyl complex of Cr(III).
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