weiming1998
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Is this hypomanganate?
Today I was trying to reduce KMnO4 with aluminium powder under alkaline conditions (the KMnO4 was alkalised by the addition of NaOH). I found out that
if NaOH(aq) was slowly added to a solution of KMnO4 with a small amount of aluminium powder suspended in it, and the resulting mixture was left to
react for a few minutes, a very deep blue (almost black) is obtained. This solution can be diluted with water, resulting in a lighter (still darker
than say, a solution of CuSO4), transparent blue solution. With the addition of a few drops of H2SO4, this blue instantly changes to the bright purple
of the MnO4- ion.
I suspect that the dark blue comes from the hypomanganate ion formed by the reduction of permanganate ions, but aqueous hypomanganates quickly
decompose in any solution other than very alkaline ones, while this solution was stable even when diluted down by water. It does quickly form
permanganate when acidified, though.
So, is the dark blue due to the hypomanganate ion, or is it due to other Mn compounds?
[Edited on 1-3-2013 by weiming1998]
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DerAlte
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@Weiming
It is very unlikely that hypomanganate is present under the conditions you describe. See the article I wrote and posted as http://www.sciencemadness.org/scipics/MnOXY.doc - specifically, look at page 3 for the Pourbaix (E0/pH) diagram. Hypomangagate is only
stable under extreme alkali conditions, such as fuzed hydroxide. What you are seeing is a combination of manganate (green) and permanganate ions
(purple, magenta) which together block out just about all but blue frequencies from the transmitted spectrum.
Hypomanagate ion is a very light blue color, not dark blue. Have a scout through the Permanganate thread, especially for a picure by Xenoid there
which shows the color well
.
Also look at Brauer, in the SM library, for a description of hypomanagante production. Made that way, it is only stable at low temperatures. It is
tricky but can be done, with care.
Der Alte
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blogfast25
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DA:
What's the OS of this supposed hypomanganate to be? Mn +II?
[Edited on 1-3-2013 by blogfast25]
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DraconicAcid
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MnO4(3-) will be Mn(V).
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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DerAlte
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Yes, As Draconic says. MnO4- = permanganate or Mn(VII); manganate MnO4(2-) or Mn (VI); Hypo = MnO4(3-), or Mn (V);
Some authorities also recognize that a brown colored ion MnO4(4-), Mn(IV), manganite, has a possibility of existence In the old Weldon process for
chlorine production, rather than the MnO3(2-). Certainly weird brown solutions are not uncommon when you mess around with Mn compounds!
Der Alte
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woelen
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The dark blue/green color is due to manganate, which contains manganese in oxidation state +6. This ion is fairly stable in solutions from pH equal to
13 and above, at lower pH it slowly disproportionates to MnO2 and MnO4(-). At acidic pH, the MnO4(2-) ion disproportionates very quickly.
Permanganate, on the other hand, is not stable at very high pH. Even without reductor, MnO4(-) ion decomposes in very strongly alkaline solutions,
such as 6 M NaOH. The solution then turns dark blue/green.
Here is a page of my website, containing pics of solutions of manganese at different oxidation states:
http://woelen.homescience.net/science/chem/solutions/mn.html
I also tried making the blue MnO4(3-) ion, but never succeeded in making this. So, indeed, this must be a very unstable ion which apparently cannot
exist in aqueous solution. It most likely immediately disproportionates to MnO2 and dark green/blue MnO4(2-).
[Edited on 1-3-13 by woelen]
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weiming1998
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The blue colour I observed was probably just a mixture of the MnO4- and MnO4(2-) ion, because I just added a small amount of MnO4- into a green
solution of manganate. It turned the exact same dark blue colour that was observed yesterday.
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TheGinginator
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I've done some work with assorted manganates, and one of my observations about Manganate(VI) is that it appears bluish under incandescent light.
Manganate (V) is a very bright blue, so what you're describing is almost certainly not that.
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