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[*] posted on 31-10-2008 at 07:46


A few more experiments.

I made a larger amount of the homemade MnO2/boiling 50 % H2SO4 mixture (3 g oxide + 50 ml acid). The pic below gives a better view of the unknown green compound (but photographed it looks more blueish than green):




I then tried to wash it to get rid of the H2SO4 but on adding water the compound hydrolyses to a brown compound. Lighter than freshly precipitated MnO2, more like Fe2O3. Could be Mn<sub>2</sub>O<sub>3</sub> or maybe Mn<sub>3</sub>O<sub>4</sub>:



So I've not been able to confirm/infirm the reaction of the green compound with HCl.


Taking a small amount of this suspension and adding 32 % HCl and it dissolves instantly to this w/o chlorine development - MnCl3?:




Heated it clears up with generation of chlorine:




What's disappointing is that the brown compound does not react at all with 1 M H2SO4, not even on heating. I tried twice with different ratios. I will try again with a slightly more concentrated H2SO4, perhaps 25 w%, instead of 1 M. If the stuff doesn't react with dilute H2SO4 then it can't be Mn2O3 or Mn3O4. For now I'm assuming it is (either) but that I'm not testing properly.


One other test I carried out was to add NaClO<sub>3</sub> (chlorate, sat.) to the purple liquid. No colour change was noted. It makes me think that the darkening I observed yesterday when adding bleach may simply have been due to the chloride content of the bleach: Mn (+III) species + 3 Cl<sup>-</sup> ---> MnCl<sub>3</sub>.

Summarising so far we've got (in all likelihood):

Homemade MnO<sub>2</sub> or unknown Mn oxide + 50 % H<sub>2</sub>SO<sub>4</sub> + heat---> soluble purple Mn (+III) species + unknown, insoluble green Mn-based species

Unknown, insoluble green Mn-based species (in 50 w % H<sub>2</sub>SO<sub>4</sub>;) + water ---> Mn<sub>2</sub>O<sub>3</sub> or Mn<sub>3</sub>O<sub>4</sub>



[Edited on 31-10-2008 by blogfast25]
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[*] posted on 31-10-2008 at 14:03


I have done some counter-experiments, starting from MnSO4 instead of MnO2. All chemicals used in this experiments are lab reagent grade chemicals.


Add MnSO4.H2O to conc. H2SO4. The solid does not dissolve. When the liquid is heated, still the solid does not dissolve.
Let cool down again and add some water, such that the acid is 60% or so. Then heat again (do not add water to hot H2SO4!!). Now the MnSO4.H2O dissolves. The liquid becomes colorless.

Add some KBrO3 to the liquid, such that the MnSO4.H2O is in large excess: At once, the liquid becomes deep purple. Some bromine is formed and oxygen is formed as well (part of the bromate decomposes in the strong acid). Boil the liquid for quite some time, so that all Br2 and possibly HBr are driven off (that's why KBrO3 is chosen as oxidizer, all remains can be driven off completely by heating). After this boiling, the liquid is turbid. When the liquid is allowed to stand for a while, a reddish brown precipitate is formed, with a deep purple liquid above it. The precipitate is somewhat like your purchased MnO2, it is fairly light brown/reddish.

Add some of the deep purple liquid to water: This results in heating up of the water (hydration of sulphuric acid), and the liquid becomes red/brown, as shown in the picture in my previous post. When more water is added, then the liquid becomes turbid and a dark brown precipitate is formed. Most likely it disproportionates in water to MnO2 and Mn(2+), or a precipitate of Mn2O3 is formed.


The experiment is repeated with 60% perchloric acid instead of sulphuric acid. Solid MnSO4.H2O is added to 60% perchloric acid. On heating, part of the solid dissolves and a white suspension is obtained. Nothing is oxidized (perchloric acid is strongly acidic, but not oxidizing at all when it is below 70%, it actually is more inert than H2SO4 of similar concentration with respect to redox reaction).
This liquid was allowed to cool down again and to this liquid, some solid KBrO3 is added. Hardly any reaction occurs. The liquid becomes very slightly brown/red.
This liquid then is heated. At a certain point, well above 100 C, a reaction occurs, which suddenly starts and is fairly vigorous. In this reaction, gaseous bromine boils away from the liquid, and the liquid becomes turbid, and a brown precipitate is formed (somewhat between the color of chocolate and red bricks). The reaction stops when all bromate is used up. No deep purple liquid is obtained.
To this liquid, some 60% H2SO4 is added. When this is done, then the liquid becomes purple, but on standing for a while, the purple color disappears again and all what remains is the brown/brick red precipitate in an almost colorless liquid.

Finally, some of the deep purple liquid from the first experiment was taken, and solid NaCl is added. Immediately, this results in formation of a dark solid, the liquid becomes almost black. When it is heated, then the liquid becomes clear again, and the color shifts from very dark brown to dark green/brown and then it gradually becomes lighter during heating, until finally the liquid is almost colorless. Some chlorine and HCl are produced in this reaction.



So, summarizing:
1) The deep purple compound almost certainly is a compound of manganese in its +3 oxidation state. It certainly is not permanganate, because when an acidic solution of permanganate is added to water, it remains purple and does not become turbid.
2) The deep purple compound depends on the presence of sulfate/sulphuric acid in high concentration. Dilution results in hydrolysis.
3) The deep purple compound is not formed in perchloric acid of similar concentration. It is specific for sulfate/sulphuric acid.
4) From the strongly acidic solutions in all cases a brown/reddish precipitate is formed, either besides the deep purple liquid, or as the only product. This almost certainly is a compound of manganese, which at least contains some of the metal in its +3 oxidation state. The precipitate has a remarkably light color.
5) With chloride, a very dark compound is formed (MnCl3 or MnCl4??), which decomposes on heating, giving chlorine and colorless Mn(2+).
6) Precipitates, formed from much more dilute solutions by mixing the purple liquid with water, are dark brown and do not have that typical somewhat reddish appearance. I have the impression that the reddish compound is not am oxide, but possibly a basic sulfate (or perchlorate in the case of perchloric acid).


Tomorrow I'll try more experimenting with MnO2 as the starting point. Up to now, I have not seen any green precipitate, I'll see if I can reproduce that part tomorrow.




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[*] posted on 31-10-2008 at 15:39


@ blogfast, woelen

From Mellors; Manganic sulphate, Mn2(SO4)3 is formed as a dark green powder (precipitate) by the action of sulphuric acid on freshly precipitated manganese dioxide. It dissolves in water to a violet solution (Mn3+) which deposits hydrated manganese dioxide on standing (dark-brown coloured).

This appears similar to the first reactions you observed in the test tubes (previous page).

Bear in mind that hydrated MnO2 will take on all sorts of brown shades depending on "grain" size, and in colloidal form will look like a brown solution.

Hmmm! A bit more reading, I found this at the 1911 Encyclopedia site;

Trimanganese tetroxide, Mn304, is produced more or less pure when the other oxides are heated. It may be obtained crystalline by heating manganese sulphate and potassium sulphate to a bright red heat (H. Debray, Comptes rendus, 1861, 52, p. 985). It is a reddish-brown powder, which when heated with hydrochloric acid yields chlorine.

If your material is reddish-brown or brownish-red, I think you might have pure Mn3O4.

There's more information on Mn compounds at the site, if you haven't already seen it!

http://www.1911encyclopedia.org/Manganese

[Edited on 31-10-2008 by Xenoid]
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[*] posted on 1-11-2008 at 07:22


@Xenoid:

You took the words right out of my mouth!

I've been pondering based on these results so far about what the 'sulfato Mn(+III) complex' could be and the lack of any cation (apart from H<sub>3</sub>O<sup>+</sup>;) makes me think that the violet colour is indeed simply Mn<sub>2</sub>(SO<sub>4</sub>;)<sub>3</sub>.

I've inadvertently obtained it by heating homemade MnO2 with 50 w% sulfuric acid, Woelen basically by dissolving MnSO4 in strong acid, followed by mild oxidation.

There remain nonetheless some questions. The nature of the blue-green compound not being the only one.

<i>"It dissolves in water to a violet solution (Mn3+) which deposits hydrated manganese dioxide on standing (dark-brown coloured)."</i>

Clearly not when the concentration of H<sub>3</sub>O<sup>+</sup> and/or SO<sub>4</sub><sup>-</sup> is great enough: my purple coloured solutions doesn't budge even on considerable dilution...

I carried out one more experiment. On standing overnight, the tube (second photo, previous post of mine) had cleared: all precipitate had collected at the bottom. After decanting off the clear supernatant liquid, I added a few centimeters of approx. 25 w% H2SO4. Even before heating, a purple solution started to form at the bottom of the tube. On heating in direct flame the tube unfortunately 'burped' and most of the content ended up splattered over my lab. The centimeter or so that was left was a clear purple solution. I need to corroborate (with steam bath heating this time) this but it seems to confirm Mellor: freshly precipitated Mn oxides dissolve readily in H2SO4 (of the right concentration?), forming Mn<sub>2</sub>(SO<sub>4</sub>;)<sub>3</sub>.

The question of what my strange looking Mn oxide really is still remains open too.

And now I'm seriously tempted to try and produce an Mn<sub>2</sub>(SO<sub>4</sub>;)<sub>3</sub> based alum, with K, Rb or Cs...

$$$$$$$$$$$

Is there no end to this weirdness??? In an attempt to corroborate the dissolution in 25 % H2SO4 of the brown hydrolysed product of the insoluble green compound, I transferred some of the green slurry (top pic, 7.46 post) into an Erlenmeyer and added a small amount of water to provoke hydrolysis. Instead the green compound dissolved - completely:



The colour is slightly more reddish that the previous purple solutions but for now I'm attributing this to concentration: in other words, it's still Mn2(SO4)3 but at a higher concentration, giving a slightly different hue. Only spectroscopy could tell that with certainly, of course.

My first thought was that it might be concentrated MnSO4 (and not Mn2(SO4)3) but adding 32 w% HCl makes it go quite dark and heating the dark solution makes it clear up to yellow, with evolution of chlorine. Mn(II) doesn't do that.

It would seem to indicate that concentration of the H2SO4 is really important: the liquid part of the green slurry was 50 w% H2SO4, diluting it a bit we're probably in the 25 % range, enough to dissolve it to Mn2(SO4)3...

I'm still nonetheless puzzled as to how a predominantly Mn (IV) compound (homemade MnO2) can be reduced to a relatively stable Mn (III) compound, seemingly in the absence of any reducing agent...

It should also be possible to determine the Mn<sup>3+</sup> concentration by adding a standardised excess of Fe<sup>2+</sup> and titrating the excess with standardised KMnO<sub>4</sub...

[Edited on 1-11-2008 by blogfast25]
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[*] posted on 1-11-2008 at 10:33


I also have done some more experimenting. I found the following:

In conc. 96% H2SO4, the purple compound cannot exist. I added some of my solid NH4[MnP2O7] to conc. H2SO4. The purple solid quickly turns brown and when the acid is heated, I obtain a reddish brown precipitate in a colorless liquid. To this, I added water, dropwise. At a certain point, the very fine precipitate dissolves and a deep purple and clear liquid is obtained. When more water is added, then the color of this solution shifts from purple (like permanganate) to reddish purple and from there to red (something like shown in the picture above of the erlenmeyer) and on addition of even more water, the solution becomes turbid and a dark brown precipitate is formed.

So, in fully concentrated H2SO4, manganese(III) seems to exist in the form of a reddish/brown solid. Is this Mn2(SO4)3? On moderate dilution of the acid, a deep purple solution is obtained. This is not plain Mn(3+) in solution, but must be some sulfato-complex of Mn(3+), a possible option could be [Mn(SO4)2](-). On even stronger dilution, the color shifts to red/brown and then aqueous Mn(3+) is obtained (as shown in my testtube picture in a previous post in this thread). When dilution is even stronger, the Mn(3+) hydrolyses to Mn2O3.

I also added some of the bright purple NH4[MnP2O7] to conc. HCl (30%). When this is done, then the liquid becomes VERY dark, almost black. But in thin layers, such as when the liquid sticks to the glass, it appears to be olive-green. On heating, the liquid becomes lighter and chlorine is released. So, the very dark brown/green compound must be a manganese(III) complex with chloride. or simply MnCl3.

Up to now, however, I have not been able to obtain the green solid material. I get an understanding of all of this chemistry, but the green material still is lacking from my experimental results. Right now, some solid MnO2 (reagent grade) is brewing in H2SO4 of medium concentration, we'll see tomorrow what this will do. I do not have such light brown manganese oxide, so that part of the experiments cannot be reproduced by me :(

[Edited on 1-11-08 by woelen]




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[*] posted on 1-11-2008 at 12:27


Quote:
Originally posted by woelen
The purple solid quickly turns brown and when the acid is
heated, I obtain a reddish brown precipitate in a colorless liquid. To this, I added water, dropwise. At a certain point, the very fine precipitate dissolves and a deep purple and clear liquid is obtained. When more water is added, then the color of this solution shifts from purple (like permanganate) to reddish purple and from there to red (something like shown in the picture above of the erlenmeyer) and on addition of even more water, the solution becomes turbid and a dark brown precipitate is formed.

So, in fully concentrated H2SO4, manganese(III) seems to exist in the form of a reddish/brown solid. Is this Mn2(SO4)3? On moderate dilution of the acid, a deep purple solution is obtained. This is not plain Mn(3+) in solution, but must be some sulfato-complex of Mn(3+), a possible option could be [Mn(SO4)2](-). On even stronger dilution, the color shifts to red/brown and then aqueous Mn(3+) is obtained (as shown in my testtube picture in a previous post in this thread). When dilution is even stronger, the Mn(3+) hydrolyses to Mn2O3.
]


These colour changes on dilution are puzzling. I'll try and reproduce that with my own purple solution. My green compound seems more in line with Xenoid's Mellor quote of a green Mn2(SO4)3, it could explain why the stuff just dissolved and not hydrolysed. But it did that only once (out of three times).

You think the purple might be Mn<sup>3+</sup> + 3 Mn(SO<sub>4</sub>;)<sub>2</sub><sup>-</sup>, huh?

Any idea how the Mn (IV) ends being reduced to Mn(II) in H2SO4 w/o the absence of an oxidiser?

Acc. 1911 Encyclopedia:
Quote:
When heated [the MnO2] with concentrated hydrochloric acid it yields chlorine, and with concentrated sulphuric acid it yields oxygen.


Obviously O [-II] ---> 1/2 O [0] + 2 e<sup>-</sup>

could provide the missing electrons. I do not recall seeing any bubbles when heating with 50 % H2SO4, though...


And strange how you don't seem to get the green compound...

If I can repeat the green-to-purple trick with a known quantity of MnO<sub>2</sub>, known volume of H2SO4 and a controlled volume of diluent water and everything dissolves I could have a good idea of the concentrations and the corresponding colours... And if I could get the unknown oxide to dissolve completely in the same way, it would open up a simply, titrimetric method for determining the Mn content of the oxide...

And adding extra H2SO4 to my reddish-purple solution could also push towards Mn(SO<sub>4</sub>;)<sub>2</sub><sup>-</sup>.

[Edited on 1-11-2008 by blogfast25]
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[*] posted on 1-11-2008 at 15:11


In the meantime I now have been able to make the green solid. I used my black reagent grade MnO2, suspended in 96% H2SO4. Only after very strong heating (the acid already gives off quite some white fumes, so it must be well over 200 C), I get slow production of gas (must be oxygen) and the color of the solid slowly changes from black to dark green. On cooling down, the color even has become a little lighter, but still dark green.

So, your quote from the 1911 encyclopedia indeed can be confirmed by me. Oxygen is produced and this makes up for the reduction:

2MnO2 --> Mn2O3 + [O], with Mn2O3 of course being converted to Mn2(SO4)3.

-----------------------------------------------

I also tried what happens with the purple solution when Na2S2O8 is added. Remarkably, no permanganate is formed, but instead, the solution becomes colorless, and bubbles of gas (must be oxygen) are produced, until the solution is totally colorless. At that point, production of oxygen stops. So, I expect the following reaction:

S2O8(2-) + 2Mn(3+) + 2H2O --> 2H2SO4 + O2 + 2Mn(2+)

This is a VERY special reaction: Here, S2O8(2-) acts as a reductor!!!

I expected formation of MnO2, or even MnO4(-), but instead, the stuff is reduced to colorless Mn(2+).

This reaction is slow, at 50 C in a water bath, it takes a few minutes.

When I add a very small pinch of KBrO3, then the liquid becomes deep purple quickly again, but this color slowly fades and after a minute or so, it is colorless again (actually light yellow, due to formation of some bromine). So, the Mn(3+) formed by quick oxidation by bromate slowly is reduced again. Altogether, this Mn(3+) species in moderately concentrated H2SO4 is really peculiar stuff...

------------------------------------------------

Finally, your different observations (sometimes material dissolving, otherwise getting a precipitate) are not that strange. If the dark green material indeed is Mn2(SO4)3, then it can either dissolve giving a red/brown solution, or hydrolyse to a flocculent precipitate. What happens depends on temperature, and the amount of acid in the water, and on the concentration of the Mn(3+). So, there are many variables, which can be adjusted, and probably there only is a relatively small range in which all of it dissolves and does not hydrolyse. So, your seemingly erratic results are more of what I expect, and are not erratic to me at all.




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[*] posted on 1-11-2008 at 21:07


@ blogfast, woelen

Hi guys, I decided to do a couple of quick experiments.

Firstly, I strongly heated my pottery grade black MnO2. Initially I used a small SS pan over a gas ring, but this was nowhere hot enough! I then transferred the material to a fire brick and heated it to red heat with a propane flame (Bernzomatic). This resulted in the formation of what I would describe as slightly orangey-brown to light chocolatey-brown material - this can only be Mn3O4. I think the colour of Mn3O4 probably varies with heating time, maximum temperature, cooling history, etc. in much the same way as "red" ferric oxide does.

Secondly, I boiled the Mn3O4 and the MnO2 with 98% sulphuric acid (SG=1.8 drain cleaner).

The brown Mn3O4 produced a greenish precipitate, overlain by a flocculant whitish layer, and a supernatant yellowish solution, when the yellowish solution was added to an approximately equal amount of water, nothing happened, and when diluted further again nothing happened. I am thinking that the green precipitate is manganic sulphate (Mn2(SO4)3) and the flocculant whitish precipitate is anhydrous manganous sulphate (MnSO4). I have some anhydrous MnSO4 which I got from a hydroponics store, it is a dirty off-white coloured powder. I'm not sure what is causing the yellowish colour of the solution!

The black MnO2 produced a dark-green precipitate overlain with greenish material in suspension. When some of this greenish suspension was added to an equal amount of water an initial intense purplish-red colour formed at the interface, when shaken, the greenish material dissolved and a purplish-red solution formed. When diluted further the solution became more of an intense rose-red colour. These reactions of MnO2 are in agreement with the observations above!

@ blogfast
I still reckon your brown material is pure Mn3O4. Let me know a specific reaction and I will try and duplicate it with my "home made" Mn3O4. You'll need to specify the acid concentrations reasonably accurately however!

[Edited on 1-11-2008 by Xenoid]
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[*] posted on 2-11-2008 at 07:21


Quote:
Originally posted by Xenoid

@ blogfast
I still reckon your brown material is pure Mn3O4. Let me know a specific reaction and I will try and duplicate it with my "home made" Mn3O4. You'll need to specify the acid concentrations reasonably accurately however!



If only I knew of such a specific reaction!

But guys, we're starting to sing from the same hymn sheet!

Having pondered my observations some more, I too am now convinced that the green compound is Mn (+III) bearing and probably simply manganic sulphate.

One thing makes me almost 100 % certain of this:

If you go back to my post of 7.20, second picture, both me and Woelen misinterpreted that. I thought MnO2 was being formed there, which then later dissolved in the extra added HCl, giving MnCl2 and Cl2. Woelen thought the heat I observed was simply due to dilution (hydration heat) of the 50 w% H2SO4. With hindsight, we were both wrong .

What happened in all likelihood is that the solid Mn (III) species reacted with the strong HCl: Mn (III) + 3 HCl + water ---> MnCl<sub>3</sub> + 3 H<sub>3</sub>O<sup>+</sup>.

The dark liquid that I (rather prejudicially) mistook for a slurry of MnO2 was in fact quite concentrated MnCl<sub>3</sub> and its formation caused the heat to be generated.

I know this almost for certain because of the very unusually fast rate this liquid cleared up on heating: the decomposition of MnCl<sub>3</sub> is much faster than the dissolution of freshly precipitated MnO2 and subsequent decomposition of the formed MnCl<sub>4</sub>. All the MnCl<sub>3</sub>-bearing solutions I've heated up show that characteristic: they decompose remarkably fast to MnCl2 and Cl2.

Ironically, what started out as a mystery regarding the reddish Mn oxide, <i>may just</i> (note cautious emphasis) have yielded the possibility of titrimetrically assaying such oxides. If I can find the right conditions (acidity and sulphate concentration) in which the green Mn (III) compound dissolves without hydrolysis to a dilut<i>ish</i> Mn<sup>3+</sup> solution, then the road is open to adding a standardised excess of Fe<sup>2+</sup>:

Mn<sup>3+</sup> + Fe<sup>2+</sup> ---> Mn<sup>2+</sup> + Fe<sup>3+</sup>

followed by back titration of the excess Fe<sup>2+</sup> with permanganate and calculation of the Mn content from sample weight, added ferrous mmol and permanganate volume. The method would of course be sensitive to the presence of chlorides.

This would really only be useful if the 'mystery oxide' behaves much in the same way but preliminary results seem to indicate that. Alternatively, if it doesn't, then the mystery oxide would have to be converted first to MnO2 via HCl and re-oxidation to Mn (IV).

Thanks for your cooperation so far!

+++++++++

And it may not even be too hard to find the window of conditions in which Mn<sup>3+</sup> can operate relatively stably. One more very small experiment was carried out. Another 3 g of homemade MnO<sub>2</sub> was boiled up with 50 W% sulphuric acid for about 15 min. The green compound appeared. It was allowed to cool down to near-RT. And this time I saw bubbles (quite small) long before the mixture came to the boil..

A small amount of approx. 1 M H2SO4 was then poured into a conical flask and to it about the same amount of green slurry (carefully stirred for homogeneity) was then added to it. The green compound dissolved effortlessly, forming a dark, beautiful wine-red solution, here seen in a test tube This must be quite concentrated Mn2(SO4)3. The photo doesn't do it justice:




My working hypothesis for now is that any reasonable amount of the manganic sulphate is probably soluble in 1 M H2SO4, without hydrolysis, reduction or oxidation.

///////////////

Here's a few dilutions using the concentrated Mn (III) solution above:



Left: 1 ml diluted about 10 times with 1 M H2SO4. Light amber/reddish yellow.

Middle: 2 ml, diluted with approx. equal amount of 1 M H2SO4. An attractive wine-red.

Right, an original MnO2 boiled up with 50 w% H2SO4, with green Mn (III) compound and violet supernatant liquid, presumably containing an Mn (III) sulphato complex, as suggested by Woelen.

No hydrolysis noted so far.

(((((((((((((

Adding deeply cooled 50 w% H2SO4 to a small amount of the liquid from the middle tube causes the colour of the liquid to shift towards the violet obtained directly from reacting solid MnO2 and 50 w% H2SO4:



Left: the middle tube from above. Wine-red.

Right: a small amount of this solution with ice cooled 50 w% H2SO4 added to it. Shifts to more purpl<i>ish</i> - violet, definitely much more like the typical purple solutions obtained by attacking MnO2 with strong H2SO4.

The mangani sulphato complex can thus also be obtained by pushing:

Mn2(SO4)3 + SO<sub>4</sub><sup>2-</sup> <---> mangani sulphato complex

to the right by increasing the SO<sub>4</sub><sup>2-</sup> concentration, not only by attacking solid MnO2 with strong H2SO4.


[Edited on 2-11-2008 by blogfast25]
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[*] posted on 3-11-2008 at 08:52


And from the middle tube (dilution about 1:1) a light brown precipitate forms when 5 M NaOH is added. In all likelihood this is Mn<sub>2</sub>O<sub>3</sub>. n H<sub>2</sub>O...

Here's at least one literature reference to the Mn (III) sulphate complex:
Quote:
Manganese III ion (or manganosulfuric acid complex) is purple, and, as it is reduced to Manganese II ion, the purple color decreases to colorless.


[Edited on 3-11-2008 by blogfast25]
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[*] posted on 3-11-2008 at 13:29


I have been reading this thread with interest from the beginning, and (unless I've missed something) no one has mentioned re-smelting the oxide with powdered carbon as a reducing agent. The melting point of manganese is between copper and iron, so equipment is a limiting factor. The only problem I see is the solubility of carbon in metallic manganese and related carbide formation. Presumably there's an acid that will dissolved the metal but not the carbide.
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[*] posted on 4-11-2008 at 08:31


@Watson.fawkes:

What would be the purpose of that? To produce the metal? Much easier to do it via thermite (see here for example) or by thermoreduction of molten, anhydrous MnCl<sub>2</sub> with magnesium. Both have their own foibles and problems though. But high purity lump Mn can be obtained by these methods.

For reduction of, say MnO<sub>2</sub> with C <i>(and not CO)</i> seriously high temperatures would be needed, I'd need to consult Ellingham diagrams on that. Just mild glowing won't reduce it to metal (but possibly to a lower oxide, yes).

Personally I won't be carrying out anymore experimentation until the weekend when hopefully I'll test my titrimetric method of Mn determination of oxides via Mn (+III)...
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[*] posted on 4-11-2008 at 22:37


Quote:
Originally posted by blogfast25. &nbsp; Re: furnace reduction
What would be the purpose of that?
To measure the oxygen content of the original material. Weigh the oxide charge beforehand, weigh the metal button afterward. Wasn't that the original question, to determine how much oxygen was in the mystery oxide?
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blogfast25
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[*] posted on 5-11-2008 at 09:40


Hmmm, I believe it will be easier and more accurate to determine Mn content either via the new proposed (and to be tested) method or via the older titrimetric analysis (oxidation to permanganate with persulphate, indicated above)...

Reduction with C would also mean inevitable contamination with C. Reduction with H2 may also be a possibility but it's hardly a <i>quick 'n easy</i> route...
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watson.fawkes
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[*] posted on 5-11-2008 at 17:41


Quote:
Originally posted by blogfast25
Hmmm, I believe it will be easier and more accurate to determine Mn content either via the new proposed (and to be tested) method or via the older titrimetric analysis (oxidation to permanganate with persulphate, indicated above)...

Reduction with C would also mean inevitable contamination with C. Reduction with H2 may also be a possibility but it's hardly a <i>quick 'n easy</i> route...
I'll make no claim as to relative accuracy. Relative ease has a huge amount to do with available equipment. As for carbide, I should imagine there's a friendly acid that will dissolve the metal but not the carbide. And then you're good to go, assuming a single carbide species, an assumption I'm not prepared to defend. But even then, if your oxide is a consistent species in an integer ratio, running multiple passes with precisely measured carbon (one per expected integer ratio) would provide a quick qualitative determination, as one batch should have minimal carbide formation.
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[*] posted on 6-11-2008 at 09:32


Quote:
Originally posted by watson.fawkes
Relative ease has a huge amount to do with available equipment.


True, but rereading the currently used titrimetric methods (there are two quite similar variants) accuracy obtained is about +/- 0.2 w% of Mn (on alloy samples). Considering obtaining three data points would take less than an hour of work, I believe the accuracy/effort ratio favours the relatively hassle-free titrations.
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