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Author: Subject: Chevreul's Salt.
MR AZIDE
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[*] posted on 13-6-2012 at 10:56
Chevreul's Salt.



Found this interesting article in another old School Science review Sep 1986.

boiling solutions of Sodium metabisulphite Na2S2O5 ( sodium disulphate iv) added to boiling CuSO4 soln , and boiled for 4 mins gives Chevreuls salt, Cu3(SO3)2.2H2O) This is a red solid , which evolves a little Sulphur Dioxide.

Copper (I) chloride will be obtained by dissolving chevreuls salt in dilute HCL, with some SO2 present.

3S2O52- + 3Cu2+ + %H2O -----> Cu3(SO3)2.2H2O + 3So2 + 2H3O+ + SO42-

Says dissolve 6g of the metabisulphite in 30ml hot water. Dissolve 7g copper sulphate in another beaker in 100 ml water, asnd heat this to boiling. With stirring as the metabisulphite is added, and boil for about 4 mins.
The precipitate if filtered off, and washed with several portions water.
This is the chevreuls salt.

Dssolving this in some dilute HCL will provide a white precip, within a greeny / blue solution. This white precip turns green slowly of exposure to air . The CuCl.
Addition of excess HCL redissolved the precip.
Addition of ammonia gives the intense blue copper tetraammine (ii) complex.


[Edited on 13-6-2012 by MR AZIDE]
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[*] posted on 13-6-2012 at 22:20


Do you have this article online? This Chevreul's salt is quite special, it must have both copper(I) and copper(II) in its lattice. Cu3(SO3)2, means that copper has charge +4 divided over 3 atoms. This means that two of the coppers must have oxidation state +1 and the last one has oxidation state +2.

I never heard of it. A nice one to try at home :P




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[*] posted on 14-6-2012 at 02:59


For crystal structure see Acta. Chem. Scand., 19, 2189-2199 (1965) (Free access)
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[*] posted on 14-6-2012 at 07:16


just did this reaction today went very smoothly and easily but since theres no info about it that i can find is there anything it is soluble in that doesn't decompose it? because i thought some nice large crystals of this substance would be pretty beautiful looking. :)
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[*] posted on 14-6-2012 at 07:52


Some interesting information on Chevreul's Salt;

http://www.scielo.br/scielo.php?script=sci_arttext&pid=S...

It looks like you can make it as red crystals, woohoo!
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[*] posted on 14-6-2012 at 11:03


I havent got this online, i dont have a image scanner, .

Further info on reaction with Dilute HCL:::


Cu3(SO4).2H2O + 4H3O+ + 2Cl- -----> 2CuCl + *H2O = 2SO2 + Cu2+

It says that the formation of Cu(II) ions favours the left side of eqilibrium.

2Cu+ <----> Cu2+ + Cu

So It can be seen that disproportionation happens.

I thought that the compound would be a precipitate, I havent done the reacion yet, I have got a small amount of the metabisulphite. So I might try this.

Usefull way to make CuCl....

Interesting to hear that this forms red crystals..........any pictures,
monoclinic, orthorombic etc... etc.......
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[*] posted on 15-6-2012 at 01:59


Quote: Originally posted by MR AZIDE  

monoclinic, orthorombic etc... etc.......

Too lazy to read the publication I posted above?
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[*] posted on 15-6-2012 at 02:09


So it seems that Cu+2 can oxidize sulfite to sulfate, if the reaction is driven by the formation of the Chevreul's Salt, which contains a portion of its copper in the +1 oxidation state. These types of reactions are not extremely unusual, for example, Cu+2 can oxidize iodide to free iodine, the reaction being driven by the formation of insoluble CuI.

Apparently Cu+1 can form strong complexes with iodide or certain sulfur-containing anions.
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[*] posted on 15-6-2012 at 05:05


Quote: Originally posted by MR AZIDE  

Usefull way to make CuCl....



Hmmm, a bit long winded, if you ask me.

Boil copper (II) sulphate solution with sodium chloride and copper wire:

Cu (II) + Cu (0) < === > 2 Cu+
Cu+ (aq) + Cl- (aq) === > CuCl (s)

White CuCl drops out.



[Edited on 15-6-2012 by blogfast25]




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[*] posted on 17-6-2012 at 23:08


I have experimented with copper sulfate and sulfite and bisulfite and I indeed can reproduce the results mentioned in this thread. I found, however, that the reaction strongly depends on the actual pH of the solutions.

When a solution of Na2S2O5 is mixed with a solution of CuSO4, then a nice grass green solution is obtained. When this solution is heated to boidling, then a brick red fine crystalline precipitate is formed and the liquid above the precipitate is (nearly) colorless.

I also made another solution of CuSO4 and excess of Na2S2O5 and put this aside at room temperature. After several hours, the bottom of the test tube is covered by many nearly black crystals (very dark red) and the liquid has become lighter green.

With sodium sulfite (Na2SO3) the behavior is very different. This produces a dark green precipitate, which quickly turns dirty brown (mustard-like). Probably first a basic copper(II)sulfite is formed, which then is converted to impure copper(I) oxide and free sulfate ion. This precipitate is slimy and becomes flocculent on shaking, while the brick red Chevreul's salt is crystalline. The latter has a much nicer appearance and is much more compact and hence can easily be separated.

I intend to do a more thorough investigation of this salt and the dependence on pH and there might follow a webpage about this interesting salt.

[Edited on 18-6-12 by woelen]




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[*] posted on 18-6-2012 at 04:06


I repeated the vague procedure available from Spectroscopic and magnetic properties of Chevreul’s salt, a mixed valence copper sulfite Cu3(SO3)2·2H2O in Inorganica Chimica Acta Volume 295, Issue 1, 15 November 1999, Pages 125–127. It states that "Chevreul’s salt was synthesized by heating a solution containing CuSO4 and NaHSO3 at a temperature between 60 and 70°C for 3 h. The concentration of CuSO4 was altered between 0.6 and 0.8 M, and the NaHSO3/CuSO4 mole ratio between 1.1 and 1.7." Unfortunately I don't have access to the original reference, which is the famous Gmelin’s Handbook, Aufl. System No. 60, vol. 8, Verlag Chemie, Weinheim/Bergstrasse, 1958, p. 484.

In 1,1 ratio and 0,6 M solution in 10 ml scale the preparation worked like charm. A quick boil led to a deep red ppt. Heating the dried product led to formation of black copper(II)oxide and volatile sulfur compounds (mostly SO2, white fumes can be seen on top of the solid) some of the sulfite also turns to sulfate, though.

[Edited on 18-6-2012 by kavu]

[Edited on 18-6-2012 by kavu]
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[*] posted on 24-6-2012 at 10:10


I made a webpage about this interesting compound, as part of a somewhat more extensive set of experiments in which I investigated the redox reaction between sulfite and copper(II) as function of pH. Quite a few interesting things were found. All of it is described in the following webpage:

http://woelen.homescience.net/science/chem/exps/copper_sulfi...




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[*] posted on 24-6-2012 at 20:59


Bravo!
What about growing larger crystals now huh? Checking their opacity to UV and IR rays maybe :o

[Edited on 6-25-2012 by Poppy]




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[*] posted on 6-10-2013 at 17:32


I just performed this synthesis tonight, and it went quite smoothly. Very simple, and the range of color changes is striking. Copper chemistry is just amazing!

I simply dissolved a small spatula of sodium metabisulfite into a few mL of water in a test tube, and did the same with copper sulfate in another test tube. This resulted in a clear solution, and a blue one. I combined the two to yield a beautiful emerald green liquid. Upon heating to boiling, the red chevruel's salt precipitated out quickly. It's a nice, compact, crystalline precipitate that settles completely within seconds. My remaining solution was still a light blue-green, indicating I had an excess of copper sulfate.

Now, I'd like to distinguish this from copper(I) oxide, which has a similar color can also be formed by addition of SO2 to solutions of Cu2+ (according to the wiki article on cuprous oxide). This is to preempt any internet nay-sayers for when I post a video on my YouTube page :)


I first took a few hundred mg of the salt and added concentrated ammonia solution, yielding the deep blue hexaaminecopper(II) complex. (proving that the compound contains copper(II) and really isn't copper(I) oxide.
I wanted to check myself by testing copper(I) with ammonia, so following the suggestion in the OP I added a similar amount of the salt to diluted hydrochloric acid. It formed a light green solution, and upon adding a bit more chevruel's salt I obtained a white precipitate of copper(I) chloride. I pipetted off the liquid, added a small amount of water to wash the precipitate, pipetted off the water, and then added some concentrated ammonia. I got the same deep blue solution! I believe this should have yielded a colorless copper(I) complex. Would slight contamination of copper(II) be enough to give the same deep blue? The copper(I) complex is also claimed to be air-sensitive - would swirling the test tube to dissolve the CuCl be enough to convert fully to the blue complex?

A very interesting and fun experiment overall.
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[*] posted on 6-10-2013 at 22:27


Very interesting inorganic chemistry, but I think one has to be careful that one indeed produces the mixed valent complex and not simply copper metal powder precipitate (complete reduction). This particularly a concern as you're heating a reducing agent with an oxidant, it can be easy to overshoot your mixed valent state and reduce fully to copper powder?

This red powder I see in some of the photos looks to me like copper powder? If I were a betting man, I would be placing my money on the @ woelen's nearly black crystals that form on standing as being the actual complex. After all Cu(I) Cu (II) charge transfer should make it VERY strongly deeply coloured.




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[*] posted on 6-10-2013 at 23:22


@MrHomeScientist: The colorless complex of copper(I) with ammonia is VERY air sensitive. The complex is colorless, but even a tiny amount of oxygen from the air makes the solution deep blue. Simply dissolving Cu2O or CuCl in ammonia does not give a colorless solution. Any air, dissolve in the ammonia or air around the ammonia causes the solution to turn blue. If you want to make a colorless solution of copper(I), then you need to have a strong reductor in solution as well, e.g. Na2S2O4 (dithionite).
Testing the Chevreuls salt can be done by taking this, together with a very small pinch of sodium dithionite and dissolving this in ammonia. You then get a blue solution. If you have a similar amount of Cu2O with the same small pinch of Na2S2O4, then the solution will be colorless and only turns blue after somewhat longer shaking with fresh air around.

Another test may be to add the Chevreuls salt to 10% HCl. That should give a clearly present smell of SO2. With Cu2O you won't get any smell.

@deltaH: I tested my red precipitate by adding acid to it. It dissolves. Fine copper powder does not dissolve. When the precipitate was added to a small amount of acid, then I could smell SO2.




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[*] posted on 7-10-2013 at 01:12


@woelen

Ok, you've convinced me it's not copper, but I'm not so convinced it's not Cu2O passivated with adsorbed surface sulfite that gives off a smell of SO2 with acid and causes the Cu2O to dissolve. This could be especially true in the case if the Cu2O particles are very small or porous, possibly even nanosized aggregates capped off by surface sulfite species?

[Edited on 7-10-2013 by deltaH]




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[*] posted on 7-10-2013 at 03:38


Yes, I understand that sulfite can be coprecipitated with Cu2O. So, the test with the smell of SO2 is not 100% sure.

If you want to do a test which is fully conclusive, then most likely you need to do a quantitative test. If you look at the formula for Chevreul's salt, then you see that there are two Cu(+) entities and two SO3(2-) entities. Fully oxidizing this to copper(II) and sulfate requires 6 electrons. If you have Cu2O, then you need 6 electrons for 3 Cu2O.

The problem with such quantitative tests is that you need to work very accurately. One mole of Chevreul's salt is 387 grams, 3 moles of cuprous oxide is 429 grams. This is fairly close. Impurities in either one of these, especially if precipitated from solution can easily be several percents (e.g. humidity, coprecipitated sodium ions, coprecipitated sulfate ions) and then the difference between the two may be swamped in inaccuracy.

Do you see a practical method for distinguishing between the two? If it is within the reach of me as a amateur chemist, then I certainly am willing to try it.




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[*] posted on 7-10-2013 at 05:06


@woelen

I hear you and very interesting challenge indeed... NOT easy!!!

Okay, I read that Cu2O is a semiconductor, so maybe that might open some avenues for testing? FYI Band gap = 2.17eV from Wiki's semiconductor data table, so that means it should absorb STRONGLY all light with wavelength shorter than 571nm... basically all green light and higher, but importantly, it should only scatter light below this. So red light is only scattered by Cu2O for example, not absorbed.

Now as for Chevreul's salt, that has broad adsorptions all over the place, see this paper on the electronic spectra of it: http://www.scielo.br/pdf/jbchs/v13n5/12815.pdf

Note that there is a broad absorbtion peak at 785nm (deep red) for Chevreul's salt, it's not the strongest, but it's there.

So I know this is not ideal, but maybe if you have can get a bright deep red LED and shine on suspect Cu2O and suspect Chevreul and compare the difference on scattered light in a darkened room?

Would be very interesting if indeed the one sample Cu2O scattered the RED light but not so much absorb it, while the Chevreul's salt absorbs it more strongly appearing darker.

What about then using ultrabright green LED, now the Cu2O should be STRONGLY absorbing, appearing black, compared to Chevreul's salt which should only be partially absorbing green light.

Then ultrabright blue LED, BOTH Chevreul's salt AND Cu2O should be absorbing that STRONGLY according to the info above.

Photo's PLEASE if this works out :D should look amazing!

[Edited on 7-10-2013 by deltaH]




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[*] posted on 7-10-2013 at 06:19


Woelen:
Thanks for the response. Unfortunately I don't have any dithionite, so I'm looking into alternatives. I think I'll try adding some of the salt to HCl to produce SO2 gas, which I will then draw into a syringe and bubble through an acidified solution of potassium dichromate. If SO2 is present, it reduces the solution and turns it from orange to green. This is the standard, specific test for sulfur dioxide. For videos, I like to be able to show visual evidence whenever possible :)

deltaH:
It's fairly obviously not copper powder, when you see it in person. Copper powder is more of a brownish color, whereas this salt is a deep red just like dark bricks. But as woelen said (and as I wrote in my original post), it is soluble in acid (and ammonia) which shows that it is indeed not copper.

If the idea of using colored LEDs is plausible, I conveniently have access to some very bright, 1 amp (3 watt) red, green, and blue LEDs. Minor obstacle is that I do not have any CuO to compare against, so I'd have to make it. I see there are multiple ways to make this compound - anyone have experience with this? Any routes more recommended than others? This sentence from wikipedia seems like the simplest to me: "Copper(I) oxide can also be prepared by reacting a copper-ammonia complex with hydrogen peroxide."
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[*] posted on 7-10-2013 at 06:58


@MrHomeScientrist

Yeah agreed, some 'standard' to compare to would be the way to go. Hope it gives nice results and pretty pictures!

Good luck

p.s. I'm jealous of you ultra brights and in all the colours too!

[Edited on 7-10-2013 by deltaH]




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[*] posted on 7-10-2013 at 12:23


A good method of making Cu2O is to use Fehling's reagent (use Google for more info on this) and add this to a solution of glucose. All chemicals needed for this are very easy to obtain.

Another method is dissolving copper sulfate in a solution of citric acid and excess sodium hydroxide or sodium carbonate. The citric acid forms a complex with the copper(II) ions and this complex reacts with glucose at high pH and in this process Cu2O is formed as red precipitate, use excess glucose. Allow standing in a warm place for a day or so (a good way is to take a bucket of hot water and place the vessel with the solution of the copper/citrate complex and glucose in this.




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[*] posted on 7-10-2013 at 23:01


Some article about preparation of this salt (and not only).

Attachment: hhhh.7z (401kB)
This file has been downloaded 525 times





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[*] posted on 8-10-2013 at 01:30


Thanks kmno4, I found this paragraph at the end of your last reference of particular importance:

"It appears from these results that the reduction of copper(II) sulfate solutions with sulfur dioxide at elevated temperatures yields Chevreul's salt, copper metal and a mixture of anions containing sulfur, in the + IV and lower oxidation states. In contrast, the sulfur dioxide reduction of an ammoniacal solution of copper(lI), i.e. a mixture of the ammine complexes of copper(IlL gives Chevreul's salt which is not contaminated with copper metal or anionic species of sulfur in oxidation states less than + IV."

N.B. if correct!




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[*] posted on 15-10-2013 at 06:23


I completed a video and writeup of this experiment, which are now posted on their respective sites. If you google Chevreul's salt, my pages are the second and third hits already! (this thread is the first)

Video: http://www.youtube.com/watch?v=CitNalVs01M
Writeup: http://thehomescientist.blogspot.com/2013/10/chevruels-salt....


I tested out deltaH's colored LED idea, and Chevreul's salt behaves just as predicted, which is pretty neat. Under red light it appears red, of course. Under green light, it is indeed partially absorbing and it still appears reddish to me. Under blue light, is looks totally black.
For doing a comparison, I tried making some CuO via a long chain of syntheses (cream of tartar -> Rochelle salt -> Fenton's reagent -> CuO), but only got a tiny amount that won't scrape off my filter paper (it seems impregnated onto it). I didn't notice anything special about it under the lights, but the paper could be interfering. I'll try making some more and this time pipette off the liquid instead of filtering, so I have a powder to look at. Once I do that, I'll post photos of the results.
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