Sciencemadness Discussion Board - Synthesis of CuBr

Synthesis of CuBr

sternman318 - 16-6-2011 at 20:04

How would I go about synthesizing Copper (I) Bromide without a sulfite salt? It appears I merely need a reducing agent, yet all I have is Mg, but that wouldnt work because they would just displace each other. From a search, I found this, but does that make sense? I have heard of using ascorbic acid as a reducing agent, can I just by vitamin C tablets?

Picric-A - 16-6-2011 at 22:49

NaBr + CuSO4 + (Vitamin C tablets crushed up and dissolved in water/filtered. )/(Sulphur dioxide)/(Copper metal) ect...

There are loads of options out there and i would be very surprised if you have a hard time finding something as simple as copper.

woelen - 17-6-2011 at 02:26

Dissolve a lot of NaBr in 40% H2SO4. Add copper sulfate and add copper wire. The copper wire reduces the copper(II) to copper(I), itself also being converted to copper(I). Allow to stand for a day or say, occasionally stirring. Assure that NO fresh air can reach the solution.
The solution will initially be dark red/purple, due to complex formation of CuBr4(2-). When copper is added it initially will become even darker, but finally it will become lighter again. After the reaction, pour the contents in not too much cold water. I expect formation of a 'snow' of white CuBr.

I actually have done this reaction quite a few times with chloride instead of bromide and that gives nice CuCl. I think that it works equally well with bromide.

LanthanumK - 17-6-2011 at 03:24

Picric-A, you can use pure Vitamin C crystals as well, not just Vitamin C tablets.

Mixell - 17-6-2011 at 03:32

Woelen, how does exactly the copper wire oxidizes the Cu22+ to Cu+? Because the electrode potentians doesn't add up:
Cu2+ + e− -> Cu+ +0.159
Cu+ + e− -> Cu(s) +0.520

Does the sulfuric acid participate in this process?

ScienceSquirrel - 17-6-2011 at 04:16

It certainly works, warming an acidic solution of copper II chloride with copper wire results in a precipitate of copper I chloride.
Sodium metabisulphite is easily obtained from home brew shops and is a useful reagent for other things.

blogfast25 - 17-6-2011 at 04:53

Quote: Originally posted by Mixell

Woelen, how does exactly the copper wire oxidizes the Cu22+ to Cu+? Because the electrode potentians doesn't add up:
Cu2+ + e− -> Cu+ +0.159
Cu+ + 2e− -> Cu(s) +0.520

Good question, Mixell.

I suspect that something similar goes on as with the classical preparation of Cu (I) chloride:

Cu2+(aq) + Cu(s) + 2 Cl-(aq) === > 2 CuCl(s)

This reaction is probably driven by the lattice energy of CuCl(s): the fact that CuCl precipitates, generates much energy from Cu+(aq) + Cl-(aq) === > CuCl(s). This makes ΔG < 0.

Mixell - 17-6-2011 at 05:00

Well, that is indeed interesting.
And what is the role of the acid in this procedure?
Also what is the recommended molar concentration of an acid (say HCl) in this procedure if I use a concentrated solution of CuCl2?

woelen - 17-6-2011 at 06:46

No, this has nothing to do with the lattice energy of CuCl. All species remain in solution in the concentrated and strongly acidic halide solutions. It is a matter of complex formation.

The redox potential of Cu(2+) + e ---> Cu(+) and the reaction Cu --> Cu(+) + e indeed does not predict reduction of copper(II) to copper(I) by copper. But things change dramatically, in concentrated halide solution. It is the copper(II) halide complex which is the active species. In chloride solution we have:

CuCl4(2-) + Cu --> 2CuCl2(-)

Here we see that the tetrachloro cuprate(II) complex is converted to a dichloro cuprate(I) complex and this reaction is favored. The redox potentials of copper(II) to copper(I) reactions are shifting strongly due to the coordination of copper. This is a common phenomenon in chemistry, an other example being that gold quickly is oxidized by plain air in the presence of cyanide.

The reaction equation above is a simplification. CuCl2(-) is colorless and probably CuBr2(-) also is. This copper(I) complex reacts in turn with excess amounts of the copper(II) complex and mixed oxidation state complexes with Cl-bridges between copper atoms are formed and these complexes have a very intense dark olivegreen color, almost black in somewhat higher concentration.

The acid probably (but that only is my educated guess) plays an important role in the formation of these mixed oxidation state complexes. In neutral solutions, there also is a redox reaction, but in that case solid material is formed and the reaction soon comes to a halt. I think that the acid prevents the hydrolysis and formation of basic insoluble compounds.

@Mixell: The higher the concentration of HCl the better it works. A beautiful solvent for copper wire is concentrated HCl in which CuCl2 or CuSO4 is dissolved.

I have done a lot of experimenting on this stuff myself and have compiled quite a few webpages about the subject. I think they are an interesting read for you:

http://woelen.homescience.net/science/chem/riddles/copperI+c...

The chemistry, described above and in the webpages, also applies for bromide instead of chloride. The experiments with chloride, however, are much more appealing, because the color changes are more striking. Copper(II) and bromide give dark red/purple color at high concentration and this dark color masks many interesting changes when copper(I) is added to the game.

[Edited on 17-6-11 by woelen]

blogfast25 - 17-6-2011 at 12:46

Hmmm… while broadly speaking I agree with you, I’ve carried out the synthesis of CuCl in the complete absence of any acid, by boiling a CuSO4 + NaCl solution with copper wire. No mixed valence complex was observed and solid white CuCl deposits swiftly on the solid copper, while the solution clears from blue to clear. That’s not to say no cuprous chlorocuprate may be present in the solution but it seems to me that the redox reaction of Cu (II) to Cu(I) with Cu (0) is 'pulled' to the right by the precipitating CuCl (thus keeping actual [Cu+] very low). That’s equivalent to saying that CuCl lattice energy is defining the equilibrium.

sternman318 - 19-6-2011 at 16:52

SO

My synthesis:

Took a test tube, put a little bit of sodium bisulphate in it. I dont have any acids, and blogfast25 mentioned success without an acidic environment, so I thought I would take the middle road and make it slightly acidic. Couldn't have hurt.

I then added a large amount of NaBr and then a small portion of CuSO4. Stripped some wire and dropped in the copper, then added enough water to cover the wire.

Again, I guess I took the middle route and heated the tube for a few minutes in a hot water bath, then I covered it and let it sit. Immediately I could see some white powder. As time passed, more and more white powder appeared. After the solution started turning light brown, I added some more NaBr and CuSO4, as there was still plenty of wire in there. Today, I decided to collect my product.

I poured the light brown solution to water, and as Woelen predicted, I got my 'snow' ( I wish I had it on video, it was pretty cool haha), and I am letting it sit so i can decant/filter it later.

In the meantime, are there any reactions or tests I can line up? I had no reason to synthesize it, but I figured I had the chemicals and it was my first reaction of any type that I could do with the chemicals I had amassed. It is exciting when things go the way you planned

woelen - 19-6-2011 at 22:49

If you try to collect the CuBr, assure that you do the workup in a reducing environment. Commercial CuBr (and also CuCl) sometimes is delivered as a fine powder under water. In air it quickly oxidizes and becomes greenish/brown. You'll see that your now beautiful bright white powder quickly becomes a dirty looking mud when it is taken out of the water. You can prevent this too some extent by rinsing the white material with a solution of SO2 in water, then taking off the water as quickly as possible and finally rinsing with acetone or (even better: ether) to get rid of the final amount of water. Once the powder is perfectly dry, it keeps somewhat better.

Instead of using a solution of SO2 you can use a solution of Na2S2O5 (metabisulfite), but the disadvantage of this is that at drying a residue of this compound remains, while a solution of SO2 completely evaporates.

blogfast25 - 20-6-2011 at 12:41

Going back a bit to the chlorides:

Cu2+ + Cu === > 2 Cu+

As Mixell pointed out, E < 0, so ΔG > 0 and K < 1

We can now distinguish between two broad cases:

Very high chloride concentration - chlorocuprate-complexes form:

CuCl42- === > Cu2+ + 4 Cl-

The equilibrium constant (right over left) here is 1 / Kf,2, with Kf,2 the complexation constant.

2 Cu+ + 4 Cl- === > 2 CuCl2-

The equilibrium constant (right over left) here is Kf,1, with Kf,1 the complexation constant.

Add up these steps:

CuCl42- + Cu === > 2 CuCl2- + 2 Cl-

The equilibrium constant (right over left) here is K x Kf,1 / Kf,2. Clearly as K < 1, then Kf,1 / Kf,2 >> 1 for the reaction to proceed.

Lower chloride concentration - precipitation of CuCl:

Basically the same but with Ks, solubility product of CuCl, instead of Kf,1 and replacing CuCl2- with CuCl.

Metabisulfite substitute

jamit - 20-6-2011 at 21:18

Is there no way to keep the copper bromide from being oxidized except to put it into a weak solution of so2. Would a weak solution of sodium metabisulfite be ok? How can I keep the powder dry and white? Can I put it into a dissector to dry and is there some thing that will absorb oxygen so that the drying will maintain the copper bromide from being oxidized...or is this not feasible for home chemist?

woelen - 20-6-2011 at 22:45

Once your powder is dry, you can keep it white quite well by putting it in a small glass vial with a very tight cap. The trouble is the drying process itself. As long as CuBr (and also CuCl) is wet or just humid, it is very easily oxidized and for the home chemist it is very difficult to handle the chemical without ever getting it in contact with air. Such a thing would require a glove box with slow inert gas or nitrogen flow and I do not think there are many home chemists who have such equipment.

So, the best thing you probably can do is keep it wet under a dilute solution of SO2 (slightly acidified sodium metabisulfite also does the job). Try to collect a lot of the white snow with just a thin layer of water above it. A nice display vial would be to have the vial completely full of water (with some SO2 or acidified metabisulfite in it) and CuBr under that water, so that no surface can be observed.

[Edited on 21-6-11 by woelen]

jamit - 20-6-2011 at 22:47

Thanks woelen. I'll try that this week.

[Edited on 21-6-2011 by jamit]

[Edited on 21-6-2011 by jamit]

woelen - 21-6-2011 at 00:07

@blogfast25: Yes, you are right with your analysis. Both complex formation of copper(I) and formation of solid CuCl can drive the reaction towards formation of copper(I). I'll try the reaction with direct formation of copper(I) chloride myself, I've never seen that happen.

blogfast25 - 21-6-2011 at 05:16

Quote: Originally posted by woelen

@blogfast25: Yes, you are right with your analysis. Both complex formation of copper(I) and formation of solid CuCl can drive the reaction towards formation of copper(I). I'll try the reaction with direct formation of copper(I) chloride myself, I've never seen that happen.

A.F. Holleman describes it (at least in my very old edition), using CuCl2. I used CuSO4 + NaCl. No acid prescribed.