Sciencemadness Discussion Board

Reduction of CuCl2 to CuCl

rstar - 5-10-2011 at 23:32

Hi Geeks,

Recently I tried to produce Copper(I) Chloride, by the Reduction of Copper(II) Chloride.
I added NaCl to a Hot solution of CuSO4, to produce CuCl2 :
2NaCl + CuSO4 = Na2SO4 + CuCl2
The solution changed from Bluish to Green color

Then I heated this solution and added K2S2O5, and there was a lot of bubbling. After bubbling stopped I got a yellow solution, but there wasn't any white precipitate of CuCl, as I expected.

I added NaOH to the solution and there was an instant red precipitate, which i believe is Cu2O, and therefore it is sure that the solution has Copper(I) or Cu<sup>+</sup> ions.

Can anyone tell me how to get CuCl out of that solution ???

kmno4 - 5-10-2011 at 23:53

Add water.
Be sure that solution is acidic, in another case your CuCl will be contaminated with Cu2O. Wet CuCl is extremely sensitive to air, so you have to act quickly.

rstar - 6-10-2011 at 00:24

It didn't work for me :(

blogfast25 - 6-10-2011 at 09:30

Search the forum for this topic: there are quite a few threads...

Nicodem - 6-10-2011 at 10:08

Quote: Originally posted by rstar  

I added NaCl to a Hot solution of CuSO4, to produce CuCl2 :
2NaCl + CuSO4 = Na2SO4 + CuCl2
The solution changed from Bluish to Green color

I'm afraid that the above equation makes no sense. First of all, you can not have NaCl, CuSO4, etc., in a monophasic aqueous solution. Instead you have solvated ions and various species derived from protonation and coordination equilibria. For example, hydrated Cu<sup>2+</sup> ions tend to exchange the H<sub>2</sub>O ligand with Cl<sup>-</sup> ligand, thus forming various chlorocuprate anions. You can observe the result of this reaction by the characteristic colour change that accompanies the ligand exchange.
Quote:
It didn't work for me

Since you did not provide the experimental data, nobody can find the origin of the unsuccessful synthesis. Though, I can make a worthless guestimate that perhaps you used a large excess of NaCl, thus preventing the formation of CuCl(s). Copper(I) cations can form chlorocuprates(I). I would expect the sodium cation present in the solution does not form insoluble chlorocuprate(I) salts, also because the complex anion is not particularly stable (fast H<sub>2</sub>O <=> Cl <sup>-</sup> ligand exchange), hence no precipitate.

nezza - 6-10-2011 at 11:20

Is bisulphite a strong enough reducing agent for this reaction ?. The method I have seen uses copper turnings in an acid solution with chloride ion present.

kmno4 - 6-10-2011 at 12:05

Take 1 drop of your yellow solution and add it to 1 cm3 of water.
Water should become cloudy and possibly you will get some white precipitate (quickly getting blue) - it is "CuCl".
If water stays clear - no Cu(I) is present.

woelen - 6-10-2011 at 12:39

Na2S2O5 definitely is a strong enough reductor. Collect a lot of the red precipitate (no need to dry, just filter and rinse with some water) and add this to concentrated HCl to which some Na2S2O5 is added. All of the precipitate will dissolve and you'll get a colorless solution in an ideal situation, in practice it will be somewhat greenish/yellow. Try to dissolve the red material in as little as possible of HCl to which Na2S2O5 is added beforehand. Then pour the solution into water. A white 'snow' will be formed. This white 'snow' is CuCl.

rstar - 6-10-2011 at 20:00

Thanx for suggestions ;)

Megamarko94 - 7-10-2011 at 02:19

this how i did it..
take some CuCO3 and add HCl to it... now you have CuCl2 in solution.
take some copper wire and make it into coil and put into the beaker now heat the solution with copper wire dont boil it dry.
get an icebath and put the beaker in it and you will see white crystals apear.(CuCl)
mine changed color to greenish due to oxidation in air and contaminants.