Sciencemadness Discussion Board - Making Copper Chloride

Making Copper Chloride

NiK - 21-11-2013 at 18:42

I'm trying to do a cool experiment that my high school chemistry teacher did years go. It was copper (II) chloride with Al foil added to demonstrate a single replacement reaction (precipitating out the Cu).

How do I make the Copper (II) Chloride solution? I watched a YouTube video that mixed HCl and H2O2 together and then added copper wire to dissolve and then evaporated off the liquid...but I still don't understand what's going on, and that's the interesting part.

So with a 2:1 ratio of HCl to H2O2 (they have about the same molar mass and the balanced equation is H2O2+ 2HCl I just mixed the same volume of some 15% H202 and ~30% HCl, that sound right??) what is your product? The balanced equation I was using says it would be chlorinated water?? or just water that will give off Cl2 gas? I have about 70mL of a murky yellow solution that fizzes when I swirl it in the beaker; what is it?

Also, from this solution what is the process with the Cu wire to make copper II chloride?? Ideally I would like the equations/stoichiometry behind it so I actually get the satisfaction of knowing/understanding what I'm doing.

Thanks!

Also if it helps at all, the solution seems to be slowly reacting like a freshly poured glass of soda; all the super tiny little bubbles jumping off the surface. Oh and I used Cloroben for my HCl and BioGuard SafeSwim for my H2O2 if this any provides insight.

[Edited on 11-22-13 by NiK]

Anddd that video was for dihydrate or something, is making the copper II chloride as easy as mixing copper sulfate (aq) with table salt?? What would the balanced equation for this looks like with the CuSO4 * 5H2O?? would it be CuSO4 * 5H2O+ NaCl---> H2O andddd?

Sorry to combine a million questions into one and to keep editing this

[Edited on 11-22-13 by NiK]

blargish - 21-11-2013 at 21:29

To answer your final question, reacting copper sulfate with sodium chloride will not work since there is no insoluble product to drive the reaction. You will just end up with a large soup of aqueous ions, namely Na⁺, Cl^-, Cu²⁺, and SO₄²⁺.

The H₂O₂ is needed in the reaction between HCl (aq) and Cu due to the fact that HCl (aq) by itself is not a strong enough oxidizer to oxidize the Cu to Cu²⁺. (It is actually the H⁺ ion that does the oxidizing with most acids, as it is reduced in turn to H₂, so I guess H⁺ is not strong enough to oxidize Cu to Cu²⁺). I am not sure exactly how the entire reaction takes place, but I believe it follows a route similar to this:

Half equations:

Cu(s) --> Cu²⁺(aq) + 2e

H₂O₂(aq) + 2H⁺(aq) + 2e --> 2H₂O(l)

Base redox equation:

Cu(s) + H₂O₂(aq) + 2H⁺(aq) --> Cu²⁺(aq) + 2H₂O(l)

Then the Cu²⁺(aq) reacts with the HCl(aq) via this reaction:

Cu²⁺(aq) + 2HCl(aq) --> CuCl₂(aq) + 2H⁺(aq)

Overall:

Cu(s) + H₂O₂(aq) + 2H⁺(aq) + 2Cl^-(aq) --> CuCl₂(aq) + 2H₂O(l)

I am not entirely sure whether I am correct, so someone please correct me if I'm wrong.

(Thanks to Draconic Acid,) The gas you see bubbling is O₂ as the copper acts as a catalyst for the decomposition of the peroxide, so the following side reaction is taking place:

2H₂O₂(aq) --> 2H₂O(l) + O₂(g)

Also, you seem to be using the coefficients as mass ratios. They are not mass ratios, but molar ratios, meaning that X moles of this will react with Y moles of that. Therefore, your mixing of equal volumes of 15% H₂O₂ and 30% HCl was not correct since the percentages are relating to mass.

I have found that the easiest way to create copper II chloride is via a two step process with the formation of copper II carbonate first. You can easily make copper II carbonate with your copper II sulfate and baking soda (sodium bicarbonate) with the following reaction:

CuSO₄(aq) + 2NaHCO₃(aq) --> CuCO₃(s) + Na₂SO4(aq) + CO₂(g) + H₂O(l)

The copper II carbonate precipitates out and is easy to filter off.
Then react the copper II carbonate with HCl(aq).

CuCO₃(s) + 2HCl(aq) --> CuCl₂(aq) + CO₂(g) + H₂O(l)

The best thing about this reaction is that all you are left with at the end is a solution of exclusively copper II chloride.
Hope I have been of some help!

[Edited on 22-11-2013 by blargish]

[Edited on 22-11-2013 by blargish]

[Edited on 22-11-2013 by blargish]

DraconicAcid - 21-11-2013 at 22:00

Quote: Originally posted by blargish

The H₂O₂ is needed in the reaction between HCl (aq) and Cu due to the fact that HCl (aq) by itself is not a strong enough oxidizer to oxidize the Cu to Cu²⁺. (It is actually the H⁺ ion that does the oxidizing with most acids, as it is reduced in turn to H₂, so I guess H⁺ is not strong enough to oxidize Cu to Cu²⁺). I am not sure exactly how the entire reaction takes place, but I believe it follows a route similar to this:

Half equations:

Cu(s) --> Cu²⁺(aq) + 2e

H₂O₂(aq) + 2H⁺(aq) + 2e --> 2H₂O(l)

Base redox equation:

Cu(s) + H₂O₂(aq) + 2H⁺(aq) --> Cu²⁺(aq) + H₂O(l)

You get 2 waters being produced.

Quote:

Then the Cu²⁺(aq) reacts with the HCl(aq) via this reaction:

Cu²⁺(aq) + 2HCl(aq) --> CuCl₂(aq) + H₂(g) , and this is the gas that you see bubbling off...

No- the bubbling of gas is because the copper also catalyzes the decomposition of hydrogen peroxide; oxygen gas is being given off.

Quote:

I have found that the easiest way to create copper II chloride is via a two step process with the formation of copper II carbonate first. You can easily make copper II carbonate with your copper II sulfate and baking soda (sodium bicarbonate) with the following reaction:

CuSO₄(aq) + NaHCO₃(aq) --> CuCO₃(s) + Na₂SO4(aq) + CO₂(g) + H₂O(l)

The copper II carbonate precipitates out and is easy to filter off.
Then react the copper II carbonate with HCl(aq).

CuCO₃(s) + HCl(aq) --> CuCl₂(aq) + CO₂(g) + H₂O(l)

The best thing about this reaction is that all you are left with at the end is a solution of exclusively copper II chloride.
Hope I have been of some help!

You need 2 HCl in this reaction.

blargish - 22-11-2013 at 05:01

Quote: Originally posted by DraconicAcid

No- the bubbling of gas is because the copper also catalyzes the decomposition of hydrogen peroxide; oxygen gas is being given off.

Ok, thanks for clarifying

Quote: Originally posted by DraconicAcid

You get 2 waters being produced

Quote: Originally posted by DraconicAcid

You need 2 HCl in this reaction

Gahh... oops

watson.fawkes - 22-11-2013 at 06:22

Quote: Originally posted by NiK

How do I make the Copper (II) Chloride solution?

Copper chloride is used as an etchant for printed circuit boards (PCB). There's plenty of online information on preparing this etchant. A page I've reference on this board before on this subject is Etching with Air Regenerated Acid Cupric Chloride. There are two methods not yet mentioned above. One is to use copper oxide or copper hydroxide and HCl. In both cases the copper is already oxidized. In this form it dissolves and complexes readily. Another way is to use atmospheric air to oxidize the copper. Reaction rate for this is limited by air exchange, so use of a bubbler is advised. This method works because any copper (II) ions formed quickly react with metallic copper to form copper (I) ions. Subsequently the copper (I) oxidizes to copper (II) in solution. (Note: all these copper ions are actually chlorine complexes; I've mentioned the simplified form.)

Copper oxide (also the carbonate) can be bought as a glaze material from ceramics suppliers. It's commonly sold in 1 lb. bags. With a little shopping you can pick up a pound for less than ten dollars. Example (no affiliation). For the purpose of a demonstration, the pottery grade is adequately pure.

NiK - 22-11-2013 at 07:33

Thank you! That clarifies it a lot; I'll try doin the method you mentioned, it sounds easier! Also I had not added the copper yet at any point so the bubbling couldn't have been from that. I'm thinking maybe there is other stuff in my H2O2 or HCl because that murky yellow solution bubbled like that for a while and was still quite warm to the touch after 45 minutes to an hour. It eventually formed some kind of white milky-looking precipitate :/

Quote: Originally posted by blargish

Also, you seem to be using the coefficients as mass ratios. They are not mass ratios, but molar ratios, meaning that X moles of this will react with Y moles of that. Therefore, your mixing of equal volumes of 15% H₂O₂ and 30% HCl was not correct since the percentages are relating to mass.

I realize I was technically using the wrong things for the ratio between HCl and H2O2 but since HCl has a molar mass of bout 35.5g/mol and H2O2 is about 34g/mol I just used an equal volume of each (except the HCl was about twice the concentration). This works out because the molar masses are almost the same, right? I'm obviously not doing a quantitative analysis haha.

watson.fawkes- Thank you for that, I will also give that a shot! Then I'll compare my results

Thanks!

[Edited on 11-22-13 by NiK]

NiK - 22-11-2013 at 07:50

Quote: Originally posted by blargish

I have found that the easiest way to create copper II chloride is via a two step process with the formation of copper II carbonate first. You can easily make copper II carbonate with your copper II sulfate and baking soda (sodium bicarbonate) with the following reaction:

CuSO₄(aq) + NaHCO₃(aq) --> CuCO₃(s) + Na₂SO4(aq) + CO₂(g) + H₂O(l)

The copper II carbonate precipitates out and is easy to filter off.
Then react the copper II carbonate with HCl(aq).

CuCO₃(s) + 2HCl(aq) --> CuCl₂(aq) + CO₂(g) + H₂O(l)

The best thing about this reaction is that all you are left with at the end is a solution of exclusively copper II chloride.
Hope I have been of some help!

On U.S. Pigment you can buy CuCO₃ for 7 bucks a pound. Sounds like more fun to just make my own though, my filtration skills need some practice anyways...or maybe it's just my patients that needs work

blargish - 22-11-2013 at 08:54

Quote: Originally posted by NiK

Quote: Originally posted by blargish

I have found that the easiest way to create copper II chloride is via a two step process with the formation of copper II carbonate first. You can easily make copper II carbonate with your copper II sulfate and baking soda (sodium bicarbonate) with the following reaction:

CuSO₄(aq) + NaHCO₃(aq) --> CuCO₃(s) + Na₂SO4(aq) + CO₂(g) + H₂O(l)

The copper II carbonate precipitates out and is easy to filter off.
Then react the copper II carbonate with HCl(aq).

CuCO₃(s) + 2HCl(aq) --> CuCl₂(aq) + CO₂(g) + H₂O(l)

The best thing about this reaction is that all you are left with at the end is a solution of exclusively copper II chloride.
Hope I have been of some help!

I made a mistake in balancing the first equation there. There should be a 2 in front of the NaHCO₃

NiK - 22-11-2013 at 09:56

I was gonna say, there arent 2 Nas on each side! But the little formal education in chemistry that I do have had be balancing it first anyways

Thank you, I will give this a shot when I get home from work TGIF--does anyone else here consider a fun Friday night chilling in my 1 person apartment doing chemistry experiments, or is that just me?? Haha

AJKOER - 13-12-2013 at 19:08

A few comments:

The hydrolysis of a dilute solution of NaHCO3 can be expressed as:

NaHCO3 (diluted) + 4Н2О = [Na(H2O)4]+ + НСO3-

НСO3- + Н2О ↔ Н2СО3 + ОН-; рKo = 7.63 (source: see http://www.allreactions.com/index.php/group-1a/natrium/sodiu... )

I suspect that a hot solution of Sodium bicarbonate would move the equilibrium to the right with the release of CO2 gas and the formation of some NaOH.

Now, per Wikipedia (http://en.wikipedia.org/wiki/Tribasic_copper_chloride ), to quote:

"Cu2(OH)3Cl can be prepared by hydrolysis of a CuCl2 solution at pH 4 ~7. A variety of bases such as sodium carbonate, ammonium, calcium, or sodium hydroxide may be used (eq. 3).[1]

2CuCl2 + 3 NaOH → Cu2(OH)3Cl + 3 NaCl (eq.3)"

So a basic copper chloride may form with a hot Sodium bicarbonate solution. Similarly, a hot aqueous reaction of CuSO4 and NaHCO3 may form a basic copper carbonate and Na2SO4. However, this is not a problem as, in any event, upon treating with HCl, for example:

Cu2(OH)3Cl + 3 HCl → 2 CuCl2 + 3 H2O
--------------------------------------------------------------

Interestingly, I believe I have easily prepared Cu2(OH)3Cl by reacting Copper in a dilute ammonia, 3% H2O2 and NaCl solution. That is, without the use of strong HCl! A greenish insoluble precipitate is visible (but, latter dissolves, there is a reference that explains why, see https://www.google.com/url?sa=t&rct=j&q=&esrc=s&... ). Also, per the same Wikipedia source:

"In the alkaline pathway, cuprammine chloride solution can be neutralized with HCl or other available acidic solutions (eq. 8).

2 [Cu(NH3)4Cl2] + 5 HCl + 3 H2O → Cu2(OH)3Cl + 8 NH4Cl (eq. 8) "

So, upon treating Cu(NH3)4Cl2 with a large excess of HCl (perhaps dilute), we may be able to form CuCl2 without strong HCl.

Alternately, form Cu(NH3)4(OH)2 by dissolving Copper in a dilute ammonia and 3% H2O2. Treat with HCl.
---------------------------------------------------

I have produced Copper oxychloride in large amounts via my so called 'bleach battery' (see http://www.sciencemadness.org/talk/viewthread.php?tid=24318&... ). It is actually a reaction of HOCl (from vinegar and bleach NaOCl) in the presence of Al and Cu metal with added NaCl to form the electrolyte. Remember to place the Copper electrode near the top of the solution (where it releases Cl2 gas) in a closed vessel. In a day, the formation of large amounts of the insoluble green copper salt is visible.

Separate out and treat with Na2CO3 to form the basic copper carbonate and proceed per above with HCl to form the CuCl2.

[Edited on 14-12-2013 by AJKOER]

macckone - 14-12-2013 at 23:45

CaCl₂ + CuSO₄ -> CaSO₄ + CuCl₂

The plaster of paris that is formed can be hard to separate unless a lot of water is used for solution.
Recrystallization is necessary if you use this method but it is basically a one pot method.

Dissolve your two chemicals in solution, mix the solutions and wait for the precipitate.

The cupric chloride will remain in solution. Boil off the water to get a crude product.
Dissolve in enough boiling water to dissolve all but a small amount of the crude product.
Filter any remaining calcium sulfate and crude curpic chloride.
Add a small amount of hydrochloric acid to the boiling hot solution.
As it cools a first crop of crystals will form. Filter and boil off more water.
Cool and more crystals will form. Repeat until the water has been reduced to
10% of the amount added.

These crystals should be fairly pure and only a small amount of hydrochloric acid is required. Ie. this is reasonably safe for a school lab.