Chemical Reaction of aluminum foil in CuCl2 solution

Question

In class we did a lab in which we first dissolved copper(II) chloride powder in water to form a blue solution. We then placed a crumbled up piece of aluminum foil in that solution. It turned brown and the (originally blue) solution that came in contact with the aluminum turned colorless. The aluminum foil turned brown.

My question here is, what is the brown substance on the aluminum foil, and how do you know it's not rust?
Thanks in advance.

Answer 1

Rust is iron oxide; you have no iron in the system so it is not rust.

The brown substance is copper dust produced by the reaction $$\ce{2Al + 3CuCl2 -> 3Cu + 2AlCl3}$$

The aluminium chloride then hydrolyses to give $\ce{AlCl3}$ hexahydrate which is colourless and acidic explanation here.

Answer 2

The aluminum foil is a thin sheet of solid aluminum. You can see that it is white colored (or silver colored, if you prefer). So solid aluminum is white in color. Aluminum salts are colorless (or you see them as white). When aluminum salts are dissolved in water, the solutions are colorless.

Now, what is the color of solid copper? Remove the plastic insulation from a piece of electric wire to see the copper wire inside, or look at the copper bottom of a frying pan. You can see that copper is a brown solid.

Copper(II) salts are blue in color. So when you dissolved Cu(II)chloride in water, you got a blue solution. The blue color indicates the presence of Cu(II) ions in the solution.

Aluminum is more reactive than copper. Therefore when aluminum foil is put into the copper salt solution, aluminum atoms on the surface of the foil (in contact with the solution) reacts and takes the place of copper(II) ions in the solution (it now becomes aluminum chloride solution). That is why the blue solution becomes colorless.

The copper(II) ions that were originally in the solution are pushed out to form solid copper powder (brown) and sticks on the surface of the aluminum foil (in chemistry, we say that the aluminum "displaced" copper from the salt). So the brown coating which appears on the aluminum foil is solid copper.

Now how do you know that it is not rust? You must know that iron objects slowly rust as time passes. This is because the iron reacts with oxygen in the atmosphere and slowly changes to iron oxide in the presence of moisture. Therefore rust is iron oxide, which is also brown in color.

Now you can infer that the brown coating on the aluminum foil is not rust, because you did not use iron anywhere in your experiment. Aluminum and copper do not rust.

Answer 3

I agree with my colleagues to the extent that it is most probably copper metal. However, given that we are starting with an acidic copper chloride solution, and likely there is some oxygen exposure either from the starting water or from air contact, a little more chemistry may be taking place, more precisely, electrochemical (or displaying a surface anode/cathode mechanism). The following reaction is known to occur, for example, with cuprous, ferrous and likely other transition metals (Mn, Co, Ce, Cr,..) in acidic solutions in the presence of oxygen:

Cu(l)/Fe(ll) + 1/4 O2 + H+ --> Cu(ll)/Fe(lll) + ½ H2O

Source for the equation above for ferrous (which is fully available as an html file at https://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/html/final-78.html and relatedly also, comments at https://pubs.acs.org/doi/10.1021/es0501058 ) and for copper, see, for example, Eq (7) at https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide.

I would rewrite the above reaction given the propensity of some transition metals (like copper) to form basic salts as follows:

4 Cu(l) + O2 + 2 H+ --> 4 Cu(ll) + 2 OH-

For college students and researchers, I have interestingly derived an underlying radical chemistry path to this reaction employing the supplement, "Impacts of aerosols on the chemistry of atmospheric trace gases: a case study of peroxides radicals"', by H. Liang1, Z. M. Chen1, D. Huang1, Y. Zhao1 and Z. Y. Li, link: https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.atmos-chem-phys.net/13/11259/2013/acp-13-11259-2013-supplement.pdf&ved=0ahUKEwj64JTH4ejMAhVCHR4KHegXCu8QFggcMAE&usg=AFQjCNGZWxTNxFPmgaT1bARYjO08w2_wIg&sig2=Gbhj5orSHmeDIV9uy-moYg :

R24 O2(aq) + Cu+ → Cu2+ + O2− ( k = 4.6xE05 )

R27 O2− + Cu+ + 2 H+ → Cu2+ + H2O2 ( k = 9.4xE09 )

R25 H2O2 + Cu+ → Cu2+ + OH + OH− ( k= 7.0 xE03 )

R23 OH + Cu+ → Cu2+ + OH− ( k = 3.0×E09 )

Net reaction again: 4 Cu+ + O2 + 2 H+ → 4 Cu2+ + 2 OH-

Now, for the experiment discussed with Al/CuCl2, I suspect elemental copper metal, especially freshly precipitated, will similarly react forming a coating of cuprous oxide per the normally slow observed oxidation of iron and copper metal in an acidic air/CO2 mix per the so called metal auto-oxidation reaction (Reaction R24 above):

O2(aq) + 4 Cu/Fe = Cu(l)/Fe(ll) + •O2−

which is also a reversible reaction. Upon acidification, the superoxide radical anion (written as O2−, •O2− or O2•− , but while the latter form is in accord convention, it can present significant issues on readability and possible misinterpretation of the nature of the active specie) can lead to H2O2 (Reaction R27) and feed a Fenton (or Fenton-type with copper) reaction (see R25 above) creating a ferric or cupric salt. As a consequence, the seemingly observed darker coloration described may actually be due to the presence of Cu2O, and at low pH and elevated oxygen levels, some much darker spots of CuO could develop.

At temperatures over 65 C, one study (see the fully available article as a pdf at https://www.sciencedirect.com/science/article/pii/0001616069900467 ) refers to “the arrival of electrons at the oxide-gas interface by thermionic emission as the rate controlling step”, where I would expect this leads to the superoxide radical anion via:

e- (aq) + O2 (aq) = •O2−

And per another study also the reaction:

e- (aq) + H+ = •H

where the hydrogen proton (derived from water vapor) apparently occupies surface Cu vacancies, formed from major defects associated with Cu2O (see discussion by Wang and Cho at https://www.jstage.jst.go.jp/article/isijinternational/49/12/49_12_1926/_pdf ).

I would also note that atmospheric superoxide plus water vapor, due to the change in the dielectric of the medium, •O2− becomes largely present as •HO2 (the proton being apparently readily donated by water). Further, the latter hydroperoxide radical is not only a slow precursor to H2O2 (and a possible fenton-type reaction noted above per R25) via the reaction:

•HO2 + •HO2 = H2O2 + O2

but also, recently recognized as an acidic radical in itself (see ‘Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic’ at https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b08081?src=recsys&journalCode=jpcafh ).