What happens when a piece of copper is placed in 1M HCl?

Question

I know that the answer is that nothing happens, but the explanation is confusing me in some places. Firstly, we are given that: \begin{align} \ce{Cl2 + 2e- &-> 2Cl-} & (E &= \pu{1.36 V})\\ \ce{Cu^{2+} + 2e- &-> Cu(s)} & (E &= \pu{0.34 V})\\ \ce{2H+ + 2e- &-> H2} & (E &= \pu{0 V}) \end{align}

What's confusing me is that in the solution, we'll have $\ce{H+}$, $\ce{Cl-}$ and $\ce{Cu(s)}$, which are all products in the equations (except for $\ce{H+}$). I get that for the reverse equation (oxidation), the oxidation potential is given by the negative of the reduction potential. What I'm wondering is, if we look at copper and chlorine first, is the only option for the $\ce{Cl-}$ in solution to be oxidised (with $E = \pu{-1.36 V}$)? Also, can the copper also only be oxidised (with $E = \pu{-0.34 V}$)? And if both of these are true, then does that mean that these two can't react at all, not because of the total $E$ being negative, but because they can both only be oxidised?

Answer 1

Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.

Your question is: What happens when a piece of copper is placed in 1M HCl?

Do a thought experiment + use some basic intuitive chemistry.

Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.

Now there are only two possibilities for each substance.

a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.

Now you have the problem well defined: Will Cu -> Cu$^{2+}$?

b) Follow the same reasoning: H$^+$, it can be oxidized to H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$

c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$

Now you can ask only two questions:

Can copper metal reduce H$^+$ to H$_2$?

or

Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.

At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation Ecell= E$_{cathode}$-E$_{anode}$

Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)

Ecell= 0.00- (+0.34) = - 0.34 V

The negative sign shows this not possible under these conditions.

Do the same for chloride ion.

Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell) Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.

In short nothing will happen to copper in HCl.

P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.

Answer 2

Without oxygen or oxidating agent, nothing happen to copper in hydrochloric acid, but dissolution of surface oxides. The only oxidant there is $\ce{H+}$, which is with $E^{\circ}=\pu{0.00 V}$ too weak oxidant to oxidize $\ce{Cu}$ with $E^{\circ}=\pu{+0.34 V}$.

If oxygen is present, then it gets slowly dissolved to tetrachlorocuprate ( similarly as to copper acetate in vinegar ):

$$\ce{2 Cu + O2 + 4 H+ + 8 Cl- -> 2 CuCl4^2- + 2 H2O}$$

The dissolution is very quick, if oxidant as hydrogen peroxide is present( it dissolves gold as well )

$$\ce{ Cu + H2O2 + 2 H+ + 4 Cl- -> CuCl4^2- + H2O}$$

Answer 3

OK, if you takes a piece of copper metal (I employ plumbing copper, high purity) in concentrated HCl the copper, in time, will be cleaned of its Cu2O coating. [EDIT] To quote a source:

Copper has excellent atmospheric corrosion resistance. It becomes naturally covered with an oxide film, changing its color to dark brown/black in normal atmospheric conditions. Later, a green patina forms on copper outdoors with a varying intensity depending on where and how the surface is exposed.

So, perform an experiment on copper metal taking before and after pictures, and see if it is noticeably different.

Underlying chemistry, per Wikipedia, to quote:

Copper(I) chloride...is a white solid sparingly soluble in water, but very soluble in concentrated hydrochloric acid. Impure samples appear green due to the presence of copper(II) chloride (CuCl2).

Also, forms a chloro-complex with concentrated HCl:

It forms complexes with halide ions, for example forming H3O+ CuCl2− in concentrated hydrochloric acid.

So, expected reactions in concentrated HCl, a cleaning off of any Cu2O coating:

$\ce{HCl + H2O <=> H3O+ + Cl-}$

$\ce{Cu2O + 2 HCl -> 2 CuCl (s) + H2O}$

$\ce{CuCl (s) + H3O+ + Cl- <=> (H3O)CuCl2 (aq)}$

I have performed experiments where one adds much NaCl creating a soluble cuprous presence (in the form of NaCuCl2). Wikipedia also cites a preparation to Cupric oxychloride based on this solubility conversion of cuprous per Eq(6) and Eq(7) here, where:

$\ce{ Cu (s) + CuCl2 (aq) -> 2 CuCl (s)}$

$\ce{ 2 CuCl (s) + 2 NaCl (aq) -> 2 NaCuCl2 (aq)}$

There is no other apparent reaction occurring with the freshly cleaned Copper.

Answer 4

The three equations you list are not enough to answer the question. The electrochemical potentials are for standard conditions, unless otherwise specified. Now when you put a piece of copper into 1 M HCl, the [H+] is 1 M (standard), the [Cl-] is 1 M (standard) and the [Cu++] is zero (nonstandard). The equation to consider is

2 H+ + Cu --> Cu++ + H2

Since the copper concentration initially is exactly zero, a few copper atoms will dissolve; the driving force can be calculated from the equation:

E = 0.34V - (0.0591V)/2 log(1.0 M/x M) = 0.34V - 0.02955V log(1/x)

0.34/0.02955 = log(1/x) = 11.5

x = 10^-11.5 = 3 x 10^-12 M

The final concentration of copper from that reaction is very small (3 trillionths of a mole per liter, 190 ppt by weight), but not zero. And I haven't taken into account that the H2 is not at standard pressure (1 atm), which will drive the voltage up a bit more, and the eventual [Cu++] also. So copper is not totally inert in 1 M HCl. And that's assuming you are talking about an inert atmosphere, like 100% N2.

In real life, this would be done in air (O2 = 0.2 atm). And oxygen will pick up electrons and give hydroxyl:

O2 + 2H2O + 4e- --> 4 OH- E = 0.40 V

Remember Cu++ + 2 e- --> Cu E = 0.34 V

Now, in 1M HCl, the [OH-] is 10^-14, so the equation for the oxygen cell voltage becomes:

E = 0.40 -0.0591V/4 log((10^-14)^4))/0.2) = 0.40 - 0.014775 - 55.3 = 1.217

Wow! All of a sudden, the cell voltage for oxygen reduction is driven high because of the acid (i.e., [OH-] = 10^-14, not 1 M). So now, copper can be oxidized and will dissolve.

You don't believe me? Drop some HCl (1 M or higher) onto a penny and watch it corrode - and when it breaks thru into the zinc interior, bubbles ahoy!