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?

  • 1
    $\begingroup$ Copper metal cannot oxidize chloride ion to chlorine gas, so forget the chloride ions. As the reduction potentials show, copper ions are spontaneously reduced to copper metal, relative to hydrogen ions being reduced to hydrogen gas. So the spontaneous reaction would be copper (II) ion plus hydrogen gas => copper metal plus hydrogen ions. But you have neither reactant: you only have the products. So the net result is nothing happens. $\endgroup$
    – Ed V
    Apr 27, 2020 at 12:30
  • $\begingroup$ One of your previous questions had a good answer by @titaniummorro, so why not do exactly what they did on this question? Simply compare standard cell potentials for $\ce{Cu (s) + 2H^+ (aq) => Cu^2+ (aq) + H_2 (g)}$ and the reverse reaction. $\endgroup$
    – Ed V
    Apr 27, 2020 at 13:05
  • 2
    $\begingroup$ @Adithya, Copper does not react with HCl in the absence of oxygen nor it will liberate hydrogen even if it is concentrated. Do you know a reference which says so? $\endgroup$
    – AChem
    Apr 27, 2020 at 16:15
  • 1
    $\begingroup$ @Adithya The link does not support what you say: it deals with copper complexes and copper compounds, not copper metal in hydrochloric acid that does not have an extrinsic oxidizing agent, e.g., dissolved oxygen from the air. $\endgroup$
    – Ed V
    Apr 27, 2020 at 16:19
  • 2
    $\begingroup$ @Adithya The reference you provided is interesting, but the question at hand is solely about clean copper metal (no oxides, patina, etc.) and 1 M HCl with no dissolved oxygen from the air or any other additional oxidizing agent, suspected or not. This is turning into the "X-Y" problem. $\endgroup$
    – Ed V
    Apr 27, 2020 at 17:32

4 Answers 4


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$_2$?


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.

  • $\begingroup$ The expansive thought-provoking answer still does not address what actually occurs in the conditions presented. The copper metal is 'cleaned' of its oxide coating (natural formed), and a little known cuprous chloro complex is created, as I noted in my answer. $\endgroup$
    – AJKOER
    Apr 29, 2020 at 16:40
  • $\begingroup$ Pardon my criticism. Note, if there is a chemist who has performed a reaction, and I get some of the details wrong, please speak up, as I, and perhaps others, should welcome the constructive advancement of the science. $\endgroup$
    – AJKOER
    Apr 29, 2020 at 16:49
  • $\begingroup$ Your point is a classical X-Y problem. You are addressing a problem which was not asked and it does not address the main problem, what happens to pure copper metal when it is dipped in 1 M HCl. You are right, a small amount of copper oxide dissolves, but the chloro complex is only formed in concentrated acid not in dilute 1 M HCl. $\endgroup$
    – AChem
    Apr 29, 2020 at 17:10
  • $\begingroup$ Thank you! That makes it clearer :) $\endgroup$
    – MatH
    Apr 30, 2020 at 0:22

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 )

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

  • $\begingroup$ Excuse me, but 2/3 of your 'answer' addresses possible reaction with O2 or H2O2 creating a cupric complex that will not be actually created as per the experiment described. The actual, little known, cuprous complex created is detailed in my response. Your not knowing this is understandable, but should not be upvoted, in my opinion. $\endgroup$
    – AJKOER
    Apr 29, 2020 at 16:25
  • $\begingroup$ @AJKOER Well, the OP has been interested in Cu itself. $\endgroup$
    – Poutnik
    Apr 29, 2020 at 17:36

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.


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!

  • 2
    $\begingroup$ You should brush up on formatting... chemistry.meta.stackexchange.com/questions/86/… $\endgroup$
    – Mithoron
    Apr 27, 2020 at 17:21
  • $\begingroup$ I guess I should do that. $\endgroup$ Apr 28, 2020 at 2:54
  • $\begingroup$ :/ defacing posts isn't welcome like anywhere. You can delete or improve it or let it be. I suggest to check out this link^ learn by practice. $\endgroup$
    – Mithoron
    Apr 28, 2020 at 16:42
  • $\begingroup$ @Mithoron: I thought I could do better on my own time, slowly. The original was rushed. $\endgroup$ Apr 28, 2020 at 17:46
  • $\begingroup$ Well, you can improve gradually, a bit here and there. However, unless you'd delete and undelete later, it should be visible. $\endgroup$
    – Mithoron
    Apr 28, 2020 at 18:16

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