Why doesn't copper react with hydrochloric acid while the other metals do?

Question

5) Which one of the following metals does not react with hydrochloric acid?

A. zinc
B. magnesium
C. iron
D. copper
E. aluminium

The right answer must be copper, but why doesn't copper react with hydrochloric acid while the other metals do?

Answer 1

You might want to look up some terms, such as

• noble and less noble metals
• reduction potential
• galvanic series

Here, reaction means that

• hydrogen gas is formed
• the metal is dissolved

In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.

$$\ce{2 H+ + 2 e- -> H2}$$

The metal serves as an electron donor and is oxidized, e.g.

$$\ce{Zn -> Zn^{2+} + 2 e- }$$

The more noble a metal is, the more reluctant it is to lose electrons. This is the case for copper, which is therefore not oxidized under these conditions.

Answer 2

In principle, non-oxidizing acids cannot directly oxidize copper since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$ or to $\ce{Cu+}$:

$$\begin{alignat}{2} \ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\ \ce{Cu^2+ + 2e- \;&<=> Cu}\quad &&E^\circ = +0.340\ \mathrm{V}\\ \ce{Cu+ + e- \;&<=> Cu}\quad &&E^\circ = +0.521\ \mathrm{V} \end{alignat}$$

Note that copper(II) is usually more stable than copper(I) in aqueous solutions.

However, the situation is slightly complicated because of the low solubility of $\ce{CuCl}$ in dilute hydrochloric acid (in concentrated hydrochloric acid, copper forms chlorido complexes such as $\ce{[CuCl2]-}$ and $\ce{[CuCl4]^2-}$):

$$K_\mathrm{sp} = \left[\ce{Cu+}\right] \cdot \left[\ce{Cl-}\right] = 1.72 \times 10^{-7}$$

The corresponding effective redox potential may be estimated as follows:

$$\begin{aligned} E&=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Cu+}\right]\\ &=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\frac{K_\mathrm{sp}}{\left[\ce{Cl-}\right]} \\ E_{\ce{CuCl}}^\circ&=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln K_\mathrm{sp}\\ &= E_{\ce{Cu+}}^\circ+0.0592\ \mathrm{V}\cdot\log K_\mathrm{sp}\\ &=0.521\ \mathrm{V}+0.0592\ \mathrm{V}\cdot\log\left(1.72 \times 10^{-7}\right)\\ &=0.121\ \mathrm{V} \end{aligned} $$

Therefore, oxidation of $\ce{Cu}$ to $\ce{Cu+}$ is favoured in dilute hydrochloric acid. Nevertheless, $\ce{H+}$ is still not strong enough to oxidize $\ce{Cu}$.

However, $\ce{Cu}$ can be oxidized by $\ce{O2}$:

$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad E^\circ = +1.229\ \mathrm{V}$$

Therefore, copper is slowly oxidized in dilute hydrochloric acid in contact with air.

Answer 3

Although you should take a look at what Klaus said, copper does in fact react with hydrochloric acid, it just takes a week until all the copper is converted into copper chloride (green) and another week or so until it forms crystals and you can dissolve them in water to form copper chloride again (but depending on the amount of chloride it has, it'll be blue or green).

$$\ce{2HCl (aq) + Cu(s) -> CuCl2(aq) + H2(g)}$$

Copper chloride reacts with aluminium to form copper (metal) and aluminium chloride.

$$\ce{3CuCl2(aq) + 2Al(s) -> 3Cu(s) + 2AlCl3(aq)}$$