The ideal answer must be copper, however why doesn"t copper react via hydrochloric acid while the other metals do?



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Here, reaction implies that

hydrogen gas is formedthe steel is dissolved

In order to form hydrogen, prolots have to be diminished to hydrogen atoms which then incorporate to $ceH2$.

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$$ce2 H+ + 2 e- -> H2$$

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

$$ceZn -> Zn^2+ + 2 e- $$

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


In principle, non-oxidizing acids cannot directly oxidize copper given that the redox potentials $E$ for $mathrmpH = 0$ present that $ceH+$ cannot oxidize $ceCu$ to $ceCu^2+$ or to $ceCu+$:

$$eginalignat2ce2H+ + 2e- ;& H2quad &&E^circ = +0.000 mathrmV\ceCu^2+ + 2e- ;& Cuquad &&E^circ = +0.340 mathrmV\ceCu+ + e- ;& Cuquad &&E^circ = +0.521 mathrmVendalignat$$

Keep in mind that copper(II) is normally even more secure than copper(I) in aqueous remedies.

However, the situation is slightly complicated because of the low solubility of $ceCuCl$ in dilute hydrochloric acid (in focused hydrochloric acid, copper forms chloricarry out complexes such as $ce-$ and also $ce^2-$):

$$K_mathrmsp = left cdot left = 1.72 imes 10^-7$$

The corresponding efficient redox potential may be estimated as follows:

$$eginalignedE&=E_ceCu+^circ+fracRTFcdotlnleft\ &=E_ceCu+^circ+fracRTFcdotlnfracK_mathrmspleft \ E_ceCuCl^circ&=E_ceCu+^circ+fracRTFcdotln K_mathrmsp\&= E_ceCu+^circ+0.0592 mathrmVcdotlog K_mathrmsp\&=0.521 mathrmV+0.0592 mathrmVcdotlogleft(1.72 imes 10^-7 ight)\&=0.121 mathrmVendaligned $$

Thus, oxidation of $ceCu$ to $ceCu+$ is favoured in dilute hydrochloric acid. Nonetheless, $ceH+$ is still not solid sufficient to oxidize $ceCu$.

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However, $ceCu$ deserve to be oxidized by $ceO2$:

$$ceO2 + 4H+ + 4e- 2H2Oquad E^circ = +1.229 mathrmV$$

As such, copper is progressively oxidized in dilute hydrochloric acid in call via air.