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$\ce{Fe}$, $\ce{Mg}$, $\ce{Ni}$, $\ce{Pb}$, $\ce{Sn}$, and $\ce{Zn}$ all react when they (in solid form) are submerged in an acid solution with the presence of a strong acid like $\ce{HCl}$, but silver and copper do not.

I have examined the electronegativity values and both $\ce{Ag}$'s and $\ce{Cu}$'s electronegativity value is only different from $\ce{Ni}$ by $\mathrm{0.1}$ on the Pauling scale. I don't see a pattern; the only thing I see which is common between copper and silver is that they both may have an oxidation number of +1.

So, why is it that $\ce{Ag}$ and $\ce{Cu}$ are immune to the H cations in the acidic solution?

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The simple answer: the electrode potential $E$ for the reaction between, e.g., solid copper and $\ce{H+}$ to yield $\ce{Cu^{+2}}$ and hydrogen gas is negative under standard conditions. Since $\Delta G_{cell} = -nFE_{cell}$, negative electrode potential implies a positive change in free energy, which indicates a nonspontaneous reaction (i.e., net work has to be done on the system for the reaction to occur, for example by application of external current). This may not be a satisfactory answer, though, since it leaves the question of why the reaction is spontaneous with other metals. – Greg E. Jun 15 '14 at 0:59
Both silver and copper react with nitric acid. So, don't say a strong acid like HCl, instead say they don't react with concentrated HCl. – LDC3 Jun 15 '14 at 0:59
@trb456, I don't think it would be a fully satisfying answer as it currently stands. Perhaps when I have a bit more time I'll consider the pertinent thermodynamic and physical parameters (i.e., things like ionization energies, enthalpies, etc.) and write up a complete answer that addresses the question at a more fundamental level. – Greg E. Jun 15 '14 at 1:50
Nitric acid is an oxidizer as well. The reaction with copper is $Cu + HNO_3 \rightharpoonup Cu^+ + H_2O + N_2O_4$ (unbalanced). – LDC3 Jun 15 '14 at 1:53
@Klik, the difference is that, with nitric acid, there are alternative redox reactions available that are more favorable, i.e., the formation of $\ce{NO2 + H2O}$. Other pathways also exist, depending on concentration and other conditions. – Greg E. Jun 15 '14 at 1:53
up vote 4 down vote accepted

This will involve some degree of hand-waving. Let me first limit the analysis to the transition metals so that more effective comparisons can be made.

The reaction between a metal and an acid requires the metal to be oxidized, i.e. lose electrons to the acid, creating positively-charged metal ions in solution. The general idea is that as you go along the transition metals in a period from left to right, there is an increase in effective nuclear charge felt by the valence $ns$ electrons (due to the additional $(n-1)d$ subshell electrons poorly shielding the $ns$ electrons from the nucleus) making the valence electrons harder to remove. This trend can be seen to some extent in the increase in ionization energies of the elements, for example, and also in the standard reduction potentials for metal cations of same charge (they become more positive, in other words making oxidation and hence attack by acid more difficult). Thus, metals at the left of the transition metals tend to be more easily attacked by acids than the metals at the right.

There is a rather interesting exception though: zinc strongly breaks the trend, being far easier to oxidize than copper, the element before it, even though zinc has a much higher ionization energy, clearly indicating its electrons are held tighter (cadmium also exhibits this anomaly, to a milder extent). What gives? The thing is that the tendency for a metal to oxidize is the result of several combined factors, one of which is also how strongly bound the solid metal is in the first place. If a metal contains atoms which are strongly bonded to each other, then oxidation tends to be more unfavourable as it would require these bonds to break. When comparing copper and zinc, it is clear that the latter has far weaker metallic bonding in the solid (their melting/boiling points are $1360\ \rm{K}$ / $2835\ \rm{K}$ and $695\ \rm{K}$ / $1180\ \rm{K}$, respectively). Thus, if you include the energy necessary to separate the atoms from the solid before ionizing them, it turns out that the process is easier for zinc than copper.

Comparing different rows of the transition metals is a bit less clear, in part because not all elements have ions of the same charge which can be directly compared. The general tendency is that the metals become tougher to oxidize as you go down the rows. For the transition metals in the sixth and seventh period this is probably a consequence of lanthanide/actinide contraction and relativistic effects, which decrease the energy of the valence $ns$ orbitals.

Now for some other elements. Magnesium is very reactive towards acids because it is both a metal containing relatively weak bonds and because its ionization energy is comparatively low, being an alkaline earth element. For tin, the metallic bonding is not weak, but its ionization energy is not too high, so it will oxidize in acids, though not as strongly. Lead is somewhat less reactive to acids, possibly because its oxidation tends to stop at +2 instead of tin's +4, creating a significantly softer cation which isn't as well solvated by water.

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So if you were given the periodic table and asked to predict another metal that wouldn't react in a solution of HCL, what element would you choose, and which figures would you look at to base your judgement on? It seems electronegativity numbers are not the be-all-end-all. Would you say melting and boiling points are a better predictor of this characteristic for redox reactions? – Klik Jun 15 '14 at 5:14
Well it's hard for me to "predict" whether a metal will react with acid or not because I already know to some extent which metals are attacked and which ones aren't. But as I said, reactivity towards acids is a combination of several factors, so looking at a single one is a bad idea. The only safe way (but even then not sure-fire) to predict reactivity towards acids would be to look at the experimental reduction potentials, as mentioned by Greg in his comment; the more positive the reduction potential, the harder it should be to dissolve by acids. – Nicolau Saker Neto Jun 15 '14 at 10:17
Experimental reduction potentials--I will definitely keep that in mind. I always thought that reactions between elements and compounds of a few elements would be easy to predict. Does the difficulty in predicting such reactions come from the geometry of the atoms? I.e. the position of the electrons and shape of the orbitals. Anyways, I appreciate the answer you've given me. Thanks! – Klik Jun 15 '14 at 18:06
Even though theoretically any problem can be solved to arbitrary accuracy, reactions are rather hard to predict from first principles (i.e. taking into account only the most fundamental assumptions) because molecular simulation calculations are computationally very expensive, and successively coarser approximations must be made as the system size increases in order to keep the problem tractable. The geometry of atoms in a system is sometimes one of the factors which contribute to complexity, though usually only in larger molecules (a few tens of atoms upwards, depending on symmetries). – Nicolau Saker Neto Jun 18 '14 at 22:24
In order to avoid getting bogged down in the dense calculations, we can sometimes isolate simpler factors which can be used on their own, after calibration against experimental data (not first principles). However, these higher-level analyses often have a reduced degree of correlation between prediction and practice, and in order to compensate, one may seek to correlate with several parameters simultaneously, again making predictions harder and problems more subtle, but increasing the range in which an analysis is applicable. – Nicolau Saker Neto Jun 18 '14 at 22:32

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