The passivity of certain metal fall in order: $\ce{Ti->Al->Cr->Be->Mo->Mg->Ni->Co->Fe->Mn->Zn->Cd->Sn->Pb->Cu}$
Can we predict the order of passivation theoretically?
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1$\begingroup$ Can you be more specific about what you mean by theoretically? $\endgroup$– hBy2PyCommented Sep 25, 2015 at 13:05
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$\begingroup$ It was a little confusing, so I just removed it. I mean this order seems quite jumble(at-least to me), and it's definitely not apparent from there positions in periodic table . So is there any criterion which must be satisfied(something which we may calculate with concentrations etc) by a metal in order to show passivity? And also can we generalise the result for any metal? $\endgroup$– TripathiCommented Sep 25, 2015 at 13:57
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2 Answers
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There is not much we can do. It is certainly not a property that is explained by a rigorously established single measurable parameter like oxidation potential. Passivation is a delicate interplay of multiple factors, some of which are not even related to the metal itself (remember, a metal may be passive in some environments, but not in the others).
However some trends can be observed. Surely, having a hard, high-melting, and chemically inert oxide does help a lot.
answered Sep 25, 2015 at 14:32
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$\begingroup$ But Aluminium Oxide's M.P is higher then Titanium Oxide's M.P yet Titanium still has a better protective layer. $\endgroup$– TripathiCommented Sep 25, 2015 at 14:46
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3$\begingroup$ That's right, oxide m.p. alone does not determine exactly the degree of passivity (neither does any other single parameter, just like I said), though has some influence on it. $\endgroup$ Commented Sep 25, 2015 at 15:00
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One of the standard tools for illustrating the tendency to passivation shown by a material is the Ellingham diagram:
An Ellingham diagram is a graph showing the temperature dependence of the stability for compounds. This analysis is usually used to evaluate the ease of reduction of metal oxides and sulphides.
Thermodynamically, the stability of the passivating oxide on the surface of the metal is directly related to the Gibbs free energy change of the oxide formation reaction:
The Gibbs free energy $\left(\Delta G\right)$ of a reaction is a measure of the thermodynamic driving force that makes a reaction occur. A negative value for $\Delta G$ indicates that a reaction can proceed spontaneously without external inputs, while a positive value indicates that it will not. ... An Ellingham diagram is a plot of $\Delta G$ versus temperature. (source PDF)
(A detailed Ellingham diagram and a more detailed description of the diagram and its construction can be found at the source link of the above quote.)
In any practical situation, some or all of the ideal assumptions made in generating an Ellingham diagram may or may not apply, but it at least provides a (semi-)quantitative means for arguing relative tendencies to passivation. In particular, as noted by Ivan Neretin, such diagrams don't consider effects of any electrochemical couples that may be active. For that, one would want to cross-reference the appropriate Pourbaix diagrams (though, these also suffer from some of the same kinds of idealized assumptions as the Ellingham diagrams).
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1$\begingroup$ You are relying on thermodynamics to explain an essentially kinetic phenomenon. This is not going to work out. Really, what the Ellingham diagram tells us that we don't know? That the oxide is stable in terms of $\Delta G$? Of course it is, nearly all of them are, so what? For similar reason, the usage of Pourbaix diagram here is downright misleading; for instance, it says that at certain conditions we should have $\ce{Fe}^{3+}$, but iron, apparently oblivious to that, would stay passive in concentrated nitric acid. $\endgroup$ Commented Sep 26, 2015 at 10:41
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2$\begingroup$ @IvanNeretin Quite true. My answer aims toward the 'broadly explanatory' level of description, though, which seemed appropriate to the question as-asked, as opposed to delving into the subtleties of interplay of kinetic/thermodynamic/mass transport/electric field effects that are, as you note, highly material- and context-specific. Ellingham/Pourbaix diagrams give a "zeroth-order" idea of what might happen; for any real system, of course one must perform experiments to determine what actually does happen. $\endgroup$– hBy2PyCommented Sep 26, 2015 at 15:16