Rusting: alternatives to water?

I have several questions about the rusting of metals (take iron as a specific case if you wish) that I have not found clear answers to:


1) Does water act as a catalyst in prototypical rusting reactions, or is it actually consumed? (keeping in mind that $H\_2$ gas produced may be oxidized back to water by ambient $O\_2$.)


2) Is dry rusting possible with just $O\_2$ and metal? Am I correct that it is thermodynamically favored but kinetically discouraged?


3) Is it realistic to rust a metal using oxygen and a dry polar solvent? If so, what solvents would be suitable?




Rust is *by definition* the transformation of iron metal to iron oxide (and iron hydroxide as an intermediate). You cannot "rust" other metals, although you *can* form oxides and degrade them etc. It's just not called rust when you convert, e.g. $\ce{Cr}$ to $\ce{CrO3}$. So there is only one rusting *process*.


1) Water is **not** a catalyst in the reaction that forms rust. A catalyst by definition is not consumed during a reaction (its bonds are not broken). Water can be and is the medium that carries $\ce{O2}$ to the iron (otherwise $\ce{O2}$ from air can react with the metal), *and* it is consumed in the process by dissociation. The fact that it is reproduced does not make it a catalyst. First iron atoms are oxidized by $\ce{O2}$ as in:


(a) $$\ce{Fe + O2\_{(aq)} \rightarrow Fe^{+2}\_{(aq)} + 2e-}$$


Then the free electrons are quickly absorbed by free protons from water by:


(b) $$\ce{2e- + 2H+\_{(aq)} \rightarrow H2}$$


It is highly unlikely for $\ce{H2}$ to react with $\ce{O2}$ to form water -- that is a combustion reaction, so high heat or pressure would have to be added. Rather, water is reproduced by a different reaction (d), below.


2) "Dry" rusting is not possible because the process requires reactions with water's ionization products $\ce{H+}\_{(aq)}$ and $\ce{OH-}\_{(aq)}$ (notice the aqueous states of all ions). The process continues by production of iron hydroxides:


(c) $$\ce{Fe^{+2}\_{(aq)} + 2OH-\_{(aq)} <=> Fe(OH)2\_{(s)}} $$
(d) $$\ce{Fe^{+2}\_{(aq)} + 4H+\_{(aq)} + O\_2\_{(aq)} <=> 4Fe^{+3}\_{(aq)} + 2H2O\_{(l)} } $$ 
$$\ce{Fe^{+3}\_{(aq)} + 3OH- <=> Fe(OH)3\_{(s)}} $$


This is why more acidic media produces rust faster; it pushes (c) and (d) to the right because (b) consumes free electrons and renders more $\ce{Fe^{+2}}$ for further reaction.


Finally the iron oxides "dry" to form the *actual rust compound*, which is a hydrated oxide: $\ce{Fe2O3\cdot nH2O }$.


3) You can see by now that water is the required solvent because the reactions of interest all involve $\ce{H+}$ and $\ce{OH-}$.


One could say there exists a "rust triangle" where "time, oxygen, and water" are required as an analog to the "fire triangle": heat, oxygen, and fuel.


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