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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?

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  • $\begingroup$ Thank you, I read that article before posting. I did not see answers to my questions there. $\endgroup$ – electronpusher Jan 28 '17 at 21:03
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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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    $\begingroup$ You are correct in that water is not a catalyst. Nevertheless, a catalyst very often participates in elementary steps (i.e, its bonds can be broken). The large regeneration is what's key. Also note that iron(III) hydroxide does not exist. The formula $\ce{Fe(OH)3}$, common in 2oth century textbooks, is quantitatively the same as the iron(III) oxide-hydroxide monohydrate. This led to confusion, but the correct compound is now known to be $\ce{FeOOH*H2O}$. $\endgroup$ – Linear Christmas Jan 28 '17 at 19:45
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    $\begingroup$ There is also no true "rust compound". As you know, it is a mixture of other oxides, hydroxides, oxide-hydroxides and their hyrdates. Making this distinction is important since $\ce{Fe2O3*nH2O}$ might not be most common compound in rust. Especially on the surface it is commonly $\ce{\alpha-FeOOH}$, followed by $\ce{Fe3O4}$. $\endgroup$ – Linear Christmas Jan 28 '17 at 19:48
  • $\begingroup$ You say dry rusting is not possible, but if you wash with an acid then workup with an alkali hydroxide? (Some dry polar solvent). Wouldn't that circumvent deliberately using water as a reactant? $\endgroup$ – electronpusher Jan 28 '17 at 21:07
  • $\begingroup$ I don't see why one can't contrive the process of oxidizing the iron, stripping electrons with acid, and reacting with NaOH or some base all with the right polar-organic solvent, but to form the product iron oxide-hydrate you'll need water molecules. The distinctive brown/red/orange substance we associate with rust is a mixture of those hydrates (and the iron hydroxides as mentioned by @LinearChristmas). $\endgroup$ – khaverim Jan 29 '17 at 4:52

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