# Why can't rust form without water?

Shouldn't iron oxide be able to form without water? It is just iron and oxygen. I don't really understand what the dot followed by the $\ce{H2O}$ means either. I was reading on wikipedia, but I have a rather terrible understanding of chemistry. What purpose does water serve in rust formation, and what does it mean for a molecule to be hydrated-how is that different then just being surrounded by water molecules?

We could explain it from an electrolytic point of view.

Iron has a certain tendency to dissolve in water according to the following equation:

$\ce{Fe (s) -> Fe^2+ (aq) + 2e^-}$

The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the region around this area becomes rich in electrons, a cathode. Now, we have to remember that water dissociates to small extents according to:

$\ce{H2O (l) <=> H^+ (aq) + OH^- (aq)}$

If it was an acid, the $\ce{H^+}$ would have quickly taken all the electrons liberated from the dissolution of iron earlier, but the concentration of hydrogen ions is not large enough in water, so that we get another reaction taking place at the cathode [The hydrogen ions actually 'gather up' near to the surface but cannot do anything, creating a thin protective layer of hydrogen ions around the cathode, but not strong enough to prevent further reaction]. This reaction involves oxygen and other water molecules:

$\ce{2H2O (l) + O2 (g) + 4e^- -> 4OH^- (aq)}$

Now this is the $\ce{OH^-}$ that reacts with the earlier liberated iron (or ferrous) ions:

$\ce{Fe^2+ (aq) + 2OH^- (aq)->Fe(OH)2 (aq)}$

Now more oxygen will react to oxidise the iron (II) hydroxide...

$\ce{Fe(OH)2 (aq) \xrightarrow[O] Fe(OH)3 (aq) \xrightarrow[O] Fe2O3 (s)}$

You probably can recognise the last product.

As the $\ce{Fe^2+}$ are consumed, more iron will get dissolved (Le Chatelier's Principle) and keep the whole process going.

Maybe a little picture to go along with this :)

• Note that water is not required if the temperature of the iron and oxygen are sufficiently high. I do not know the precise temperature, but from experience with welding I can say that noticeable amounts of orange-colored iron oxide begin to form as steel approaches its melting point, which is around 1370 degrees C (2500°F). This high-temperature oxidation in air, without water, is similar to what we usually call charring or burning. – Ralph Dratman Feb 21 '19 at 18:22
• @RalphDratman That sounds accurate. Heat basically prompts iron to react with oxygen, first giving FeO, and with more oxygen in the fray, Fe2O3 (rust). We could say that's the actual way that rust is formed, and water acts as catalyst (makes the same reaction possible with lower energy (heat) requirement. – Jerry Feb 21 '19 at 18:35

The dot means that the compound is a hydrate of the parent chemical; the parent compound forms a weakly-bonded structure with one or more water molecules per parent molecule, because either the hydrogen atoms of the water, or the partial positively-charged region of the water molecule caused by its "bent" molecular shape, is attracted to partially negative regions of the parent molecule. It's not a "full" traditional ionic bond but it works in somewhat of a similar way. In solid form, this mixture of the parent compound and water molecules forms a crystal structure like ice (it doesn't have to be a well-ordered crystal like you typically get with salts), with properties that differ from a form of the substance that has more, less or no water in its structure.

An everyday example of the difference between a dehydrated and hydrated compound is Portland cement, used in concrete. In its raw form, it's "dehydrated" calcium carbonate along with a few other components, made by heating limestone until the water is released from the rock's structure. The remaining solid forms a fine powder that can't hold much of a shape. Add the water back in while mixing concrete, and the water molecules are re-incorporated into a solid structure with the calcium carbonate to essentially reform a rock in the desired shape.

As far as why rust only forms in the presence of water, water's "polar" nature is at work here as well. Being a polar solvent, it has some affinity for electrons, which are easily given up by the iron as a transition metal, and are attracted to the hydrogen atoms in the water. If a hydrogen atom successfully "captures" an electron, it balances its own electric charge, and is "liberated" from the bond it has with the oxygen in the water molecule, instead pairing with another liberated hydrogen to form a diatomic gas. Now, this hydrogen gas molecule won't get far, because water usually has some "dissolved" oxygen gas in it, and there's more oxygen just beyond the water, waiting to re-oxidize this hydrogen gas back into water. The net result is a positively-charged iron atom (typically in the +2 or +3 oxidation states, having given up 2 or 3 electrons to the water, respectively), negatively-charged hydroxide ions ($\ce{OH-}$), and water. These form a mixture of iron hydroxides:

$$\ce{2(Fe - 2e^{-}) + 4(H_2O + e^+) + O_2 \\ \to 2Fe^{2+} + 4OH^- + 2H_2 + O_2 \\ \to 2Fe(OH)_2 + 2H_2O}\\$$ $$\ce{4(Fe - 3e^{-}) + 12(H_2O + e^+) + 3O_2 \\ \to 4Fe^{3+} + 12OH^- + 6H_2 + 3O_2 \\ \to 4Fe(OH)_3 + 6H_2O}$$

That second equation typically occurs "stepwise", as you'll notice that's quite a lot of electrons and ions floating around at once. Typically, iron(II) hydroxide is formed first, and the iron will then readily oxidize further to its +3 state, either taking on an additional hydroxide to form iron(III) hydroxide, or liberating a hydrogen that will oxidize to form water, resulting in the monohydrate of iron(III) oxide-hydroxide:

$$\ce{4(Fe(OH)2 - e^{-}) + 2H_2O + O_2 \to 4FeO(OH)*H_2O}$$

Iron(III) hydroxide will readily rearrange itself into much the same state (it's not so much decomposition, as the water stays weakly bonded to the oxide-hydroxide to form the monohydrate):

$$\ce{Fe(OH)3 \to FeO(OH)*H_2O}$$

Finally, two of these iron oxide-hydroxide molecules are generally considered to "share" an oxygen from one of the hydroxide ions between them, with the hydrogen instead bonding to the oxygen of the other hydroxide, reforming water as a hydrate (possibly with some of it being released):

$$\ce{2FeO(OH)*H_2O \to Fe_2O_3*xH_2O + yH_2O}$$

Various hydrate structures of this iron oxide exist, which result in different colors of the compound from dark brown to deep red to red-orange. Other oxides are also possible by more direct action of the oxygen on the iron itself, such as iron(II) oxide ($\ce{FeO}$). Heating rust releases the water and forces the iron oxides to rearrange, which can also produce the third oxide of iron, iron (II,III) oxide (known in mineral form as magnetite).

You'll notice that the water participates in the overall reaction without really being consumed by it to form the product; it's destroyed and then re-formed in equal amounts, creating intermediates in the process. As such, water is a catalyst for these reactions. You'll also hopefully notice that this "ideal" reaction chain is just one of the possibilities; "rust" is an ill-defined term in chemistry, with the everyday compound consisting of a mixture of various hydrates of the three iron oxides, all of which are produced depending on subtle localized differences in the temperature and ratios of water, oxygen and iron.

By definition, rust includes water in its structure. Its not rust if it doesn't contain water.
Rust is porous- it has holes so corrosion of iron keeps going a lot longer if it rusts rather it forms a straight oxide.

Iron forms oxides instantly which are more protective than rust- not a lot more. Hot iron in oxygen environments corrodes by forming iron oxide.

What purpose does water serve in rust formation?"

First, one should ask what is the temperature of the water and the source of the water? Cold from a faucet implying possible oxygen content along with say free $$\ce{Cl2}$$ or even $$\ce{NH2Cl}$$, or natural aerated waters containing transition metal salts (including those of $$\ce{Fe, Mn}$$, some $$\ce{Cu}$$, …), along with possibly gases other than $$\ce{O2}$$ like $$\ce{N2O}$$ from nitrate (present in well water) decomposition, or perhaps boiled distilled water without any dissolved gases or minerals. The composition of the iron alloy, I will ignore and assume pure $$\ce{Fe}$$, along with a pH range for the $$\ce{H2O}$$ of between 6 to 8.

I will attempt a brief depiction of the chemistry which could actually occur in a few scenarios to give a closer perspective of real world chemistry. For natural waters, as a starter, the chemistry described below is certainly relevant. However, with the distilled water in open air contact with submerged powdered iron, I will expect a comparatively much longer induction period till any rust is observed per the reactions outlined above. This is due to the underlying electrochemical nature of the corrosion process and no electrolyte (from dissolved salts, but instead awaiting dust particles) and no dissolved $$\ce{O2}$$ or an acid (source of $$\ce{H+}$$ could arrive via air containing $$\ce{CO2}$$). So iron rusting proceeds much faster in those cases containing dissolved oxygen, an acid source (free chlorine), and high mineral content (electrolyte).

There is also a possible direct action of water on iron via the following path without any oxygen presence:

\begin{align}\ce{2 [ H2O &<=> H+ + OH- ]}\\[1.3em] \ce{Fe + 2 OH- &-> Fe(OH)2 + 2 e-}\\[1.3em] \ce{2 [ H+ + e- &<=> H^. ]}\\[1.3em] \ce{H^. + H^. &-> H2}\end{align}

\begin{align}\text{Net:} &&\ce{Fe + 2 H2O -> Fe(OH)2 + H2}&&\end{align}

This could then be followed by the so called Schikorr reaction, which details the conversion of iron(II) hydroxide into the mixed oxide iron(II,III), also written as Fe3O4 (see Wikipedia https://en.wikipedia.org/wiki/Schikorr_reaction ).

Interestingly, there is also a possible action on any formed ferrous in water containing $$\ce{N2O}$$ in the further presence of cupric (see Chemical Reduction of Nitrite and Nitrous Oxide by Ferrous Iron, by J. T. Moraghan and R. J. Buresh,in SSSAJ, Vol. 41 No. 1, p. 47–50, 1976.).

The reason for mentioning an acid source is due to the following reaction converting ferrous to ferric in the presence of oxygen:

$$\ce{4 Fe^{II} (aq) + 4 H+ + O2 -> 4 Fe^{III} + 2 H2O}$$

So, depending on various sources of the water, in addition to the mechanism with water and oxygen described above, the chemistry could be a bit more complex.

Water is the catalyst in the process of rusting, therefore you cannot get rust without water.

Taken from http://www.chemicalformula.org/rust:

The overall chemical equation for the formation of rust is

Iron + water + oxygen --> rust

$$\ce{4 Fe(s) + 6 H2O(l) + 3 O2(g) -> 4 Fe(OH)3(s)}$$

Iron(III) hydroxide, $\ce{Fe(OH)3}$ then dehydrates to produce $\ce{Fe2O3.nH2O(s)}$ or rust

• That doesn't help my understanding much-All you did was introduce a new word. – user1638 May 29 '13 at 5:36
• Let me put it this way - the reaction that causes rust requires iron, oxygen, and water. – Dustin L. May 29 '13 at 5:43
• The OP already knew that; the question was why. – KeithS May 29 '13 at 17:23

"Rust" can form without water, if by rust you mean an iron oxide. Fe2 O3 ( the red oxide) can form in air at ambient temperatures. The only example I know is fretting corrosion where the characteristic red powder forms around two rubbing steel/iron surfaces. Of course you can make all sorts of iron oxides at elevated temperatures.