Why $\ce{Fe}$ with $\ce{H2O}$ gives $\ce{Fe3O4}$ not $\ce{Fe2O3}$?

Like, $\ce{Zn + H2O \rightarrow ZnO + H2}$, why $\ce{Fe}$ doesn't simply gives $\ce{Fe2O3}$ when it reacts with steam?

  • 1
    $\begingroup$ Another similar question is : why does a strange compound like $\ce{Fe3O4}$ exist ? Why is Iron (plus one or two other metals) able to make an oxide containing the same element at two different oxidation states ? Why is it not the case for other metals like copper for example ? $\endgroup$
    – Maurice
    Commented Oct 15, 2023 at 17:07
  • 2
    $\begingroup$ An idea : did you compute the Gibbs enthalpy for either one of the two reactions? $\endgroup$
    – Buttonwood
    Commented Oct 15, 2023 at 17:44
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    $\begingroup$ Five words: Steam is not oxidizing enough. $\endgroup$ Commented Oct 15, 2023 at 18:01
  • $\begingroup$ And chemistry.stackexchange.com/questions/73275/… $\endgroup$ Commented Oct 16, 2023 at 2:13
  • $\begingroup$ @NilayGhosh It seems to me neither of linked answers addresses the reason why Fe2O3 is not formed. $\endgroup$
    – Poutnik
    Commented Oct 16, 2023 at 3:07

2 Answers 2


Quoted picture from Researchgate.net for Ellingham diagram for iron oxides and hydrogen:

Ellingham diagram for iron oxides and hydrogen

On the diagram above, you can see equilibrium partial pressures (expressed as oxygen chemical potential} of oxygen for particular redox systems, depending on temperature.

Each system has a line $\mu(\ce{O2})=f(T)$. A system is able to oxidize systems with their respective lines below and vice versa.

We can conclude these consequences:

This reaction is thermodynamically favorable at all temperatures for the given range:

$$\ce{3 Fe2O3 + H2 -> 2 Fe3O4 + H2O}$$

This reaction is thermodynamically favorable below $\pu{900 K}$.

$$\ce{3 FeO + H2O-> Fe3O4 + H2}$$

The reaction is thermodynamically favorable below roughly $\pu{1700 K}$.

$$\ce{Fe + H2O -> FeO + H2}$$

It should be obvious that

  • The reduction of water by iron stops at $\ce{Fe3O4}$.
  • Above $\pu{900 K}$, it stops even at $\ce{FeO}$.
  • Above $\pu{1700 K}$, hydrogen reduces $\ce{FeO}$ to iron.

Note that the respective temperatures are related to equilibrium states. The open system with the flow of steam reactant and removal of hydrogen somewhat shifts the reactions toward hydrogen as the product, so the effective decisive temperatures would be higher.

OTOH, for reduction of oxides by hydrogen, it is the opposite case and needed temperature would be lower.

I assume for $\ce{Fe + H2O <=> FeO + H2}$, for temperature near $\ce{1200 K}$ and above, the reaction would go in large extend forward if hot water steam goes over hot iron, or backward, if hydrogen goes over iron oxides.


Consider the Schikorr reaction, proceeding in two stages: "[A]naerobic oxidation of two Fe(II) into Fe(III) by the protons of water... [then] loss of two water molecules from the iron(II) and iron(III) hydroxides giving rise to... the formation of a thermodynamically more stable phase iron(II,III) oxide. Ma et al state, that crystalline $\ce{Fe3O4}$ (magnetite) is thermodynamically more stable than amorphous $\ce{Fe(OH)2}$.

BTW, the stability of magnetite, and its adherence to iron, is used to protect steel from rusting. "Blueing... creates a layer of magnetite over the metal to prevent rust."

See also this question.

  • $\begingroup$ If protons of water were able to oxidize Fe(II) into Fe(III), then all of it would end up in Fe(III), either as Fe(OH)3 or as Fe2O3. $\endgroup$ Commented Oct 15, 2023 at 18:06
  • $\begingroup$ I think the OP has in mind rather high T chemistry without hydrated forms of iron oxides, $\endgroup$
    – Poutnik
    Commented Oct 15, 2023 at 19:14

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