# Can Iron(III) oxide oxidize elemental iron in a nitrogen environment under room temperature without any catalysis? [closed]

Will there be electron transfer to allow this process to happen: $$\ce{2 Fe(III) + Fe(0) -> 3 Fe(II)}$$

## closed as unclear what you're asking by airhuff, Mithoron, M.A.R. ಠ_ಠ, Todd Minehardt, Mathew MahindaratneJun 3 at 2:19

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• In what situation? Also what speed? – Mithoron Jun 1 at 21:45
• What are the counter ions? – Karsten Theis Jun 2 at 11:02

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions. One mechanism of atmospheric rusting involves the $$\ce{Fe(III)}$$ bearing rust reacting with fresh iron to form $$\ce{Fe3O4}$$, which then oxidizes in air to form more rust. See Section 1.6 here.

We can capture the $$\ce{Fe3O4}$$ stage by merely placing the rusty material in water with a little electrolyte, such as rinse water after an acid pickling process, where air cannot easily penetrate. The rust, being converted to $$\ce{Fe3O4}$$, turns black. This emphasizes the fact that the iron reacts with the $$\ce{Fe(III)}$$ in the rust without oxygen, once the rust has first formed.

I've actually seen the $$\ce{Fe3O4}$$ formation happen on steel products during rinsing after pickling, giving black stains or stains bearing a black component. To prevent it, we had to design and maintain the rinsing (and drying) processes to prevent the rusting.

• Thanks for answering this question! I guess I should rephrase my question: if solid Fe2O3 and elemental iron were mixed in powder form, and stored under anhydrous environment, will there be electron transfer from iron(0) Fe2O3, leading to the formation of Fe(II) species? – Z-BO Jun 2 at 1:02

Reactions between solids are generally slow because diffusion only goes so fast. However, if you produced the two materials in nano size to increase surface area and contact, mixed thoroughly (in absence of air) and then perhaps under pressure (density of $$\ce{FeO}$$ is 5.745, density of $$\ce{Fe2O3}$$ is 5.25) with an increase in temperature ($$\ce{FeO}$$ is stable above 575 °C: https://en.wikipedia.org/wiki/Iron%28II%29_oxide) after some time (how long depends on temperature and pressure), you could get complete transformation of $$\ce{Fe2O3 + Fe}$$ to $$\ce{3 FeO}$$.

If magnetite ($$\ce{Fe3O4)}$$ forms as an intermediate product, its low density (5.175) suggests it would not survive.

But on the other hand, the heats of formation do not favor $$\ce{FeO}$$: for $$\ce{FeO}$$, 65 kcal/mol; for $$\ce{Fe2O3}$$, 200 kcal/mol; for $$\ce{Fe3O4}$$, 268 kcal/mol (CRC Handbook).

It's hard to say. The electrochemical data do not take oxygen or solid states into consideration. But if you squeeze the mixture hard enough, I'm sure it will proceed.

Theoretically, yes. Just look at the following redox reaction: \begin{align} \ce{Fe &<=> Fe^2+ + 2e-} &\quad \mathrm{E^\circ} &= \pu{0.447 V} \tag{Oxidation}\\ \ce{Fe^3+ + e- &<=> Fe^2+} &\quad \mathrm{E^\circ} &= \pu{0.771 V} \tag{Reduction}\\ \ce{2Fe^3+ + Fe &<=> 3Fe^2+ } &\quad \mathrm{E_{cell}^\circ} &= \pu{1.218 V} \tag{Total redox} \end{align}

Accordingly, $$\mathrm{E_{cell}^\circ}$$ is a positive value, and therefore, the reaction is spontaneous.

• Thanks! But I thought this is enthalpy calculation, am I wrong? I am not familiar with the notation E0 here. Do you have a reference article about this process? Or probably a book chapter I can read? – Z-BO Jun 2 at 1:09
• The values I used are from sites.chem.colostate.edu/diverdi/all_courses/…. Yet, some certain values you can find in any General Chemistry book. – Mathew Mahindaratne Jun 2 at 4:04