$$\ce{Fe2O3 + 6HI -> 2FeI2 + I2 + 3H2O}$$

Why don't we get $\ce{FeI3}$? After all, iron's oxidation state is $+3$ in the reagent.

Should one just memorize that up to bromine, it's $\ce{FeX3}$, and below it's $\ce{FeX2}$?

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    $\begingroup$ Because the iodide will reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$. $\endgroup$ – bon May 23 '16 at 10:08
  • $\begingroup$ @bon - the iodide (-3) ion will give an electron to Fe(+3)? Because their electronegativity difference is low? $\endgroup$ – CowperKettle May 23 '16 at 10:10
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    $\begingroup$ There is no $\ce{I^{3-}}$ ion. $\endgroup$ – bon May 23 '16 at 10:11
  • $\begingroup$ @bon - I see. So the I(-1) ion will give it's electron to Fe(3+), even though I(-1) is more electronegative. $\endgroup$ – CowperKettle May 23 '16 at 10:15
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    $\begingroup$ Look at the redox potentials for iodide, bromide and chloride here. $\endgroup$ – bon May 23 '16 at 10:17

The standard reduction potentials for the following half reactions can be found here.

$$ \begin{align} \ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\ \ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\ \ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\ \ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V} \end{align} $$

You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.

The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.


Actually, some have voiced an opinion that FeI3 can apparently be created with difficulty, and is reputedly very unstable, decomposing into FeI2 and I2. Similarly, with CuI2, which is likewise unstable. To quote a source reference. $$ FeCl3 + 3KI = FeI3 + 3KCl ; FeI3 = FeI2 + I2 $$


Large anions diminish the lattice energy of the higher halide to such an extent that the higher halide may be thermodynamically unstable.

Sample Solution-

Iodide ion is a good reducing agent and hence reduces metal in higher oxidation state into lower Ones.

Even, more interesting is the fact that FeI3 is extremely photosensitive! See this ebook, which describes FeI3 as "very difficult-to-prepare" and as "intensely black" in appearance, which is somewhat definite for a compound arguably that some claim does NOT exist.

So, in light, I would expect:

I- + hv -> .I + e-(aq)

Fe(III) (aq) + e- (aq) -> Fe(II)

.I + .I = I2

which means, in light, an even shorter half-life with potential transient atomic iodine presence and creation of elemental iodine.

So per reference above, one may be able to prepare (albeit briefly) some FeI3 (aq) by adding FeCl3 to a source of iodide (like aqueous KI). Here also is an interesting (but still speculative) supporting source, to quote from the abstract:

This study first reports that ferric chloride (FeCl3) can lead to the formation of iodinated coagulation byproducts (I-CBPs) from iodide-containing resorcinol solution or natural waters. The unwanted I-CBP formation involved the oxidation of iodide by ferric ions to generate various reactive iodine species, which further oxidize organic compounds. Although the oxidation rate of iodide by FeCl3 was several orders of magnitude slower than that by chlorine or chloramine, most of the converted iodide under the ferric/iodide system was transformed into iodine and iodinated organic compounds rather than iodate. Formation of four aliphatic I-CBPs was observed, and four aromatic I-CBPs were identified by gas chromatography mass-spectrometry and theoretical calculation. Coagulation of iodide-containing waters with FeCl3 also produced I-CBPs ranging from 12.5 ± 0.8 to 32.5 ± 0.2 μg/L as I. These findings call for careful consideration of the formation of I-CBPs from coagulation of iodide-containing waters with ferric salts.

The liberation of iodine together with the formation of iodinated coagulation byproducts (supportive of possible atomic iodine formation with any light exposure from the highly photo-sensitive FeI3) possibly suggests a photo-assisted breakdown of FeI3 (aq), in my speculation.

  • $\begingroup$ To teachers, I apology for voicing an opinion that FeI3 may indeed exist. But, I see interesting experimental evidence that it may indeed behave as expected. Further, the possible existence of a highly photosensitive FeI3 and associated active iodine radical presence is, in my opinion, noteworthy, in itself. For those, like myself, interested in recent scientific thought, with potential patent applications, welcome. $\endgroup$ – AJKOER Jan 16 '20 at 2:18

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