# Why doesn't iron (III) iodide exist?

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

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 $\ce{FeX2}$?

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

$$\ce{Fe^3+ (aq) + e- -> Fe^2+ (aq)}\mathrm{~~~~~E^{o} = +0.77V}$$ $$\ce{I2 (s) + 2e- -> 2I- (aq)}\mathrm{~~~~~E^{o} = +0.54V}$$ $$\ce{Br2 (l) + 2e- -> 2Br- (aq)}\mathrm{~~~~~E^{o} = +1.07V}$$ $$\ce{Cl2 (g) + 2e- -> 2Cl- (aq)}\mathrm{~~~~~E^{o} = +1.36V}$$
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.