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

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

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

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.

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.

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