[mrjdfield]

Hi All,

I am actually a chemistry teacher, and was setting some redox problems for my class when I came across a problem of my own so....

My question is essentially: Why does silver iodide form?

My normal approach to this would be to look at the standard electrode potentials of the relevant half-equations, which are:
 
1/2 I₂ + e^-  I^-  (E^o = +0.54 V)
Ag⁺ + e^-  Ag   (E^o = +0.80 V)

My understanding, is that, since the Ag⁺/Ag half-equation is more positive, that Ag⁺ ions will be able to oxidise iodide ions to produce silver and iodine. Clearly however this doesn't happen though, as we see by the heavy yellow precipitate produced whenever I^- and Ag⁺ get a chance to meet.

Is my reasoning just wrong, or is it the case that whilst unfavoured energetically, this reaction proceeds for other reasons, and if so what are those reasons?

On the face of it it would be entropically unfavourable too since you are producing a highly ordered solid, when previously you had a highly disordered mixture. My thinking is that perhaps the insolublity of the product simply drives an equilibrium all the way to the right.

Any thoughts on this would be much appreciated.

Jon   

[mjc123]

That heavy yellow precipitate is the key, isn't it. You are correct, the insolubility of the product drives the equilibrium to the right.
Ag(s) + 1/2 I₂(s)  Ag⁺(aq) + I^-(aq)
E° for this reaction equals E°(I₂/I^-) - E°(Ag⁺/Ag), which is negative. But we can't achieve standard conditions because of the insolubility of AgI. For the real situation, assuming unit activities of Ag and I₂, the Nernst equation reduces to
E = E°(I₂/I^-) - E°(Ag⁺/Ag) - RT/F*ln K_sp(AgI)
Using a value of K_sp = 8.3e-17 M², E = +0.69V
Remember, standard EPs only tell us what should happen in molar solutions - at different concentrations, different reactions may occur.
Another example - why does aqua regia dissolve gold, when the SEPs suggest it shouldn't?

[Corribus]

Just as a point of elaboration to add onto the previous (very nice) response, you can also look at it in terms of Gibbs energy changes. You essentially have two processes going on - an oxidation/reduction between silver and iodide/iodine, and an precipitation between the products of the oxidation/reduction. The over all product will be determined by the over all (summed) thermodynamic driving force between the two processes.

I.e.,

(1) Ag(s) + 1/2 I₂(s) Ag⁺(aq) + I^-(aq)
(2) Ag⁺(aq) + I^-(aq) AgI(s)

Summing to

Ag(s) + 1/2 I₂(s) AgI(s)

Each of (1) and (2) is associated with ΔG. Since Gibbs energy is a state function, ΔG of the total process can be determined by simply summing the ΔG's of the two processes (Hess Law).

For (1), if I did my quick math right, I determined ΔG to be +25.09 kJ/mol, reflecting what you've already figured out about the direction this redox reaction should go.  However, the formation of AgI is so favorable (ΔG = -91.73 kJ/mol, from the Ksp value), that the overall ΔG = -66.63 kJ/mol. Meaning that AgI(s) will spontaneously form (after equilibrium is reached), even though the first step (oxidation of silver) is thermodynamically unfavorable as written.

Entropy and enthalpy will both be factors here, of course, and ΔG incorporates both of these. If you wanted to find out the relative impact of enthalpy and entropy in determining why formation of AgI is so favorable, you could look up standard reference values for the various species involved to calculate the ΔH and ΔS components of ΔG. Alternatively you could do some experiments - I suggest varying the temperature. This will give you some information about enthalpy, which could in turn be used to determine the entropy.

[mrjdfield]

Thanks both, your answers are very helpful.