Redox and redox equilibria

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I'm doing my background chemistry for a redox titration to determine the concentration of copper ions using potassium iodide. I know the ionic equation I need is:
 
2Cu²⁺(aq) + 4I^-(aq) -> 2CuI(s) + I₂(aq)
 
I'm having problems explaining why this reaction occurs though because my two half-equations are:
 
Cu²⁺(aq) + e^- ->Cu⁺(aq) - E^o = 0.15 V
I₂(aq) + 2e^- -> 2I^-(aq) - E^o = 0.54 V
 
The E^o values would normally mean I would reverse the Cu half-cell as it has the more negative E^o value. But to get my overall ionic equation I actually need to reverse the iodine half-cell. So I'm confused why the reaction occurs.
 
Vikki

Corrie writes ....
 
Let's look at the two half-reactions involved here again:
 
Cu²⁺(aq) + e <=> Cu⁺(aq) - E^o = +0.15 V
 
I₂(aq) + 2e^- <=> 2I^-(aq) - E^o = +0.54 V
 
On this basis I₂ is a stronger oxidising agent than Cu²⁺ (more positive E^o value). Therefore the reaction should go in the direction: 2Cu⁺ + I₂ -> 2Cu²⁺ + 2I^-, as you have noted.
 
Instead it actually goes, as we know: 2Cu²⁺ + 2I^- -> 2Cu⁺ + I₂
 
The reason for this is that one of the products is insoluble CuI. So the [Cu⁺(aq)] is very low indeed, certainly no where near the 1 M value that the standard E^o values above are based on.
 
Lowering the [Cu⁺(aq)] will have the effect of making the Cu²⁺ + e <=> Cu⁺ halfcell potential much more positive, i.e. much more oxidising. So much so that in fact Cu²⁺ becomes a stronger oxidising agent than I₂, and the reaction goes in a direction not predicted by just considering standard electrode potentials - a not uncommon situation in test-tube reactions where conditions are not standard.

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updated: 29 November 2007