[LucaZombini]

determinate the direction of the reaction 2Fe³⁺ + 2I^-  I₂ + 2Fe²⁺ when every specie is 1 mol/dm^-3 and find the concentration of every ionic specie when the equilibrium is reached. T = 25 °C

E_Fe³⁺/_Fe²⁺ = 0,771V
E_I2/I^- = 0,536V

given the Nernst's formula i find the Keq
E = RT/zF * logKeq
Keq = 10^(zE/0,05916) = 8,8*10⁷

[Fe³⁺] = [I^-] = 1-2x
[Fe²⁺] = 1+2x

then i don't know how to proceed. i've tried to substitute the results above in the Keq equation but i wasn't able to find the results.
 
for example

Keq = 1+2x/(1-2x)²

[mjc123]

What is the expression for K_eq for the reaction
2Fe³⁺ + 2I^-  I₂ + 2Fe²⁺ ?

[LucaZombini]

Keq = [Fe²⁺]² [I₂]/ [Fe³⁺ ]² [I^-]²

I₂ should be solid, i think.
 
EDIT:

I forgot to write the solutions

Fe³⁺ = I^- = 1,44E^-2
Fe²⁺ = 2-1,44E^-2

[mjc123]

Yes (leaving out [I₂] which is a solid). That is not (1+2x)/(1-2x)², is it?

You could simplify it by writing
Fe³⁺ + I^-    1/2 I₂ + Fe²⁺
for which z = 1 in the Nernst equation. Then
K_eq = [Fe²⁺]/[Fe³⁺][I^-]
but this K_eq has a different value from the one you calculated.

[LucaZombini]

I've managed to solve it by making an approximation which eliminated some terms!! 

i counted z as 2, that's how i found that value for Keq