Setting up a galvanic cell (1.25V)

[nautica]

This is what I have chosen

I2 (s) + 2e- = 2I- (aq) reduction potential is +0.53
Zn2+ (aq) + 2e- = Zn (s) reduction potential is -0.76

using nerntz I get

E = 1.29 - (0.0257/2)(ln (2.24/0.1)
E= 1.2500

What would could I use to get the I and Zn ions? Also, would the Zn be the solution that should be 2.24 molar or the I?

Would KCl be a good salt bridge?

Thanks
Nautica

 

[nautica]

I am also finding E not, Delta G not, and K

I calulated
Enot=1.29V
DeltaGnot = -nFEcell = -248,944J
and for K I used 1.29V=0.0257V/2e- lnK and came up with 3.97 x 10^43 but this sounds like an extremely large number.

thanks
nautica