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

 

[ravenprp]

Thank you for sharing your chosen setup for the galvanic cell. It seems like you have chosen a reaction between solid iodine (I2) and zinc ions (Zn2+) to generate a voltage of 1.25V. Your calculations using the Nernst equation look correct, and the resulting voltage is within the expected range for this type of cell.

To get the ions for your cell, you can use solid iodine (I2) and zinc sulfate (ZnSO4) as your reactants. The zinc sulfate will dissociate in water to release Zn2+ ions, while the solid iodine will react with the Zn2+ ions to form I- ions.

As for the concentrations, it is up to you to decide. You can use a 2.24 M solution of either zinc sulfate or iodine, or you can use different concentrations for each reactant. Just make sure to keep the ratio of reactants the same as the balanced equation to maintain the correct voltage.

KCl can be a good choice for a salt bridge in this cell, as it is a common and inert electrolyte. It will allow for the transfer of ions between the two half-cells without reacting with them. Other options for a salt bridge could include potassium nitrate (KNO3) or sodium sulfate (Na2SO4).

I hope this helps answer your questions. Happy experimenting!