Question

The solubility product constant for CuI(s) is 1.1 x 10^-12. Calculate the value of E knot for the half-reactoin: CuI + e- ---> Cu + I-.

For Cu+ + e- ---> Cu, E knot = 0.52V

Answer 1

The solubility product constant for CuI(s) is 1.1 x 10^-12. Calculate the value of E knot for the half-reactoin: CuI + e- ---> Cu + I-.

For Cu+ + e- ---> Cu, E knot = 0.52V

Given that;

CuI + e- => Cu + I- (at the anode)

Cu+ + e- => Cu (at the cathode) - SRP = + 0.52V

We know that;

Eºcell = Eºreduction (Cathode) - Eºreduction (Anode)

here given that K as Ksp which is is 1.1 x 10^-12. Now we calculate the value of free energy as follows:

ΔGº = - RT ln Ksp
= - (8.3145 J K/mol * 298 K) ln (1*10^-12)
= 6.846e4 J

now we can calculate the Eºcell
ΔGº = -nFEºcell


6.846e4 J = -1 x 96485 Coloumbs/mol x Eºcell

Eºcell = 6.846e4J / -96485 Coloumbs/mol
= -0.7095 V

Eºcell = Eºreduction (Cathode) - Eºreduction (Anode)

Now we fill in the blanks

-0.7095V = Eºreduction (Cathode) - 0.52V (this being the standard reduction potential for the anode reaction)

-0.7095V + 0.52V = -0.1895V

= -0.19V