Question

What is the Ecell at 298K of the spontaneous cell made from Ag+/Ag (0.8V) and Cl2/Cl-(1.36V) half cells if the [Ag+]= 0.01M and [Cl^-]= .0001M with Pressure of Cl2 = 1.1atm? I know the answer is 0.92V but i am having trouble getting there. Thanks.

Answer 1

oxidation : anode

2Ag ----------------> 2Ag+ + 2e- , Eo = 0.80 V

reduction : cathode

Cl2 + 2e- ----------------> 2 Cl- . Eo = 1.36 V

Eocell = Eo cathode -Eo anode = 1.36 - 0.80 = 0.56 V

overall reaction :

2 Ag + Cl2 ------------> 2 Ag+ + 2 Cl-

Q = [Ag+]^2 [Cl-]^2 / P_Cl2

Q = (0.01)^2 (0.0001)^2 / 1.1

Q = 9.09 x 10^-13

Nernest equation:

E_cell = E^ocell - 0.05916 / n * log Q (n = number of electrons = 2)

=  0.56 - 0.05916 / 2 * log (9.09 x 10^-13 )

= 0.92 V

E_cell  = 0.92 V