Electrochemistry

Question

Given the Ag+/Ag=0.8V, Cu2+/Cu=0.34V Calculate the standard emf of the cell and the equilibrium constant

Answer 1

Ag⁺ + e- ==> Ag Eº = 0.8 V (cathode)

Cu²⁺ + 2e- ==> Cu Eº = 0.34 V (anode)

2Ag⁺ + Cu ==> 2Ag + Cu²⁺ ... overall reaction

Standard emf = cathode - anode = 0.80 V - 0.34 V = 0.46 V

To find the Keq, we can use

-nFEº = -RT ln K

n = 2 mols electrons

F = 96,500

Eº = 0.46

R = 8,314

T = 298K (assumed)

ln K = ?

Plugging in, we have...

-(2)(96500)(0.46) = -(8.314)(298) ln K

ln K = -(2)(96500)(0.46) / -(8.314)(298)

ln K = 35.8

K = 3.5x10¹⁵