Cd | Cd 2+ (0.05 M) || Cl - (0.10 M ) | Cl 2(1 atm ) | Pt,
E o = 1.76 V. Calculate a) the standard electrode potential of the cadmium electrode, b) the equilibrium constant for the cell reaction and c) the emf at 25 oC.
Solution:
The cell reaction is
Cd (s) + Cl 2 (g, 1 atm) Cd 2+ (0.05 M) + 2 Cl - (0.1M)
a)
The value of E o for the cell is 1.76V. The standard electrode potential for the Cl 2| Cl - | Pt cell is 1.3595 V. Therefore the electrode (reduction) potential for the Cd 2+ | Cd cell is: 1.76 V = E o = E oCl-/ C l 2 - E oCd 2+/ Cd
E oCd 2+/ Cd = 1.3595 – 1.76 = - 0.4005 V
b)
The equilibrium constant for the reaction (n = 2) is
K = exp [2 x 96500 x 1.76 / 8.314 x 298] = exp(137.1) = 3.49 x 10 59
c)
The reaction quotient for the reaction is
Q = [Cd 2+ ] / [Cd (s)] x [Cl - ] 2 / Cl 2(g)]
Both the denominators Cd(s) and Cl2(g) at 1 atm pressure are in their standard states.
Cd 2+ (0.05 M) + 2 Cl - (0.1M) 
Q = 0.05 x (0.1) 2 = 0.0005 
