Question

Consider the following reactions:

Zn^2+ + 4H3 <----> Zn(NH3)4^2+ beta 4 = 5.01 x 10^8

Zn^2+ + 2e^- <-----> Zn(s) E degree = -0.762 V

Assuming there is negligible current and nearly all Zn^2+ is in the form Zn(NH3)4^2+. what cathode potential (vs. S.H.E.) would be required to reduce 99.99% of the Zn^2+ from a solution containing 0.20 M Zn^2+ in 1.4 M ammonia? (Assume T= 298 K)

________V

Answer 1

For the half cell

Zn²⁺+ + 2e- <-----> Zn(s)            E^o = -0.762 V

E^o = (0.0592 V/2)logK= 0.0296V x logK

logK = - 0.762 V/0.0296 V = - 25.74

K = 10^-25.74= 1.82x10^-26 = 1/ [Zn2+]

For the complex formation

Zn²⁺ + 4NH3 <----> Zn(NH3)₄²⁺          Kf=β4 = 5.01 x 10⁸

Kf=β4 = [Zn(NH3)₄²⁺] / ([Zn2+][NH3]⁴)

[Zn(NH3)₄²⁺] is reduced from 0.2 M to

(0.01/100)x 0.2M = 2x10^-5 M

Using Kf expression:

[Zn2+] = [Zn(NH3)₄²⁺] / (Kf [NH3]⁴)=

          = 2x10^-5 M / (5.01 x 10⁸ x 1.4⁴)=

           = 1.04x10^-14 M

Again for the half cell

The quotient Q = 1/[Zn2+] = 1/1.04x10^-14 M = 9.615x10¹³

logQ = 13.98

E_cell = E^o - (0.0592 V/2)logQ= -0.762 V - 0.0296V x 13.98

       = -0.762 V - 0.414 V = - 1.176 V

If the applied potential is more negative the reduction of Zn²⁺ will continue.