Question

4. Calculate the equilibrium constant at standard temperature, 25oC, for the redox reaction between Pb2+(aq) and Zn(s). HINT: Write a balanced chemical reaction first. a) 1.17x1030 b) 1.92x1021 c) 4.38x1010 d) 1.08x1015 e) 1.75x109

Answer 1

The reaction taking place at cathode:

Pb²⁺ + 2e Pb [E_c= -0.126] [Hint: look at standarad emf series for E values]

The reaction taking place at anode:

Zn Zn + 2e [E_a= +0.763] [Hint: Look at standarad emf series for E values]

E^o_cell = E_c - E_a

E^o_cell = -0.126 - 0.763

E^o_cell = - o.889

Using the relation;

E^o_cell = 0.0591 log(K) / n

where K is equilibrium constant & n is no. of moles of electron tranferred in balanced equation.

Here n = 2;

-0.889 = 0.0591 X log(K) / 2

-0.889 X 2 = 0.0591 X log(K)

-1.778 = 0.0591 X log(K)

-1.778 / 0.0591 = log(K)

-30.08 = log(K)

Antilog(-30.08) = K

K= 1.17 X 10³⁰

Hence option (a) is correct