Question

3. (10 marks) Consider the electrochemical cell: voltmeter Pb wire It bridg Ag wire M Ag 1.8 M Pb2+ Ag,SO, (s) Using this data, calculate the Kap for Ag SO,(s). Note that to obtain ions in the right com- partment, excess silver sulphate solid was added and a small amount of it dissolved. Ag (aq) +e Ag(s)E0.80 V

Answer 1

here we have reduction potentials

reduction potential of Ag+ is greater than Pb+2

so CATHODE = Ag

ANODE= Pb

now we will write reduction half reaction for both anode and cathode

CATHODE = 0.80 - (0.059/1)Log (1/[Ag+]) Ag+ + e- ----------> Ag

ANODE = - 0.13 - (0.059/2)Log (1/[Pb+2]) Pb+2 + 2e- -----------> Pb

E = E cathode - E anode

0.83 = 0.93 + (0.059/2)Log ([Ag+]^2 /[Pb+2])

solving eqn concentration of Ag+ is 0.026457M

Ag2SO4 -------------> 2Ag+ + SO4 -2

2s s

here s is solubality

Ksp = (2s)^2 * s = 4s^3

as we know 2s = 0.026457

s = 0.1322877

Ksp = 4s^3 = 9.26 * 10^ -6