Question

Devise a cell for which the full cell reaction is:   Ag⁺(aq) + Br⁻(aq) → AgBr (s)

a. What is the cell emf at 25°C?

b. Calculate DG˚rxn from both E° and standard Gibbs energies of formation.

c. Calculate the equilibrium constant.

d. Calculate the solubility, S, of AgBr.

Answer 1

reduction half reaction

Ag+(aq) + e-    -----------------> Ag(s) E = 0.80 V

oxidation half reaction

Ag(s) + Br-(aq) -----------------> AgBr(s) + e- E = 0.07 V

overall reaction :

Ag⁺(aq) + Br⁻(aq) ----------------> AgBr (s)

a)

E° cell = E°reduction - E° oxidation

           = 0.80 - 0.07

           = 0.73 V

b)

DG˚rxn = - n F E^ocell

             = - 1 x 96500 x 0.73 x 10^3

             = - 70.4 kJ/mol

c)

DG˚rxn = - RTlnK

-70.4 = - 8.314 x 10^-3 x 298 x ln K

K = 2.18 x 10^12

d)

E = E^ocell - (0.0591 / 1 ) log [1/[Ag+][Br-]]   (at equilibrium E = 0)

E^ocell = (0.0591 / 1 ) log [1/[Ag+][Br-]]

0.73 = (0.0591 / 1 ) log [1/[Ag+][Br-]]

log [1/[Ag+][Br-]] = 12.35

[Ag+][Br-] = 4.47 x 10^-13

Ksp = 4.47 x 10^-13