Astudent does an experiment to determine the molar solubility of lead(ii) bromide. she constructs a voltaic cell at 298 k consisting of 0.733 m lead nitrate solution and a lead electrode in the cathode compartment, and a saturated lead bromide solution and a lead electrode in the anode compartment. if the cell potential is measured to be 5.45×10-2 v, what is the molar solubility of lead bromide at 298 k determined in this experiment?

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.