Question

A student 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?

Answer 1

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

Step-by-step explanation:

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

`E = E^(o) - (0.0591)/(n) log([ox])/([red])`

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:

`E^(o)  = (0.0591)/(n) log ([ox])/([red]) \\\\log ([ox])/([red]) = (nE^(o) )/(0.0591) \\\\log[red] =  log[ox] -  (nE^(o) )/(0.0591)\\\\[red] = 10^{ log[ox] -  (nE^(o) )/(0.0591)} \\\\[red] = 10^{ log0.733 -  (2x5.45x10^(-2)  )/(0.0591)}\\\\`

[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.