Question

A student does an experiment to determine the molar solubility of lead(II) bromide. He constructs a voltaic cell at 298 K consisting of 0.821 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 5.60*10-2 V, what is the molar solubility of lead bromide at 298 K determined in this experiment? mol/L

Answer 1

1.In order to solve this problem,we need to use Nernst equation:

  

where E is the cell potential at a instant.E0 refers to the cell potential,n is the number of electron takes part in the overall reaction. is the concentration of the oxidated species and is the concentration of the reduced species.

the cell reaction at equilibrium,E=0,therefore

  

The reduction will happen in the anode compartment.So the concentration of the reduced species is equivalent to the molar solubility of lead bromide.

Molar solubility of lead bromide at 298K is 0.010 mol/L