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.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).