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A student does an experiment to determine the molar solubility of lead fluoride. He constructs a voltaic cell at 298 K consisting of a 0.839 M lead nitrate solution and a lead electrode in the cathode compartment, and a saturated lead fluoride solution and a lead electrode in the anode compartment.

If the cell potential is measured to be 7.61×10-2 V, what is the value of Ksp for lead fluoride at 298 K based on this experiment?

1 Answer

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The balanced chemical equation for the dissolution of lead fluoride is:

PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)

The solubility product expression for lead fluoride is:

Ksp = [Pb2+][F-]^2

At equilibrium, the cell potential (Ecell) is related to the Gibbs free energy change (ΔG) by the equation:

ΔG = -nFEcell

where n is the number of moles of electrons transferred in the cell reaction, and F is the Faraday constant (96,485 C/mol).

In this case, the cell reaction is:

Pb(s) + PbF2(s) ⇌ 2Pb2+(aq) + 2F-(aq)

Two moles of electrons are transferred, so n = 2.

The free energy change can be related to the equilibrium constant (K) by the equation:

ΔG = -RT ln(K)

where R is the gas constant (8.314 J/K·mol), and T is the temperature in Kelvin.

We can combine these equations to obtain an expression for Ksp:

Ksp = e^(-ΔG/RT) = e^(nFEcell/RT)

Substituting the given values:

n = 2
F = 96,485 C/mol
Ecell = 7.61×10^-2 V
R = 8.314 J/K·mol
T = 298 K

Ksp = e^(2 * 96,485 C/mol * 7.61×10^-2 V / (8.314 J/K·mol * 298 K))

Ksp = 3.01×10^-8

Therefore, the value of Ksp for lead fluoride at 298 K based on this experiment is 3.01×10^-8.
User Alexander Gladysh
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