A cell was constructed with two lead electrodes. The electrolyte compartment is 1 M Pb(NO3)2 (aq). In the other compartment, NaI has been added to a Pb(NO3)2 solution until a yellow precipitate forms and the concentration of I- ions is .5 M. The potential of the cell is .155 V at 298 K. Calculate the concentration of Pb2+ ions in the second compartment. Calculate the solubility produce of PbI2.

To calculate the concentration of Pb2+ ions in the second compartment, we can use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the species involved in the cell reaction.

The general form of the Nernst equation is:

Ecell = E°cell - (RT / nF) * ln(Q)

where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons transferred in the cell reaction
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient

In this case, the cell reaction is given by:

Pb2+(aq) + 2I-(aq) → PbI2(s)

Based on the cell diagram, the half-reaction occurring at the anode is:

Pb(s) → Pb2+(aq) + 2e-

And the half-reaction occurring at the cathode is:

2I-(aq) → I2(s) + 2e-

First, we need to calculate the reaction quotient Q using the concentrations of the species involved in the reaction. Q is given by:

Q = [Pb2+][I-]^2

We are given that [I-] = 0.5 M. However, we need to determine [Pb2+] in the second compartment.

Now we will use the Nernst equation. The equation can be rearranged to solve for Q:

ln(Q) = (Ecell - E°cell) / ((RT / nF))

Substituting the given values: Ecell = 0.155 V, E°cell = 0 V (since the cell is non-standard), R = 8.314 J/(mol·K), T = 298 K, n = 2, and F = 96485 C/mol, we can calculate ln(Q).

ln(Q) = (0.155 - 0) / ((8.314 * 298) / (2 * 96485))

Using ln(Q) = ln([Pb2+][I-]^2):

ln([Pb2+][0.5]^2) = (0.155 - 0) / ((8.314 * 298) / (2 * 96485))

Now we can solve for [Pb2+]. First, rearrange the equation:

[Pb2+] = e^(((0.155 - 0) / ((8.314 * 298) / (2 * 96485))) / (0.5^2))

Calculate the value of the exponent and solve for [Pb2+].

Once you have the [Pb2+] concentration, you can use it to calculate the solubility product of PbI2. The solubility product constant (Ksp) is the product of the concentrations of the Pb2+ and I- ion in a saturated solution of the solid compound.

The balanced equation for the dissolution of PbI2 is:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

The Ksp expression for PbI2 is:

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

Now that you have the [Pb2+] concentration from the previous calculation and [I-] = 0.5 M, you can substitute these values into the Ksp expression and solve for the solubility product of PbI2.