Cant figure this out, I feel like the answer I keep getting is way too low.

A voltaic cell is constructed that is based on the following reaction:

Sn2+ (aq) +Pb(s) = Sn(s) + Pb2+

(a) If the concentration of Sn2+ in the cathode compartment is 1.00M and the cell generates an emf of 0.22V, what is the concentration of Pb2+ in the anode compartment?

(b) If the anode compartment contains [SO4^2-] = 1.00M in equilibrium with PbSO4, what is the Ksp of PbSO4?

To solve this problem, we can use the Nernst equation:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:
Ecell = the cell potential
E°cell = the standard cell potential
R = the gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
n = number of moles of electrons transferred in the balanced equation
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient

(a) Let's first find the value of Q using the given concentration of Sn2+ in the cathode compartment (1.00M) and the emf generated by the cell (0.22V):

Ecell = E°cell - (RT/nF) * ln(Q)

0.22V = E°cell - (RT/nF) * ln(Q)

Since the reaction is based on the transfer of one mole of electrons (n = 1), we can simplify the equation:

0.22V = E°cell - (RT/F) * ln(Q)

We can rearrange this equation to solve for ln(Q):

ln(Q) = (E°cell - 0.22V) * (F/RT)

Next, we need the standard cell potential (E°cell). To find it, we can consult a table of standard reduction potentials or calculate it using the half-cell potentials of the redox reaction:

E°cell = E°cathode - E°anode

From the given equation:
Sn2+ (aq) + Pb(s) = Sn(s) + Pb2+

The half-cell reactions are:
Sn2+(aq) + 2e- = Sn(s) (reduction)
Pb(s) = Pb2+(aq) + 2e- (oxidation)

The standard reduction potentials for these half-cell reactions can be found in reference tables:
E°cathode = +0.14 V
E°anode = -0.13 V

Substituting these values into the equation for E°cell:

E°cell = 0.14 V - (-0.13 V)
= 0.27 V

Now, we can substitute the value of E°cell, T (which you didn't provide), R, and F into the equation for ln(Q) to find Q:

ln(Q) = (0.27 V - 0.22 V) * (96485 C/mol) / (8.314 J/(mol·K) * T)

To find the concentration of Pb2+ in the anode compartment, we need to relate Q to the concentrations of the reactants and products. The balanced equation gives us a 1:1 stoichiometric ratio between Sn2+ and Pb2+. Therefore, when Sn2+ is 1.00M in the cathode compartment, Pb2+ in the anode compartment must also be 1.00M to ensure that Q is 1.00.