Problem: Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K.

We are asked to find the cell potential (E_cell) for the following lead concentration cell given. 

We will use the Nernst Equation to calculate the cell potential. The Nernst Equation relates the concentrations of compounds and cell potential.

$\boxed{{\mathbf{E}}_{\mathbf{cell}}\mathbf{=}\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{-}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{ }\mathbf{V}}{\mathbf{n}}\mathbf{\right)}\mathbf{log}\mathbf{ }\mathbf{Q}}$

E_cell = cell potential under non-standard conditions
E°_cell = standard cell potential
n = number of e^- transferred
Q= reaction quotient = [products]/[reactants] 

In the Nernst Equation, the E°_cell is needed but only E_cell was given. We have to determine the E°_cell first and as well as the anode (oxidation) and cathode (reduction) in the concentration cell and the number of electrons transferred (n).