Consider the cell described below: $$\text { Al }\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the [Al \(^{3+} ]\) to have changed by 0.60 \(\mathrm{mol} / \mathrm{L}\) . (Assume \(T=25^{\circ} \mathrm{C} .\)

First, we need to determine the initial cell potential of the galvanic cell. To do this, we'll need to use the standard reduction potentials for the half-cell reactions. The standard reduction potentials are: $$E^0_{Al^{3+}/Al} = -1.66 \ V$$ $$E^0_{Pb^{2+}/Pb} = -0.13 \ V$$ Now we can calculate the initial cell potential using the equation: $$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$$ $$E_{cell}^{0} = (-0.13V) - (-1.66V)$$

We are given that the change in the concentration of Al^{3+} is 0.60 mol/L, and we need to find the new concentrations after the reaction has operated long enough. First, calculate the moles of electrons, \(n_e\), that would be required by the change in the concentration of \(Al^{3+}\): $$n_e = 3 \times \Delta [Al^{3+}] = 3 \times 0.60 \ mol$$ Now we can calculate the new concentration of \(Pb^{2+}\) by considering the stoichiometry of the reaction: $$\Delta[Pb^{2+}] = \frac{2}{3} \times \Delta[Al^{3+}] = \frac{2}{3} \times 0.60 \ mol$$ Lastly, we can calculate the new concentrations of \(Al^{3+}\) and \(Pb^{2+}\): $$[Al^{3+}'] = 1.00 \ M - 0.60 \ M$$ $$[Pb^{2+}'] = 1.00 \ M + \frac{0.60 \ M}{3} \times 2$$

Now we'll apply the Nernst equation to calculate the cell potential after the reaction has operated long enough: $$E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q$$ where R is the ideal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred per reaction, F is Faraday's constant, and Q is the reaction quotient given by: $$Q = \frac{[Pb^{2+}'][Al^{3+}]^3}{[Al^{3+}'][Pb^{2+}]^2}$$ Plug in the values of the constants and the new concentrations of \(Al^{3+}\) and \(Pb^{2+}\), and calculate the new cell potential, \(E_{cell}\).