Consider the cell described below: $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M) \| \mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{~mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

To calculate the cell potential, we need to identify the half-reactions occurring at the anode (Zn) and cathode (Cu). The balanced half-reactions are given below: At the anode (oxidation): \(Zn \rightarrow Zn^{2+} + 2e^-\) At the cathode (reduction): \(Cu^{2+} + 2e^- \rightarrow Cu\)

The overall cell reaction is obtained by adding the balanced oxidation and reduction half-reactions: \(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu\)

To calculate the standard cell potential, we need to consult a table of standard electrode potentials and find the values for Zn²⁺/Zn and Cu²⁺/Cu half-cells: The standard reduction potentials are: \(E^0_{Zn^{2+}/Zn} = -0.76 V\) \(E^0_{Cu^{2+}/Cu} = +0.34 V\) Now, we can calculate the standard cell potential (\(E^0_{cell}\)): \(E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 V - (-0.76 V) = 1.10 V\)

The Nernst equation relates the cell potential, standard cell potential, temperature, and the concentrations of the species involved in the reaction: \(E_{cell} = E^0_{cell} - \frac{0.05916}{n} \cdot \log{Q}\) where: - \(E_{cell}\) is the cell potential - \(E^0_{cell}\) is the standard cell potential - n is the number of moles of electrons transferred - Q is the reaction quotient For this particular cell, the reaction quotient Q is: \(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\) Since [Zn²⁺] has changed by 0.20 mol/L and [Cu²⁺] has changed by the same value but in the opposite direction, we can calculate the new concentrations: \([\mathrm{Zn}^{2+}]_\mathrm{final} \ = \ [\mathrm{Zn}^{2+}]_\mathrm{initial} \ + 0.20 \mathrm{~mol/L} \ = \ 1.00 \mathrm{~mol/L} \ + 0.20 \mathrm{~mol/L} \ = \ 1.20 \mathrm{~mol/L}\) \([\mathrm{Cu}^{2+}]_\mathrm{final} \ = \ [\mathrm{Cu}^{2+}]_\mathrm{initial} \ - 0.20 \mathrm{~mol/L} \ = \ 1.00 \mathrm{~mol/L} \ - 0.20 \mathrm{~mol/L} \ = \ 0.80 \mathrm{~mol/L}\) Now, we can substitute the values into the Nernst equation: \(E_{cell} = 1.10 V - \frac{0.05916}{2} \cdot \log{\frac{1.20}{0.80}}\)

Now we can calculate the cell potential for the new concentrations using the Nernst equation: \(E_{cell} \approx 1.10 V - \frac{0.05916}{2} \cdot \log{1.5}\) \(E_{cell} \approx 1.10 V - 0.02117 V \approx 1.07883 V\) The cell potential after the reaction has operated long enough for the [Zn²⁺] to have changed by 0.20 mol/L is approximately \(1.079 V\).