Consider the following galvanic cell at \(25^{\circ} \mathrm{C} :\) $$\text { Pt }\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \rightleftharpoons_{2 \mathrm{Cr}^{3+}}(a q)+\mathrm{Co}(s) \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E},\) for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

First, let's identify the half-reactions involved. The anode half-reaction has a Cr²⁺/Cr³⁺ couple, and the cathode half-reaction has a Co/Co²⁺ couple. Anode: \(Cr^{2+} \to Cr^{3+} + e^-\) Cathode: \(Co^{2+} + 2e^- \to Co\)

To obtain the overall reaction, we need to balance the electrons in both half-reactions. \(2\ [Cr^{2+} \to Cr^{3+} + e^-]\) \(1\ [Co^{2+} + 2e^- \to Co]\) Now, we can write the overall cell reaction: \(2\ Cr^{2+}(aq) + Co^{2+}(aq) \to 2\ Cr^{3+}(aq) + Co(s)\)

Now, let's use the Nernst equation: \(\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{zF} \ln{Q}\) where \(\mathscr{E}^\circ\) is the standard cell potential, R is the gas constant, T is the absolute temperature, z is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient. Given the equilibrium constant, we can calculate the standard cell potential using: \(\Delta G^\circ = -RT \ln{K}\) \(\mathscr{E}^\circ = \frac{-\Delta G^\circ}{zF}\) First, we need to find \(\Delta G^\circ\): From the information given, K = 2.79 x 10⁷, T = 298 K, n = 2, F = 96,485 C/mol \(\Delta G^\circ = -RT \ln{K}\) \(\Delta G^\circ = -(8.314\ \mathrm{J/K \cdot mol}) (298\ \mathrm{K}) \ln{(2.79 \times 10^{7})}\) \(\Delta G^\circ = -3.66 \times 10^5\ \mathrm{J/mol}\) Now, let's find \(\mathscr{E}^\circ\): \(\mathscr{E}^\circ = \frac{-\Delta G^\circ}{zF}\) \(\mathscr{E}^\circ = \frac{-(-3.66 \times 10^5\ \mathrm{J/mol})}{(2)(96,485\ \mathrm{C/mol})}\) \(\mathscr{E}^\circ = 1.90\ \mathrm{V}\) Next, we will calculate the reaction quotient, Q: \(Q = \frac{[\mathrm{Cr}^{3+}]^2}{ [\mathrm{Cr}^{2+}]^2 [\mathrm{Co}^{2+}]} = \frac{(2.0\ \mathrm{M})^2}{(0.30\ \mathrm{M})^2(0.20\ \mathrm{M})}\) Now, we can find the cell potential \(\mathscr{E}\): \(\mathscr{E} = 1.90\ \mathrm{V} - \frac{(8.314\ \mathrm{J/K \cdot mol}) (298\ \mathrm{K})}{(2)(96,485\ \mathrm{C/mol})} \ln{Q}\) \(\mathscr{E} = 1.90\ \mathrm{V} - 0.198\ \mathrm{V}\) \(\mathscr{E} = 1.70\ \mathrm{V}\)

Finally, we can calculate the change in Gibbs free energy, \(\Delta G\), using the cell potential \(\mathscr{E}\): \(\Delta G = -zF\mathscr{E}\) \(\Delta G = -(2)(96,485\ \mathrm{C/mol})(1.70\ \mathrm{V})\) \(\Delta G = -3.28 \times 10^5\ \mathrm J/mol\) Therefore, the cell potential for this galvanic cell is 1.70 V, and the change in Gibbs free energy for the cell reaction is -3.28 x 10⁵ J/mol.