Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\) . b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\). c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

In this situation, both Iodine and Chlorine are halogens, which have a similar tendency to gain electrons. However, due to the lower electronegativity of Iodine compared to Chlorine, there is no redox reaction, and the substances do not react. b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\).

In this situation, the \(\mathrm{Cl}_{2}\) gas will oxidize the iodide ions in the Nal solution: \[\mathrm{2I^- + Cl_2 → I_2 + 2Cl^-}\]

From the reduction potential table, we find the standard reduction potentials: \(\mathrm{Cl_2}\) + 2e⁻ → 2\(\mathrm{Cl^-}\): \(E_{Cl}^{\circ} = +1.36\,V\) \(\mathrm{I_2}\) + 2e⁻ → 2\(\mathrm{I^-}\): \(E_{I}^{\circ} = +0.54\,V\) Now, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{Cl}^{\circ} - E_{I}^{\circ} = +1.36\,V - (+0.54\,V) = +0.82\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 2 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(2)(96485\,C/mol)(0.82\,V) \approx -158\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{158\,000\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 1.5 × 10^{18}\) c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\)

In this situation, copper ions in the CuCl₂ solution will reduce and deposit on the silver wire: \[\mathrm{2Ag + Cu^{2+} → Cu + 2Ag^{+}}\]

Using the standard reduction potential table, we find the reduction potentials: \(\mathrm{Cu^{2+}}\) + 2e⁻ → \(\mathrm{Cu}\): \(E_{Cu}^{\circ} = +0.34\,V\) \(\mathrm{Ag^{+}}\) + e⁻ → \(\mathrm{Ag}\): \(E_{Ag}^{\circ} = +0.80\,V\) Then, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{Ag}^{\circ} - E_{Cu}^{\circ} = +0.80\,V - (+0.34\,V) = +0.46\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 2 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(2)(96485\,C/mol)(0.46\,V) \approx -89.2\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{89\,200\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 4.5 × 10^{14}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air.

In this situation, the iron ions in the FeSO₄ solution will oxidize in the air (exposed to O₂) in an acidic medium: \[\mathrm{4Fe^{2+} + O_2 + 4H^+ → 4Fe^{3+} + 2H_2O}\]

From the standard reduction potential table, we find the reduction potentials: \(\mathrm{O_2}\) + 4e⁻ + 4H⁺ → 2\(\mathrm{H_2O}\): \(E_{O_2}^{\circ} = +1.23\,V\) \(\mathrm{Fe^{3+}}\) + e⁻ → \(\mathrm{Fe^{2+}}\): \(E_{Fe}^{\circ} = -0.77\,V\) Then, we calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): \(\mathscr{E}_{cell}^{\circ} = E_{O_2}^{\circ} - E_{Fe}^{\circ} = +1.23\,V - (-0.77\,V) = +2.00\,V\) Next, we find the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the equation: \(\Delta G^{\circ} = -nFE_{cell}^{\circ}\) n = 4 (electrons transferred) F = 96485 C/mol (Faraday's constant) \(\Delta G^{\circ} = -(4)(96485\,C/mol)(2.00\,V) \approx -774\,kJ/mol\) Finally, we calculate the equilibrium constant (K) using the equation: \(K = e^{-\frac{\Delta G^{\circ}}{RT}}\) R = 8.314 J/mol·K (gas constant) T = 298 K \(K = e^{\frac{774\,000\,J/mol}{(8.314\,J/mol·K)(298\,K)}} \approx 2.2 × 10^{80}\)