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 NaI. 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 \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) at \(25^{\circ} \mathrm{C}\).

In this case, we have iodine (\(\mathrm{I}_{2}\)) and sodium chloride (\(\mathrm{NaCl}\)). As there is no known reaction between iodine and sodium chloride, no reaction occurs. ### Operation b: \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI.###

Here, we have chlorine gas (\(\mathrm{Cl}_{2}\)) and sodium iodide (\(\mathrm{NaI}\)). Chlorine is more electronegative than Iodine and will displace it from the salt in a single-displacement reaction.

The balanced equation for the halogen displacement reaction between \(\mathrm{Cl}_{2}\) and \(\mathrm{NaI}\) is: \[\mathrm{Cl}_2 + 2\mathrm{NaI} \rightarrow 2\mathrm{NaCl} + \mathrm{I}_2\]

To calculate \(\mathscr{8}^{\circ}\), we need to know the standard reduction potentials: \[\mathrm{Cl}_2 + 2\mathrm{e}^- \rightarrow 2\mathrm{Cl}^- \hspace{1cm} E^{\circ}_1 = +1.36 \, \mathrm{V}\] \[\mathrm{I}_2 + 2\mathrm{e}^- \rightarrow 2\mathrm{I}^- \hspace{1cm} E^{\circ}_2 = +0.54 \, \mathrm{V}\] The \(\mathscr{8}^{\circ}\) for this cell will be \( E^{\circ}_1 - E^{\circ}_2 = +1.36 - 0.54 = 0.82 \, \mathrm{V}\). To calculate \(\Delta G^{\circ}\), we can use the formula \(\Delta G^{\circ} = -nFE^{\circ}\) where \(n\) is the number of moles of transferred electrons and \(F\) is the Faraday constant (\(F \approx 96,485 \, \frac{\mathrm{C}}{\mathrm{mol}}\)). In this case, we have the transfer of 2 moles of electrons: \[\Delta G^{\circ} = -2 (96,485 \, \frac{\mathrm{C}}{\mathrm{mol}})(0.82 \, \mathrm{V}) = -158,555.8 \, \frac{\mathrm{J}}{\mathrm{mol}} \approx -158.56 \, \frac{\mathrm{kJ}}{\mathrm{mol}}\] Finally, we can calculate the equilibrium constant \(K\) using the relationship \(\Delta G^{\circ} = -RT \ln K\), where \(R\) is the gas constant (\(8.314 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\)) and \(T\) is the temperature in Kelvin (\(25^{\circ} \mathrm{C} = 298.15 \, \mathrm{K}\)). Solving for \(K\) gives: \[K = e^{\frac{-\Delta G^{\circ}}{RT}} = e^{\frac{158,555.8 \, \frac{\mathrm{J}}{\mathrm{mol}}}{(8.314 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}})(298.15 \, \mathrm{K})}} \approx 2.54 \times 10^{21}\] ### Operation c: A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\). ###

In this case, we have a silver wire (\(\mathrm{Ag}\)) and copper(II) chloride (\(\mathrm{CuCl}_{2}\)). Silver can displace copper from the \(\mathrm{CuCl}_{2}\) in a single-displacement reaction since it is higher in the reactivity series.

The balanced equation for the displacement reaction between \(\mathrm{Ag}\) and \(\mathrm{CuCl}_{2}\) is: \[2\mathrm{Ag} + \mathrm{CuCl}_2 \rightarrow \mathrm{Cu} + 2\mathrm{AgCl}\]

To be continued in a similar manner as operation b. ### Operation d: An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. ###

In this case, we have an acidic solution of \(\mathrm{FeSO}_{4}\) being exposed to air, which contains \(\mathrm{O}_{2}\) gas. The \(\mathrm{Fe}^{2+}\) ions in the \(\mathrm{FeSO}_{4}\) can be oxidized by the oxygen in the air to form \(\mathrm{Fe}^{3+}\) ions via a redox reaction.

The balanced equation for the redox reaction between \(\mathrm{Fe}^{2+}\) ions and \(\mathrm{O}_{2}\) gas is: \[4\mathrm{Fe}^{2+} + \mathrm{O}_{2} + 4\mathrm{H}^{+} \rightarrow 4\mathrm{Fe}^{3+} + 2\mathrm{H}_{2}\mathrm{O}\]

To be continued in a similar manner as operation b.