For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\).

First, we need to write the corresponding half-cell reactions for both the oxidation and reduction processes: Oxidation half-reaction: \(\mathrm{2I}^{-} \rightarrow \mathrm{I}_{2} + 2e^-\) Reduction half-reaction: \(\mathrm{Hg}_{2}^{2+} + 2e^- \rightarrow \mathrm{2Hg}\)

Now, we combine the two half-cell equations to form a balanced redox equation: \(\mathrm{Hg}_{2}^{2+} + 2I^- \rightarrow \mathrm{I}_{2} + 2Hg\)

Using the standard reduction potentials for the given half-cell reactions (which can be found in a reference table), we can calculate the standard emf as follows: \(E^{\circ}_{cell} = E^{\circ}_{reduction} - E^{\circ}_{oxidation}\) For this reaction: \(E^{\circ}_{cell} = (+0.79~V) - (+0.54~V) = +0.25~V\)

Now, we can calculate the standard Gibbs free energy change using the following relation: \(\Delta G^{\circ} = -nFE^{\circ}_{cell}\) Where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday's constant (\(96485~C~mol^{-1}\)), and \(E^{\circ}_{cell}\) is the standard emf. For this reaction: \(n=2\) (2 moles of electrons are transferred) \(\Delta G^{\circ} = -2 \times 96485~C~mol^{-1}\times 0.25~V = -48242.5~\text{J}~\text{mol}^{-1} \)

Lastly, we can find the equilibrium constant \(K\) at \(298~K\) using the standard Gibbs free energy change and the following equation: \(\Delta G^{\circ} = -RT \ln{K}\) Where \(R\) is the gas constant (\(8.314~\text{J}~\text{K}^{-1}\text{mol}^{-1}\)) and \(T\) is the temperature (in Kelvin). Solving for \(K\): \( K = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{\frac{48242.5}{8.314\times298}} \approx 7.8 \times 10^{10}\) So, for Part (a), we have a balanced redox equation, a standard emf of \(0.25~V\), a standard Gibbs free energy change of \(-48242.5~\text{J}~\text{mol}^{-1}\), and an equilibrium constant of about \(7.8 \times 10^{10}\) at \(298~K\). Now, we'll proceed to Part (b) and (c) similarly. ## Part (b) ## (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. ## Part (c) ## (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\).