Consider the following data: $$\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{~V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}){ }^{3+} & K &=2.0 \times 10^{47} \end{aligned}$$ where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}$$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

We can use the Nernst equation, which relates the standard cell potential, equilibrium constant, and stoichiometric coefficients of the half-reactions, as follows: $$\mathscr{E}^{\circ}_{\text{cell}} = \frac{RT}{nF} \ln K$$ The Nernst equation for the given half-reactions would look like this: $$\mathscr{E}^{\circ}_1 = \frac{RT}{n_1F} \ln K_1$$ $$\mathscr{E}^{\circ}_2 = \frac{RT}{n_2F} \ln K_2$$ We know that \(\mathscr{E}^{\circ}_1=1.82 \thinspace\mathrm{V}\), \(K_1=1.5 \times 10^{12}\), and \(K_2=2.0 \times 10^{47}\). We're given that the first half-reaction involves 1 electron transfer (\(n_1=1\)). To find \(\mathscr{E}^{\circ}\) for the desired half-reaction, we need to add the potentials of the known half-reactions: $$\mathscr{E}^{\circ} = \mathscr{E}^{\circ}_1 + \mathscr{E}^{\circ}_2$$ Since the total balanced redox reaction for parts 2 and 3 should involve 1 electron transfer like part 1, we can assume that \(n_2=1\). The potential for the second half reaction can be found as follows: $$\mathscr{E}^{\circ}_2 = \frac{RT}{F} \ln K_2$$ Using the gas constant (R) and Faraday's constant (F) values, we have: $$\mathscr{E}^{\circ}_2 = \frac{(8.314 \thinspace\mathrm{J \thinspace mol^{-1} K^{-1}})(298\thinspace\mathrm{K})}{(1)(96485 \thinspace\mathrm{C \thinspace mol^{-1}})} \ln (2.0 \times 10^{47}) = 3.07\thinspace\mathrm{V}$$ Now we can add the potentials to find the desired standard reduction potential: $$\mathscr{E}^{\circ} = 1.82 + 3.07 = 4.89\thinspace\mathrm{V}$$ The standard reduction potential for the given half-reaction is \(4.89\thinspace\mathrm{V}\).

To determine the stronger oxidizing agent, we'll compare the standard reduction potentials. A higher standard reduction potential corresponds to a stronger oxidizing agent. \(\mathrm{Co}^{3+}\) has a standard reduction potential of \(1.82\thinspace\mathrm{V}\), while \(\mathrm{Co}(\mathrm{en})_{3}^{3+}\) has a standard reduction potential of \(4.89\thinspace\mathrm{V}\). Since \(4.89\thinspace\mathrm{V} > 1.82\thinspace\mathrm{V}\), \(\mathrm{Co}(\mathrm{en})_{3}^{3+}\) is the stronger oxidizing agent.

In the crystal field model, ligands are assumed to interact with the central metal ion by creating an electric field that can stabilize or destabilize the energy levels of the d electrons in the metal ion. The stronger the ligand field, the greater the splitting of d-orbitals into high-energy and low-energy sets. In this case, the ethylenediamine (en) ligand forms strong coordination bonds with cobalt, increasing the ligand field strength and subsequently stabilizing the low-energy d-orbitals. Thus, it becomes more difficult for electrons to be removed from these orbitals, which in turn results in higher reduction potentials. This stabilization could explain why \(\mathrm{Co}(\mathrm{en})_{3}^{3+}\) has a higher standard reduction potential than \(\mathrm{Co}^{3+}\), making it a stronger oxidizing agent.