Question

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.

Short answer

The standard reduction potential for the half-reaction \(\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}\) is \(4.89\thinspace\mathrm{V}\). Comparing the reduction potentials, \(\mathrm{Co}(\mathrm{en})_{3}^{3+}\) is the stronger oxidizing agent. In the crystal field model, the ethylenediamine (en) ligand increases the ligand field strength, stabilizing the low-energy d-orbitals in cobalt, which makes it difficult for electrons to be removed and results in a higher reduction potential.

Step-by-step solution

a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction.

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 \(4.89\thinspace\mathrm{V}\).

b. Determine the stronger oxidizing agent.

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.

c. Rationalize the result using the crystal field model.

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.

Key concepts

Nernst Equation

The Nernst equation plays a crucial role in electrochemistry. It helps us relate the standard electrode potential of a reaction to the concentration of the reacting ions and their activities. It is given by: \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \] where \( \mathscr{E} \) is the cell potential, \( \mathscr{E}^{\circ} \) is the standard cell potential, \( R \) is the gas constant (8.314 J mol\(^{-1}\) K\(^{-1}\)), \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant (96485 C mol\(^{-1}\)), and \( Q \) is the reaction quotient. In the exercise, the students calculated the standard reduction potential \( \mathscr{E}^{\circ} \) for the cobalt-ethylenediamine complex reaction using the Nernst equation. They found that \( \mathscr{E}^{\circ} \) was significantly higher when ethylenediamine was bound to the cobalt ions, showing the effect of ligand stabilization.

Oxidizing Agent

An oxidizing agent is a substance that gains electrons in a chemical reaction. This means it is reduced while causing another substance to be oxidized. In the world of electrochemistry, oxidizing agents possess a high standard reduction potential. The higher the potential, the stronger the oxidizing capability. For instance,

• If a substance has a higher reduction potential, it is more favorable for it to gain electrons.
• In our example, \( \mathrm{Co}^3+ \) has a standard reduction potential of 1.82 V, whereas \( \mathrm{Co}( ext{en})_3^{3+} \) holds a potential of 4.89 V.
• Thus, \( \mathrm{Co}( ext{en})_3^{3+} \) is a stronger oxidizing agent.

The strong oxidizing nature of \( \mathrm{Co}( ext{en})_3^{3+} \) arises from the substantial energy needed to remove electrons from its stable complex.

Crystal Field Theory

Crystal Field Theory (CFT) provides a model explaining how metal-ligand bonds affect the energy of a d orbital in transition metal complexes. According to CFT, when ligands approach a metal ion, they alter the electron distribution among the d-orbitals due to electrostatic interactions.

• The theory explains the splitting of degenerate d-orbitals into different energy levels. This splitting depends on the strength of the ligand field.
• Strong field ligands, like ethylenediamine, result in greater splitting. This is because they raise the energy of the high-energy orbitals more than weaker ligands.
• In the exercise, ethylenediamine stabilizes the \( \mathrm{Co} \) complex, making it harder for electrons to be removed, which leads to a high reduction potential.

By understanding the interactions within Crystal Field Theory, students can comprehend why the \( \mathrm{Co}( ext{en})_3^{3+} \) complex presents enhanced oxidizing strength. Such insight into electron configurations and energetics between metal ions and ligands enables deeper appreciation of complex formation dynamics.