Problem 91 - Chemistry An Atoms First Approach

Question

Consider the following data: $$ \begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & & K=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{3+} & & K=2.0 \times 10^{47} \end{aligned} $$ 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 oxi- $$ \text { dizing agent, } \mathrm{Co}^{3+} \text { or } \mathrm{Co}(\mathrm{en})_{3}^{3+} ? $$ c. Use the crystal field model to rationalize the result in part b.

Step-by-Step Solution

Answer

In summary, we calculated the standard reduction potential for the half-reaction involving Co(en)_3^3+ and Co(en)_3^2+, which was found to be 0.068 V. By comparing the reduction potentials, we determined that Co^3+ is the stronger oxidizing agent. The crystal field model helps rationalize this result as the strong field ligand, ethylenediamine, causes a large splitting of d-orbitals, thus making Co(en)_3^3+ more stable and less prone to accept an electron compared to Co^3+.

Step 1: a. Calculate the standard reduction potential for the half-reaction

We will use the Nernst equation to calculate the standard reduction potential for the half-reaction: \(\mathscr{E}^{\circ} = \mathrm{RT}/\mathrm{nF} \cdot \mathrm{ln}\,K\) For the overall reaction: \(\mathrm{Co}^{3+}+\mathrm{e}^{-}+\mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}^{3+}+\mathrm{e}^{-}+\mathrm{Co}(\mathrm{en})_{3}^{2+}\) The equilibrium constant is given by: \(K = \frac{[\mathrm{Co}(\mathrm{en})_{3}^{2+}]}{[\mathrm{Co}^{2+}][\mathrm{en}]^3} = 1.5 \times 10^{12}\) Now, applying the Nernst Equation to the half-reaction: \(\mathscr{E}_{\text{Co(en)}_3^3+/\text{Co(en)}_3^2+}^{\circ} = \frac{RT}{nF} \cdot \mathrm{ln}\,K\) Replacing the values and using \(\mathscr{E}_{\text{Co}^{3+}/\text{Co}^{2+}}^{\circ}=1.82\,V\), \(\mathscr{E}_{\text{Co(en)}_3^3+/\text{Co(en)}_3^2+}^{\circ} = \frac{RT}{F} \cdot \mathrm{ln}\,(1.5 \times 10^{12})\) Calculating the value, we get: \(\mathscr{E}_{\text{Co(en)}_3^3+/\text{Co(en)}_3^2+}^{\circ} = 0.068\,V\)

Step 2: b. Determine the stronger oxidizing agent

In order to compare the oxidizing strength of Co^3+ and Co(en)_3^3+, we need to compare their reduction potentials. The higher the reduction potential, the stronger the oxidizing agent. \(\mathscr{E}^{\circ}_{\text{Co}^{3+}/\text{Co}^{2+}} = 1.82\,V\) \(\mathscr{E}^{\circ}_{\text{Co(en)}_3^3+/\text{Co(en)}_3^2+} = 0.068\,V\) Since \(\mathscr{E}^{\circ}_{\text{Co}^{3+}/\text{Co}^{2+}} > \mathscr{E}^{\circ}_{\text{Co(en)}_3^3+/\text{Co(en)}_3^2+}\), Co^3+ is the stronger oxidizing agent.

Step 3: c. Crystal field model rationalization

According to the crystal field model, the energy of d-orbitals in the presence of ligands depends on the geometry and strength of the ligand field. In this case, en (ethylenediamine) is a strong field ligand that causes a large splitting of d-orbitals. This results in Co(en)_3^3+ having less energy available to accept an electron than Co^3+. As a result, Co(en)_3^3+ has a lower standard reduction potential than Co^3+, making the latter a stronger oxidizing agent. The higher stability of Co(en)_3^3+ in the presence of the strong field ligand can be attributed to the greater crystal field stabilization energy.

Key Concepts

Nernst Equation

The Nernst equation is a relationship used in electrochemistry to determine the reduction potential of a half-cell in an electrochemical cell, or to find the equilibrium constant for a reversible reaction. This equation makes it possible to calculate the electrical potential of a cell under non-standard conditions by accounting for the temperature and concentrations of the reactants and products.

For a reaction like \( \mathrm{A}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{A} \), the Nernst equation is given by:\[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \cdot \ln{Q} \]
where \( \mathscr{E} \) is the cell potential, \( \mathscr{E}^{\circ} \) is the standard cell potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient. In the textbook solution, the Nernst equation was utilized to calculate the standard reduction potential for \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \) by incorporating the equilibrium constant \( K \) for the formation of the complex.

Oxidizing Agent

An oxidizing agent, also known as an oxidant, is a substance that has the ability to oxidize other substances — in other words, it gains electrons in a chemical reaction. In our exercise, we're comparing the oxidizing strength of \( \mathrm{Co}^{3+} \) and \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \). The one with a higher standard reduction potential (\( \mathscr{E}^{\circ} \)) is considered a stronger oxidizing agent because it has a greater tendency to gain electrons and be reduced.

A higher \( \mathscr{E}^{\circ} \) value indicates the species has a stronger attraction for electrons, making it a more formidable oxidizing agent. In the solution, \( \mathrm{Co}^{3+} \) has a higher standard reduction potential than \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \) and is thus the stronger oxidant, favoring the reduction process more strongly and thereby oxidizing other substances more effectively.

Crystal Field Model

The crystal field model is a theory that describes the breaking of degeneracies of electron orbital states, usually d or f orbitals, due to a set of surrounding anions or ligands. It explains how the field created by a ligand affects the energies of the d-orbitals of the metal ion. Strong field ligands, like ethylenediamine (en) used in the complex \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \), create a large energy gap between the high-spin and low-spin states.

These ligands can cause a pairing of electrons within the lower energy d-orbitals, leading to a more stable configuration due to the high crystal field stabilization energy (CFSE). In the given example, the presence of the strong field ligand en stabilizes the \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \) complex, as reflected in its lower reduction potential. This lower potential indicates that it is less likely to gain an electron and highlights why \( \mathrm{Co}^{3+} \) is a better oxidizer, given that it is not as stabilized by crystal field effects as the \( \mathrm{Co}(\mathrm{en})_{3}^{3+} \) complex is.