Problem 95 - Chemistry: The Central Science in SI Units, Global Edition

Question

The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: $$ \begin{aligned} \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons \\ \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+}(a q) & E^{\circ}=1.12 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons & \\ &\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36 \mathrm{~V} \end{aligned} $$ (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, \(\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}\) or \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\) is more difficult to reduce? (c) Suggest an explanation for your answer to (b).

Step-by-Step Solution

Answer

(a) Yes, it is thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs since their E° values are positive: \(E^{\circ} \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}/[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+} = 1.12\mathrm{~V}\); \(E^{\circ} \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}/[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} = 0.36\mathrm{~V}\). (b) The complex \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is more difficult to reduce since it has a lower E° value compared to \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}\). (c) The structural differences in ligands cause the reduction difficulty: \(o-\mathrm{phen}\) ligands in \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}\) have a stronger overlap with the metal ion, making it easier to reduce, while the cyanide ions in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) do not have the same strong stabilizing effect.

Step 1: (a) Determine the Thermodynamic Favorability

To determine the thermodynamic favorability of reducing the two iron complexes, we will look at the E° values provided in the exercise: \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+}(a q)\); \(E^{\circ}=1.12\mathrm{~V}\) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36\mathrm{~V}\) Both E° values are positive, which implies that the reduction of both Fe(III) complexes to their Fe(II) counterparts is thermodynamically favorable.

Step 2: (b) Compare Reduction Potentials

To see which complex is more difficult to reduce, we will compare their E° values: \(E°\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}/[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+} = 1.12\mathrm{~V}\) \(E°\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}/[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} = 0.36\mathrm{~V}\) Since the E° value for the reduction of the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is lower, it is more difficult to reduce.

Step 3: (c) Suggest an Explanation for Reduction Difficulty

An explanation for the difference in reduction difficulty can be found in the ligands coordinating the central iron atom. The \(o- phen\) ligands in the \(\left[\mathrm{Fe}(o- \mathrm{phen})_{3}\right]^{3+}\) complex are large, planar, and aromatic ligands, which results in a stronger overlap between the metal ion and the ligand orbitals, stabilizing the reduced Ｆｅ(II) state and making it easier to reduce. On the other hand, the ligands in the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex are cyanide ions. While cyanide ions have strong σ-donor and π-acceptor properties, their overall stabilizing effect on the Fe(II) state is not as strong as that of the larger and more conjugated \(o - phen\) ligands. Consequently, the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is more difficult to reduce than the \(\left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}\) complex.

Key Concepts

Reduction Potentials

Reduction potentials, often symbolized as E°, help us understand how likely a molecule is to gain electrons, or be reduced. When you see a positive E° value, it indicates a high likelihood that reduction will occur because it means the reaction is energetically favorable. The higher the E° value, the easier it is for the reaction to take place. In the exercise, we see that the complexes have different E° values:

• For \[\left[\text{Fe}(o-\text{phen})_3\right]^{3+} \rightarrow \left[\text{Fe}(o-\text{phen})_3\right]^{2+}, E^{\circ} = 1.12 \text{ V}.\]
• For \[\left[\text{Fe}(\text{CN})_6\right]^{3-} \rightarrow \left[\text{Fe}(\text{CN})_6\right]^{4-}, E^{\circ} = 0.36 \text{ V}.\]

Clearly, the higher E° value of the first complex means it has a greater tendency to be reduced compared to the second complex. This is a quick way to judge the likelihood of a reduction reaction for both complexes. Comparing these potentials, for the given reaction, tells us which reactant is more willing to be reduced.

Thermodynamic Favorability

Thermodynamic favorability refers to whether or not a chemical reaction will likely occur under normal conditions. It's rooted in the concept of Gibbs free energy, but in redox reactions, we often use reduction potentials as a surrogate to predict favorability. When both E° values are positive, as in this case for both complexes shown in the exercise, we learn that the reactions are favorable because a positive reduction potential indicates the reaction releases energy, driving it spontaneously:

• A higher E° value means more energy is released, and the reaction is even more favorable.
• If E° had been negative, it would indicate a reaction that is not spontaneous and requires an input of energy.

Therefore, \[\left[\text{Fe}(o-\text{phen})_3\right]^{3+} \text{ and } \left[\text{Fe}(\text{CN})_6\right]^{3-}\]being reduced are both favorable reactions, although one is substantially more favorable than the other due to a higher reduction potential.

Ligand Effects in Coordination Chemistry

In coordination chemistry, ligands play a critical role in defining the properties of the central metal atom. This can include influencing reduction potentials, as seen in our exercise. The \[o-\text{phen}\] and \[\text{CN}^{-}\] ligands interact differently with the iron center due to their electronic properties and structure:

• The \[o-\text{phen}\] ligand, being large, planar, and aromatic, offers strong overlap with the metal orbitals, leading to a stable reduced state for the central iron ion. More overlap stabilizes the reduced form, increasing the likelihood of reduction and resulting in a higher E° value.
• The \[\text{CN}^{-}\] ligand, while a strong σ-donor and π-acceptor, does not stabilize the reduced state as effectively as the aromatic \[o-\text{phen}\] ligands do. This poorer stabilization results in a lower E° value, indicating a lesser tendency to undergo reduction.

Such differences are key in predicting and rationalizing the behavior of complexes in chemical reactions. The stability conferred by their ligands governs the complex's reactivity and is central to coordination chemistry.