Problem 95
Question
The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: \(\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons\) $$ \left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{2+}(a q) \quad E^{\circ}=1.12 \mathrm{~V} $$ \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons\) $$ \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36 \mathrm{~V} $$ (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, [Fe(o-phen) \(\left._{3}\right]^{3+}\) 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
VerifiedKey Concepts
Reduction Potential
- \([\mathrm{Fe}(\text{o - phen})_{3}]^{3+}\) has a reduction potential of 1.12 V.
- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) has a reduction potential of 0.36 V.
Thermodynamic Favorability
- The reduction of \([\mathrm{Fe}(\text{o - phen})_{3}]^{3+}\) results in a \(E^{\circ}\) of 1.12 V, indicating a strong favorability.
- Similarly, \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) shows favorability with a \(E^{\circ}\) of 0.36 V, though less than the o-phen complex.
Ligand Field Strength
- CN^- is a strong field ligand, increasing the stabilization of the \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) complex, thus making its reduction less favorable.
- o-phen is a weaker field ligand than CN^-, which results in a smaller energy gap, making the \([\mathrm{Fe}(\text{o - phen})_{3}]^{3+}\) complex more easily reducible.
Complex Stability
- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) is more stable due to the strong field effect of the cyanide, anchoring the Fe(III) state.
- Conversely, \([\mathrm{Fe}(\text{o - phen})_{3}]^{3+}\), while stable, allows for easier reduction due to weaker ligand fields allowing greater electron flexibility.