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).

(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.