Question #44863

Consider the first-row transition metal iron (Fe). The standard reduction potential for its hexaaqua complex is given as: $\ce{[Fe(H2O)6]^{3+} + e^- \rightleftharpoons [Fe(H2O)6]^{2+}} \quad E^\circ = +0.77 \text{ V}$ Both the $+2$ and $+3$ hexaaqua complexes are high-spin. When excess $\ce{Cl^-}$ is added, the $+2$ ion undergoes ligand substitution and a geometry change to form a high-spin tetrahedral complex: $\ce{[Fe(H2O)6]^{2+} + 4Cl^- \rightleftharpoons [FeCl4]^{2-} + 6H2O}$ Assume the standard free energy change ($\Delta G^\circ$) for this specific substitution is governed entirely by the change in Crystal Field Stabilization Energy (CFSE). Given the crystal field splitting parameters: $\Delta_o \text{ for } \ce{[Fe(H2O)6]^{2+}} = 10,400 \text{ cm}^{-1}$ $\Delta_t \text{ for } \ce{[FeCl4]^{2-}} = 4,000 \text{ cm}^{-1}$ Use the conversion factor $1 \text{ eV} \approx 8,000 \text{ cm}^{-1}$. (Note: $1 \text{ eV}$ per molecule corresponds to $96.5 \text{ kJ/mol}$, meaning $\Delta G^\circ$ in eV is numerically equal to the potential in Volts for a 1-electron transfer). Calculate the standard reduction potential ($E^\circ_{complex}$) for the coupled process: $\ce{[Fe(H2O)6]^{3+} + 4Cl^- + e^- \rightleftharpoons [FeCl4]^{2-} + 6H2O}$