When analyzing standard electrode potentials (), the redox couple displays an exceptionally more pos — d and f Block Elements Chemistry Question
Question
When analyzing standard electrode potentials ($E^\circ$), the $Mn^{3+}/Mn^{2+}$ redox couple displays an exceptionally more positive potential compared to the analogous $Cr^{3+}/Cr^{2+}$ couple. What is the fundamental, underlying electronic mechanism forcing this thermodynamic disparity?
💡 Solution & Explanation
A highly positive standard electrode potential ($E^\circ$) indicates a very strong tendency to undergo reduction (gain an electron). For $Mn$, gaining an electron transitions it from a $3d^4$ state ($Mn^{3+}$) to a spherically symmetric, half-filled $3d^5$ state ($Mn^{2+}$), which is extremely stable due to maximum exchange energy. For $Cr$, the $Cr^{3+}$ state ($t_{2g}^3$) is highly stable in an aqueous field, so it resists reduction to $Cr^{2+}$ ($d^4$), making its $E^\circ$ much less positive.