Question #44892

The actinoid series exhibits complex redox chemistry due to the relativistic stabilization of the 7s orbital and the varying energy of the 5f orbitals. Consider the Latimer diagram for Americium (Am) in aqueous acidic medium ($\ce{pH} = 0$) at $298\text{ K}$: $\ce{AmO2^{2+}(aq) ->[+1.60\text{ V}] AmO2+(aq) ->[+1.04\text{ V}] Am^{4+}(aq) ->[+2.60\text{ V}] Am^{3+}(aq) ->[-2.07\text{ V}] Am(s)}$ Unlike early actinoids (e.g., U, Np) where the $+4$ state is highly stable, the increasing effective nuclear charge in Am significantly lowers the 5f orbital energy, making $\ce{Am^{4+}}$ a powerful oxidant. A chemist investigates the thermodynamic stability of the $\ce{AmO2+(aq)}$ ion. It undergoes spontaneous disproportionation to yield $\ce{AmO2^{2+}(aq)}$ and the most thermodynamically stable lower oxidation state under these conditions. Calculate the standard Gibbs free energy change ($\Delta G^\circ$) for the balanced disproportionation reaction involving $3\text{ moles}$ of $\ce{AmO2+(aq)}$. (Given: $F = 96500\text{ C mol}^{-1}$)