The relationship between the standard Gibbs free energy change ($\Delta G^{\circ}$) and standard cell potential ($E^{\circ}_{cell}$) is given by: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$

For Cell 1: $Fe|Fe^{2+}||Fe^{3+}|Fe$ Oxidation at anode: $Fe \rightarrow Fe^{2+} + 2e^-$ Reduction at cathode: $Fe^{3+} + 3e^- \rightarrow Fe$ To balance the electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. Total electrons transferred, $n = 6$. $\Delta G^{\circ}_1 = -6 \times F \times 0.404 = -2.424\text{ F}$

For Cell 2: $Fe|Fe^{2+}||Fe^{3+}, Fe^{2+}|Pt$ Oxidation at anode: $Fe \rightarrow Fe^{2+} + 2e^-$ Reduction at cathode: $Fe^{3+} + e^- \rightarrow Fe^{2+}$ To balance the electrons, we multiply the reduction half-reaction by 2. Total electrons transferred, $n = 2$. $\Delta G^{\circ}_2 = -2 \times F \times 1.211 = -2.422\text{ F}$

For Cell 3: $Fe|Fe^{3+}||Fe^{3+}, Fe^{2+}|Pt$ Oxidation at anode: $Fe \rightarrow Fe^{3+} + 3e^-$ Reduction at cathode: $Fe^{3+} + e^- \rightarrow Fe^{2+}$ To balance the electrons, we multiply the reduction half-reaction by 3. Total electrons transferred, $n = 3$. $\Delta G^{\circ}_3 = -3 \times F \times 0.807 = -2.421\text{ F}$

The respective values are $-2.424\text{ F}$, $-2.422\text{ F}$, and $-2.421\text{ F}$.