Using the standard electrode potentials,predict if the reaction between the following is feasible:
$(a) Fe_{(aq)}^{3+} \text{ and } I_{(aq)}^{-}$
$(b) Ag_{(aq)}^{+} \text{ and } Cu_{(s)}$
$(c) Fe_{(aq)}^{3+} \text{ and } Cu_{(s)}$
$(d) Ag_{(s)} \text{ and } Fe_{(aq)}^{3+}$
$(e) Br_{2(aq)} \text{ and } Fe_{(aq)}^{2+}$

(N/A) reaction is feasible if the standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ is positive.
$(a) 2Fe^{3+}_{(aq)} + 2I^{-}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + I_{2(s)}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{I_{2}/I^{-}} = 0.77 - 0.54 = +0.23 \text{ V}$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(b) 2Ag^{+}_{(aq)} + Cu_{(s)} \rightarrow 2Ag_{(s)} + Cu^{2+}_{(aq)}$
$E^{\circ}_{cell} = E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Cu^{2+}/Cu} = 0.80 - 0.34 = +0.46 \text{ V}$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(c) 2Fe^{3+}_{(aq)} + Cu_{(s)} \rightarrow 2Fe^{2+}_{(aq)} + Cu^{2+}_{(aq)}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Cu^{2+}/Cu} = 0.77 - 0.34 = +0.43 \text{ V}$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(d) Ag_{(s)} + Fe^{3+}_{(aq)} \rightarrow Ag^{+}_{(aq)} + Fe^{2+}_{(aq)}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Ag^{+}/Ag} = 0.77 - 0.80 = -0.03 \text{ V}$. Since $E^{\circ}_{cell} < 0$,the reaction is not feasible.
$(e) Br_{2(aq)} + 2Fe^{2+}_{(aq)} \rightarrow 2Br^{-}_{(aq)} + 2Fe^{3+}_{(aq)}$
$E^{\circ}_{cell} = E^{\circ}_{Br_{2}/Br^{-}} - E^{\circ}_{Fe^{3+}/Fe^{2+}} = 1.09 - 0.77 = +0.32 \text{ V}$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.