Using the standard electrode potentials given in Table $3.1$ of the $NCERT$ textbook,predict if the reaction between the following is feasible:
$(i)$ $Fe^{3+}_{(aq)}$ and $I^{-}_{(aq)}$
$(ii)$ $Ag^{+}_{(aq)}$ and $Cu_{(s)}$
$(iii)$ $Fe^{3+}_{(aq)}$ and $Br^{-}_{(aq)}$
$(iv)$ $Ag_{(s)}$ and $Fe^{3+}_{(aq)}$
$(v)$ $Br_{2(aq)}$ and $Fe^{2+}_{(aq)}$

(N/A) reaction is feasible if the standard cell potential $E^{\circ}_{cell}$ is positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
$(i)$ $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$: $E^{\circ}_{cell} = 0.77V - 0.54V = +0.23V$ (Feasible).
$(ii)$ $2Ag^+ + Cu \rightarrow 2Ag + Cu^{2+}$: $E^{\circ}_{cell} = 0.80V - 0.34V = +0.46V$ (Feasible).
$(iii)$ $2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2$: $E^{\circ}_{cell} = 0.77V - 1.09V = -0.32V$ (Not feasible).
$(iv)$ $Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}$: $E^{\circ}_{cell} = 0.77V - 0.80V = -0.03V$ (Not feasible).
$(v)$ $Br_2 + 2Fe^{2+} \rightarrow 2Br^- + 2Fe^{3+}$: $E^{\circ}_{cell} = 1.09V - 0.77V = +0.32V$ (Feasible).