To determine which reaction cannot occur, we need to analyze the standard electrode potentials.$\\$Given standard electrode potentials at 298 K:$\\$• $$Cu$^{2+}/$Cu$: 0.34 \,$V$\\$• $$Zn$^{2+}/$Zn$: -0.76 \,$V$\\$• $$Fe$^{2+}/$Fe$: -0.44 \,$V$\\$• $$Ag$^{+}/$Ag$: 0.80 \,$V$\\$The feasibility of a redox reaction can be predicted by calculating the cell potential $E_{cell}. \\$A positive $E_{cell}$ indicates a spontaneous reaction.$\\$Let's analyze each option:$\\$Option 1: $$CuSO$_4($aq$) +$Zn$($s$) \rightarrow$ZnSO$_4($aq$) +$Cu$($s$) \\$• Oxidation: $$Zn$\rightarrow$Zn$^{2+} + 2e^- \quad (E^\circ = 0.76 \,$V$) \\$• Reduction: $$Cu$^{2+} + 2e^- \rightarrow$Cu$\quad (E^\circ = 0.34 \,$V$) \\ E_{cell} = E_{cathode} - E_{anode} = 0.34 - (-0.76) = 1.10 \,$V$\\$Since $E_{cell} > 0,$ the reaction is feasible.$\\$Option 2: $$CuSO$_4($aq$) +$Fe$($s$) \rightarrow$FeSO$_4($aq$) +$Cu$($s$) \\$• Oxidation: $$Fe$\rightarrow$Fe$^{2+} + 2e^- \quad (E^\circ = 0.44 \,$V$) \\$• Reduction: $$Cu$^{2+} + 2e^- \rightarrow$Cu$\quad (E^\circ = 0.34 \,$V$) \\ E_{cell} = 0.34 - (-0.44) = 0.78 \,$V$\\$Since $E_{cell} > 0,$ the reaction is feasible.$\\$Option 3: $$FeSO$_4($aq$) +$Zn$($s$) \rightarrow$ZnSO$_4($aq$) +$Fe$($s$) \\$• Oxidation: $$Zn$\rightarrow$Zn$^{2+} + 2e^- \quad (E^\circ = 0.76 \,$V$) \\$• Reduction: $$Fe$^{2+} + 2e^- \rightarrow$Fe$\quad (E^\circ = 0.44 \,$V$) \\ E_{cell} = 0.44 - (-0.76) = 1.20 \,$V$\\$Since $E_{cell} > 0,$ the reaction is feasible.$\\$Option 4: $2$CuSO$_4($aq$) + 2$Ag$($s$) \rightarrow 2$Cu$($s$) +$Ag$_2$SO$_4($aq$) \\$• Oxidation: $2$Ag$\rightarrow 2$Ag$^{+} + 2e^- \quad (E^\circ = -0.80 \,$V$) \\$• Reduction: $$Cu$^{2+} + 2e^- \rightarrow$Cu$\quad (E^\circ = 0.34 \,$V$) \\ E_{cell} = 0.34 - 0.80 = -0.46 \,$V$\\$Since $E_{cell} < 0,$ the reaction is not feasible.$\\$Therefore, the reaction in Option 4 cannot occur.