To balance the given redox reaction, we need to follow these steps:$\\$1. Identify the oxidation and reduction half-reactions.$\\$• The oxidation half-reaction involves $$C$_2$O$_4^{2-}$ being oxidized to $$CO$_2. \\$• The reduction half-reaction involves $$MnO$_4^-$ being reduced to $$Mn$^{2+}. \\$2. Write the half-reactions:$\\$• Oxidation: $$C$_2$O$_4^{2-} \rightarrow 2$CO$_2 + 2e^- \\$• Reduction: $$MnO$_4^- + 8$H$^+ + 5e^- \rightarrow$Mn$^{2+} + 4$H$_2$O$\\$3. Balance the electrons in the half-reactions.$\\$• Multiply the oxidation half-reaction by 5:$\\ 5($C$_2$O$_4^{2-} \rightarrow 2$CO$_2 + 2e^-) \\$• Multiply the reduction half-reaction by 2:$\\ 2($MnO$_4^- + 8$H$^+ + 5e^- \rightarrow$Mn$^{2+} + 4$H$_2$O$) \\$4. Combine the balanced half-reactions:$\\ 5$C$_2$O$_4^{2-} + 2$MnO$_4^- + 16$H$^+ \rightarrow 10$CO$_2 + 2$Mn$^{2+} + 8$H$_2$O$\\$5. Verify the balance of atoms and charges:$\\$• Carbon: $5 \times 2 = 10$ on both sides$\\$• Manganese: $2$ on both sides$\\$• Oxygen: $5 \times 4 + 2 \times 4 = 28$ on both sides$\\$• Hydrogen: $16$ on both sides$\\$• Charge: $-10 + 2 \times (-1) + 16 = +4$ on both sides$\\$Thus, the balanced equation is:$\\ 2$MnO$_4^- + 5$C$_2$O$_4^{2-} + 16$H$^+ \rightarrow 2$Mn$^{2+} + 10$CO$_2 + 8$H$_2$O$\\$Therefore, the correct coefficients for $$MnO$_4^-,$C$_2$O$_4^{2-},$H$^+$ are $2, 5, 16. \\$This corresponds to Option 2.