To balance the redox reaction between potassium permanganate KMnO4 and oxalic acid H2C2O4 in acidic medium, we need to follow these steps:1. Write down the unbalanced equation:KMnO4 + H2C2O4 Mn^2+ + CO22. Separate the equation into two half-reactions:Oxidation half-reaction: H2C2O4 CO2Reduction half-reaction: KMnO4 Mn^2+3. Balance the atoms in each half-reaction:Oxidation half-reaction: H2C2O4 2 CO2 + 2 H^+Reduction half-reaction: MnO4^- + 8 H^+ + 5 e^- Mn^2+ + 4 H2O4. Balance the electrons in both half-reactions:To balance the electrons, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:Oxidation half-reaction: 5 H2C2O4 10 CO2 + 10 H^+ + 10 e^-Reduction half-reaction: 2 MnO4^- + 16 H^+ + 10 e^- 2 Mn^2+ + 8 H2O5. Add the two balanced half-reactions together and simplify:5 H2C2O4 + 2 MnO4^- + 16 H^+ 10 CO2 + 10 H^+ + 2 Mn^2+ + 8 H2OSimplify by canceling out the 10 H^+ on both sides:5 H2C2O4 + 2 MnO4^- + 6 H^+ 10 CO2 + 2 Mn^2+ + 8 H2O6. Write the balanced redox reaction:5 H2C2O4 + 2 KMnO4 + 6 H^+ 10 CO2 + 2 Mn^2+ + 8 H2ONote: In the final balanced equation, we replaced MnO4^- with KMnO4, as potassium K^+ is a spectator ion and does not participate in the redox reaction.