a C ₂ O ₄²⁻ + b MnO ₄^- + c H ^+ x Mn ²⁺ + y H ₂ O + z CO ₂ a and x respectively are

Step-by-step solution

The given reaction is a redox reaction in an acidic medium. The oxidation half-reaction is: C ₂ O ₄²⁻ 2 CO ₂ + 2e^- The reduction half-reaction is: MnO ₄^- + 8 H ^+ + 5e^- Mn ²⁺ + 4 H ₂ O To balance the number of electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 : 5 C ₂ O ₄²⁻ 10 CO ₂ + 10e^- 2 MnO ₄^- + 16 H ^+ + 10e^- 2 Mn ²⁺ + 8 H ₂ O Adding the two balanced half-reactions, we get: 5 C ₂ O ₄²⁻ + 2 MnO ₄^- + 16 H ^+ 2 Mn ²⁺ + 8 H ₂ O + 10 CO ₂ Comparing this with the given e