Consider the following redox reaction in an acidic medium: x MnO ₄⁻ + y C ₂ O ₄²⁻ + z H ⁺ x Mn ²⁺ + 2y CO ₂ + z 2 H ₂ O When the equation is correctly balanced using the smallest possible whole numbers, the values of the stoichiometric coefficients x , y , and z are respectively:

Step-by-step solution

The given redox reaction can be split into two half-reactions. Oxidation half-reaction: C ₂ O ₄²⁻ 2 CO ₂ + 2e⁻ Reduction half-reaction: 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 gives the overall balanced equation: 2 MnO ₄⁻ + 5 C ₂ O ₄²⁻ + 16 H ⁺ 2 Mn ²⁺ + 10 CO ₂ + 8 H ₂ O Comparing this with the