In the reaction 2MnO4⁻ + 5C2O4²⁻ + 16H⁺ → 2Mn²⁺ + 10CO2 + 8H2O, which species is

In the reaction 2MnO4⁻ + 5C2O4²⁻ + 16H⁺ → 2Mn²⁺ + 10CO2 + 8H2O, which species is reduced? (2019)

In the reaction 2MnO4⁻ + 5C2O4²⁻ + 16H⁺ → 2Mn²⁺ + 10CO2 + 8H2O, which species is reduced? (2019)

Step 1: Identify the reactants and products in the reaction.
Step 2: Look for the oxidation states of the elements in the reactants and products.
Step 3: Determine the oxidation state of manganese (Mn) in MnO4⁻. It is +7.
Step 4: Determine the oxidation state of manganese (Mn) in Mn²⁺. It is +2.
Step 5: Compare the oxidation states. Mn goes from +7 to +2, which means it gains electrons.
Step 6: Since gaining electrons indicates reduction, MnO4⁻ is the species that is reduced.

No concepts available.