In the reaction 2MnO4- + 5C2O4^2- + 16H+ → 2Mn^2+ + 10CO2 + 8H2O, what is being

In the reaction 2MnO4- + 5C2O4^2- + 16H+ → 2Mn^2+ + 10CO2 + 8H2O, what is being oxidized? (2019)

In the reaction 2MnO4- + 5C2O4^2- + 16H+ → 2Mn^2+ + 10CO2 + 8H2O, what is being oxidized? (2019)

Step 1: Identify the reactants in the reaction: 2MnO4-, 5C2O4^2-, and 16H+.
Step 2: Determine the oxidation states of the elements in the reactants.
Step 3: Focus on the C2O4^2- (oxalate ion) and find its oxidation state. Each carbon in C2O4^2- has an oxidation state of +3.
Step 4: Look at the products of the reaction, which include CO2. In CO2, the oxidation state of carbon is +4.
Step 5: Compare the oxidation states: the carbon in C2O4^2- goes from +3 to +4, indicating it is losing electrons.
Step 6: Conclude that since C2O4^2- is losing electrons, it is being oxidized to CO2.

No concepts available.