In the balanced redox example, what is the oxidation half-reaction?

• A. MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O
• B. CO2 → CO3^2- + e-
• C. C2O4^2- → 2 CO2 + 2 e-
• D. MnO4- → Mn2+ + 5 e-

Answer: (C)

The main idea is to identify the half-reaction that releases electrons. In a redox balancing process, the oxidation half shows a species losing electrons, so its oxidation state must increase.

Here, the oxidation occurs when oxalate (C2O4^2−) is converted to carbon dioxide (CO2). The carbon atoms go from an oxidation state of +3 in C2O4^2− to +4 in CO2, which is an increase, meaning electrons are released. That half-reaction is written with electrons on the product side: C2O4^2− → 2 CO2 + 2 e−. It shows two electrons produced, consistent with the oxidation of oxalate.

The other options don’t fit as the oxidation half: one shows a species gaining electrons (reduction), another would imply no change in oxidation state, and another depicts oxidation of permanganate in a way that isn’t the oxidation half for this balance.