【Daily Question】Theoretical Foundation 2020.09.02
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Multiple-choice question: When the following substances come into contact with an H2O2 aqueous solution, which one can cause H2O2 to exhibit reducitive properties? (A) KMnO4 (acidic) (B) SnCl2 (C) Fe2+ (D) NaOH. Answer: A. Hint: Extra rewards will be given to those who can explain the reasoning behind the answer; the answer can be found in the response
Answer: A. The order of the electrode potentials is: EΘ(H2O2/H2O) > EΘ(MnO4–/MnO42+) > EΘ(O2/H2O2) > EΘ(Fe3+/Fe2+) > EΘ(O2/OH–) > EΘ(Sn4+/Sn2+). It can be seen that MnO4–, as an oxidizing agent, can oxidize H2O2, which is in the reduced form in the O2/H2O2 redox pair; thus, KMnO4 can make H2O2 exhibit reducing properties. In the H2O2/H2O couple, H2O2 is in the oxidized state.