Consider the following reaction: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. In this reaction, the number of electrons transferred per mole of MnO₄⁻ is:
(A) 1
(B) 3
(C) 5
(D) 8
Correct: C
MnO₄⁻ (Mn in +7 oxidation state) is reduced to Mn²⁺ (oxidation state +2). The change in oxidation state per Mn atom is 7 − 2 = 5 electrons gained. Since the balanced equation shows 2 moles of MnO₄⁻, the total electrons transferred in the reaction is 10. However, the question asks per mole of MnO₄⁻, so it is 5 electrons. The oxidation half-reaction: C₂O₄²⁻ → 2CO₂ + 2e⁻ (each oxalate loses 2 electrons). For 5 oxalate ions, 10 electrons are released, which match the 10 electrons accepted by 2 MnO₄⁻ ions. Per mole of MnO₄⁻, 5 electrons are transferred. Answer: C.