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Number of moles of KMnO₄ required to oxidize 1 mole of Fe²⁺ in acidic solution:

A2
B5
C1/5 (Mn(VII) → Mn(II): 5e⁻ accepted; Fe²⁺ → Fe³⁺: 1e⁻ lost)
D1
Answer & Solution
Correct answer: C. 1/5 (Mn(VII) → Mn(II): 5e⁻ accepted; Fe²⁺ → Fe³⁺: 1e⁻ lost)
MnO₄⁻ + 5 e⁻ → Mn²⁺. Fe²⁺ → Fe³⁺ + e⁻. **So 1 MnO₄⁻ oxidizes 5 Fe²⁺ → 1/5 mole KMnO₄ per mole Fe²⁺.**
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