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JEE Main 2021 · Chemistry · Redox Reactions and Titrations · medium

20 mL of 0.50 M FeC2O4 (ferrous oxalate) is titrated in acidic medium with 0.10 M KMnO4. The volume of KMnO4 required for complete oxidation is

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20 mL20\ \mathrm{mL} of 0.50 M0.50\ \mathrm{M} FeC2O4\mathrm{FeC_2O_4} (ferrous oxalate) is titrated in acidic medium with 0.10 M0.10\ \mathrm{M} KMnO4\mathrm{KMnO_4}. The volume of KMnO4\mathrm{KMnO_4} required for complete oxidation is
  1. A. 20 mL20\ \mathrm{mL}
  2. B. 40 mL40\ \mathrm{mL}
  3. C. 60 mL60\ \mathrm{mL}
  4. D. 100 mL100\ \mathrm{mL}
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Answer: (C) 60 mL60\ \mathrm{mL}

Both Fe2+\mathrm{Fe^{2+}} and oxalate are oxidised. Per formula unit: Fe2+→Fe3+\mathrm{Fe^{2+}}\to\mathrm{Fe^{3+}} gives 1 e−1\,e^- and C2O42−→2CO2\mathrm{C_2O_4^{2-}}\to2\mathrm{CO_2} gives 2 e−2\,e^-, so the n-factor of FeC2O4\mathrm{FeC_2O_4} is 33. For MnO4−→Mn2+\mathrm{MnO_4^-}\to\mathrm{Mn^{2+}} the n-factor is 55. Millimoles of FeC2O4=20×0.5=10\mathrm{FeC_2O_4}=20\times0.5=10. Equivalents: 10×3=3010\times3=30 meq =V×0.10×5=V\times0.10\times5, so V=60 mLV=60\ \mathrm{mL}. Counting only Fe2+\mathrm{Fe^{2+}} gives 20 mL20\ \mathrm{mL} and counting only oxalate gives 40 mL40\ \mathrm{mL}.
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