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JEE Main 2022 · Chemistry · d- and f-Block Elements and Metallurgy · medium

The volume of 0.020 M KMnO4 required in acidic medium to completely oxidise 0.72 g of FeC2O4 (take Fe=56, C=12, O=16) is

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The volume of 0.020 M KMnO40.020\ \mathrm M\ \mathrm{KMnO_4} required in acidic medium to completely oxidise 0.72 g0.72\ \mathrm g of FeC2O4\mathrm{FeC_2O_4} (take Fe=56\mathrm{Fe}=56, C=12\mathrm{C}=12, O=16\mathrm{O}=16) is
  1. A. 100 mL100\ \mathrm{mL}
  2. B. 150 mL150\ \mathrm{mL}
  3. C. 250 mL250\ \mathrm{mL}
  4. D. 50 mL50\ \mathrm{mL}
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Answer: (B) 150 mL150\ \mathrm{mL}

Both ions are oxidised by permanganate: Fe2+→Fe3+\mathrm{Fe^{2+}\to Fe^{3+}} (1 e−^-) and C2O42−→2CO2\mathrm{C_2O_4^{2-}\to2CO_2} (2 e−^-), so each formula unit supplies 33 e−^-. Moles of FeC2O4=0.72/144=5.0\mathrm{FeC_2O_4}=0.72/144=5.0 mmol, giving 1515 mmol of electrons. MnO4−→Mn2+\mathrm{MnO_4^-}\to\mathrm{Mn^{2+}} takes 55 e−^-, so KMnO4=3.0\mathrm{KMnO_4}=3.0 mmol and V=3.0/0.020=150 mLV=3.0/0.020=150\ \mathrm{mL}. Counting only oxalate gives 100100 mL; only iron gives 5050 mL.
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