Redox Titration (permanganometry)
In an acidic medium permanganometric titration, 20 mL of 0.05 M oxalic acid solution was titrated against potassium permanganate of concentration 0.02 M. What volume of permanganate is required to reach the end point?
Select the correct option:
Solution
20 mL
Permanganate titrations are redox reactions in which the equivalents of oxidant equal the equivalents of reductant at the end point. In acidic medium the permanganate ion gains five electrons, MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O, so its n-factor is 5. Oxalic acid is oxidised at both carbon atoms, C_2O_4^{2-} → 2CO_2 + 2e^-, giving it an n-factor of 2. The milliequivalents of oxalic acid = molarity × volume × n-factor = 0.05 × 20 × 2 = 2.0 meq. Setting this equal to the permanganate milliequivalents, 0.02 × V × 5 = 2.0, gives 0.1 × V = 2.0, so V = 20 mL. Option 10 mL would result from using an n-factor of 5 for oxalic acid as well, ignoring its true value of 2. Option 50 mL arises from treating permanganate with an n-factor of 2. Option 8 mL comes from omitting the oxalic acid n-factor entirely. This equivalence approach reflects the NCERT redox-titration method. Plausibility check: because permanganate transfers more electrons per mole than oxalate, the equal equivalents balance out here to give a coincidentally equal volume, which is dimensionally consistent.
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- redox titration (permanganometry)
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
20 mL
Permanganate titrations are redox reactions in which the equivalents of oxidant equal the equivalents of reductant at the end point. In acidic medium the permanganate ion gains five electrons, MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O, so its n-factor is 5. Oxalic acid is oxidised at both carbon atoms, C_2O_4^{2-} → 2CO_2 + 2e^-, giving it an n-factor of 2. The milliequivalents of oxalic acid = molarity × volume × n-factor = 0.05 × 20 × 2 = 2.0 meq. Setting this equal to the permanganate milliequivalents, 0.02 × V × 5 = 2.0, gives 0.1 × V = 2.0, so V = 20 mL. Option 10 mL would result from using an n-factor of 5 for oxalic acid as well, ignoring its true value of 2. Option 50 mL arises from treating permanganate with an n-factor of 2. Option 8 mL comes from omitting the oxalic acid n-factor entirely. This equivalence approach reflects the NCERT redox-titration method. Plausibility check: because permanganate transfers more electrons per mole than oxalate, the equal equivalents balance out here to give a coincidentally equal volume, which is dimensionally consistent.
This hard difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of redox titration (permanganometry). It appeared in the 2025 exam.
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