JEE Advanced level
A JEE Advanced-level example on d- and f-Block Elements
Question. Potassium permanganate (KMnO₄) is used to titrate 25 mL of a solution containing Fe²⁺ ions in acidic medium. The titration requires 20 mL of 0.02 M KMnO₄ solution. (a) What is the oxidation state change of Mn during the reaction? (b) Calculate the molarity of Fe²⁺ in the original solution. (c) If the same Fe²⁺ solution is titrated with K₂Cr₂O₇ in acidic medium, what volume of 0.01 M K₂Cr₂O₇ would be required?
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- In acidic medium, MnO₄⁻ (Mn in +7) is reduced to Mn²⁺ (Mn in +2). Oxidation state change = 7 − 2 = 5 units decrease.
- Balanced equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. So 1 mol KMnO₄ reacts with 5 mol Fe²⁺.
- Moles of KMnO₄ used = 0.02 × (20/1000) = 4 × 10⁻⁴ mol. Moles of Fe²⁺ = 5 × 4 × 10⁻⁴ = 2 × 10⁻³ mol.
- Molarity of Fe²⁺ = (2 × 10⁻³)/(25 × 10⁻³) = 0.08 M.
- For dichromate: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. So 1 mol K₂Cr₂O₇ reacts with 6 mol Fe²⁺.
- Moles of K₂Cr₂O₇ needed = (2 × 10⁻³)/6 = 3.33 × 10⁻⁴ mol. Volume = (3.33 × 10⁻⁴)/0.01 = 0.0333 L = 33.3 mL.
Answer: (a) Mn changes from +7 to +2 (decrease of 5 units); (b) [Fe²⁺] = 0.08 M; (c) Volume of K₂Cr₂O₇ required = 33.3 mL
This example was solved twice independently and both solutions agree.