Question

An equal volume of a reducing agent is titrated separately with 1M KMnO₄ in acid neutral and alkaline media. The volumes of KMnO₄ required are 20 ml. in acid, 33.4 ml. neutral and 100 ml. in alkaline media. Find out the oxidation state of manganese in each reduction product. Give the balanced equations for all the three half reactions. Find out the volume of 1M K₂Cr₂O₇ consumed; if the same volume of the reducing agent is titrated in acid medium.

Answer 1

. Let the n-factor of KMnO4 in acid, neutral and alkaline media are N1, and N3 respectively. Also, same volumes of reducing agent is used everytime, same number of equivalents of KMnO4 would be required every time. ⇒ 20N1 = 100/3N2 = 100 N3 ⇒ N1 = 5/3N2 = N3 Also, n-factors are all integer and greater than or equal to one but less than six, N3 must be 1. ⇒ N1 = 5, N2 = 3 ∴ In acid medium : 〖MnO〗_4^- → Mn2+ In neutral medium : 〖MnO〗_4^- → Mn4+ In alkaline medium : 〖MnO〗_4^- → Mn6+ For K2Cr2O7 Volume : In acidic medium meq of 〖MnO〗_4^- used = 20 × 5 = 100 = meq of R – A ⇒ meq of K2Cr2O7 required = 100 ⇒ 100 = 1 × 6 × V (n-factor = 6) ⇒V = 100/6 = 16.67 mL.

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