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`        An equal volume of a reducing agent is titrated separately with 1 M KMnO4 in acid, neutral and alkaline medium. The volumes of KMnO4 required are 20 mL in acid, 33.3 mL in 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 reaction. Find out the volume of 1 M K2Cr2O7 consumed, if the same volume of the reducing agent is titrated in acid medium. `
5 years ago

Jitender Pal
365 Points
```							Sol. 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

```
5 years ago
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