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The equivalent mass of KMnO4 in alkaline medium is a.31.6 b.52.7 c.79.d.158

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Answer: (b)

In acidic medium the number of electrons involved are 5.

Mno4 + 8H+ + 5e– → Mn2+ + 4H2O

Eq. mass of KMnO4 = 158/ 5 =31.6

In basic medium the number of electrons involved are 3.

Mno4 + 2H2O+ + 3e– → MnO2 + 4OH-

Eq. mass of KMnO4 = 158/ 3 = 52.7

Equivalent mass of KMnO4 in acidic and alkaline medium is 31.6 and 52.7 respectively.

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