1.

Calculate equivalent weight of KMnO_(4) in Acidic, Basic and Neutral medium.

Answer»

SOLUTION :Acidic medium : `MnO_(4)^(-)+5e^(-)toMn^(+2)" "thereforeM/5`
Basic medium : `MnO_(4)^(-)+3E^(-)toMnO_(2)" "thereforeM/3`
NEUTRAL medium : `MnO_(4)^(-)+E^(-)toMnO_(4)^(-2)" "thereforeM/1`


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