For decolourization of 1 mole of KMnO_(4) the moles fo H_(2)O_(2) requ

1.	For decolourization of 1 mole of KMnO_(4) the moles fo H_(2)O_(2) required is
Answer» `1//2` `3//21` `5//2` `7//2` Solution :Red half reaction `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` Oxid half reaction `H_(2)O_(2)rarrO_(2)+2H^(+)+2e^(-)` since 1 MOLE of `H_(2)O_(2)` requires `5e^(-)` therefore REDUCTION of 1 mole of `KMnO_(4)` will REQUIRE `5//2` molesof `H_(2)O_(2)`

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