When 0.1 mol MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO41- is

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answer is 3.

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Detailed Solution

The oxidation reaction isMn+6O42-0.1 mol→ Mn+7O4- +e-0.1 molQ=0.1×F=0.1×96500C=9650C

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