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What is the quantity of charge needed fo...

What is the quantity of charge needed for the reduction of `1` mole of `MnO_(4)^(-)` ion to `Mn^(2+)` ion ?

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The correct Answer is:
5
`underset(("Oxidation NO. of Mn" = +7))(MnO_(4)^(-)) + 5e rarr underset(("Oxidation no. of "Mn = 2))(Mn^(2+))`
`:. "Charge" = 5 xx 96500C = 5F`
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