How many electrons are required for the reduction of 1 mole of MnO_(4)

1.	How many electrons are required for the reduction of 1 mole of MnO_(4)^(-)" to "Mn^(2+) ?
Answer» `3.011xx10^(24)` `6.022xx10^(24)` `1.2044xx10^(24)` `1.8066xx10^(24)` SOLUTION : `therefore` Reduction half reaction : `MnO_(4)^(-)+5bare+8H^(+)toMn^(2+)+4H_(2)O` Thus reduction of 1 mole of `MnO_(4)^(-)` requires 5 mole of electrons But 1 mole electron = `6.022xx10^(23)` electron `therefore5" mole electron"=(5xx6.022xx10^(23))/1` = `30.10xx10^(23)` = `3.011xx10^(24)` electrons

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