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The charge required for the reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` isA. 1FB. 3FC. 5FD. 6F |
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Answer» Correct Answer - B `overset(+7)(MnO_(4)^(-)rarroverset(+4)(MnO_(2))` `i.e. MnO_(4)^(-)+3e^(-)rarrMnO_(2)` `MnO_(4)^(-)+2H_(2)O+3e^(-) rarrMnO_(2)+4OH^(-)` Thus for reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` charge required =3F |
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