1.

The charge required for the reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` isA. 1FB. 3FC. 5FD. 6F

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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