InterviewSolution
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InterviewSolution
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`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. ` 2//3` mol of `MnO_4^(-)` and `1//3` mol of `MnO_2`B. ` 1//3` mol of `MnO_4^(-)` and `2//3` mol of `MnO_2`C. ` 1//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2`D. ` 2//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2` |
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Answer» Correct Answer - A ` MnO_4^(2-)` is quite strongly oxdizing and is only stable in very strong alkali. In dilute alkali, water, or acidic solutions, it disproportionates : ` MnO_4^(2-) rarr MnO_4^(-) MnO_2` Oxidation half-reaction: `overset(+6)(M)nO_(4)^(2-) rarr overset(+7)(M)nO_(4)^(-) +e^(-)` Reduction half-reaction: `overset(+6)(M)nO_(4)^(2-)+2e^(-) rarr overset(+4)(M)nO_(2)` Balance ionic charges by adding `H^(+)` ions: `MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)` `MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)` Balance H and O by adding `H_(2)O` molecules `MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)` `MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)+2H_(2)O` To equalize electron transfer, we multiply oxidation half-reaction by 2 and add them, cancelling electrons on both sides: `3MnO_(4)^(2-) +4H^(+) rarr 3MnO_(4)^(-)+MnO_(2)+2H_(2)O` Similarly we can balance it by means of `OH^(-)` ions: `3MnO_(4)^(2-)+2H_(2)O rarr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)` Thus, 3 mol of `MnO_(4)^(2-)` yield 2 mol of `MnO_(4)^(-)` and 1 mole of `MnO_(2)`. Consequently, 1 mol of `MnO_(4)^(2-)` disproportionates to yield `2//3` mol of `MnO_(4)^(2-)` and `1//3` mol of `MnO_(2)`. Note that in neutral aqueous meadium, we can balance the charge either using `H^(+)` ions or using `OH^(-)` ions. |
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