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Permanagante ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion write the balaced chemical equation for the reaction |
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Answer» Solution :Step 1 write the skeletal equaton the skeletal equation for the given reaction is `MnO_(@)^(-)(aq)+Br^(-)rarrMnO_(2)(s)BrO_(3)^(-)+BrO_(3)^(-)`(aq) step 2 find out the elements which undergo a CHANGE in oxidatoin number (O.N)  here O.N of Br increases from -1 in `Br^(-)` to +5in `BrO_(3)^(-)therefore Br^(-)` ACTS as reductant further O.N of Mn decreases form +7 in `MnO_(4)^(-)` to +4 in `MnO_(2) THEREFOR MnO_(4)^(-)` acts as OXIDANT step 3 find out total increase / decrease in O.N since there is only one Br atom on either side therefore total in O.N of Br is 6 further since there is only one Mn atom on either side therefore total decrease O.N of Mn is 3 step 4 balance increase / decrease in O.N since the total increase in O.N is 6 and decrease in O.N is 3 therefore multiply `MnO_(4)^(-)` by 2 combing steps 2 and 3 we have `2MnO_(5)^(-)(aq)Br^(-)(aq)rarrMnO_(2)(s)+BrO_(3)^(-)(aq)` step 5 balance all atoms other than O and H to balance Mn on either side of Eq (ii) multiply `MnO_(2)` by 2 we have `2MnO_(4)^(-)(aq)+Br^(-)(aq)rarr2MnO_(2)(s)+BrO_(3)^(-)(aq)+H_(2)O(l)` step 7balance H atoms by adding `H_(2)O` and `OH^(-)` ions since the reactin occurs in basic medium since there are tow H atoms on R.H.S and none on L.H.S of equ(IV) therefore add `2H_(2)O` to L.H.S and `2OH^(-)` to R.H.S of eq(iv) we have `2MnO_(4)^(-)(aq)+Br^(-)(aq)+2H_(2)O(l)rarr2MnO_(2)(s)+BrO_(3)^(-)(aq)+H_(2)O(l)+2OH^(-)(aq)` |
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