1.

Permanganate ion reacts with bromide ion is a basic medium, to give manganese dioxide and bromate ion. Write the balance chemical equation for the reaction.

Answer»

Solution :Step- 1 Write the skeletal equation. The skeletal equation for the given reaction is:
`MnO_4^(-)(aq)+Br^(-) (aq) toMnO_2(s)+BrO_3(aq)`
Step -2 Find out the elements which undergo a change in oxidation number (O.N.)

O.N DECREASES by 3 PER Mn atom.......(i)
Here, O.N. of Br increases from -1 Br to +5 in `BrO_3^(-)` and therefore, `Br^(-)` acts as the reactant. Further O.N. of Mn decreases from +7 in `MnO_4^(-)` to +4 in `MnO_2`, therefore `MnO_4^(-)` acts as the oxidant.
Step -3: Balance O atoms by ADDING `H_2O` molecules.
`2MnO_4^(-)(s)+Br^(-) (aq)to2MnO_2 (s)+BrO_3^(-)(aq)+H_2O(l)`.....(iv)
Step -4: Balance H atoms by adding `H_2O` and OH IONS since the reaction occurs in basic medium, therefore, add `2H_2O` to L.H.S and 2OH to R.H.S of Eq. (iv), we have,
`2MnO_4^(-)(aq)+Br^(-)(aq)+2H_2O(l)to2MnO_2+BrO_3^(-)(aq)+H_2O(l)+2OH^(-) (aq)`


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