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Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction. |
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Answer» Solution :Step-1 : The skeletal ionic equation is : `MnO_(4(AQ))^(-)+Br_((aq))^(-)toMnO_(2(s))+BrO_(3(aq))^(-)` Step-2 : Assign oxidation numbers for Mn and Br. `overset(+7)(MnO_(4(aq))^(-))+overset(-1)(Br_((aq))^(-))tooverset(+4)(MnO_(2(s)))+overset(+5)(BrO_(3(aq))^(-))` This indicates that permanganate ion is the oxidant and bromide ion is the reductant. Step-3 : Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease. `overset(+7)(2MnO_(4(aq))^(-))+overset(-1)(Br_((aq))^(-))tooverset(+4)(2MnO_(2(s)))+overset(+5)(BrO_(3(aq))^(-))` Step-4 : As the reaction OCCURS in the basic medium, and the ionic charges are not equal on both sides, add `2OH^(-)` ions on the right to make ionic charges equal. `2MnO_(4(aq))^(-)+Br_((aq))^(-)to2MnO_(2(s))+BrO_(3(aq))^(-)+2OH_((aq))^(-)` Step-5 : Finally, count the hydrogen atoms and ass APPROPRIATE number of water molecules (i.e. one `H_(2)O` molecule) on the left side to achieve BALANCED redox change. `2MnO_(4(aq))^(-)+Br_((aq))^(-)+H_(2)O_((l))to2MnO_(2(s))+BrO_(3(aq))^(-)+2OH_((aq))^(-)` |
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