Permanganate (VII) ion, `MnO_(4)^(-)` in basic solution oxidises iodide ion, `I^(-)` to produce molecular iodine `(I_(2))` and manganese (IV) oxide `(MnO_(2))`. Write a balanced ionic equation to represent this redox reaction.

Explanation

Step -1 :
First we write the skeletal ionic equation, which is
`MnO_(4)^(-)(aq)+1(aq)to MnO_(2)(s)+I_(2)(s)`
Step -2 :
The two half-reactions are :
-1 0
Oxidation half : `1^(-)(aq)toI_(2)(s)`
+7 +4
Reduction half : `MnO_(4)^(-)(aq)to MnO2(s)`
Step -3 :
To balance the I atoms in the oxidation half reaction,we rewrite it as :
`2I^(-)(aq)to I_(2)(s)`
Step -4: As the reaction takes place to basic medium, to balance the O atoms in the reduction half reaction,we add `OH^(-)` ions in required number.
`MnO_(4)^(-)(aq)to MnO_(2)(s)+2HO^(-)(I)` To balance the H atoms, we add two `H_(2)O` molecules on the left.
`Mn_(4)^(-)(aq)+2H_(2)O(aq)to MnO_(2)(s)+2HO^(-)(I)`
Balance H and O atoms by bit and trial method even if it requires morethan once we have to do it.
The resultant equation is :
`MnO_(4)^(-)(aq)+2H_(2)O(I)to MnO_(2)(s)+4OH^(-)(aq)`
Note : While balanceing H atom O atoms do not disturb the coefficients of other species (oxidant and reductant and the products) .
Step -5: In this step we balance the charges of the two half-reactions in the manner depi-cted as :
`2I^(-)(aq)to I_(2)(s)+2e^(-)`
`Mn_(4)^(-)(aq)+2H_(2)O(I)+3e^(-)to MnO_(2)(s)+4OH^(-)(aq)`
Now to equalise the number of electrons, in the two half reactions we multiply the oxidation half-reaction by 3 and the reductiion half-reaction by 2.
`6I^(-)(aq)to 3I_(2)(s)+6e^(-)`
`2MnO_(4)(aq)+4H_(2)O(I)+6e^(-)to 2MnO_(2)(s)+8OH^(-)(aq)`
Step -6 :
Add two half- reactions to obtain the net reaction and canceel the electrons on both sides.
`6I^(-)(aq)+2MnO_(4)^(-)(aq)+4H_(2)O(I)to 3I_(2)(s)+2MnO_(2)(s)+8OH^(-)(aq)`
Step -7 :
A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.