Write a balanced ionic equation to describe the oxidation of iodide `(I^-)` in by permanganate `(MnO_(4)^(-))` ion in basic solution to yield molecular iodine `(l_2)` and manganese (IV) oxide `(MnO_2)`.
Strategy : We are given the formulas for two reactants and two prodcts. We use these to write the skeletal ionic equatin. We construct and balance the appropriate half-reactions using the rules just described. Then we add the half -reactions and eliminate common terms.

Explanation

Step 1: First we write the skeletal ionic equation, which is
`MnO_(4)^(-) (aq) + I^(-) (aq) to MnO_(2)(s) + I_(2)(s)`
Step 2: The two half-reactions are:
Oxidation half : `overset(-1)(I^(-))(aq) to overset(0)(I_(2))(s)`
Reduction half : `overset(+7)(MnO_(4)^(-)) (aq) to MnO_(2)(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 : To balance the O atoms in the reduction half reaction , we add two water molecules on the right :
`MnO_(4)^(-) (aq) + 4 H^(+) (aq) to MnO_(2)(s) + 2H_(2)O(1)` To balance the H atoms , we add four `H^(+)` ions on the left :
`MnO_(4)^(-) (aq) + 4 H^(+) (aq) to MnO_(2)(s) + 2H_(2)O (1)`
As the reaction takes place in a basic solution , therefore , for four `H^(+)` ions , we add four `OH^(-)` ions to both sides of the equation :
`MnO_(4)^(-) (aq) + 4 H^(+) (aq) + 4 OH^(-) (aq) to MnO_(2) (s) + 2H_(2)O (1) + 4 OH^(-) (aq)`
Replacing the `H^(+)` and `OH^(-)` ions with the water , the resultant equation is : `MnO_(4)^(-)(aq) + 2H_(2)O(1) to MnO_(2)(s) + 4 OH^(-)(aq)`
Step 5 : In this step we balance the charges of the two half - reactions in the manner depicted as :
`2I^(-) (aq) to I_(2) (s) + 2e^(-)`
`MnO_(4)^(-) + 2H_(2)O (1) + 3 e^(-) to MnO_(2)(s) + 4 OH^(-) (aq)`
Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.
`6I^(-) (aq) to 3 I_(2) (s) + 6 e^(-)`
`2 MnO_(4)^(-) (aq) + 4 H_(2)O(1) + 6 e^(-) to 2 MnO_(2) (s) + 8 OH^(-) (aq)`
Step 6 : Add two half-reactions to obtain the net reactions after cancelling electrons on both sides.
`6I^(-)(aq) + 2MnO_(4)^(-) (aq) + 4H_(2)O(1) to 3I_(2)(s) + 2 MnO_(2) (s) + 8 OH^(-) (aq)`
Step 7 : : A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.