Give the balanced ionic equation for the oxidation of `I^(-)` by `MnO_4^(-)` ions in basic medium to produce molecular iodine `(I_2)` and manganese (IV) oxide `(MnO_2)`.

Explanation

Step 1: The unbalanced ionic equation: ` overset(+7)(MnO_(4)^(-) + overset(+7)(I^(-)) to overset(+4)(MnO_(2))+overset(0)(I_2)`
Step 2: Separating the equation into two half-reactions, the oxidation half and the reduction half,
` underset("Oxidation half-reaction")overset(-1 " " 0)( I^(-) to I_2) , underset("Reduction half-reaction ") overset(+7 " " +4)(MnO_(4)^(-) to MnO_(2))`
Step 3: To balance the number I atoms, the oxidation half reaction is multiplied by 2,
` 2I^(-) to I_(2) , MnO_(4)^(-) to MnO_(2)`
Step 4: To balance O atoms in the reduction half reaction, water molecules are added on the right:
`2I^(-) to I_(2) , MnO_(4)^(-) to MnO_(2) + 2H_2 O`
To balance H atoms, four `H^(+)` ions are added on the left:
`2I^(-) to I_2 , MnO_(4)^(-) + 4H^(-) to MnO_(2) + 2H_2 O`
As the reaction takes place in basic medium, four `H^(+)` ions, four OH-ions are added to both sides of the reduction half-reaction:
` MnO_(4)^(-) + 4H^(+) + 4OH^(-) to MnO_2 + 2 H_2 O + 4 OH^(-)`
Replacing the Hand `OH^(-)` ions with water,
` 2 I^(-) to I_2 + 2e^(-) , MnO_(4)^(-) + 2H_2 O + 3 e^(-) to MnO_(2) + 4 OH^(-)`
Step 5: On balancing the charges of the two half-reactions as given below:
`2I^(-) to I_2 + 2e^(-) , MnO_(4)^(-) + 2H_2 O + 2e^(-) to MnO_(2) + 4Oh^(-)`
Equalising the number of electrons in both the equations by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2
` 6 I^(-) to 3 I_2 + 6e^(-) , 2 MnO_(4)^(-) + 4H_2 O + 6e^(-) to 2MnO_(2) + 8 OH^(-)`
Step 6: Adding the half reactions to get the overall reaction and cancelling electrons on both sides.
` 6 I^(-) + 2 MnO_(4)^(-) + 4H_2 O to 3 I_2 + 2MnO_(2) + 8OH^(-)`