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)(Mn)O_4^(-) + overset(-1)(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)(I^-) to overset(0)(I_2)), underset("Reduction half-reaction")(overset(+7)(MnO_4^(-)) to overset(+4)(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_2O`
To balance Hatoms, four `H^+` ions are added on the left:
`2I^(-) to I_2 , MnO_4^(-) + 4H^(+) to MnO_2 + 2H_2O`
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 + 2H_2O + 4OH^(-)`
Replacing the `H^(+)` and `OH^(-)` ions with water,
`2I^(-) to I_2 , MnO_4^(-) + 2H_2O to MnO_2 + 4OH^(-)`
Step 5: On balancing the charges of the two half-reactions as given below:
`2I^(-) to I_2 + 2e^(-) , MnO_4^(-) + 2H_2O + 3e^(-) to MnO_2 + 4OH^(-)`
Equalising the number of electrons in both the equations by multiplying the oxidation halfreaction by 3 and the reduction half reaction by 2 .
`6I^(-) to 3I_2 + 6e^(-) , 2MnO_4^(-) + 4H_2O + 6e^(-) to 2MnO_2 + 8OH^(-)`
Step 6: Adding the half reactions to get the overall reaction and cancelling electrons on both sides.
`6I^(-) + 2MnO_4^(-) + 4H_2O to 3I_2 + 2MnO_2 + 8OH^(-)`