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Permanganate (VII) ion, MnO(4)^(-) in ba...

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.

Text Solution

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Step 1 : First we write the skeletal ionic equation, which is
`MnO_(4)^(-) (aq) + I^(-) (aq) rarr MnO_(2) (s) + I_(2) (s)`
Step 2 : The two half-reactions are :
Oxidation half : `overset(-1)(I^(-))(aq) rarr overset(0)(I_(2))(s)`
Reduction half : `overset(+7)(Mn)O_(4)^(-) (aq) rarr overset(+4)(Mn)O_(2) (s)`
Step 3 : To balance the I atoms in the oxidation half reaction, we rewrite it as :
`2I^(-) (aq) rarr 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) rarr MnO_(2) (s) + 2H_(2)O (I)`
To balance the H atoms, we add four `H^(+)` ions on the left :
`MnO_(4)^(-) (aq) + 4 H^(+) (aq) rarr MnO_(2) (s) + 2H_(2)O (I)`
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) + 4H^(+) (aq) + 4 OH^(-) (aq) rarr MnO_(2) (s) + 2H_(2)O (I) + 4 OH^(-) (aq)`
Replacing the `H^(+) and OH^(-)` ions with water, the resultant equation is :
`MnO_(4)^(-) (aq) + 2H_(2)O (I) rarr MnO_(2)(s) + 4OH^(-) (aq)`
Step 5 : In this step we balance the charges of the two half-reactions in the manner depicted as :
`2I^(-) (aq) rarr I_(2) (s) + 2e^(-) MnO_(4)^(-) (aq) + 2H_(2)O(I) + 3e^(-) rarr MnO_(2)(s) + 4OH^(-) (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) rarr 3I_(2) (s) + 6e^(-) 2 MnO_(4)^(-) (aq) + 4H_(2)O (I) + 6e^(-) rarr 2MnO_(2) (s) + 8OH^(-) (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(I) rarr 3I_(2)(s) + 2MnO_(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.
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