Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :Step-1 : First we WRITE the skeletal ionic EQUATION, which is `MnO_(4(aq))^(-)+I_((aq))^(-)toMnO_(2(s))+I_(2(s))` Step-2 : The two half-reactions are : Oxidation half : `overset(-1)(I_((aq))^(-))tooverset(0)(I_(2(s)))` Reduction half : `overset(+7)(MnO_(4(aq))^(-))tooverset(4)(MnO_(2(s)))` Step-3 : To balance the I atoms in the oxidation half reaction, we rewrite it as : `2I_((aq))^(-)toI_(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))^(-)toMnO_(2(s))+2H_(2)O_((l))` To balance the H atoms, we add four `H^(+)` ions on the left : `MnO_(4(aq))^(-)+4H_((aq))^(+)toMnO_(2(s))+2H_(2)O_((l))` 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))^(+)4OH_((aq))^(-)toMnO_(2(s))+2H_(2)O_((l))+4OH_((aq))^(-)` Replacing the `H^(+)andOH^(-)` ions with water, the resultant equation is : `MnO_(4(aq))^(-)+2H_(2)O_((l))toMnO_(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))^(-)toI_(2(s))+2e^(-)` `MnO_(4(aq))^(-)+2H_(2)O_((l))+3e^(-)toMnO_(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))^(-)to3I_(2(s))+6e^(-)` `MnO_(4(aq))^(-)+4H_(2)O_((l))+6e^(-)to2MnO_(2(s))+8OH_((aq))^(-)` Step-6 : Add two half-reactions to obatin the net reactions after CANCELLING electrons on both sides. `6I_((aq))^(-)+2MnO_(4(aq))^(-)+4H_(2)O_((l))to3I_(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. |
|