InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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. |
|
Answer» 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. |
|