Question

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 1

Step 1: Write the unbalanced (skeletal) ionic equation :
`MnO_4^(-) (aq.) + I^(-) (aq.) rarr MnO_2(s) + I_2(s)`
Step 2: Assign the oxidation numbers :
`overset(-7)(Mn) overset(-2)(O_2^(-)) (aq.) + overset(-1)(I^(-))(aq.) rarr overset(+4) (Mn) overset (-2) (O_2) (s) + overset (0) (I_2)(s)`
Step 3: Ideintify the oxidizing and reducing agents to construct the unbalanced half-reaction :
Oxidation half-reaction :
`overset(-1)(I^(-)) (aq.) rarr overset(0)(I_2)(s)`
Reduction half-reaction:
`overset (-7) (Mn) O_4^(-) (aq.) rarr overset (+4) (Mn)O_2 (s)`
Step 4: (a) To balance the I atoms in the oxidation half-reaction, we write
`2I^(-)(aq.) rarr I_2(s)`
(b) To account for the change in oxidation numbers, we add two electrons to the right-hand side fo oxidation half-reaction and three electrons to the left-hand side of reduction half-reaction:
`2I^(-) (aq.) rarr I_2(s) +2e^(-)`
`MnO_4^(-) (aq.) +3e^(-) rarr MnO_2(s)`
( c ) Balance the ionic charges in each half-reaction by adding `OH^(-)` ions as the reaction carried out in basic medium. There is no need of this in oxidation half-reaction. In rduction half-reaction we add four ` OH^(-)` ions to the right -hand side :
`MnO_4^(-) (aq.) + 3e^(-) rarr MnO_2(s) +4OH^(-) (aq.)`
Add `H_2O` molecules to balance the H and O atoms . There is no such requirement in oxidation half-reaction. In reduction half-reaction, we add two `H_(2)O` molecules on the left.
`MnO_(4)^(-) (aq.)+2H_(2)O(1)+3e^(-) rarr MnO_(2)(s)+4OH^(-) (aq.)`
Step 5: Balance the electron transfer. To equalize the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
`6 I^(-) (aq.) rarr 3 I_(2) (s) +6 e^(-)`
`2MnO_(4)^(-) (aq.)+4H_(2)O(1)+6 e^(-) rarr 2MnO_(2)(s)+8OH^(-) (aq.)`
Step 6: Add the two half-reactions to give
`MnO_(4)^(-)(aq.) +6I^(-) (aq.)+4H_(2)O(1) +6e^(-) rarr 2MnO_(2)(s)+3I_(2)(s)+8OH^(-) (aq.)+6 e^(-)`
After cancelling the electrons on both sides, we obtain
`MnO_(4)^(-)(aq.)+6I^(-) (aq.)+4H_(2)O(1) rarr 2MnO_(2)(s)+3I_(2)(s)+8OH^(-)(aq.)`
Step 7: A final check shows that the equation is balanced with respect to both atoms and charges.