Question

The product of oxidation of ${{I}^{-}}$ with $Mn{{O}_{4}}^{-}$ in alkaline medium is:
A)$I{{O}_{3}}^{-}$
B)${{I}_{2}}$
C)$I{{O}^{-}}$
D)$I{{O}_{4}}^{-}$

Answer 1

The answer to this question is based on the concept of reduction of $Mn{{O}_{4}}^{-}$ and oxidation of ${{I}^{-}}$ by balancing the charge of the equation and the product obtained here thus gives you the correct answer.

Complete answer:
We are familiar with the concepts of balancing a chemical equation when a reactant reacts with other reactants to give products and also the basic reactions like oxidation reaction, reduction reactions etc.
Now, here we are dealing with the oxidation reaction that takes place when iodide ion is treated with the permanganate ion.
Now, let us see the reaction taking place and the balancing of that particular reaction which will lead us to the required answer.
- We know that potassium permanganate is a good reducing agent and this reagent in acidic medium oxidises the double bond and replaces it by the two carbonyl groups.
Now, here the iodide ion that is ${{I}^{-}}$ in the form of KI is treated with the alkaline potassium permanganate and the reaction taking place is as shown below,
$2KMn{{O}_{4}}+KI+{{H}_{2}}O\to KI{{O}_{3}}+2Mn{{O}_{2}}+2KOH$
Here, the oxidation state of manganese changes from +5 to +2, that is it is undergoing reduction and in turn oxidises iodide ion of potassium iodide to iodate ion of potassium iodate.

Therefore, the correct answer is option A) $I{{O}_{3}}^{-}$

Note: Note that acidic potassium permanganate is a very strong oxidizing agent compared to that of alkaline potassium permanganate. In the strong acid medium, the potassium permanganate oxidises the compounds strongly and in neutral and alkaline medium it acts as weak oxidant.