Question

For $KMn{O_4}$ acts as an oxidizing agent in alkaline medium, when alkaline $KMn{O_4}$ is treated with $KI$, iodine ion oxidized to:
$\left( A \right)$ ${I_2}$
$\left( B \right)$ $I{O^ - }$
$\left( C \right)$ $IO_3^ - $
$\left( D \right)$ $IO_4^ - $

Answer 1

Hint: In this reaction $KMn{O_4}$ acts as an oxidizing agent (Oxidizing agent is an agent which oxidizes the other substance and reduces itself). It can oxidize other substances by accepting the electron.

Complete step by step answer:
In acidic solution, it oxidises to molecular iodine.
$2KMn{O_4} + 10KI + 8{H_2}S{O_4} \to 6{K_2}S{O_4} + 2KMnS{O_4} + 5{I_2} + 8{H_2}O$
While in an alkaline or neutral solution, the iodides are oxidized to iodates.
\[\]$2KMn{O_4} + {H_2}O + KI \to 2Mn{O_2} + 2KOH + \mathop {KI{O_3}}\limits_{\left( {Iodate} \right)} $
Valance factor or n-factor of $KI$ is in acidic medium as iodine charges from $ - 1$ (in $KI$ ) to zero (in${I_2}$ ).
So, the correct option is $\left( C \right)$ .

Additional information: In the above reaction $KMn{O_4}$ is an oxidizing agent because it oxidized Iodide ion $\left( {{I^ - }} \right)$ to Iodine molecule and potassium iodide $\left( {KI} \right)$ is an reducing agent because it reduced potassium permanganate $\left( {KMn{O_4}} \right)$ to magnesium sulphate $\left( {MnS{O_4}} \right)$ .Potassium permanganate $\left( {KMn{O_4}} \right)$ : Potassium permanganate is an inorganic compound in which the central atom is manganese attached to four oxygen atoms. It is also a secondary solution used in the reaction of acid base titration. In acid base titration it acts as an auto catalyst which speeds the rate of reaction and it is also used as an indicator in acid base titration because in basic condition color of potassium permanganate is changed pink to colorless. The change of color indicates the end point in acid base titration.

Note:
Reaction of acidified $KMn{O_4}$ with $KI$ is a type of redox reaction (A redox reaction is a reaction in which oxidation and reduction takes place). This reaction is a type of redox reaction because oxidation of manganese in $KMN{O_4}$ is $ + 7$ reduced to $ + 2$ in manganese sulphate and oxidation state of Iodine atom in $\left( {{I^ - }} \right)$ is $ - 1$ oxidized to $0$. Thus, oxidation of iodine takes place and reduction of manganese takes place. Therefore, we can say that it is a type of redox reaction.