Question

The number of moles of $KMn{O}_4$ reduced by one mole of $KI$ in alkaline medium is:

• A. 2.00
• B. 2
• C. 2.0

Answer 1

Answer: (A), (B), (C)

The reaction can be represented as
$2KMn{O}_4+KI+H_{2}O\underset{}{\overset{}{\to }}KI{O}_3+2Mn{O}_2+2KOH$
The oxidation number of iodine increases from $-1($ in $KI)$ to $+5($in $KI{O}_3)$.
The total increase in oxidation number of $I$ is $6$.
The oxidation number of $Mn$ decreases from $+7($in $KMn{O}_4)$ to $+4($in $Mn{O}_2)$.
The decrease in the oxidation number of one $Mn$ is $3$.
For $2Mn$ atoms, the decrease in oxidation number is $6$.
$2Mn$ atoms balance the charge of $1$ Iodine atom.
Thus $2$ moles of $KMn{O}_4$ is reduced during the reaction.
Answer $=2$ moles.