Question

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

• A. one-fifth
• B. five
• C. one
• D. two

Answer 1

Answer: (D)

two

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

$2KMn{O}_4+H_{2}O⟶2Mn{O}_2+2KOH+3\left[O\right]$

$KI+3\left[O\right]⟶KI{O}_3$

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$2KMn{O}_4+H_{2}O+KI⟶2Mn{O}_2+2KOH+KI{O}_3$

The oxidation number of iodine increases from $-1$ (in KI) to $+5$ (in $KI{O}_3$).
The total increase in the oxidation number of iodine is 6.

The oxidation number of Mn decreases from $+7$ (in $KMn{O}_4$) to $+4$ ( im $Mn{O}_2$).
The decrease in the oxidation number of one Mn atom is 3.
For 2 Mn atoms, the decrease in oxidation number is 6.

Thus, the increase in the oxidation number of iodine is balanced with a decrease in the oxidation number of Mn if the mole ratio $KMn{O}_4:KI=2:1$.