The number of moles of $K M n O_{4}$ reduced by one mole of $K I$ in alkaline medium is:

one-fifth

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five

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one

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two

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Solution

The correct option is C two
The number of moles of $K M n O_{4}$ reduced by one mole of $K I$ in alkaline medium is two.

$2 K M n O_{4} + H_{2} O ⟶ 2 M n O_{2} + 2 K O H + 3 [O]$
$K I + 3 [O] ⟶ K I O_{3}$
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$2 K M n O_{4} + H_{2} O + K I ⟶ 2 M n O_{2} + 2 K O H + K I O_{3}$

The oxidation number of iodine increases from $- 1$ (in KI) to $+ 5$ (in $K I O_{3}$ ).
The total increase in the oxidation number of iodine is 6.
The oxidation number of Mn decreases from $+ 7$ (in $K M n O_{4}$ ) to $+ 4$ ( im $M n O_{2}$ ).
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 $K M n O_{4} : K I = 2 : 1$ .

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