In the chemical reaction between stoichiometric quantities of $K M n O_{4}$ and $K I$ in weakly basic solution, what is the number of moles of $I_{2}$ released for $4$ mol of $K M n O_{4}$ consumed?

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Solution

The ionic form of reaction of $K M n O_{4}$ and KI in weakly basic solution is given as:
$2 7 M n O_{4}^{-} + 6 {- 1 I}^{-} + 4 H_{2} O \to 2 + 4 M n O_{2} + 3_{2} + 8 O H^{-}$
Here, oxidation state of $M n$ decreases from $+ 7$ in $K M n O_{4}$ to $+ 4$ in $M n O_{2}$
n- factor of $K M n O_{4} = change in oxidation number per formula unit = 3$
oxidation state of $I$ increases from $- 1$ in $I^{-}$ to $0$ in $I_{2}$
now, n- factor of $I_{2} = change in oxidation number per I_{2} molecule = 2$
$Since, n-factor \times Number of mol(n) = number of g-equivalents$
and at completion of reaction,
g-equivalents of $K M n O_{4}$ consumed = g-equivalents of $I_{2}$ formed
So, $3 \times m o l o f K M n O_{4} = 2 \times m o l o f I_{2}$
$3 \times 4 = 2 \times m o l o f I_{2}$
Amount of $I_{2} = 6$ mol

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K M n O_{4}

K I

I_{2}

M n O_{2}

K I

K M n O_{4}

I_{2}

K M n O_{4}

E

K M n O_{4}

H_{2} O_{2} + K M n O_{4} + H_{2} S O_{4} \to M n S O_{4} + H_{2} O + O_{2} + K_{2} S O_{4}

K M n O_{4}

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M n^{2 +}

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= 158

250 m L

0.04 N

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