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Molarity

If equal volu...

Question

If equal volumes of 0.1 $M K M n O_{4}$ and 0.1 $M K_{2} C r_{2} O_{7}$ solutions are allowed to oxidise $F e^{2 +}$ to $F e^{3 +}$ in acidic medium, then $F e^{2 +}$ oxidised will be:

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K M n O_{4}

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K_{2} C r_{2} O_{7}

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Equal in both cases

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Cannot be determined

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Solution

The correct option is A More by $K_{2} C r_{2} O_{7}$
$K_{2} C r_{2} O_{7}$ In acidic medium $C r^{+ 6}$ goes to $C r^{+ 3}$ state and hence there is a net gain of 3 electrons. As there are 2 moles of $C r$ in $K_{2} C r_{2} O_{7}$
The equivalent moles of $K_{2} C r_{2} O_{7} = 6$
Normality $N_{1} = 0.1 M \times 6 = 0.6 N$
$K M n O_{4}$ In Acidic medium, $M n^{+ 7}$ goes to $M n^{+ 2}$ state and hence there is a net gain of 5 electrons
The equivalent moles of $K M n O_{4} = 5$
Normality $N_{2} = 0.1 M \times 5 = 0.5 N$

If the equal volumes of both are used that means more of $F e^{2 +}$ will be neutralized by $K_{2} C r_{2} O_{7}$ as it is having $\frac{6}{5}$ more electrons to gain then $K M n O_{4}$

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If equal volumes of 0.1 $M K M n O_{4}$ and 0.1 $M K_{2} C r_{2} O_{7}$ solutions are allowed to oxidise $F e^{2 +}$ to $F e^{3 +}$ in acidic medium, then $F e^{2 +}$ oxidised will be: