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Types of Redox Reactions

The ratio of ...

Question

The ratio of number of moles of $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ required to oxidise $0.1$ mol $S n^{+ 4}$ in acidic medium :

6 : 5

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5 : 6

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1 : 2

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2 : 1

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Solution

The correct option is A $6 : 5$
$+ 7 K M n O_{4} a c i d i c m e d i u m - ------------ \to + 2 M n^{+ 2}$
$+ 6 K_{2} C r_{2} O_{7} a c i d i c m e d i u m - ------------ \to + 3 C r^{+ 3}$
$+ 4 S n^{+ 4} r e d u c i n g a g e n t - ------------ \to + 2 S n^{+ 2}$
Number of equivalents $= m o l e s \times n f$
$n (S n^{+ 4}) = 0.1$
$n f (S n^{+ 4}) = 2$
Equating number of equivalents of $K M n O_{4}$ and $S n^{+ 4}$
$n_{1} {n f}_{1} = n_{2} {n f}_{2}$
$0.1 \times 2 = 5 n_{2} n_{21} = 0.04$
Equating number of equivalents $K_{2}$ $C r_{2}$ $O_{7}$ and $S n^{+ 4}$
$n_{1} f_{1} = n_{2} 2 f_{2} 2$
$0.1 \times 2 = 6 n_{22}$
$n_{22} = \frac{0.1}{3}$
Ratio of moles of $K M n O_{3}$ and $K_{2} C r_{2} O_{7} = \frac{0.04}{0.1} = \frac{6}{5}$

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