Ratio of moles of Fe (II) oxidized by equal volumes of equimolar $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ solutions in acidic medium will be :

5 : 3

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1 : 1

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1 : 2

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5 : 6

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Solution

The correct option is B 1 : 2
$F e^{+ 2}$ in acidic medium $M u^{+ 7} \to M u^{+ 4}$
and $C r^{+ 6} \to C r^{+ 3}$
So Ratio $K M n O_{4} : K_{2} C r_{2} O_{7}$
$(7 - 4) n : 2 \times (6 - 3) n$
$3 n : 6 n$
$1 : 2$

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Ratio of moles of Fe (II) oxidized by equal volumes of equimolar KMnO4 and K2Cr2O7 solutions in acidic medium will be :

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The correct option is B 1 : 2
$F e^{+ 2}$ in acidic medium $M u^{+ 7} \to M u^{+ 4}$
and $C r^{+ 6} \to C r^{+ 3}$
So Ratio $K M n O_{4} : K_{2} C r_{2} O_{7}$
$(7 - 4) n : 2 \times (6 - 3) n$
$3 n : 6 n$
$1 : 2$