In the reaction of $F e S O_{4}$ with $K_{2} C r_{2} O_{7}$ in the presence of an acid, the changes in the formal oxidation numbers of Cr and Fe are _________ respectively.

1 and 5

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3 and 1

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5 and 1

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1 and 3

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Solution

The correct option is C 3 and 1
$C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 6 F e^{3 +} + 2 C r^{3 +} + 7 H_{2} O$
Change in oxidation number of Cr is $= 6 - 3 = 3$
Change in oxidation number of Fe is $= 3 - 2 = 1$

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F e C r_{2} O_{4}

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C r_{2} O_{7}^{2 -}

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The oxidation state of Cu and Fe in $C u F e S_{2}$ are $+ 1$ and___respectively

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