A given amount of Fe^{2+} is oxidized by x m | vedclass

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(B) According to the Law of Equivalence, the number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing a

Vedclass · Accepted Answer

$0.83x$

Vedclass · Answer

(B) According to the Law of Equivalence, the number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent.Number of equivalents of $Fe^{2+}$ = Number of equivalents of $MnO_{4}^{-}$ = Number of equivalents of $Cr_{2}O_{7}^{2-}$.For $MnO_{4}^{-}$ in acidic medium: $MnO_{4}^{-} + 8H^{+} + 5e^{-} ightarrow Mn^{2+} + 4H_{2}O$. The $n$-factor is $5$.Number of equivalents of $MnO_{4}^{-} = x 	imes 5 = 5x$.For $Cr_{2}O_{7}^{2-}$ in acidic medium: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} ightarrow 2Cr^{3+} + 7H_{2}O$. The $n$-factor is $6$.Let the number of moles of $Cr_{2}O_{7}^{2-}$ be $y$.Number of equivalents of $Cr_{2}O_{7}^{2-} = y 	imes 6 = 6y$.Equating the equivalents: $6y = 5x$.Therefore, $y = \frac{5}{6}x = 0.833x \approx 0.83x$.

$A$ given amount of $Fe^{2+}$ is oxidized by $x \ mol$ of $MnO_{4}^{-}$ in acidic medium. The number of moles of $Cr_{2}O_{7}^{2-}$ required to oxidize the same amount of $Fe^{2+}$ in acidic medium is

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