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

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$0.83x$

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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 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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$A$ 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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