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A given amount of $\mathrm{Fe}^{2+}$ is oxidized by $\mathrm{x}$ mol of $\mathrm{MnO}_{4}$ in acidic medium. The number of moles of $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ required to oxidize the same amount of $\mathrm{Fe}^{2+}$ in acidic medium is
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The correct answer is:
$0.83 \mathrm{x}$
As per Law of Equivalence
no. of eq. of $\mathrm{Fe}^{+2}=$ no. of eq. of $\mathrm{KMnO}_{4}=$ no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}$
$\begin{array}{l}
\mathrm{MnO}_{4}^{-}(\mathrm{O} \cdot \text { Agent })+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{+2}+4\mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=5) \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}(\mathrm{O} \cdot \mathrm{A})+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{+3}+7 \mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=6)
\end{array}$
no. of eq. of $\mathrm{KMnO}_{4}=$ no. of moles $\times \mathrm{n}$. factor $=\mathrm{x} \times 5$
no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}=$ no. of moles $\times n . \mathrm{f}=\mathrm{y} \times 6$
From Law of equivalence
$\begin{array}{l}
6 y=5 x \\
\text { or } y=5 / 6 x=0.83 x
\end{array}$
no. of eq. of $\mathrm{Fe}^{+2}=$ no. of eq. of $\mathrm{KMnO}_{4}=$ no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}$
$\begin{array}{l}
\mathrm{MnO}_{4}^{-}(\mathrm{O} \cdot \text { Agent })+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{+2}+4\mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=5) \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}(\mathrm{O} \cdot \mathrm{A})+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{+3}+7 \mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=6)
\end{array}$
no. of eq. of $\mathrm{KMnO}_{4}=$ no. of moles $\times \mathrm{n}$. factor $=\mathrm{x} \times 5$
no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}=$ no. of moles $\times n . \mathrm{f}=\mathrm{y} \times 6$
From Law of equivalence
$\begin{array}{l}
6 y=5 x \\
\text { or } y=5 / 6 x=0.83 x
\end{array}$
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