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Question:
Answered & Verified by Expert
Given :
$$
\begin{gathered}
\mathrm{E}_{\frac{1}{2} \mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V} \\
\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\mathrm{o}}=1.51 \mathrm{~V}
\end{gathered}
$$
The correct order of reducing power of the species $\left(\mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}\right.$ and $\left.\mathrm{Cl}^{-}\right)$will be:
Options:
$$
\begin{gathered}
\mathrm{E}_{\frac{1}{2} \mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V} \\
\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\mathrm{o}}=1.51 \mathrm{~V}
\end{gathered}
$$
The correct order of reducing power of the species $\left(\mathrm{Cr}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}\right.$ and $\left.\mathrm{Cl}^{-}\right)$will be:
Solution:
1910 Upvotes
Verified Answer
The correct answer is:
$\mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr}$
$\mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr}$
Lower the value of reduction potential higher will be reducing power hence the correct order will be
$$
\mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr}
$$
$$
\mathrm{Mn}^{2+} < \mathrm{Cl}^{-} < \mathrm{Cr}^{3+} < \mathrm{Cr}
$$
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