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The standard reduction potential data at \( 25^{\circ} \mathrm{C} \) is given below.
\[
\begin{array}{l}
\mathrm{E}^{\circ}\left(\mathrm{Fe}^{3} \cdot \mathrm{Fe}^{2}\right)=+0.77 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{Fe}^{2} \cdot \mathrm{Fe}\right)=-0.44 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{Cu}^{2} \cdot \mathrm{Cu}\right)=-0.34 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{Cu}^{\prime} \cdot \mathrm{Cu}\right)=+0.52 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{*}+4 \mathrm{c}^{\mathrm{c}} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\right)-+1.23 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}\right)=+0.40 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{Cr}^{3} \cdot \mathrm{Cr}\right)=-0.74 \mathrm{~V} \\
\mathrm{E}^{\circ}\left(\mathrm{Cr}^{2} \cdot \mathrm{Cr}\right)=-0.91 \mathrm{~V}
\end{array}
\]
[JEE Advanced-2013]
Match \( \mathrm{E}^{\circ} \) of the rebox pair in List I with the values given in List II and select the correct answer using the code given below the lists :
\( \begin{array}{llll} & \text { List I } & & \text { List II } \\ \text { P. } & \mathrm{E}^{\prime}\left(\mathrm{Fe}^{\prime} \cdot \mathrm{Fe}\right) & \text { 1. } & -0.36 \mathrm{~V} \\ \text { Q. } & \mathrm{E}^{\prime}\left(4 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 4 \mathrm{H}^{*}+4 \mathrm{OH}^{-}\right) & 2 . & -0.4 \mathrm{~V} \\ \mathrm{R} & \mathrm{E}^{\prime}\left(\mathrm{Cu}^{2-}+\mathrm{Cu} \rightarrow 2 \mathrm{Cu}^{*}\right) & 3 . & -0.04 \mathrm{~V} \\ \text { S. } & \mathrm{E}^{\prime}\left(\mathrm{Cr}^{3}, \mathrm{Cr}^{-2}\right) & \text { 4. } & -0.83 \mathrm{~V}\end{array} \)
Codes
(A) \( 4 \quad 4 \quad 1 \quad 3 \)
(B) \( 2 \quad 3 \quad 4 \quad 1 \)
(C) \( 1 \quad 2 \quad 3 \quad 4 \)
(D) \( 3 \quad 4 \quad 1 \quad 2 \)
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