Based on the data given below :

$$\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \end{array}$$

the strongest reducing agent is :

<p>To determine the strongest reducing agent among the options, we need to consider which species most readily donates electrons (i.e., is most easily oxidized). One standard method is to compare the standard reduction potentials of the corresponding half-reactions. A lower (more negative) standard reduction potential indicates that the reduced species is less stable and more prone to oxidation.</p> <p>Here’s the data provided:</p> <p><p>$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7H_2O,\quad E^\circ = +1.33\,V$$ </p></p> <p><p>$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V$ </p></p> <p><p>$$\mathrm{MnO}_4^- + 8H^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4H_2O,\quad E^\circ = +1.51\,V$$ </p></p> <p><p>$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V$ </p></p> <p>Now, let’s consider the species from the Options:</p> <p><p><strong>Option A: Cl⁻</strong> </p> <p>The relevant half-reaction is: </p> <p>$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V.$ </p> <p>The reduced form here is Cl⁻. Its corresponding oxidation (the reverse reaction) would have an electrode potential of </p> <p>$E^\circ_{\text{oxidation}} = -1.36\,V.$ </p></p> <p><p><strong>Option B: MnO₄⁻</strong> </p> <p>In the given half-reaction, MnO₄⁻ acts as the oxidizing agent (it is reduced to Mn²⁺). Therefore, MnO₄⁻ is not a good reducing agent.</p></p> <p><p><strong>Option C: Cr (metallic chromium)</strong> </p> <p>The relevant half-reaction is: </p> <p>$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V.$ </p> <p>Here, Cr is the reduced species. Reversing this reaction gives: </p> <p>$\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^-,$ </p> <p>with an effective oxidation potential of </p> <p>$E^\circ_{\text{oxidation}} = +0.74\,V.$</p></p> <p><p><strong>Option D: Mn²⁺</strong> </p> <p>In the MnO₄⁻/Mn²⁺ couple, Mn²⁺ is the reduced form. Its reverse (oxidation) potential would be: </p> <p>$E^\circ_{\text{oxidation}} = -1.51\,V.$ </p></p> <p>When comparing the oxidation potentials (remember, a higher oxidation potential means the species is more willing to lose electrons), we have:</p> <p><p>Cl⁻: $E^\circ_{\text{ox}} = -1.36\,V$ </p></p> <p><p>Cr: $E^\circ_{\text{ox}} = +0.74\,V$ </p></p> <p><p>Mn²⁺: $E^\circ_{\text{ox}} = -1.51\,V$ </p></p> <p>Among these, metallic Cr stands out because its corresponding reduction half-reaction has the most negative potential ($-0.74\,V$). This negative value indicates that Cr (in its elemental form) is unstable relative to its oxidized form ($\mathrm{Cr}^{3+}$) and, therefore, is the most eager to lose electrons. In other words, Cr is the strongest reducing agent.</p> <p>Thus, the answer is:</p> <p>Option C: $\mathrm{Cr}$</p>