Given :
$$E^\circ_{\frac{1}{2}Cl_2/Cl^-} = 1.36$$ V, $$E^\circ_{Cr^{3+}/Cr} = -0.74$$ V
$$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33$$ V, $$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51$$ V
The correct order of reducing power of the species (Cr, Cr$$^{3+}$$, Mn$$^{2+}$$ and Cl$$^-$$) will be:

The reducing power of a species depends on its tendency to lose electrons, which is indicated by the standard reduction potential of its corresponding oxidation half-reaction. A more negative reduction potential means the species is a stronger reducing agent because it is harder to reduce and easier to oxidize.

Given the standard reduction potentials:

• $$E^\circ_{\frac{1}{2}Cl_2/Cl^-} = 1.36$$ V (for $$\frac{1}{2}Cl_2 + e^- \rightarrow Cl^-$$)
• $$E^\circ_{Cr^{3+}/Cr} = -0.74$$ V (for $$Cr^{3+} + 3e^- \rightarrow Cr$$)
• $$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33$$ V (for $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$)
• $$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51$$ V (for $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$)

We need the reduction potentials for the couples where the given species are oxidized:

• For Cr (oxidized to Cr³⁺): The reduction potential for $$Cr^{3+} + 3e^- \rightarrow Cr$$ is $$-0.74$$ V.
• For Cr³⁺ (oxidized to Cr₂O₇²⁻): The reduction potential for $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$ is $$1.33$$ V.
• For Mn²⁺ (oxidized to MnO₄⁻): The reduction potential for $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$ is $$1.51$$ V.
• For Cl⁻ (oxidized to Cl₂): The reduction potential for $$\frac{1}{2}Cl_2 + e^- \rightarrow Cl^-$$ is $$1.36$$ V.

Now, compare these reduction potentials:

• Cr: $$-0.74$$ V (most negative)
• Cr³⁺: $$1.33$$ V
• Cl⁻: $$1.36$$ V
• Mn²⁺: $$1.51$$ V (most positive)

Since reducing power decreases as the reduction potential becomes more positive, the order of reducing power (strongest to weakest) is:

Cr > Cr³⁺ > Cl⁻ > Mn²⁺

This can be written as:

Mn²⁺ < Cl⁻ < Cr³⁺ < Cr

Comparing with the options:

• Option A: Mn²⁺ < Cl⁻ < Cr³⁺ < Cr (matches)
• Option B: Mn²⁺ < Cl³⁺ < Cl⁻ < Cr (incorrect, Cl³⁺ is not a given species)
• Option C: Cr³⁺ < Cl⁻ < Mn²⁺ < Cr (incorrect order)
• Option D: Cr³⁺ < Cl⁻ < Cr < Mn²⁺ (incorrect order)

Hence, the correct answer is Option A.