Based on the data given below: $$\begin{aligned}E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} &= 1.33 \text{V} &E^\circ_{\text{Cl}_2/\text{Cl}^-} &= 1.36\ \text{V} \\E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} &= 1.51 \text{V} &E^\circ_{\text{Cr}^{3+}/\text{Cr}} &= -0.74\ \text{V}\end{aligned}$$ the strongest reducing agent is:

We need to identify the strongest reducing agent from the given standard reduction potentials.

$$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33$$ V

$$E^\circ_{Cl_2/Cl^-} = 1.36$$ V

$$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51$$ V

$$E^\circ_{Cr^{3+}/Cr} = -0.74$$ V

Key Concept: A reducing agent is a species that gets oxidized (loses electrons). The strongest reducing agent is the species whose oxidation is most thermodynamically favorable, i.e., it has the most negative (or least positive) standard reduction potential.

Each species listed in the options acts as a reducing agent in the reverse of the given half-reactions:

- $$Cr$$ gets oxidized: $$Cr \rightarrow Cr^{3+} + 3e^-$$; $$E^\circ_{red} = -0.74$$ V (most negative)

- $$Cl^-$$ gets oxidized: $$2Cl^- \rightarrow Cl_2 + 2e^-$$; $$E^\circ_{red} = +1.36$$ V

- $$Mn^{2+}$$ gets oxidized: $$Mn^{2+} \rightarrow MnO_4^-$$; $$E^\circ_{red} = +1.51$$ V

- $$MnO_4^-$$ is already in a high oxidation state and acts as an oxidizing agent, not a reducing agent.

The lower the standard reduction potential of a half-cell, the greater the tendency of that species to get oxidized (act as a reducing agent). $$Cr^{3+}/Cr$$ has $$E^\circ = -0.74$$ V, which is the most negative value among all the half-cells listed.

Therefore, $$Cr$$ is the strongest reducing agent.

The correct answer is Option 1: $$Cr$$.