Concept: The strongest reducing agent is the species that is most easily oxidized.

Equivalently, among the given species, the best reducing agent will correspond to the most negative reduction potential for its conjugate reduction half-reaction.

Examine the options one by one

We are given standard reduction potentials:

$E^\circ_{Cl_2/Cl^-}=+1.36\,\text{V}$ $E^\circ_{Cr^{3+}/Cr}=-0.74\,\text{V}$ $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}}=+1.33\,\text{V}$ $E^\circ_{MnO_4^-/Mn^{2+}}=+1.51\,\text{V}$

Relate each option to its oxidation tendency

• Option A: $Mn^{2+}$

  From $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O, \quad E^\circ=+1.51\,\text{V}$ the reverse oxidation is $Mn^{2+} \to MnO_4^-$ with $E^\circ_{ox}=-1.51\,\text{V}$ Very unfavorable. So $Mn^{2+}$ is not a strong reducing agent.

• Option B: $Cr^{3+}$

  From $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O, \quad E^\circ=+1.33\,\text{V}$ reverse oxidation of $Cr^{3+}$ to dichromate is unfavorable: $E^\circ_{ox}=-1.33\,\text{V}$ Also from $Cr^{3+}+3e^-\to Cr, \quad E^\circ=-0.74\,\text{V}$ this shows $Cr^{3+}$ itself prefers reduction rather than oxidation. So $Cr^{3+}$ is not a strong reducing agent.

• Option C: $Cl^-$

  From $Cl_2 + 2e^- \to 2Cl^-, \quad E^\circ=+1.36\,\text{V}$ oxidation is $2Cl^- \to Cl_2 + 2e^-$ with $E^\circ_{ox}=-1.36\,\text{V}$ Again unfavorable. So $Cl^-$ is not the strongest reducing agent.

• Option D: $Cr$

  Given $Cr^{3+}+3e^-\to Cr, \quad E^\circ=-0.74\,\text{V}$ therefore oxidation of chromium is $Cr \to Cr^{3+}+3e^-$ with $E^\circ_{ox}=+0.74\,\text{V}$ This is favorable compared with the other options.

Comparison

The oxidation tendencies are:

• $Mn^{2+}$: $-1.51\,\text{V}$
• $Cr^{3+}$: very unfavorable as reducing agent
• $Cl^-$: $-1.36\,\text{V}$
• $Cr$: $+0.74\,\text{V}$

The species with the greatest tendency to get oxidized is $Cr$.

Final answer

$\boxed{Cr}$

So the strongest reducing agent is Option D.