Given
E°$$_{Cl_{2}/Cl^{-}}$$ = 1.36 V, E°$$_{Cr^{3+}/Cr}$$ = -0.74 V
E°$$_{Cr_{2}O_{7}^{2-}/Cr^{3+}}$$ = 1.33 V, E°$$_{MnO_{4}^{-}/Mn^{2+}}$$ = 1.51 V
Among the following, the strongest reducing agent is:

We begin with the definition of standard reduction potential. For any redox couple written in the reduction direction, for example

$$X^{n+}+ne^- \rightarrow X$$

the listed value $$E^\circ_{X^{n+}/X}$$ is the reduction potential. If we wish to look at the reverse (oxidation) direction, we use the relation

$$E^\circ_{\text{oxidation}}=-\,E^\circ_{\text{reduction}}.$$

A species behaves as a reducing agent when it itself is oxidised, i.e. when the reverse reaction is favoured. The larger (more positive) the oxidation potential, the greater the reducing power. Equivalently, the smaller (more negative) the reduction potential, the greater the reducing power. So, in practice, we simply compare the given reduction potentials and pick the most negative one.

The data supplied are

$$E^\circ_{Cl_{2}/Cl^-}=+1.36\ \text{V},$$

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

$$E^\circ_{Cr_{2}O_{7}^{2-}/Cr^{3+}}=+1.33\ \text{V},$$

$$E^\circ_{MnO_{4}^- /Mn^{2+}}=+1.51\ \text{V}.$$

Now we match every option with its relevant couple and note the corresponding reduction potential:

(A) $$Mn^{2+}$$ comes from the couple $$MnO_{4}^- /Mn^{2+}$$ with $$E^\circ=+1.51\ \text{V}.$$

(B) $$Cr^{3+}$$ is part of the couple $$Cr_{2}O_{7}^{2-}/Cr^{3+}$$ with $$E^\circ=+1.33\ \text{V}.$$

(C) $$Cl^-$$ belongs to the couple $$Cl_{2}/Cl^-$$ with $$E^\circ=+1.36\ \text{V}.$$

(D) $$Cr$$ is linked with the couple $$Cr^{3+}/Cr$$ whose reduction potential is $$E^\circ=-0.74\ \text{V}.$$

Among the four numbers  +1.51 V, +1.36 V, +1.33 V and -0.74 V, the smallest (most negative) value is $$-0.74\ \text{V}.$$

Because a more negative reduction potential means a more positive oxidation potential (by $$E^\circ_{\text{oxidation}}=-E^\circ_{\text{reduction}}$$), the couple with $$E^\circ=-0.74\ \text{V}$$ provides the strongest tendency for the reverse (oxidation) reaction. That reverse reaction is

$$Cr \rightarrow Cr^{3+}+3e^-,$$

showing that metallic chromium, $$Cr$$, donates electrons most readily. Therefore $$Cr$$ is the strongest reducing agent among the given choices.

Hence, the correct answer is Option 4.