Given:
$$E^\circ_{Cr^{3+}/Cr} = -0.74$$ V; $$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51$$ V
$$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33$$ V; $$E^\circ_{Cl_2/Cl^-} = 1.36$$ V
Based on the data given above, strongest oxidising agent will be:

We begin by recalling the basic electrochemical principle that the larger (more positive) the standard reduction potential $$E^\circ$$ of a half-reaction, the greater is the tendency of the reactant side of that half-reaction to accept electrons, that is, to get reduced. A substance that readily gets reduced will, in turn, oxidise other substances; hence it acts as a strong oxidising agent.

For every $$E^\circ$$ value supplied, we first write the corresponding reduction half-reaction in the conventional form “oxidised species $$+$$ electrons $$\rightarrow$$ reduced species”, identify the reactant (the species on the left), and note its ability to serve as an oxidising agent.

1. For $$E^\circ_{Cr^{3+}/Cr} = -0.74 \text{ V}$$ we have
$$Cr^{3+} + 3e^- \rightarrow Cr$$.

The reactant here is $$Cr^{3+}$$. Because its $$E^\circ$$ is negative, its tendency to get reduced is low, so $$Cr^{3+}$$ is a weak oxidising agent.

2. For $$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$$ we write
$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$.

The reactant side contains $$MnO_4^-$$. The very positive value $$1.51 \text{ V}$$ means $$MnO_4^-$$ has a very strong tendency to accept electrons; therefore it is a very strong oxidising agent.

3. For $$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$$ we would have
$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$.

The reactant $$Cr_2O_7^{2-}$$ has a high positive potential, but this species does not appear in the given options, so we only note its strength for comparison.

4. For $$E^\circ_{Cl_2/Cl^-} = 1.36 \text{ V}$$ the reduction half-reaction is
$$Cl_2 + 2e^- \rightarrow 2Cl^-$$.

The oxidising species here is $$Cl_2$$, not $$Cl^-$$. Since the option list contains $$Cl^-$$ rather than $$Cl_2$$, $$Cl^-$$ itself is on the product side of the reduction and therefore is not an oxidising agent at all; it is actually a possible reducing agent.

Now we examine the four species that actually appear in the options:

• $$Mn^{2+}$$ (Option A) is the reduced form in reaction 2; its $$E^\circ$$ for being reduced further is not given and, being already in a low oxidation state, it cannot serve as an effective oxidiser.

• $$MnO_4^-$$ (Option B) possesses $$E^\circ = 1.51 \text{ V}$$ and is an extremely strong oxidising agent.

• $$Cl^-$$ (Option C) is a product of the reduction of $$Cl_2$$, hence it cannot oxidise anything; it is not an oxidising agent.

• $$Cr^{3+}$$ (Option D) corresponds to $$E^\circ = -0.74 \text{ V}$$ and therefore is a very weak oxidising agent.

Comparing all these, $$MnO_4^-$$ with the highest positive reduction potential $$1.51 \text{ V}$$ has the greatest tendency to get reduced and consequently is the strongest oxidising agent among the choices.

Hence, the correct answer is Option B.