1. Identify what determines oxidising strength

An oxidising agent is the species that gets reduced.
So, the larger the standard reduction potential E^c, the stronger the oxidising agent.

We compare the species given in the options based on whether they can act as oxidising agents in the listed reduction half-reactions.

2. Write the relevant reduction half-reactions

Given:

$Cr^{2+} + 2e^- \rightarrow Cr \qquad E^\circ = -0.74\,\text{V}$

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \qquad E^\circ = 1.51\,\text{V}$

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \qquad E^\circ = 1.33\,\text{V}$

$Cl_2 + 2e^- \rightarrow 2Cl^- \qquad E^\circ = 1.36\,\text{V}$

3. Check the options

Option A: $Cr^{3+}$

The given chromium reduction potential involving $Cr^{3+}$ is not directly $Cr^{3+} \to Cr$; instead, we are given:

$Cr_2O_7^{2-} \rightarrow Cr^{3+}$

Here $Cr^{3+}$ is the reduced form, so it is not the strongest oxidising agent.

Option B: $Mn^{2+}$

From

$MnO_4^- \rightarrow Mn^{2+}$

$Mn^{2+}$ is the reduced form, so it is not an oxidising agent here.

Option C: $MnO_4^-$

This is reduced according to

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

with

$E^\circ = 1.51\,\text{V}$

This is very high, so $MnO_4^-$ is a very strong oxidising agent.

Option D: $Cl^-$

From

$Cl_2 + 2e^- \rightarrow 2Cl^-$

$Cl^-$ is the reduced form. It cannot be the oxidising agent; rather $Cl_2$ would be.

4. Compare oxidising agents by $E^\circ$

Among the species listed, the only clear strong oxidising species is $MnO_4^-$, and its reduction potential is highest:

$1.51\,\text{V} > 1.36\,\text{V} > 1.33\,\text{V}$

Hence, the strongest oxidising agent is:

$\boxed{MnO_4^-}$

5. Compare with stored correct answer

Stored correct answer: C
Derived answer: C

They match.