1. We need to compare the standard reduction potentials

$E^\circ_{M^{3+}/M^{2+}}$

for the given first-row transition metals: Cr, Mn, Fe, and Co.

This corresponds to the reduction:

$M^{3+} + e^- \rightarrow M^{2+}$

A higher value of $E^\circ$ means that $M^{3+}$ is more easily reduced to $M^{2+}$, or equivalently, $M^{3+}$ is relatively less stable than $M^{2+}$.

2. Write the electronic configurations of the relevant ions.

• $\mathrm{Cr}$: $\mathrm{Cr}: [Ar]3d^5 4s^1$ $\mathrm{Cr}^{2+}: [Ar]3d^4$ $\mathrm{Cr}^{3+}: [Ar]3d^3$

• $\mathrm{Mn}$: $\mathrm{Mn}: [Ar]3d^5 4s^2$ $\mathrm{Mn}^{2+}: [Ar]3d^5$ $\mathrm{Mn}^{3+}: [Ar]3d^4$

• $\mathrm{Fe}$: $\mathrm{Fe}: [Ar]3d^6 4s^2$ $\mathrm{Fe}^{2+}: [Ar]3d^6$ $\mathrm{Fe}^{3+}: [Ar]3d^5$

• $\mathrm{Co}$: $\mathrm{Co}: [Ar]3d^7 4s^2$ $\mathrm{Co}^{2+}: [Ar]3d^7$ $\mathrm{Co}^{3+}: [Ar]3d^6$

3. Use stability of $d$-electron configurations.

Specially stable configurations are:

• half-filled: $d^5$
• sometimes extra stability also for $d^3$ in octahedral environments, but for aqueous ions the strongest simple argument here is $d^5$ stability.

Now compare each pair:

• For $\mathrm{Cr}$: $\mathrm{Cr}^{3+} = d^3, \quad \mathrm{Cr}^{2+} = d^4$ Here $\mathrm{Cr}^{3+}$ is more stable than $\mathrm{Cr}^{2+}$, so reduction $\mathrm{Cr}^{3+} \to \mathrm{Cr}^{2+}$ is not especially favorable. Hence $E^\circ$ is relatively low.

• For $\mathrm{Mn}$: $\mathrm{Mn}^{3+} = d^4, \quad \mathrm{Mn}^{2+} = d^5$ Since $d^5$ is highly stable, $\mathrm{Mn}^{2+}$ is much more stable than $\mathrm{Mn}^{3+}$. Therefore, $\mathrm{Mn}^{3+} + e^- \rightarrow \mathrm{Mn}^{2+}$ is highly favorable, so $E^\circ$ should be very high.

• For $\mathrm{Fe}$: $\mathrm{Fe}^{3+} = d^5, \quad \mathrm{Fe}^{2+} = d^6$ Here $\mathrm{Fe}^{3+}$ has the extra-stable $d^5$ configuration, so reduction to $\mathrm{Fe}^{2+}$ is less favorable. Thus $E^\circ$ is comparatively smaller.

• For $\mathrm{Co}$: $\mathrm{Co}^{3+} = d^6, \quad \mathrm{Co}^{2+} = d^7$ There is no special half-filled stabilization in $\mathrm{Co}^{2+}$. So this reduction is favorable to some extent, but not as strongly as in Mn.

4. Known trend / standard values in aqueous medium support this:

Approximate values are:

$E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}) \approx -0.4\,\text{V}$ $E^\circ(\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}) \approx +1.5\,\text{V}$ $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) \approx +0.77\,\text{V}$ $E^\circ(\mathrm{Co}^{3+}/\mathrm{Co}^{2+}) \approx +1.8\,\text{V}$

Thus, among the given options, the highest value is for cobalt.

5. Therefore, the correct option is

$\boxed{\text{C: Co }(Z=27)}$

6. Comparison with stored correct answer:

Stored correct answer = C.

My derived answer = C.

So they agree.