1. We are given standard reduction potentials of the type

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

for the metals:

• $\mathrm{Cr^{3+}/Cr^{2+}} : E^\circ = -0.41\,\text{V}$
• $\mathrm{Mn^{3+}/Mn^{2+}} : E^\circ = +1.57\,\text{V}$
• $\mathrm{Fe^{3+}/Fe^{2+}} : E^\circ = +0.77\,\text{V}$
• $\mathrm{Co^{3+}/Co^{2+}} : E^\circ = +1.97\,\text{V}$

2. The question asks: for which metal is the change in oxidation state from $+2$ to $+3$ easiest?

That corresponds to the oxidation:

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

3. The given $E^\circ$ values are for the reverse process (reduction):

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

So the oxidation potential for

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

will be the negative of the given reduction potential.

Thus, oxidation is easiest when the reduction potential $E^\circ(M^{3+}/M^{2+})$ is most negative.

4. Compare the given values:

$-0.41,\quad +1.57,\quad +0.77,\quad +1.97$

The most negative value is:

$-0.41\,\text{V for Cr}$

5. Therefore, $\mathrm{Cr^{2+}}$ is most easily oxidized to $\mathrm{Cr^{3+}}$.

Hence, the correct option is:

$\boxed{\text{C: Cr}}$