The $E_{M^{3+} / M^{2+}}^o$ values for $\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}$ and Co are $-0.41,+1.57,+0.77$ and +1.97 V respectively. For which of these metals, the change in oxidation state from +2 to +3 is the easiest?

The oxidation of a metal ion from $+2$ to $+3$ state is represented by the half-reaction $M^{2+} \rightarrow M^{3+} + e^{-}$.

The standard electrode potential for this process is given by $E_{M^{2+} / M^{3+}}^o = -E_{M^{3+} / M^{2+}}^o$.

A process is easiest when the standard reduction potential $E_{M^{3+} / M^{2+}}^o$ is the most negative, which corresponds to the most positive oxidation potential $E_{M^{2+} / M^{3+}}^o$.

Comparing the given values of $E_{M^{3+} / M^{2+}}^o$:
For Cr: $-0.41$ V
For Mn: $+1.57$ V
For Fe: $+0.77$ V
For Co: $+1.97$ V

The value for Cr is the most negative ($-0.41$ V), which implies that the oxidation potential for $\mathrm{Cr}^{2+} \rightarrow \mathrm{Cr}^{3+} + e^{-}$ is the most positive ($+0.41$ V).

Therefore, the change in oxidation state from $+2$ to $+3$ is easiest for Chromium.

Answer: Cr