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

The given values are standard reduction potentials $$(E^\circ_{M^{3+}/M^{2+}})$$, which measure the tendency of a metal ion to be reduced from +3 to +2:
$$M^{3+}+e^{-}\rightarrow M^{2+}$$ The question asks for the ease of oxidation from +2 to +3 $$(M^{2+} \rightarrow M^{3+} + e^-)$$.
The standard oxidation potential is simply the negative of the reduction potential:

• Cr: $$E^\circ_{\text{ox}} = -(-0.41\text{ V}) = \mathbf{+0.41\text{ V}}$$
• Fe: $$E^\circ_{\text{ox}} = -(+0.77\text{ V}) = -0.77\text{ V}$$
• Mn: $$E^\circ_{\text{ox}} = -(+1.57\text{ V}) = -1.57\text{ V}$$
• Co: $$E^\circ_{\text{ox}} = -(+1.97\text{ V}) = -1.97\text{ V}$$

A more positive oxidation potential (or a more negative reduction potential) means the species loses an electron more readily. Because Chromium has a positive oxidation potential $$(+0.41\text{ V})$$, $$Cr^{2+}$$ is highly unstable and easily oxidises to the more stable $$Cr^{3+}$$ state (which is stabilized by a half-filled $$t_{2g}^{3}$$ d-orbital configuration)