The correct order of $E^0{\mathrm{M}}^{2+}\mid \mathrm{M}$ values with negative sign for the four successive elements $\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}$ and $\mathrm{Co}$  is

The $E^{\circ }$ values are as follows.

 $$\begin{gathered} E^{\circ }\left({\mathrm{Cr}}^{2+}\mid \mathrm{Cr}\right)=-0.91\mathrm{V};E^{\circ }\left({\mathrm{Mn}}^{2+}\mid \mathrm{Mn}\right)=1.18\mathrm{V}; \\ {E}^{\circ }\left({\mathrm{Fe}}^{2+}\mid \mathrm{Fe}\right)=-0.44\mathrm{V};E^{\circ }\left({\mathrm{Co}}^{2+}\mid \mathrm{Co}\right)=-0.28\mathrm{V} \end{gathered}$$

Electrode potential becomes less negative as one goes from left to right across 3d transition elements. Chromium seems to be exceptional case since its (first+ second) ionization energy is larger than that of manganese. In general, larger the (first+ second) ionization energy, lesser negative is the electrode potential.