Among $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$, the stronger reducing agent in aqueous solution is:

Standard electrode potentials: $E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}) = -0.41\,\text{V}$ (very negative, easily oxidised, strong reductant), $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\,\text{V}$ (positive, less easily oxidised). $\mathrm{Cr}^{2+}$ ($d^4$) loses one electron to give $\mathrm{Cr}^{3+}$ ($d^3$, half-filled $t_{2g}$ in an octahedral field), which is extra stable. By contrast, $\mathrm{Fe}^{2+}$ ($d^6$) loses one electron to give $\mathrm{Fe}^{3+}$ ($d^5$), also stable but the change is energetically less favourable from $d^6$ than from $d^4$.