Question:
Which is a stronger reducing agent Cr2+ or Fe2+ and why?
Solution:
The following reactions are involved when Cr2+ and Fe2+ act as reducing agents.
$\mathrm{Cr}^{2+} \longrightarrow \mathrm{Cr}^{3+} \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$
The $E_{\mathrm{Cr}^{2} / \mathrm{Cr}^{2}}^{\circ}$ value is $-0.41 \mathrm{~V}$ and $E_{\mathrm{Fe}^{2} / \mathrm{Fe}^{2}}^{\circ}$ is $+0.77 \mathrm{~V}$. This means that $\mathrm{Cr}^{2+}$ can be easily oxidized to $\mathrm{Cr}^{3+}$, but $\mathrm{Fe}^{2+}$ does not get oxidized to $\mathrm{Fe}^{3+}$ easily. Therefore, $\mathrm{Cr}^{2+}$ is a better reducing agent that $\mathrm{Fe}^{3+}$.