Explain the trends in $E_{{M^{3 + }}/{M^{2 + }}}^ o $ electrode potential.

The scandium has low value of $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ potentials because of its noble gas configuration.

The highest value of $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ is for zinc because $\mathrm{Zn}^{2+}$ has $\mathrm{d}^{10}$ configuration which is extra stable. High value of $\mathrm{Mn}^{3+} \mid \mathrm{Mn}^{2+}$ is due to half-filled $\left(d^{5}\right)$ orbitals in $\mathrm{Mn}^{2+}$ while low values of $\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}$ shows extra stability of $\mathrm{Fe}^{3+}\left(d^{5}\right)$. The vanadium has comparatively less negative values because of higher stability of $\mathrm{V}^{2+}$ that has half filled configuration $\left(\mathrm{t}_{2 \mathrm{~g}}^{3}\right)$.

The high value of $\mathrm{Co}^{3+} / \mathrm{Co}^{2+}$ is because of high values of $\mathrm{CFSE}$ of $\mathrm{t}_{2 \mathrm{~g}}^{5} \mathrm{e}^{2}\left[\mathrm{Co}^{2+}\right]$ system in a aqueous medium.

The high values of $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ of $\mathrm{Mn}^{3+}$ and $\mathrm{Co}^{3+}$ indicates that they are good oxidizing agents while $\mathrm{Ti}^{2+}, \mathrm{V}^{2+} \mathrm{Cr}^{2+}$ etc. are good reducing agents and thus these liberates $\mathrm{H}_{2}$ from acids.

Ex. : $2 \mathrm{Cr}_{\text {(aq) }}^{2+}+2 \mathrm{H}_{\text {(aq) }}^{+} \rightarrow 2 \mathrm{Cr}_{\text {(aq) }}^{3+}+\mathrm{H}_{2 \text { (aq) }} \uparrow$

Elements	Electrode potentials$\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)$
$Ti$	$-0.37$
$V$	$-0.26$
$\mathrm{Cr}$	$-0.41$
$Mn$	$+1.57$
$\mathrm{Fe}$	$+0.77$
$Co$	$+1.97$

In presence of oxidizing agents such as $\mathrm{HNO}_{3}$, the chromium forms layer of oxide $\mathrm{Cr}_{2} \mathrm{O}_{3}$ on its surface and hence it is passive towards oxidation.