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

(N/A) The $E_{M^{3+}/M^{2+}}^{o}$ values are determined by the stability of the $M^{3+}$ and $M^{2+}$ oxidation states.
$1$. $Sc^{3+}$ has a noble gas configuration,making it very stable,which results in a very low (highly negative) $E_{M^{3+}/M^{2+}}^{o}$ value.
$2$. The high value for $Mn^{3+}/Mn^{2+}$ $(+1.57 \ V)$ is due to the extra stability of the half-filled $d^5$ configuration in $Mn^{2+}$.
$3$. The low value for $Fe^{3+}/Fe^{2+}$ $(+0.77 \ V)$ is due to the extra stability of the $Fe^{3+}$ $(d^5)$ ion.
$4$. The high value for $Co^{3+}/Co^{2+}$ $(+1.97 \ V)$ is due to the high Crystal Field Stabilization Energy $(CFSE)$ of the $Co^{2+}$ $(d^7)$ system in an aqueous medium.
$5$. Elements with high positive $E_{M^{3+}/M^{2+}}^{o}$ values (like $Mn^{3+}$ and $Co^{3+}$) act as strong oxidizing agents. Conversely,elements with low or negative values (like $Ti^{2+}, V^{2+}, Cr^{2+}$) act as strong reducing agents and can liberate $H_2$ from acids.
Example: $2Cr^{2+}_{(aq)} + 2H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + H_{2(g)} \uparrow$

Element	Electrode potential ($E_{M^{3+}/M^{2+}}^{o}$ in $V$)
$Ti$	$-0.37$
$V$	$-0.26$
$Cr$	$-0.41$
$Mn$	$+1.57$
$Fe$	$+0.77$
$Co$	$+1.97$