Chemistry - Complete d-block Question with Solution | TestHub

The standard electrode potential ($E^{\circ }$) for the $M{n}^{+3}/M{n}^{+2}$ couple is much more positive than for $C{r}^{+3}/C{r}^{+2}$ primarily due to the electronic configurations. $M{n}^{+3}$ has a $d^4$ configuration, which readily accepts an electron to achieve the more stable half-filled $d^5$ configuration of $M{n}^{+2}$. This strong tendency for reduction makes the $E^{\circ }$ value highly positive. The third ionization enthalpy of Mn is also significantly higher than Cr, making the formation of $M{n}^{+3}$ less favorable and thus favoring its reduction.

The relevant half-reactions are:

$M{n}^{+3}(aq)+e^{-}\to M{n}^{+2}(aq)$

$C{r}^{+3}(aq)+e^{-}\to C{r}^{+2}(aq)$

Electronic configurations:

$M{n}^{+3}$: $[Ar]3d^4$

$M{n}^{+2}$: $[Ar]3d^5$ (half-filled, stable)

$C{r}^{+3}$: $[Ar]3d^3$ (stable, $t_{2g}^3$ in octahedral field)

$C{r}^{+2}$: $[Ar]3d^4$

The stability of the $d^5$ configuration in $M{n}^{+2}$ drives the reduction of $M{n}^{+3}$, leading to a very positive $E^{\circ }$ value. In contrast, $C{r}^{+3}$ ($d^3$) is relatively stable due to crystal field stabilization energy, making its reduction to $C{r}^{+2}$ ($d^4$) less favorable.