Question

Account for: ${E_0}$ value for the $\dfrac{{M{n^{3 + }}}}{{M{n^{2 + }}}}$ couple is highly positive (+1.57V) as compare to $\dfrac{{C{r^3}}}{{C{r^{2 + }}}}$ which is negative (-0.4V). Why?

Answer 1

As it is known that, of the reaction is always equal to $- nFE$ where n represents no of electrons and F and E represent Faraday's constant and electrode potential simultaneously. As we know, for a reaction to be feasible thermodynamically $\Delta G$ should be less than 0, so apparently E will be greater than 0 and also more positive.

Complete step by step answer:
The outer electronic configuration of $M{n^{3 + }}$ is $3{d^4}$ and $M{n^{2 + }}$ has an outer electronic configuration of $3{d^5}$. So the conversion of $M{n^{3 + }}$ to $M{n^{2 + }}$ will be a favorable reaction since $3{d^5}$ is very much stable configuration because of its half-filled configuration. Therefore, ${E_0}$ value for $\dfrac{{M{n^{3 + }}}}{{M{n^{2 + }}}}$ together is + (positive). Similarly, $C{r^3}$ to $C{r^{2 + }}$ undergoes a change in outer electronic configuration that is from $3{d^3}$ to $3{d^4}$. $F{e^{3 + }}$ to $F{e^{2 + }}$ undergoes a change in the outer electronic configuration that is from $3{d^5}$ to $3{d^6}$. Both of these resultant configurations must not be stable and so have lower ${E_0}$ value.

Note: The more thermodynamically feasible the reaction and so greater is the electrode potential. So, for $M{n^{3 + }}$ to $M{n^{2 + }}$, configuration of $M{n^{3 + }}$ is $3{d^4}$ and similarly configuration of $M{n^{2 + }}$ is $3{d^5}$. For $M{n^{2 + }}$ the configuration is a stable half-filled configuration that is $3{d^5}$. So this is more feasible whereas for reaction from $C{r^{3 + }}$ to $C{r^{2 + }}$, the configuration for $C{r^{3 + }}$ is $3{d^3}$ and $C{r^{2 + }}$ is $3{d^4}$. Since the reactant is even more stable, the reaction is unfavorable or we can also say E is less than 0. Similarly, the same things are seen with $F{e^{3 + }}$ to $F{e^{2 + }}$ because to their $3{d^5}$ and $3{d^6}$ configuration of $F{e^{3 + }}$ and $F{e^{2 + }}$ simultaneously.