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

Hint: 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.