Question

Explain why
1-${{\text{E}}^ \circ }$ for ${\text{M}}{{\text{n}}^{3 + }}{\text{/M}}{{\text{n}}^{2 + }}$ couple is more positive than that for ${\text{F}}{{\text{e}}^{3 + }}{\text{/F}}{{\text{e}}^{2 + }}$ (At. no. of ${\text{Mn}} = 25$, ${\text{Fe}} = 26$).
2-${\text{C}}{{\text{e}}^{3 + }}$ can be easily oxidised to ${\text{C}}{{\text{e}}^{4 + }}$ (At. no. of ${\text{Ce}} = 58$).

Answer 1

Hint:The ${{\text{E}}^ \circ }$ for the metal depends on the valence shell electronic configuration. The more stable valence shell electronic configuration gives a more positive value of the standard potential ${{\text{E}}^ \circ }$. To solve this we must know the valence shell electronic configuration of the given ions.

Complete step by step answer:
1) We are given that ${{\text{E}}^ \circ }$ for ${\text{M}}{{\text{n}}^{3 + }}{\text{/M}}{{\text{n}}^{2 + }}$ couple is more positive than that for ${\text{F}}{{\text{e}}^{3 + }}{\text{/F}}{{\text{e}}^{2 + }}$.
The atomic number of ${\text{Mn}}$ is 25. Thus, the electronic configuration of ${\text{Mn}}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^2}$
The valence shell electronic configuration of ${\text{M}}{{\text{n}}^{3 + }}$ ion is $3{d^4}$ and the valence shell electronic configuration of ${\text{M}}{{\text{n}}^{2 + }}$ is $3{d^5}$.
When ${\text{M}}{{\text{n}}^{3 + }}$ is converted to ${\text{M}}{{\text{n}}^{2 + }}$ the valence shell electronic configuration changes from $3{d^4}$ to $3{d^5}$.
The $3{d^5}$ electronic configuration is very stable as it is a half-filled configuration.
The atomic number of ${\text{Fe}}$ is 26. Thus, the electronic configuration of ${\text{Fe}}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}4{s^2}$
The valence shell electronic configuration of ${\text{F}}{{\text{e}}^{3 + }}$ ion is $3{d^5}$ and the valence shell electronic configuration of ${\text{F}}{{\text{e}}^{2 + }}$ is $3{d^6}$.
When ${\text{F}}{{\text{e}}^{3 + }}$ is converted to ${\text{F}}{{\text{e}}^{2 + }}$ the valence shell electronic configuration changes from $3{d^5}$ to $3{d^6}$.
The $3{d^6}$ electronic configuration is not stable.
We know that the more stable valence shell electronic configuration gives a more positive value of the standard potential ${{\text{E}}^ \circ }$.
Thus, ${{\text{E}}^ \circ }$ for ${\text{M}}{{\text{n}}^{3 + }}{\text{/M}}{{\text{n}}^{2 + }}$ couple is more positive than that for ${\text{F}}{{\text{e}}^{3 + }}{\text{/F}}{{\text{e}}^{2 + }}$.
2) We are given that ${\text{C}}{{\text{e}}^{3 + }}$ can be easily oxidised to ${\text{C}}{{\text{e}}^{4 + }}$.
The atomic number of ${\text{Ce}}$ is 58. Thus, the electronic configuration of ${\text{Ce}}$ is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}4{d^{10}}5{s^2}5{p^6}4{f^1}5{d^1}6{s^2}$
The valence shell electronic configuration of ${\text{C}}{{\text{e}}^{3 + }}$ is $4{f^1}5{d^0}6{s^0}$. The ${\text{C}}{{\text{e}}^{3 + }}$ ion can easily lose the one electron in 4f and form ${\text{C}}{{\text{e}}^{4 + }}$.
The electronic configuration of ${\text{C}}{{\text{e}}^{4 + }}$ is $4{f^0}5{d^0}6{s^0}$ which is a stable configuration.
Thus, ${\text{C}}{{\text{e}}^{3 + }}$ can be easily oxidised to ${\text{C}}{{\text{e}}^{4 + }}$.

Note:The difference in the potential of two half cells is known as the electrode potential. Electrode potential is denoted by E. when the concentration of all the reactant and product species in the reaction is unity is known as the standard electrode potential. Standard electrode potential is denoted by ${{\text{E}}^0}$.