Chapter 21, Problem 91CP

(a)

Interpretation Introduction

Interpretation: The standard electrode potential $\left(E^{\text{ο}}\right)$ for the reaction and stronger oxidizing agent among ${\text{Co}}^{\text{+3}}$ and $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ is to be stated. An explanation on basis of the crystal field model corresponding to the fact that one of the given two ions is the stronger oxidizing agent is to be stated.

Concept introduction: The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it. The energy difference between these two levels depends upon the properties of both metal and the ligands. If the ligand is strong, then splitting will be high and the complex will be low spin. If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: The standard electrode potential $\left(E^{\text{ο}}\right)$ for the given half reaction.

Answer to Problem 91CP

Answer

The standard electrode potential $\left(E^{\text{ο}}\right)$ for the given reaction is $\underset{\_}{-0.254V}$.

Explanation of Solution

Explanation

The given equations are,

${\text{Co}}^{\text{+3}}+{\text{e}}^{-}\to {\text{Co}}^{\text{+2}}{E}^{\text{ο}}=1.82\text{V}$ (1)

${\text{Co}}^{2+}+3\text{en}\to \text{Co}{\left(\text{en}\right)}_3{}^{2+}K=1.5\times {10}^{12}$ (2)

${\text{Co}}^{3+}+3\text{en}\to \text{Co}{\left(\text{en}\right)}_3{}^{3+}{K}_2=2.0\times {10}^{47}$ (3)

Where,

• $E^{\text{ο}}$ is the standard electrode potential for equation (1).

• $K$ is the equilibrium constant of equation (2).

• $K_2$ is the equilibrium constant of equation (3).

The equilibrium constant $\left(K_1\right)$ for equation (1) is calculated by the formula,

$\log K_1=\frac{n{E}^{\text{ο}}}{0.0591}$

Where,

• $K_1$ is the equilibrium constant of equation (1).

• $n$ is the number of electrons.

The value of $n$ is $1$.

Substitute the value of $n$ and $E^{\text{ο}}$ in the above formula.

$\begin{array}{c}\log K_1=\frac{1\times 1.82}{0.0591}\\ \log K_1=30.79\\ K_1=6.165\times {10}^{30}\end{array}$

Now the equation (1) becomes,

${\text{Co}}^{\text{+3}}+{\text{e}}^{-}\to {\text{Co}}^{\text{+2}}{K}_1=6.165\times {10}^{30}$ (4)

To obtain the required half cell the equation (3) is reversed and added with equation (2) and equation (4). The overall reaction becomes,

$\text{Co}{\left(\text{en}\right)}_3{}^{3+}+{\text{e}}^{-}\to \text{Co}{\left(\text{en}\right)}_3{}^{2+}$ (5)

The equilibrium constant $\left(K_3\right)$ for the above reaction is calculated as,

$K_3=\frac{1}{K_2}\times K\times K_1$

Substitute the values of $K$, $K_1$ and $K_2$ in the above formula.

$\begin{array}{c}{K}_3=\frac{1}{2\times {10}^{47}}\times 1.5\times {10}^{12}\times 6.165\times {10}^{30}\\ =4.7\times {10}^{-5}\end{array}$

The equlibrium constant for theequation (5) is calculated by the formula,

$E^{\text{ο}}=\frac{0.0591}{n}\log K_3$

The value of $n$ is $1$.

Substitue the values of $n$ and $K_1$ in the above equation.

$\begin{array}{c}{E}^{\text{ο}}=\frac{0.0591}{1}\log 4.7\times {10}^{-5}\\ =\underset{\_}{-0.254V}\end{array}$

Therefore, the standard electrode potential $\left(E^{\text{ο}}\right)$ for the given reaction is $\underset{\_}{-0.254V}$.

(b)

Interpretation Introduction

Interpretation: The standard electrode potential $\left(E^{\text{ο}}\right)$ for the reaction and stronger oxidizing agent among ${\text{Co}}^{\text{+3}}$ and $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ is to be stated. An explanation on basis of the crystal field model corresponding to the fact that one of the given two ions is the stronger oxidizing agent is to be stated.

Concept introduction: The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it. The energy difference between these two levels depends upon the properties of both metal and the ligands. If the ligand is strong, then splitting will be high and the complex will be low spin. If the ligand is weak, then splitting will be less and the complex will be high spin.

Answer to Problem 91CP

Answer

The ${\text{Co}}^{\text{+3}}$ is stronger oxidizing agent than $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$.

Explanation of Solution

Explanation

The standard reduction potential, $E^{\text{ο}}{}_{{\text{Co}}^{\text{3+}}{\text{/Co}}^{\text{2+}}}$ is $1.82\text{V}$ and the standard reduction potential, $E^{\text{ο}}{}_{\text{Co}{\left(\text{en}\right)}_3{}^{\text{3+}}\text{/Co}{\left(\text{en}\right)}_3{}^{\text{2+}}}$ is $-0.254\text{V}$.

Since reduction potential of ${\text{Co}}^{\text{+3}}$ is higher than $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$, therefore it is stronger oxidizing agent than $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$.

(c)

Interpretation Introduction

Interpretation: The standard electrode potential $\left(E^{\text{ο}}\right)$ for the reaction and stronger oxidizing agent among ${\text{Co}}^{\text{+3}}$ and $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ is to be stated. An explanation on basis of the crystal field model corresponding to the fact that one of the given two ions is the stronger oxidizing agent is to be stated.

Concept introduction: The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it. The energy difference between these two levels depends upon the properties of both metal and the ligands. If the ligand is strong, then splitting will be high and the complex will be low spin. If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: The use to crystal field model to rationalise the result in part b.

Answer to Problem 91CP

Answer

An explanation on basis of the crystal field model corresponding to the fact that ${\text{Co}}^{\text{+3}}$ is the stronger oxidizing agent has been rightfully stated.

Explanation of Solution

Explanation

The oxidation state of both $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$ and $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ is $+3$; since, ehtlenediammine is a neutral ligand.

The electronic configuration of $\text{Co}$ is $\left[\text{Ar}\right]3d^{7}4s^2$. With the loss of three electrons, it comes in oxidation state of $+3$ with electronic configuration of $\left[\text{Ar}\right]3d^6$.

Cobalt ion form $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$ with water as follows:

${\text{Co}}^{3+}{\text{+6H}}_2\text{O}\to \text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$

Both $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$ and $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ are octahedral complex but $\text{en}$ is a strong field ligand while water is a weak field ligand. The six electrons will occupy ${\text{t}}_{\text{2}}\text{g}$ set of orbitals in $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ complex while in case of $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$ complex, they will occupy both $\text{eg}$ and ${\text{t}}_{\text{2}}\text{g}$ set of orbitals. $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$ complex have higher crystal field energy than $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$, therefore $\text{Co}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{6}}{}^{\text{3+}}$ can gain electrons more easily than $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$. Therefore, ${\text{Co}}^{\text{+3}}$ is a stronger oxidizing agent $\text{Co}{\left(\text{en}\right)}_{\text{3}}{}^{\text{3+}}$.