Chapter 19, Problem 101CP

(a)

Interpretation Introduction

Interpretation: The value of $\xi °$ for the given half-reaction needs to be determined.

  ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

Given:

  $\begin{array}{l}{\text{Co}}_{}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}\text{ ξ°=1}\text{.82V}\\ {\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{ = K}}_{\text{f}}\text{ = 2}{\text{.0×10}}^{\text{47}}\\ {\text{ Co(en)}}_{\text{3}}^{\text{2+}}{\text{= K}}_{\text{f}}\text{ = 1}{\text{.5×10}}^{\text{12}}\end{array}$

Concept Introduction:

Crystal field theory is the theory given to explain the bonding in the coordination complexes. As ligand approaches towards the metal ion, the d-orbital of metal ion divide according to the energy of metal ion. On the basis of energy and degeneracy, the d-orbital can be classified as ${\text{t}}_{\text{2g}}$ and ${\text{e}}_{\text{g}}$ orbitals. The electrons will fill according to their energy levels.

In octahedral complex, the ${\text{t}}_{\text{2g}}$ orbitals are lower energy orbitals in d-orbital splitting whereas ${\text{e}}_{\text{g}}$ orbitals are higher energy orbitals.

Answer to Problem 101CP

  ${\text{E}}_{\text{a}}^{\text{°}}\text{= -0}\text{.26V}$

Explanation of Solution

  $\begin{array}{l}{\text{Co}}_{}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}\text{ ξ°=1}\text{.82V}\\ {\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{ = K}}_{\text{f}}\text{ = 2}{\text{.0×10}}^{\text{47}}\\ {\text{ Co(en)}}_{\text{3}}^{\text{2+}}{\text{= K}}_{\text{f}}\text{ = 1}{\text{.5×10}}^{\text{12}}\end{array}$

For ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

  $\begin{array}{l}{\text{Co}}_{}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}{\text{ ξ}}_c^{°}\text{ =1}\text{.82V}\\ {\text{ Co(en)}}_{\text{3}}^{\text{2+}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{ + e}}^{\text{-}}{\text{ -ξ}}_a^{°}\text{= ?}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{Co}}_{}^{\text{3+}}{\text{+ Co(en)}}_{\text{3}}^{\text{2+}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}{\text{ +Co(en)}}_{\text{3}}^{\text{3+}}{\text{ ξ}}_{cell}^{°}=1.82-{\text{ξ}}_a^{°}\end{array}$

  $\begin{array}{l}{\text{Co}}_{}^{\text{3+}}\text{+ 3en }\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{ K}}_{\text{1}}\text{=2}{\text{.0×10}}^{\text{47}}\\ {\text{ Co(en)}}_{\text{3}}^{\text{2+}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}{\text{ + 3en K}}_{\text{2}}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\text{1}{\text{.5×10}}^{\text{12}}$}\right.\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{Co}}_{}^{\text{3+}}{\text{+ Co(en)}}_{\text{3}}^{\text{2+}}\stackrel{}{\to }{\text{Co}}_{}^{\text{2+}}{\text{ +Co(en)}}_{\text{3}}^{\text{3+}}{\text{ K = K}}_{\text{1}}\times {\text{ K}}_{\text{2}}\text{ =}\raisebox{1ex}{$\text{2}{\text{.0×10}}^{\text{47}}$}\!\left/ \!\raisebox{-1ex}{$\text{1}{\text{.5×10}}^{\text{12}}$}\right.\text{= 1}{\text{.3×10}}^{35}\end{array}$

Calculate E_cell from Nernst equation:

  $\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}\text{ =}\frac{\text{0}\text{.0591}}{\text{n}}\text{logK}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{ = }\frac{\text{0}\text{.0591}}{\text{1}}\text{log (1}{\text{.3×10}}^{\text{3}}{}^{\text{5}}\text{)}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{ = 2}\text{.08V}\\ {\text{E}}_{\text{cell}}^{\text{°}}\text{=1}{\text{.82-E}}_{\text{a}}^{\text{°}}\text{=2}\text{.08V}\\ {\text{-E}}_{\text{a}}^{\text{°}}\text{=2}\text{.08-1}\text{.82=0}\text{.26V}\\ {\text{E}}_{\text{a}}^{\text{°}}\text{= -0}\text{.26V}\end{array}$

(b)

Interpretation Introduction

Interpretation: The stronger oxidizing agent out of ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ or ${\text{Co}}_{}^{\text{3+}}$ needs to be determined, if ${\text{E}}_{\text{a}}^{\text{°}}\text{= -0}\text{.26V}$ for the given reaction.

  ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

Concept Introduction:

Crystal field theory is the theory given to explain the bonding in the coordination complexes. As ligand approaches towards the metal ion, the d-orbital of metal ion divide according to the energy of metal ion. On the basis of energy and degeneracy, the d-orbital can be classified as ${\text{t}}_{\text{2g}}$ and ${\text{e}}_{\text{g}}$ orbitals. The electrons will fill according to their energy levels.

In octahedral complex, the ${\text{t}}_{\text{2g}}$ orbitals are lower energy orbitals in d-orbital splitting whereas ${\text{e}}_{\text{g}}$ orbitals are higher energy orbitals.

Answer to Problem 101CP

  ${\text{Co}}_{}^{\text{3+}}$ is much stronger oxidizing agent ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ .

Explanation of Solution

For the given reaction; ${\text{E}}_{\text{a}}^{\text{°}}\text{= -0}\text{.26V}$

  ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

A stronger oxidizing agent will be the more easily reduced species therefore the more positive standard reduction potential will be associated with strong oxidizing agent. Hence ${\text{Co}}_{}^{\text{3+}}$ is much stronger oxidizing agent ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ as ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ hasE° = -0.26 V.

(c)

Interpretation Introduction

Interpretation: The crystal field model for the stronger oxidizing agent out of ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ or ${\text{Co}}_{}^{\text{3+}}$ needs to be determined, if ${\text{E}}_{\text{a}}^{\text{°}}\text{= -0}\text{.26V}$ for the given reaction.

  ${\text{Co(en)}}_{\text{3}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Co(en)}}_{\text{3}}^{\text{2+}}$

Concept Introduction:

Crystal field theory is the theory given to explain the bonding in the coordination complexes. As ligand approaches towards the metal ion, the d-orbital of metal ion divide according to the energy of metal ion. On the basis of energy and degeneracy, the d-orbital can be classified as ${\text{t}}_{\text{2g}}$ and ${\text{e}}_{\text{g}}$ orbitals. The electrons will fill according to their energy levels.

In octahedral complex, the ${\text{t}}_{\text{2g}}$ orbitals are lower energy orbitals in d-orbital splitting whereas ${\text{e}}_{\text{g}}$ orbitals are higher energy orbitals.

Answer to Problem 101CP

Since en is a stronger-field ligand than H2O, the d-orbital splitting is larger for ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ thus it needs more energy than to ${\text{Co(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ . Hence ${\text{Co(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ will gain electron more easily than for ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ .

Explanation of Solution

In aqueous solution, ${\text{Co}}_{}^{\text{3+}}$ forms hydrate complex ${\text{Co(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ and ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ the metal ion exists as ${\text{Co}}_{}^{\text{3+}}$ with 3d⁶ with six d electrons. In a strong-field case for each complex ion, the d-orbital splitting diagram will be

  

Since, en is a stronger-field ligand than H2O, the d-orbital splitting is larger for ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ thus it needs more energy than to ${\text{Co(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ . Hence ${\text{Co(H}}_{\text{2}}{\text{O)}}_6^{\text{3+}}$ will gain electron more easily than for ${\text{Co(en)}}_{\text{3}}^{\text{3+}}$ .