Chapter 17, Problem 17.21P

(a)

Interpretation Introduction

Interpretation:

The current density and overpotential of reduction of azobenzene has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Overpotential: The activation energy of a reaction at an electrode can be overcome by voltage.  The required voltage to apply is called overpotential.

Ohmic potential:  In electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage.  The required voltage to apply is called ohmic potential.

${\text{E}}_{\text{ohmic}}\text{=}\text{IR}$

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.21P

The overpotential of reduction of azobenzene is $\text{0}\text{.85}\text{V}$

Explanation of Solution

To determine: The current density and overpotential of reduction of azobenzene.

The reduction of azobenzene is given as

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{N=NC}}_{\text{6}}{\text{H}}_{\text{5}}\text{+}{\text{4H}}^{\text{+}}\text{+}{\text{4e}}^{\text{-}}\stackrel{}{\to }{\text{2C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$

$\begin{array}{l}\text{Electron flow=}\left(\text{4}\frac{{\text{e}}^{\text{-}}\text{s}}{{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{N=NC}}_{\text{6}}{\text{H}}_{\text{5}}}\right)\left(\text{25}\text{.9}\frac{\text{nmol}}{\text{s}}\right)\left(\text{96485}\text{C/mol}\right)\\ \\ \text{=}\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{C/s}\end{array}$

The current density of reduction of azobenzene is

$\text{=}\frac{\text{1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{A}}{\text{1}\text{.00}\text{×}{\text{10}}^{\text{-4}}{\text{m}}^{\text{2}}}\text{=}\text{100}{\text{A/m}}^{\text{2}}$

For smooth platinum electrode current density of $\text{100}{\text{A/m}}^{\text{2}}$ is equal to $\text{0}\text{.85}\text{V}$.

Over potential is $\text{0}\text{.85}\text{V}$

(b)

Interpretation Introduction

Interpretation:

The cathode potential at which azobenzene get reduced has to be calculated

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Overpotential: The activation energy of a reaction at an electrode can be overcome by voltage.  The required voltage to apply is called overpotential.

Ohmic potential:  In electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage.  The required voltage to apply is called ohmic potential.

${\text{E}}_{\text{ohmic}}\text{=}\text{IR}$

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.21P

The cathode potential at which azobenzene get reduced is $\text{-}\text{0}\text{.036V}$

Explanation of Solution

To determine: The cathode potential at which azobenzene get reduced.

$\begin{array}{l}\text{Using}\text{Nernst}\text{eqn}\text{the}\text{Cathode}\text{potential}\text{is}\text{calculated}\text{as}\\ \\ \text{E(cathode}\text{) =}\text{0}\text{.100 -}\text{0}\text{.05916}\text{log}\frac{{{\text{[Ti}}^{\text{3+}}\text{]}}_{\text{s}}}{{{\text{[TiO}}^{\text{2+}}\text{]}}_{\text{s}}{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}\\ \\ \text{=}\text{0}\text{.100 -}\text{0}\text{.05916}\text{log}\frac{\text{[0}\text{.10]}}{\text{[0}\text{.050][0}{\text{.10]}}^{\text{2}}}\\ \\ \text{=}\text{0}\text{.100}\text{V-}\text{0}\text{.136}\text{V}\text{=-}\text{0}\text{.036V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The anode potential at which oxygen get reduced to water has to be calculated

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Overpotential: The activation energy of a reaction at an electrode can be overcome by voltage.  The required voltage to apply is called overpotential.

Ohmic potential:  In electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage.  The required voltage to apply is called ohmic potential.

${\text{E}}_{\text{ohmic}}\text{=}\text{IR}$

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.21P

The anode potential at which oxygen get reduced is $1.160\text{V}$

Explanation of Solution

To determine: The anode potential at which azobenzene get reduced.

$\begin{array}{l}\text{Using}\text{Nernst}\text{eqn}\text{the}\text{anode}\text{potential}\text{is}\text{calculated}\text{as}\\ \\ \text{E(anode}\text{) =}1.229\text{ -}\frac{\text{0}\text{.05916}}{4}\text{log}\frac{1}{{\text{P}}_{\text{O}}{}_{{}_{\text{2}}}{{\text{[H}}^{\text{+}}\text{]}}^4}\\ \\ \text{=}1.229\text{ -}\frac{\text{0}\text{.05916}}{4}\text{log}\frac{1}{(0.20){(0.10)}^4}\\ \\ \text{=}1.229\text{V-}0.069\text{V}\text{=}1.160\text{V}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The voltage at which azobenzene reduced to aniline has to be calculated

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Overpotential: The activation energy of a reaction at an electrode can be overcome by voltage.  The required voltage to apply is called overpotential.

Ohmic potential:  In electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage.  The required voltage to apply is called ohmic potential.

${\text{E}}_{\text{ohmic}}\text{=}\text{IR}$

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.21P

The voltage at which azobenzene reduced to aniline is $\text{-2}\text{.57}\text{V}$

Explanation of Solution

To determine: The voltage at which azobenzene reduced to aniline.

The reduction of azobenzene is given as

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{N=NC}}_{\text{6}}{\text{H}}_{\text{5}}\text{+}{\text{4H}}^{\text{+}}\text{+}{\text{4e}}^{\text{-}}\stackrel{}{\to }{\text{2C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$

$\begin{array}{l}\text{E =}\text{E(cathode)}\text{-}\text{E(anode)}\text{-}\text{IR}\text{-}\text{Overpotential}\\ \\ \text{=}\text{-0}\text{.036}\text{-1}\text{.160}\text{-(1}\text{.00}\text{×}{\text{10}}^{\text{-2}}\text{A)(52}\text{.4}\text{Ω)}\text{-0}\text{.85}\\ \\ \text{=}\text{-2}\text{.57}\text{V}\end{array}$