Chapter 17, Problem 17.4P

(a)

Interpretation Introduction

Interpretation:

The voltage required to complete the given net reaction 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 17.4P

The voltage required to complete the given net reaction is $\text{-1}\text{.906}\text{V}$

Explanation of Solution

To determine: The voltage required to complete the given net reaction.

When the electric current is negligible, the voltage of net reaction is given as

$\begin{array}{l}\text{E = E(cathode)}\text{-}\text{E(anode)}\\ \\ \text{=}{\text{{E}}^{\circ }\text{(cathode)-}\text{0}\text{.05916}\text{log}{\text{P}}_{{\text{H}}_{\text{2}}}{}^{\text{0}\text{.5}}{\text{[OH}}^{\text{-}}\text{]}-}\\ {\text{{E}}^{\circ }\text{(anode)-}\text{0}\text{.05916}\text{log}{\text{[Br}}^{\text{-}}\text{]}}\\ \\ \text{=}\text{{0}\text{.828 -}\text{0}\text{.059 16}\text{log}{\text{(1}}^{\text{.0)}}\text{[0}\text{.10]-}\\ \text{{1}\text{.078}\text{-0}\text{.05916}\text{log}\text{[0}\text{.10]}}\text{=}\text{-1}\text{.906}\text{V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ohmic potential which is voltage required to overcome the resistance of cell has to be calculated.

Concept Introduction:

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.

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

Answer to Problem 17.4P

The ohmic potential which is voltage required to overcome the resistance of cell is $\text{0}\text{.20V}$.

Explanation of Solution

To determine: The ohmic potential which is voltage required to overcome the resistance of cell.

The ohmic potential of the cell is calculated as follows

$\begin{array}{l}{\text{E}}_{\text{ohmic}}\text{=}\text{IR}\\ \\ \text{=}\text{(0}\text{.100A)(2}\text{.0}\text{Ω)}\\ \\ \text{=}\text{0}\text{.20V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The voltage required to overcome the ohmic potential effect and electrolysis effect 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 17.4P

The voltage required to overcome the ohmic potential effect and electrolysis effect is $\text{-2}\text{.17}\text{}\text{V}$.

Explanation of Solution

To determine: The voltage required to overcome the ohmic potential effect and electrolysis effect.

$\begin{array}{l}\text{E = E(cathode)}\text{-}\text{E(anode)}\text{-}\text{I}\text{.R}\text{-}\text{overpotential}\\ \\ \text{=}\text{-1}\text{.906}\text{-}\text{0}\text{.20}\text{-(0}\text{.20}\text{+0}\text{.40)}\\ \\ \text{=}\text{-2}\text{.17}\text{}\text{V}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The voltage required to overcome the ohmic potential effect and anodic reaction overpotential 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 17.4P

The voltage required to overcome the ohmic potential effect and electrolysis effect is $\text{-2}\text{.82}\text{}\text{V}$.

Explanation of Solution

To determine: The voltage required to overcome the ohmic potential effect and anodic reaction overpotential.

$\begin{array}{l}\text{E}{\text{(cathode) = E}}^{\text{o}}\text{(cathode)}\text{-}\text{0}\text{.05916}\text{log}{\text{P}}_{{\text{H}}_{\text{2}}}{}^{\text{0}\text{.5}}{\text{[OH}}^{\text{-}}\text{]}\\ \\ \text{=}\text{0}\text{.828 -}\text{0}\text{.059 16}\text{log}{\text{(1}}^{\text{.0)}}\text{[1}\text{.0]}\\ \\ \text{=}\text{-0}\text{.828}\text{V}\\ \\ \text{E}{\text{(anode) = E}}^{\text{o}}\text{(anode)}\text{-}\text{0}\text{.05916}{\text{log[Br}}^{\text{-}}{\text{]}}_{\text{s}}\\ \\ \text{=}1.078\text{ -}\text{0}\text{.059 16}\text{log}\text{(0}\text{.010)}\\ \\ \text{=}\text{-1}\text{.196}\text{V}\end{array}$

From the anode voltage and cathode voltage, the voltage required to overcome the ohmic potential effect and anodic reaction overpotential is calculated as

$\begin{array}{l}\text{E = E(cathode)}\text{-}\text{E(anode)}\text{-}\text{I}\text{.R}\text{-}\text{overpotential}\\ \\ \text{=}\text{-0}\text{.828}\text{-}\text{1}\text{.196}\text{-}\text{0}\text{.20}\text{-}\text{(0}\text{.20}\text{+0}\text{.40)}\\ \\ \text{=}\text{-2}\text{.82}\text{}\text{V}\end{array}$