Chapter 20, Problem 68A

(a)

Interpretation Introduction

Interpretation:

The voltage of cell is to be determined.

Concept introduction:

The total voltage of the cell is the sum of the electric potential of its respective half cells.

Answer to Problem 68A

Voltage of the cell is 3.45V

Explanation of Solution

Since Ag half cell has positive potential it will act as reduction half cell.

Reaction for Ag|Ag+ half cell(reduction):

$A{g}^{+}+e^{-}\to Ag\text{ E}°\text{=0}\text{.8V}$

Reaction for Be2+|Be half cell(oxidation):

$Be\to B{e}^{2+}+2e^{-}\text{ E}°\text{=+1}\text{.85V}$

Cell potential:

$\begin{array}{l}E{°}_{oxidation}=1.85V\\ E{°}_{reduction}=0.8V\times 2=1.6V\\ E{°}_{cell}=1.85+1.6\\ E{°}_{cell}=3.45V\end{array}$

(b)

Interpretation Introduction

Interpretation:

The voltage of cell is to be determined.

Concept introduction:

The total voltage of the cell is the sum of the electric potential of its respective half cells.

Answer to Problem 68A

Voltage of the cell is 0.89V

Explanation of Solution

Since Ag half cell has less positive reduction potential it will act as oxidation half cell.

Reaction for Au+|Au half cell(reduction):

$A{u}^{+}+e^{-}\to Au\text{ E}°\text{=1}\text{.69V}$

Reaction for Be2+|Be half cell(oxidation):

$Ag\to A{g}^{+}+e^{-}\text{ E}°\text{=-0}\text{.8V}$

Cell potential:

$\begin{array}{l}E{°}_{oxidation}=-0.8V\\ E{°}_{reduction}=1.69V\\ E{°}_{cell}=1.69-0.8\\ E{°}_{cell}=0.89V\end{array}$

Interpretation Introduction

(c)

Interpretation:

The voltage of cell is to be determined.

Concept introduction : The total voltage of the cell is the sum of the electric potential of its respective half cells.

Answer to Problem 68A

Voltage of the cell is 2.08V

Explanation of Solution

Since Ag half cell has positive potential it will act as reduction half cell.

Reaction for Ag|Ag+ half cell(reduction):

$2A{g}^{+}+2e^{-}\to 2Ag\text{ E}°\text{=1}\text{.6V}$

Reaction for S|S²- half cell(oxidation):

$S^{2-}\to S+2e^{-}\text{ E}°\text{=+0}\text{.48V}$

Cell potential:

$\begin{array}{l}E{°}_{oxidation}=0.48V\\ E{°}_{reduction}=1.6V\\ E{°}_{cell}=0.48+1.6\\ E{°}_{cell}=2.08V\end{array}$

Interpretation Introduction

(d)

Interpretation:

The voltage of cell is to be determined.

Concept introduction:

The total voltage of the cell is the sum of the electric potential of its respective half cells.

Answer to Problem 68A

Voltage of the cell is 1.06V

Explanation of Solution

Since Ag half cell has positive potential it will act as reduction half cell.

Reaction for Ag|Ag+ half cell(reduction):

$2A{g}^{+}+2e^{-}\to 2Ag\text{ E}°\text{=1}\text{.6V}$

Reaction for I₂|I- half cell(oxidation):

$2I^{-}\to I_2+2e^{-}\text{ E}°\text{=-0}\text{.54V}$

Cell potential:

$\begin{array}{l}E{°}_{oxidation}=-0.54V\\ E{°}_{reduction}=1.6V\\ E{°}_{cell}=1.6-0.54\\ E{°}_{cell}=1.06V\end{array}$