Chapter 14, Problem 14.CE

(a)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

${\text{aA+ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{bB}$

The Nernst equation results in the half-cell potential E as,

$\text{E=E°-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

Here,

$\text{E°}$ = standard reduction potential ( ${\text{A}}_{\text{A}}{\text{=A}}_{\text{B}}\text{=1}$ )

$\text{R}$ = gas constant $\left(\text{8}\text{.314}\text{J/}\left(\text{K}\text{.mol}\right)\right)\text{=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)$

T =Temperature (in K)

N = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$ )

A = Activity of species, i

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode( ${\text{E}}_{\text{+}}$) – potential of left hand electrode( ${\text{E}}_{-}$).

Answer to Problem 14.CE

The cell voltage  of the battery is $\text{1}\text{.65 V}$.  Electrons flow from left hand electrode to right hand electrode that is iron to platinum.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

$\begin{array}{l}\text{Right hand cell:}{\text{Br}}_{\text{(l)}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{2Br}}^{\text{-}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{1}\text{.078 V}\\ \text{-}\\ \text{Left hand cell:}{\text{Fe}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Fe}}_{\text{(s)}}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{- 0}\text{.44 V}\\ \\ \text{Net reaction:}{\text{Br}}_{\text{2}}{}_{\text{(l)}}{\text{ + Fe}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ 2Br}}^{\text{-}}{\text{ + Fe}}^{\text{2+}}\end{array}$

Applying the electrode potential in Nernst equation, the cell voltage is calculated as

$\begin{array}{l}{\text{E}}_{\text{+}}\text{=}\left(\text{1}\text{.078}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}{\text{(0}\text{.050)}}^{\text{2}}\right)\text{-}\left(\text{-0}\text{.44}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{\text{1}}{\text{0}\text{.010}}\right)\\ \\ \text{=}\text{1}\text{.155 - (- 0}\text{.50) = 1}\text{.65 V}\end{array}$

The cell potential is positive one.  Thus, electrons flow from left hand electrode to right hand electrode that is iron to platinum.

(b)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

${\text{aA+ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{bB}$

The Nernst equation results in the half-cell potential E as,

$\text{E=E°-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

Here,

$\text{E°}$ = standard reduction potential ( ${\text{A}}_{\text{A}}{\text{=A}}_{\text{B}}\text{=1}$ )

$\text{R}$ = gas constant $\left(\text{8}\text{.314}\text{J/}\left(\text{K}\text{.mol}\right)\right)\text{=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)$

T =Temperature (in K)

N = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$ )

A = Activity of species, i

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode ( ${\text{E}}_{\text{+}}$) – potential of left hand electrode( ${\text{E}}_{-}$).

Answer to Problem 14.CE

The cell voltage of the battery is $\text{- 0}\text{.77 V}$.  Electrons flow from right hand electrode to left hand electrode that is iron to copper.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

$\begin{array}{l}\text{Right hand cell:}{\text{Fe}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{2Fe}}_{\text{(s)}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{-0}\text{.44 V}\\ \text{-}\\ \text{Left hand cell:}{\text{Cu}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Cu}}_{\text{(s)}}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{0}\text{.339 V}\\ \\ \text{Net reaction:}{\text{Fe}}^{\text{2+}}{\text{ + Cu}}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ Fe}}_{\text{(s)}}{\text{ + Cu}}^{\text{2+}}\end{array}$

Applying the electrode potential in Nernst equation, the cell voltage is calculated as

$\begin{array}{l}{\text{E}}_{\text{+}}\text{=}\left(\text{-0}\text{.44}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{\text{1}}{\text{0}\text{.050}}\right)\text{-}\left(\text{0}\text{.339}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{\text{1}}{\text{0}\text{.020}}\right)\\ \\ \text{=}\text{- 0}\text{.48 - (0}\text{.289) = - 0}\text{.77 V}\end{array}$

The cell potential is negative one.  Thus, electrons flow from right hand electrode to left hand electrode that is iron to copper.

(c)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

${\text{aA+ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{bB}$

The Nernst equation results in the half-cell potential E as,

$\text{E=E°-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

Here,

$\text{E°}$ = standard reduction potential ( ${\text{A}}_{\text{A}}{\text{=A}}_{\text{B}}\text{=1}$ )

$\text{R}$ = gas constant $\left(\text{8}\text{.314}\text{J/}\left(\text{K}\text{.mol}\right)\right)\text{=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)$

T =Temperature (in K)

N = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$ )

A = Activity of species, i

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode( ${\text{E}}_{\text{+}}$) – potential of left hand electrode( ${\text{E}}_{-}$).

Answer to Problem 14.CE

The cell voltage  of the battery is $\text{- 0}\text{.77 V}$.  Electrons flow from right hand electrode to left hand electrode that is iron to copper.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

$\begin{array}{l}\text{Right hand cell:}{\text{Cl}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{2Cl }}^{\text{-}}{}_{\text{(s)}}{\text{E}}_{\text{+}}{}^{\text{0}}\text{=}\text{1}\text{.360V}\\ \text{-}\\ \text{Left hand cell:}{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{2Hg}}_{\text{(s)}}\text{+}{\text{2Cl}}^{\text{-}}{\text{E}}_{\text{-}}{}^{\text{0}}\text{=}\text{0}\text{.268V}\\ \\ \text{Net reaction:}{\text{Cl}}_{\text{2}}{}_{\text{(g)}}{\text{ + 2Hg}}_{\text{(l)}}\stackrel{}{\rightleftharpoons }{\text{ Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}{}_{\text{(s)}}\end{array}$

Applying the electrode potential in Nernst equation, the cell voltage is calculated as

$\begin{array}{l}{\text{E}}_{\text{+}}\text{=}\left(\text{1}\text{.360}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}\frac{{\text{(0}\text{.040)}}^{\text{2}}}{\text{0}\text{.050}}\right)\text{-}\left(\text{0}\text{.268}\text{-}\frac{\text{0}\text{.05916}}{\text{2}}\text{log}{\text{(0}\text{.060)}}^{\text{2}}\right)\\ \\ \text{= 1}\text{.434 - (0}\text{.340) = 1}\text{.094V}\end{array}$

The cell potential is positive one.  Thus, electrons flow from left hand electrode to right hand electrode that is mercury to platinum.