Chapter 14, Problem 14.38P

(a)

Interpretation Introduction

Interpretation:

The half-reaction and Nernst equation for each half-cell has to be written.

Concept introduction:

Nernst equation can be used to determine the cell potential at any instant, the difference from the standard state.

${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{ - }\left(\frac{\text{RT}}{\text{nF}}\right)\text{lnQ}$

Where:

${\text{E}}_{\text{cell}}$ = cell potential under nonstandard conditions

$\text{E}{}_{\text{cell}}^{\text{o}}$ = cell potential under standard conditions

R = gas constant, which is $\text{8}\text{.31 }\left(\text{volt-coulomb}\right)\text{/}\left(\text{mol-K}\right)$

T = temperature (kelvin), which is generally $\text{298°K }\left(\text{77°F/25°C}\right)$

n = number of moles of electrons exchanged in the electrochemical reaction

F = Faraday's constant, $\text{96500 coulombs/mol}$

Q = reaction quotient, which is the equilibrium expression with initial concentrations rather than equilibrium concentrations

Explanation of Solution

The half-reaction and Nernst equation for each half-cell can be calculated using below formula

$\text{E}\text{(right)}\text{=}{\text{E}}^{\text{o}}\text{(right) }\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{A}}_{{\text{F}}^{\text{-}}}^{\text{2}}\text{(right)}}{{\text{P}}_{{\text{O}}_{\text{2}}}^{\text{1/2}}\text{(right)}}$

$\text{E}\text{(left)}\text{=}{\text{E}}^{\text{o}}\text{(left) }\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{A}}_{{\text{F}}^{\text{-}}}^{\text{2}}\text{(left)}}{{\text{P}}_{{\text{O}}_{\text{2}}}^{\text{1/2}}\text{(left)}}$

Both half left and right are added

$\begin{array}{l}{\text{MgF}}_{\text{2(s)}}\text{+}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\text{+}{\text{2e}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{MgAl}}_{\text{2}}{\text{O}}_{\text{4(s)}}\text{+}{\text{2F}}^{\text{-}}\\ \underset{\_}{{\text{MgO}}_{\text{(s)}}\text{+}{\text{2F}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{MgF}}_{\text{2}}{}_{\text{(s)}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2(g)}}\text{+}{\text{2e}}^{\text{-}}}\\ \text{net}\text{reaction}\text{:}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{+MgO}}_{\text{(s)}}\underset{}{\rightleftharpoons }{\text{MgAl}}_{\text{2}}{\text{O}}_{\text{4(s)}}\end{array}$

Nernst equation for net reaction

$\text{E}\text{(right)-E}\text{(left)}\text{=}{\text{E}}^{\text{o}}{\text{(right)-E}}^{\text{o}}\text{(left) }\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{A}}_{{\text{F}}^{\text{-}}}^{\text{2}}\text{(right)}}{{\text{P}}_{{\text{O}}_{\text{2}}}^{\text{1/2}}\text{(right)}}\text{+}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{{\text{A}}_{{\text{F}}^{\text{-}}}^{\text{2}}\text{(left)}}{{\text{P}}_{{\text{O}}_{\text{2}}}^{\text{1/2}}\text{(left)}}$

${\text{F}}^{\text{-}}$ have both side same activities of ${\text{O}}_{\text{2}}$.  The ln terms cancel, leaving

$\text{E}\text{(cell)}\text{=}{\text{E}}^{\text{o}}\text{(right)}{\text{-E}}^{\text{o}}\text{(left)=}{\text{E}}^{\text{o}}\text{(cell)}$

(b)

Interpretation Introduction

Interpretation:

The value of ${\text{ΔG}}^{\text{o}}$ has to be calculated.

Concept introduction:

The value of ${\text{ΔG}}^{\text{o}}$ can be calculated as,

${\text{E}}^{\text{o}}\text{=}\frac{{\text{-ΔG}}^{\text{o}}}{\text{nF}}$

Where

${\text{E}}^{\text{o}}$ = The standard reduction potential

n = number of electrons

The equilibrium constant for the reaction can be calculated as,

Finding $\text{K}$ from ${\text{E}}^{\text{o}}$

$\text{K}\text{=}\frac{{\text{10}}^{{\text{nE}}^{\text{o}}}}{\text{0}\text{.05916}}\text{at}{\text{25}}^{\text{o}}\text{C}$

Where $\text{K}$ is equilibrium constant

Answer to Problem 14.38P

The value of ${\text{ΔG}}^{\text{o}}$ is $\text{-29}\text{.51}\text{kJ/mol}$

Explanation of Solution

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}\text{-}{\text{nFE}}^{\text{o}}\\ \text{=}\text{-}\text{(2)}\text{(9}\text{.6485×}{\text{10}}^{\text{4}}\text{C/mol)}\text{(0}\text{.1529}\text{J/C)}\text{=}\text{-29}\text{.51}\text{kJ/mol}\end{array}$

The fact that a volt is equivalent to one joule/coulomb

(c)

Interpretation Introduction

Interpretation:

The electron flow direction has to be found.

Concept introduction:

The value of ${\text{ΔG}}^{\text{o}}$ can be calculated as,

${\text{E}}^{\text{o}}\text{=}\frac{{\text{-ΔG}}^{\text{o}}}{\text{nF}}$

Where

${\text{E}}^{\text{o}}$ = The standard reduction potential

n = number of electrons

$\begin{array}{l}{\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}\\ \text{where,}\\ \text{T}\text{is}\text{temperature}\\ {\text{ΔG}}^{\text{o}}\text{is}\text{stand}\text{free}\text{energy}\text{change}\\ {\text{ΔH}}^{\text{o}}\text{is}\text{stand}\text{enthalpy}\text{change}\end{array}$

Explanation of Solution

$\begin{array}{l}{\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}\\ {\text{-nFE}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}\\ \text{=}\text{-nF(0}\text{.1223}\text{V+3}{\text{.06×10}}^{\text{-5}}\text{T)}\\ \text{=}\text{-nF(0}\text{.1223}\text{V)}\text{-}\text{nF(3}{\text{.06×10}}^{\text{-5}}\text{T)}\\ \text{=}\text{-}\underset{{\text{ΔH}}^{\text{o}}}{\underbrace{\text{nF(0}\text{.1223}\text{V}}}\text{)}\text{-}\text{T}\{\underset{{\text{ΔS}}^{\text{o}}}{\underbrace{\text{nF(3}{\text{.06×10}}^{\text{-5}}\text{V/K}}}\text{)}\}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}\text{-nF}\text{(0}\text{.1223}\text{V)}\\ \text{=}\text{-}\text{(2)(9}\text{.6485}\text{×}{\text{10}}^{\text{4}}\text{C/mol)}\text{(0}\text{.1223J/C)}\\ \text{=}\text{-23}\text{.60}\text{kJ/mol}\end{array}$

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}\text{-}\text{nF}\text{(3}\text{.06×}{\text{10}}^{\text{-5}}\text{V/K)}\\ \text{=}\text{-}\text{(2)(9}\text{.6485}\text{×}{\text{10}}^{\text{4}}\text{C/mol)}\text{(3}\text{.06×}{\text{10}}^{\text{-5}}\text{V/K)}\\ \text{=}\text{5}\text{.90}\text{C}\text{.V/(K}\text{.mol)=}\text{5}\text{.90}\text{J/(K}\text{.mol)}\end{array}$

We made use of the conversion coulomb volt =joule