Chapter 14, Problem 14.BE

(a)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=-0.112V$

The equilibrium constant for the given reaction is $K=1\times 10^{-19}$

Explanation of Solution

Given data:

The given reaction is ${\text{I}}_{\text{2}}(s)+{\text{ 5Br}}_{\text{2}}{\text{(aq) + 6H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }{\text{ 2IO}}_{\text{3}}{}^{-}{\text{ +10Br}}^{-}+12{\text{H}}^{\text{+}}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ 5Br}}_{\text{2}}{\text{(aq) + 10e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ 10Br}}^{-}{\text{E}}_{+}^{°}\text{ = 1}\text{.098 V}\\ \frac{-2{\text{IO}}_{\text{3}}{}^{-}{\text{ + 12H}}^{+}{\text{+10e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ I}}_{\text{2}}(s)+6{\text{H}}_{\text{2}}\text{O}}{{\text{I}}_{\text{2}}{\text{(s) + 5Br}}_{\text{2}}\text{(aq)}+6{\text{H}}_{\text{2}}\text{O }\stackrel{}{\rightleftharpoons }2{\text{IO}}_{\text{3}}{}^{-}{\text{ +10Br}}^{-}+{\text{ 12H}}^{+}}{\text{E}}_{-}^{°}\text{ = 1}\text{.201 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =1}\text{.098}-\text{ 1}\text{.210}\\ \text{ =}-\text{0}\text{.112 V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $-0.112V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(10)(-0.112)}{}^{/0.05916}\\ \text{ =1}\times {\text{10}}^{-19}\end{array}$

Therefore, the equilibrium constant is $1\times 10^{-19}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=-0.112V$

The equilibrium constant for the given reaction is calculated as $K=1\times 10^{-19}$

(b)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=-0.45V$

The equilibrium constant for the given reaction is $K=6\times 10^{-16}$

Explanation of Solution

Given data:

The given reaction is ${\text{Cr}}^{\text{2+}}\text{ + Fe(s) }\stackrel{}{\rightleftharpoons }{\text{ Fe}}^{2+}\text{ + Cr(s)}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ Cr}}^{2+}{\text{ + 2e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Cr(s)}{\text{E}}_{+}^{°}\text{ = -0}\text{.89 V}\\ \frac{-{\text{ Fe}}^{\text{2+}}{\text{ + 2e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Fe}(s)}{{\text{Cr}}^{2+}\text{ +Fe(s) }\stackrel{}{\rightleftharpoons }\text{ Cr(s)}+{\text{ Fe}}^{2+}}{\text{E}}_{-}^{°}\text{ = -0}\text{.44 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =}-\text{0}\text{.89}-\text{ (}-0.44)\\ \text{ =}-0.45\text{ V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $-0.45V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(2)(-0.45)}{}^{/0.05916}\\ \text{ =6}\times {\text{10}}^{-16}\end{array}$

Therefore, the equilibrium constant is $6\times 10^{-16}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=-0.45V$

The equilibrium constant for the given reaction is calculated as $K=6\times 10^{-16}$

(c)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=3.720V$

The equilibrium constant for the given reaction is $K=6\times 10^{125}$

Explanation of Solution

Given data:

The given reaction is ${\text{Mg(s) + Cl}}_{\text{2}}\text{(g) }\stackrel{}{\rightleftharpoons }{\text{ Mg}}^{2+}{\text{ + 2Cl}}^{-}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ Cl}}_{\text{2}}(g){\text{ + 2e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ 2Cl}}^{-}{\text{E}}_{+}^{°}\text{ = 1}\text{.360 V}\\ \frac{-{\text{ Mg}}^{\text{2+}}{\text{ + 2e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Mg}(s)}{{\text{Mg(s) +Cl}}_{\text{2}}\text{(g) }\stackrel{}{\rightleftharpoons }{\text{ Mg}}^{2+}+{\text{ 2Cl}}^{-}}{\text{E}}_{-}^{°}\text{ = -2}\text{.360 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =1}\text{.360}-\text{ (}-2.360)\\ \text{ =3}\text{.720 V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $3.720V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(2)(3.720)}{}^{/0.05916}\\ \text{ =6}\times {\text{10}}^{125}\end{array}$

Therefore, the equilibrium constant is $6\times 10^{125}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=3.720V$

The equilibrium constant for the given reaction is calculated as $K=6\times 10^{125}$

(d)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=-0.462V$

The equilibrium constant for the given reaction is $K=1\times 10^{-47}$

Explanation of Solution

Given data:

The given reaction is ${\text{5MnO}}_{\text{2}}{\text{(s) + 4H}}^{+}\stackrel{}{\rightleftharpoons }{\text{ 3Mn}}^{2+}{\text{ + 2MnO}}_{\text{4}}{}^{-}+2{\text{H}}_{\text{2}}\text{O}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ 3[MNO}}_{\text{2}}(s){\text{ + 4H}}^{+}{\text{ + 2e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ Mn}}^{2+}{\text{ + 2H}}_{\text{2}}\text{O}]{\text{E}}_{+}^{°}\text{ = 1}\text{.230 V}\\ \frac{-{\text{ 2[MnO}}_{\text{4}}{}^{-}{\text{ + 4H}}^{+}{\text{ + 3e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ MnO}}_{\text{2}}(s){\text{+ 2H}}_{\text{2}}\text{O}]}{{\text{5MnO}}_{\text{2}}{\text{(s) + 4H}}^{+}\stackrel{}{\rightleftharpoons }{\text{ 2MnO}}_{\text{4}}{}^{-}+{\text{ 3Mn}}^{2+}{\text{ + 2H}}_{\text{2}}\text{O}}{\text{E}}_{-}^{°}\text{ = 1}\text{.692 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =1}\text{.230}-1.692\\ \text{ =}-0.462\text{ V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $-0.462V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(6)(-0.462)}{}^{/0.05916}\\ \text{ =1}\times {\text{10}}^{-47}\end{array}$

Therefore, the equilibrium constant is $1\times 10^{-47}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=-0.462V$

The equilibrium constant for the given reaction is calculated as $K=1\times 10^{-47}$

(e)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=0.782V$

The equilibrium constant for the given reaction is $K=2\times 10^{13}$

Explanation of Solution

Given data:

The given reaction is ${\text{Ag}}^{+}{\text{ + 2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{2-}\stackrel{}{\rightleftharpoons }{\text{ Ag(S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{)}}_{\text{2}}{}^{3-}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ Ag}}^{+}{\text{ + e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Ag(s)}{\text{E}}_{+}^{°}\text{ = 0}\text{.799 V}\\ \frac{-{\text{Ag(S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{)}}_{\text{2}}{}^{3-}{\text{ + e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Ag}(s){\text{+ 2S}}_2{\text{O}}_{\text{3}}{}^{2-}}{{\text{Ag}}^{+}{\text{ + 2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{2-}\stackrel{}{\rightleftharpoons }{\text{ Ag(S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{)}}_{\text{2}}{}^{3-}}{\text{E}}_{-}^{°}\text{ = 0}\text{.017 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =0}\text{.799}-0.017\\ \text{ =0}\text{.782 V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $0.782V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(1)(0.782)}{}^{/0.05916}\\ \text{ =2}\times {\text{10}}^{13}\end{array}$

Therefore, the equilibrium constant is $2\times 10^{13}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=0.782V$

The equilibrium constant for the given reaction is calculated as $K=2\times 10^{13}$

(f)

Interpretation Introduction

Interpretation:

The $\text{E}°$ and $\text{K}$ for the given reaction has to be calculated.

Concept Introduction:

Finding $E°$ from $K$:

The equation used for finding $\text{E}°$ from $\text{K}$ is:

$\text{E° =}\frac{0.05196}{n}\log \text{ K}\text{(at }25°\text{C})$

Finding $K$ from $E°$:

The equation used for finding $\text{K}$ from $\text{E°}$ is:

$\text{K =}{10}^{nE°}{}^{/0.05916}\text{(at }25°\text{C})$

To Calculate: The $\text{E}°$ and $\text{K}$ for the given reaction

Answer to Problem 14.BE

The $\text{E}°$ value for the given reaction is $E°=-0.703V$

The equilibrium constant for the given reaction is $K=1\times 10^{-12}$

Explanation of Solution

Given data:

The given reaction is $\text{CuI(s) }\stackrel{}{\rightleftharpoons }{\text{ Cu}}^{+}{\text{ + I}}^{-}$

Calculation of equilibrium constant:

First divide the given reaction into two half-reactions with their $\text{E}°$ value found from Appendix H.

$\begin{array}{l}{\text{ CuI(s) + e}}^{-}\stackrel{}{\rightleftharpoons }{\text{ Cu(s) + I}}^{-}{\text{E}}_{+}^{°}\text{ =}-0.185\text{ V}\\ \frac{-{\text{Cu}}^{+}{\text{ + e}}^{-}\stackrel{}{\rightleftharpoons }\text{ Cu}(s)}{\text{CuI(s) }\stackrel{}{\rightleftharpoons }{\text{ Cu}}^{+}{\text{ + I}}^{-}}{\text{E}}_{-}^{°}\text{ = 0}\text{.518 V}\end{array}$

Find the $\text{E}°$ for the net reaction as follows,

$\begin{array}{l}\text{E}°{\text{ = E}}_{+}^{°}-{\text{ E}}_{-}^{°}\\ \text{ =}-0.185-0.518\\ \text{ =}-0.703\text{ V}\end{array}$

Therefore, $\text{E}°$ for the net reaction is $-0.703V$

Substitute obtained $\text{E}°$ value and calculated the equilibrium constant as follows,

$\begin{array}{l}\text{K =}{10}^{nE°}{}^{/0.05916}\\ \text{ =}{10}^{(1)(-0.703)}{}^{/0.05916}\\ \text{ =1}\times {\text{10}}^{-12}\end{array}$

Therefore, the equilibrium constant is $1\times 10^{-12}$

Conclusion

The $\text{E}°$ value for the given reaction is calculated as $E°=-0.703V$

The equilibrium constant for the given reaction is calculated as $K=1\times 10^{-12}$