Chapter 19, Problem 23QP

Use the standard reduction potentials to find the equilibrium constant for each of the following reactions at $\text{25°C:}$

$\begin{array}{l}(a)\\ B{r}_2(l)+{21}^{-}(aq)\rightleftarrows 2B{r}^{-}(aq)+I_2(s)\end{array}$

$\begin{array}{l}(b)\\ 2C{c}^{4+}(aq)+2{\text{CI}}^{-}(aq)\rightleftarrows C{l}_2(g)+2C{c}^{3+}(aq)\end{array}$

$\begin{array}{l}(c)\\ 5F{e}^{2+}(aq)+Mn{o}_4^{-}(aq)+8H^{+}(aq)\rightleftarrows M{n}^{2+}(aq)+4H_2\text{O}(l)+5F{e}^{3+}(aq)\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium constant for each of the given chemical reactions is to be determined.

Concept introduction:

The standard cell potential of a particular cell is given as:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Here, ${\text{E}}^{\text{0}}{}_{\text{cathode}}{\text{and E}}^{\text{0}}{}_{\text{anode}}$ is the standard reduction potential occurring at the cathode and anode respectively.

The relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given as:

$E^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}$

Here, $K$ is the equilibrium constant, $n$ is the transfer of electrons in the half-cell, and $E_{cell}^0$ is the standard cell potential.

Answer to Problem 23QP

Solution:

(a) $K\text{ = 2 }\times {\text{10}}^{\text{18}}$

(b) $K\text{ = 3}\times {\text{10}}^{\text{8}}$

(c) $K\text{ = 3 }\times {\text{10}}^{\text{62}}$

Explanation of Solution

a) ${\text{Br}}_{\text{2}}(l){\text{ + 2I}}^{-}\text{(}aq\text{) }\rightleftharpoons {\text{2Br}}^{-}\text{(}aq{\text{) + I}}_{\text{2}}\text{(}s\text{) }$

This reaction can be written as two half-reactions, as shown below:

$\begin{array}{l}\text{At cathode (reduction):}\\ {\text{Br}}_{\text{2}}(l){\text{ + 2e}}^{-}\to 2{\text{Br}}^{-}(aq)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ {\text{2I}}^{-}\text{(}aq\text{) }\to {\text{ I}}_{\text{2}}(s){\text{ + 2e}}^{-}\end{array}$

From table 19.1, the reduction potential of bromine is $1.07\text{ V}$ and the reduction potential of iodine is $0.53\text{ V}$.

$\begin{array}{l}{E}^o{}_{cathode}\text{= 1}\text{.07 V}\\ \text{and }{E}^{\text{o}}{}_{\text{anode}}\text{ = 0}\text{.53 V}\end{array}$

The standard cell potential of the cell is given by the expression shown below:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Substitute the values:

$\begin{array}{c}{E}^o{}_{cell}\text{ = 1}\text{.07 V }-\text{0}\text{.53 V}\\ \text{ = 0}\text{.54 V}\end{array}$

Now, the relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given by the following expression:

$\begin{array}{l}{E}^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}\\ \Rightarrow {\text{ K = 10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Here, the number of exchange of electrons is two;

Thus, the value of $n\text{ = 2}$.

Substitute the values of $n$ and $E_{cell}^0$ in the above expression and solve,

$\begin{array}{l}{\text{K = 10}}^{\frac{\text{2}\times \text{ 0}\text{.54 V}}{0.0592}}\\ \Rightarrow {\text{ K = 10}}^{\text{18}\text{.2}}=\text{ 2 }\times {\text{ 10}}^{\text{18}}\end{array}$

Thus, the equilibrium constant of this reaction is $2\times {\text{ 10}}^{\text{18}}$.

b) $2{\text{Ce}}^{4+}(aq){\text{ + 2Cl}}^{-}\text{(}aq\text{) }\rightleftharpoons {\text{Cl}}_2\text{(}g\text{)}+{\text{2Ce}}^{\text{3+}}\text{(}aq\text{) }$

This reaction can be written as two half-reactions, as shown below:

$\begin{array}{l}\text{At cathode (reduction):}\\ {\text{2Ce}}^{\text{4+}}(aq)+{\text{2e}}^{-}\to 2{\text{Ce}}^{3+}(aq)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ {\text{2Cl}}^{-}\text{(}aq\text{) }\to {\text{ Cl}}_{\text{2}}(g){\text{ + 2e}}^{-}\end{array}$

Fromtable 19.1, the reduction potential of cerium is $\text{1}\text{.61 V}$ and the reduction potential of chlorine is $\text{1}\text{.36 V}$.

$\begin{array}{l}{E}^o{}_{cathode}\text{= 1}\text{.61 V}\\ E^{\text{o}}{}_{\text{anode}}\text{ = 1}\text{.36 V}\end{array}$

Now, the standard cell potential of a cell is given by the expression shown below:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Substitute the values of half-cell potentials to the above expression,

$\begin{array}{c}{E}^o{}_{cell}=\text{1}\text{.61 V }-\text{1}\text{.36 V}\\ \text{ = 0}\text{.25 V}\end{array}$

Now, the relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given by the following expression:

$\begin{array}{l}{E}^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}\\ \Rightarrow {\text{ K = 10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Here, the number of exchange of electrons is two;

Thus, the value of $n\text{ = 2}$.

Substitute the values of $n$ and cell potential in the above expression and solve,

$\begin{array}{l}{\text{K = 10}}^{\frac{\text{2}\times \text{ 0}\text{.25 V}}{0.0592}}\\ \Rightarrow {\text{ K = 10}}^{\text{8}\text{.44}}=\text{ 3 }\times {\text{ 10}}^{\text{8}}\end{array}$

Thus, the equilibrium constant of this reaction is $\text{3 }\times {\text{ 10}}^{\text{8}}$.

c)

$\begin{array}{l}5{\text{Fe}}^{\text{2+}}(aq){\text{ + MnO}}_4{}^{-}\text{(}aq{\text{) + 8H}}^{\text{+}}\text{(}aq\text{)}\rightleftharpoons {\text{Mn}}^{2+}\text{(}aq{\text{) + 4H}}_{\text{2}}\text{O(}l{\text{) + 5Fe}}^{\text{3+}}\text{(}aq\text{)}\\ \end{array}$

This reaction can be written as two halves reactions as follows:

$\begin{array}{l}\text{At cathode (reduction):}\\ {\text{MnO}}_4{}^{-}(aq){\text{ + 8 H}}^{\text{+}}\text{(}aq{\text{) + 5e}}^{-}\to {\text{Mn}}^{\text{2+}}(aq)+4{\text{H}}_{\text{2}}\text{O}(l)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ {\text{5Fe}}^{\text{2+}}\text{(}aq\text{) }\to {\text{ 5Fe}}^{\text{3+}}\text{(}aq{\text{) + 5e}}^{-}\end{array}$

From table 19.1, the reduction potential of manganese is $\text{1}\text{.51 V}$ and the reduction potential of iron is $\text{0}\text{.77 V}$.

$\begin{array}{l}{E}^o{}_{cathode}\text{= 1}\text{.51 V}\\ E^{\text{o}}{}_{anode}\text{ = 0}\text{.77 V}\end{array}$

Now, the standard cell potential of a cell is given by the expression shown below:

$E^0{}_{cell}={\text{E}}^{\text{0}}{}_{cathode}-{\text{E}}^0{}_{anode}$

Substitute the values of half-cell potential in the above expression,

$\begin{array}{c}{E}^o{}_{cell}\text{ = 1}\text{.51 V }-\text{ 0}\text{.77 V}\\ \text{ = 0}\text{.74 V}\end{array}$

Now, the relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given by the following expression:

$\begin{array}{l}{E}^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}\\ \Rightarrow {\text{ K = 10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Here, the number of exchange of electrons is five;

Thus, the value of $n\text{ = 5}$.

Substitute the values of $n$ and standard cell potential in the above expression.

$\begin{array}{l}{\text{K = 10}}^{\frac{5\times \text{ 0}\text{.74 V}}{0.0592}}\\ \Rightarrow {\text{ K = 10}}^{62.5}=\text{ 3 }\times {\text{ 10}}^{62}\end{array}$

Thus, the equilibrium constant of this reaction is $\text{3 }\times {\text{ 10}}^{62}$.