Chapter 11, Problem 55E

(a)

Interpretation Introduction

Interpretation:Overall cell reaction from half cell reactions of $\text{Au}$ and $\text{Tl}$ should be determined. Also, standard electrode potential of reaction is to be determined.

Concept introduction:Study of interchange between electrical and chemical energy falls under branch of chemistry called electrochemistry. It includes occurrence of oxidation and reduction reactions. This includes production of electric current from chemical reaction and vice-versa.

Reactions that occur in galvanic cells can be broken down into two half-cell reactions. One half-cell reaction is that of reduction whereas another half-cell reaction is that of oxidation.

Answer to Problem 55E

Value of $E{°}_{\text{cell}}$ is $\text{1}\text{.84 V}$ .

Explanation of Solution

The reduction half cell reaction for $\text{Au}$ is as follows:

  ${\text{Au}}^{+2}+2{\text{e}}^{-}\to \text{Au }E°=1.50\text{ V}$

The reduction half cell reaction for $\text{Tl}$ is as follows:

  ${\text{Tl}}^{+}+{\text{e}}^{-}\to \text{Tl }E°=-0.34\text{ V}$

Reduction potential is the ability of species to get reduced. More value of reduction electrode potential determines the tendency of species to get reduced. Since, reduction electrode potential of $\text{Au}$ is higher than $\text{Tl}$ . Therefore, $\text{Au}$ undergoes reduction and $\text{Tl}$ undergoes oxidation.

The half cell reaction at anode is as follows:

  $\text{Tl}\to {\text{Tl}}^{+}+{\text{e}}^{-}$

The half cell reaction at cathode is as follows:

  ${\text{Au}}^{+3}+3{\text{e}}^{-}\to \text{Au}$

To equalize number of electrons half cell reaction of $\text{Tl}$ is multiplied by 3. Hence, half cell reaction for $\text{Tl}$ at anode is as follows:

  $\text{3Tl}\to 3{\text{Tl}}^{+}+3{\text{e}}^{-}$

On addition of above two reactions overall cell reaction is obtained and is as follows:

  ${\text{Au}}^{+3}+\text{3Tl}\to \text{Au}+3{\text{Tl}}^{+}$

Expression for $E{°}_{\text{cell}}$ is as follows:

  $E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E°\left(\text{cathode}\right)$ is standard reduction potential for cathode.
• $E°\left(\text{anode}\right)$ is standard oxidation potential for anode.

Value of $E°\left(\text{cathode}\right)$ is $1.50\text{ V}$ .

Value of $E°\left(\text{anode}\right)$ is $-0.34\text{ V}$ .

Substitute the values in above equation.

  $\begin{array}{c}E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)\\ =1.50\text{ V}-\left(-0.34\text{ V}\right)\\ =\text{1}\text{.84 V}\end{array}$

Hence, value of $E{°}_{\text{cell}}$ is $\text{1}\text{.84 V}$ .

(b)

Interpretation Introduction

Interpretation:Value of $\Delta G°$ and $K$ for galvanic cell is to be calculated.

Concept introduction:Study of interchange between electrical and chemical energy falls under branch of chemistry called electrochemistry. It includes occurrence of oxidation and reduction reactions. This includes production of electric current from chemical reaction and vice-versa.

Reactions that occur in galvanic cells can be broken down into two half-cell reactions. One half-cell reaction is that of reduction whereas another half-cell reaction is that of oxidation.

Answer to Problem 55E

Value of $\Delta G°$ is $-533\text{ kJ}$ and that of $K$ is $2.52\times {10}^{93}$ .

Explanation of Solution

The reduction half cell reaction for $\text{Au}$ is as follows:

  ${\text{Au}}^{+2}+2{\text{e}}^{-}\to \text{Au }E°=1.50\text{ V}$

The reduction half cell reaction for $\text{Tl}$ is as follows:

  ${\text{Tl}}^{+}+{\text{e}}^{-}\to \text{Tl }E°=-0.34\text{ V}$

Reduction potential is the ability of species to get reduced. More value of reduction electrode potential determines the tendency of species to get reduced. Since, reduction electrode potential of $\text{Au}$ is higher than $\text{Tl}$ . Therefore, $\text{Au}$ undergoes reduction and $\text{Tl}$ undergoes oxidation.

The half cell reaction at anode is as follows:

  $\text{Tl}\to {\text{Tl}}^{+}+{\text{e}}^{-}$

The half cell reaction at cathode is as follows:

  ${\text{Au}}^{+3}+3{\text{e}}^{-}\to \text{Au}$

To equalize number of electrons half cell reaction of $\text{Tl}$ is multiplied by 3. Hence, half cell reaction for $\text{Tl}$ at anode is as follows:

  $\text{3Tl}\to 3{\text{Tl}}^{+}+3{\text{e}}^{-}$

On addition of above two reactions overall cell reaction is obtained and is as follows:

  ${\text{Au}}^{+3}+\text{3Tl}\to \text{Au}+3{\text{Tl}}^{+}$

Expression for $E{°}_{\text{cell}}$ is as follows:

  $E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E°\left(\text{cathode}\right)$ is standard reduction potential for cathode.
• $E°\left(\text{anode}\right)$ is standard oxidation potential for anode.

Value of $E°\left(\text{cathode}\right)$ is $1.50\text{ V}$ .

Value of $E°\left(\text{anode}\right)$ is $-0.34\text{ V}$ .

Substitute the values in above equation.

  $\begin{array}{c}E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)\\ =1.50\text{ V}-\left(-0.34\text{ V}\right)\\ =\text{1}\text{.84 V}\end{array}$

Expression for $\Delta G°$ is as follows:

  $\Delta G°=-nFE{°}_{\text{cell}}$

Where,

• $\Delta G°$ is standard Gibbs free energy change.
• $n$ is moles of electrons involved in reaction.
• $E{°}_{\text{cell}}$ is standard electrode potential for cell.

Value of $n$ is $3\text{ mol }{e}^{-}$ .

Value of $E{°}_{\text{cell}}$ is $\text{1}\text{.84 V}$ .

Value of $F$ is $96485\text{ C/mol }{e}^{-}$ .

Substitute the values in above equation.

  $\begin{array}{c}\Delta G°=-nFE{°}_{\text{cell}}\\ =-\left(\text{3 mol }{e}^{-}\right)\left(96485\text{ C/mol }{e}^{-}\right)\left(\text{1}\text{.84 V}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\\ =-533\text{ kJ}\end{array}$

Expression for $E{°}_{\text{cell}}$ is as follows:

  $\Delta G°=-nRT\ln K$

Where,

• $\Delta G°$ is standard Gibbs free energy change.
• $R$ isgas constant.
• $T$ is temperature of reaction.
• $K$ is equilibrium constant for reaction.

Rearrange above equation for $\ln K$ .

  $\ln K=-\frac{\Delta G°}{nRT}$

Value of $R$ is $\text{8}\text{.314 J /K mol}$ .

Value of $T$ is $298\text{ K}$ .

Value of $\Delta G°$ is $-533\text{ kJ}$ .

Value of $n$ is $1\text{ mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\ln K=-\frac{\Delta G°}{nRT}\\ =-\frac{\left(-533\text{ kJ}\right)}{\left(1\text{ mol}\right)\left(\text{8}\text{.314 J /K mol}\right)\left(298\text{ K}\right)}\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)\\ =215.13\end{array}$

Value of $K$ is calculated as follows:

  $K=2.52\times {10}^{93}$

Hence, value of $\Delta G°$ is $-533\text{ kJ}$ and that of $K$ is $2.52\times {10}^{93}$ .

(c)

Interpretation Introduction

Interpretation:Value of $E_{\text{cell}}$ at $25°\text{C}$ is to be calculated.

Concept introduction: Nernst equation represents relation between potential of electrochemical reaction, standard cell potential, activities of species and temperature.

Expression of Nernst equation at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(Q\right)$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Answer to Problem 55E

Value of $E_{\text{cell}}$ of electrochemical cell is $2.04\text{ V}$ .

Explanation of Solution

Given Information:Value of $\left[{\text{Au}}^{+3}\right]$ is $1\times {10}^{-2}M$ , that of $\left[{\text{Tl}}^{+}\right]$ is $1\times {10}^{-4}M$ and that of temperature is $25°\text{C}$ .

Given equilibrium reaction is as follows:

  ${\text{Au}}^{+3}+\text{3Tl}\to \text{Au}+3{\text{Tl}}^{+}$

Expression of reaction quotient for above chemical equation is as follows:

  $Q=\frac{{\left[{\text{Tl}}^{+}\right]}^3}{\left[{\text{Au}}^{\text{+3}}\right]}$

Where,

$\left[{\text{Tl}}^{+}\right]$ is concentration of ${\text{Tl}}^{+}$ .

  $\left[{\text{Au}}^{+3}\right]$ is concentration of ${\text{Au}}^{\text{+3}}$ .

Value of $\left[{\text{Tl}}^{+}\right]$ is $1\times {10}^{-4}M$ .

Value of $\left[{\text{Au}}^{+3}\right]$ is $1\times {10}^{-2}M$ .

Substitute the values in above equation.

  $\begin{array}{c}Q=\frac{{\left[{\text{Tl}}^{+}\right]}^3}{\left[{\text{Au}}^{\text{+3}}\right]}\\ =\frac{{\left(1\times {10}^{-4}\right)}^3}{\left(1\times {10}^{-2}\right)}\\ ={10}^{-10}\end{array}$

The half cell reaction at anode is as follows:

  $\text{3Tl}\to 3{\text{Tl}}^{+}+3{\text{e}}^{-}$

The half cell reaction at cathode is as follows:

  ${\text{Au}}^{+3}+3{\text{e}}^{-}\to \text{Au}$

On addition of above two reactions overall cell reaction is obtained and is as follows:

  ${\text{Au}}^{+3}+\text{3Tl}\to \text{Au}+3{\text{Tl}}^{+}$

Expression for $E{°}_{\text{cell}}$ is as follows:

  $E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E°\left(\text{cathode}\right)$ is standard reduction potential for cathode.
• $E°\left(\text{anode}\right)$ is standard oxidation potential for anode.

Value of $E°\left(\text{cathode}\right)$ is $1.50\text{ V}$ .

Value of $E°\left(\text{anode}\right)$ is $-0.34\text{ V}$ .

Substitute the values in above equation.

  $\begin{array}{c}E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)\\ =1.50\text{ V}-\left(-0.34\text{ V}\right)\\ =\text{1}\text{.84 V}\end{array}$

Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log K$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Value of $E_{\text{cell}}^0$ is $1.84\text{ V}$ .

Value of $n$ is $3$ .

Value of $Q$ is ${10}^{-10}$ .

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log K\\ =1.84-\frac{0.0591}{3}\log \left({10}^{-10}\right)\\ =1.84-0.020\left(-10\log 10\right)\\ =2.04\text{ V}\end{array}$

Hence, $E_{\text{cell}}$ of electrochemical cell is $2.04\text{ V}$ .