Chapter 11, Problem 61E

(a)

Interpretation Introduction

Interpretation:Electrode potential for below reaction is to be determined.

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 61E

Value of $E_{\text{cell}}$ for cell reaction is $2.12\text{ V}$ .

Explanation of Solution

The reduction half cell reaction at anode is as follows:

  $\text{Pb}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}$

The reduction half cell reaction at cathode is as follows:

  ${\text{PbO}}_2\left(s\right)+3{\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4^{-}\left(aq\right)+2{\text{e}}^{-}\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Overall reaction is as follows:

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{{\left[{\text{HSO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^2}\right)$

Where,

• $E_{\text{cell}}$ iselectrode potential for cell.
• $E_{\text{cell}}^{°}$ is standard electrode potential for cell.
• $n$ is total electrons transferred.
• $\left[{\text{HSO}}_4^{-}\right]$ is concentration of ${\text{HSO}}_4^{-}$ .
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ .

Value of $E{°}_{\text{cell}}$ is 2.04.

Value of $n$ is 2.

Value of $\left[{\text{HSO}}_4^{-}\right]$ is 4.5.

Value of $\left[{\text{H}}^{+}\right]$ is 4.5.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{{\left[{\text{HSO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^2}\right)\\ =2.04-\frac{0.0591}{2}\log \left(\frac{1}{{\left[4.5\right]}^2{\left[4.5\right]}^2}\right)\\ =2.04+0.077\\ =2.12\text{ V}\end{array}$

Hence, $E_{\text{cell}}$ for cell reaction is $2.12\text{ V}$ .

(b)

Interpretation Introduction

Interpretation:Value of $E°$ for below reaction at $-20°\text{C}$ is to be determined.

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 61E

Value of $E{°}_{\text{cell}}$ is $1.98\text{ V}$ .

Explanation of Solution

The formula to change temperature from $°\text{C}$ to K is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Where,

$T\left(\text{K}\right)$ is temperature in K.

  $T\left(°\text{C}\right)$ is temperature in $°\text{C}$ .

Value of $T\left(°\text{C}\right)$ is $-20°\text{C}$ .

Substitute value in above equation.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =-20+273.15\\ =253.15\text{ K}\end{array}$

Overall reaction is as follows:

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The expression to calculate $\Delta G°$ from value of $\Delta H°$ and $\Delta S°$ is as follows:

  $\Delta G°=\Delta H°-T\Delta S°$

Where,

$\Delta G°$ is change in standard gibbs free energy.

  $\Delta H°$ is change in standard enthalpy.

  $T$ is temperature in kelvin.

  $\Delta S°$ is change in standard entropy.

Value of $\Delta H°$ is $-315.9\text{ kJ}$ .

Value of $\Delta S°$ is $263.5\text{ J/K}$ .

Value of $T$ is $253.15\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}\Delta G°=\Delta H°-T\Delta S°\\ =\left(-315.9\text{ kJ}\right)-\left(253.15\text{ K}\right)\left(263.5\text{ J/K}\right)\left(\frac{{10}^{-3}\text{ kJ }}{1\text{ J}}\right)\\ =-315.9\text{ kJ}-66.70\text{ kJ}\\ =-\text{382}\text{.6 kJ}\end{array}$

Expression for $\Delta G°$ to calculate $E{°}_{\text{cell}}$ is as follows:

  $\Delta G°=-nFE{°}_{\text{cell}}$

Rearrange above equation for $E{°}_{\text{cell}}$ .

  $E{°}_{\text{cell}}=-\frac{\Delta G°}{nF}$

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 $2\text{ mol }{e}^{-}$ .

Value of $\Delta G°$ is $-\text{382}\text{.6 kJ}$ .

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

Substitute the values in above equation.

  $\begin{array}{c}E{°}_{\text{cell}}=-\frac{\Delta G°}{nF}\\ =\frac{-\left(-\text{382}\text{.6 kJ}\right)}{2\left(96485\text{ C/mol }{e}^{-}\right)}\left(\frac{{10}^{-3}\text{ J}}{1\text{ kJ}}\right)\\ =1.98\text{ V}\end{array}$

Hence, value of $E{°}_{\text{cell}}$ is $1.98\text{ V}$ .

(c)

Interpretation Introduction

Interpretation:Electrode potential for below reaction at $-20°\text{C}$ is to be determined.

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 61E

Value of $E_{\text{cell}}$ for cell reaction is $2.05\text{ V}$ .

Explanation of Solution

The formula to change temperature from $°\text{C}$ to K is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Where,

$T\left(\text{K}\right)$ is temperature in K.

  $T\left(°\text{C}\right)$ is temperature in $°\text{C}$ .

Value of $T\left(°\text{C}\right)$ is $-20°\text{C}$ .

Substitute value in above equation.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =-20+273.15\\ =253.15\text{ K}\end{array}$

The reduction half cell reaction at anode is as follows:

  $\text{Pb}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}$

The reduction half cell reaction at cathode is as follows:

  ${\text{PbO}}_2\left(s\right)+3{\text{H}}^{+}\left(aq\right)+{\text{HSO}}_4^{-}\left(aq\right)+2{\text{e}}^{-}\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Overall reaction is as follows:

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{HSO}}_4^{-}\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{RT}{nF}\log \left(\frac{1}{{\left[{\text{HSO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^2}\right)$

Where,

• $E_{\text{cell}}$ is electrode potential for cell.
• $E_{\text{cell}}^{°}$ is standard electrode potential for cell.
• $n$ is total electrons transferred.
• $R$ is universal gas constant.
• $T$ is temperature at Kelvin.
• $F$ is faraday constant.
• $\left[{\text{HSO}}_4^{-}\right]$ is concentration of ${\text{HSO}}_4^{-}$ .
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ .

Value of $E{°}_{\text{cell}}$ is 2.04.

Value of $n$ is 2.

Value of $R$ is $8.314\text{ J/K mol}$ .

Value of $T$ is $253.15\text{ K}$ .

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

Value of $\left[{\text{HSO}}_4^{-}\right]$ is 4.5.

Value of $\left[{\text{H}}^{+}\right]$ is 4.5.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{°}-\frac{RT}{nF}\log \left(\frac{1}{{\left[{\text{HSO}}_4^{-}\right]}^2{\left[{\text{H}}^{+}\right]}^2}\right)\\ =2.04-\frac{\left(8.314\text{ J/K mol}\right)\left(253.15\text{ K}\right)}{2\left(96485\text{ C/mol }{e}^{-}\right)}\log \left(\frac{1}{{\left(4.5\right)}^2{\left(4.5\right)}^2}\right)\\ =2.04-0.011\left(-2.61\right)\\ =2.05\end{array}$

Hence, $E_{\text{cell}}$ for cell reaction is $2.05\text{ V}$ .

(d)

Interpretation Introduction

Interpretation:Reason for fail of batteries in cold days than in warm days should be determined.

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

Expression of Nernst equation is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{RT}{nF}\log Q$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $R$ is universal gas constant.
• $T$ is temperature at Kelvin.
• $F$ is faraday constant.
• $Q$ is reaction quotient.

Explanation of Solution

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

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{RT}{nF}\log Q$

In accordance with above expression with increase in electrode potential or cell potential temperature decreases as cell potential is inversely related to temperature. On cold days temperature lowers down thus this increase the cell potential and hence, oil present in battery becomes viscous and current does not flow. Thus, batteries fail on colder days than on warm days.