Chapter 11, Problem 62E

(a)

Interpretation Introduction

Interpretation:Value of $E_{\text{cell}}$ and $\Delta G$ 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.

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 62E

Value of $E_{\text{cell}}$ is $0.204\text{ V}$ and $\Delta G$ is $-3.93\times {10}^4\text{ J}$ .

Explanation of Solution

Given equilibrium reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+2{\text{e}}^{-}\to 2\text{Ag}+{\text{CrO}}_4^{2-}$

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

  $Q=\left[{\text{CrO}}_4^{2-}\right]$

Where,

• $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

  Value of $\left[{\text{CrO}}_4^{2-}\right]$ is 1.

  Substitute the values in above equation.

    $\begin{array}{c}Q=\left[{\text{CrO}}_4^{2-}\right]\\ =1\end{array}$

  The half cell reaction at anode is calomel electrode and is as follows:

    $\text{Hg}\to {\text{Hg}}^{+2}+2{\text{e}}^{-}$

  The half cell reaction at cathode is as follows:

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

  To equalize number of electrons half cell reaction of $\text{Ag}$ is multiplied by 2. Hence, half cell reaction for $\text{Ag}$ at cathode is as follows:

    ${\text{2Ag}}^{+}+2{\text{e}}^{-}\to 2\text{Ag}$

  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 $0.446\text{ V}$ .

  Value of $E°\left(\text{anode}\right)$ is $0.242\text{ V}$ .

  Substitute the values in above equation.

    $\begin{array}{c}E{°}_{\text{cell}}=E°\left(\text{cathode}\right)-E°\left(\text{anode}\right)\\ =0.446\text{ V}-0.242\text{ V}\\ =\text{0}\text{.204 V}\end{array}$

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

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

  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 $0.204\text{ V}$ .

  Value of $n$ is 2.

  Value of $Q$ is 1.

  Substitute the value in above equation.

    $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log Q\\ =0.204-\frac{0.0591}{2}\log 1\\ =0.204\end{array}$

  Hence, $E_{\text{cell}}$ of electrochemical cell is $0.204\text{ V}$ .

  Expression for $\Delta G$ is as follows:

    $\Delta G=-nF{E}_{\text{cell}}$

  Where,

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

  Value of $n$ is $2\text{ mol }{e}^{-}$ .

  Value of $E_{\text{cell}}$ is $\text{0}\text{.204 V}$ .

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

  Substitute the values in above equation.

    $\begin{array}{c}\Delta G=-nF{E}_{\text{cell}}\\ =-\left(\text{2 mol }{e}^{-}\right)\left(96485\text{ C/mol }{e}^{-}\right)\left(\text{0}\text{.204 V}\right)\\ =-39325.08\text{ CV}\left(\frac{\text{J}}{\text{CV}}\right)\\ =-3.93\times {10}^4\text{ J}\end{array}$

  Hence, value of $E_{\text{cell}}$ is $0.204\text{ V}$ and $\Delta G$ is $-3.93\times {10}^4\text{ J}$ .

(b)

Interpretation Introduction

Interpretation:Nernst equation with concentration of SCE kept constant 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.

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 62E

Nernst equation for reaction is as follows:

  $E_{\text{cell}}=0.242-\frac{0.0591}{2}\log \left[{\text{CrO}}_4^{2-}\right]$

Explanation of Solution

Given equilibrium reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+2{\text{e}}^{-}\to 2\text{Ag}+{\text{CrO}}_4^{2-}$

The half cell reaction at anode is calomel electrode and is as follows:

  $\text{Hg}\to {\text{Hg}}^{+2}+2{\text{e}}^{-}$

Since, concentration of SCE is kept constant, therefore, electrode oxidation potential of calomel is taken. Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{SCE}}^0-\frac{0.0591}{n}\log \left[{\text{CrO}}_4^{2-}\right]$

Where,

• $E_{\text{cell}}$is the cell potential.
• $E_{\text{SCE}}^0$is the standard oxidation potential of SCE.
• $n$is total electrons transferred.
• $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

Value of $E_{\text{SCE}}^0$ is $0.242\text{ V}$ .

Value of $n$ is 2.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{SCE}}^0-\frac{0.0591}{n}\log \left[{\text{CrO}}_4^{2-}\right]\\ =0.242-\frac{0.0591}{2}\log \left[{\text{CrO}}_4^{2-}\right]\end{array}$

Hence, Nernst equation for react is as follows:

  $E_{\text{cell}}=0.242-\frac{0.0591}{2}\log \left[{\text{CrO}}_4^{2-}\right]$

(c)

Interpretation Introduction

Interpretation:Cell potential of reaction if concentration of ${\text{CrO}}_4^{2-}$ is $1.00\times {10}^{-5}M$ 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.

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 62E

  $E_{\text{cell}}$ of electrochemical cell is $0.352\text{ V}$ .

Explanation of Solution

Given equilibrium reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+2{\text{e}}^{-}\to 2\text{Ag}+{\text{CrO}}_4^{2-}$

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

  $Q=\left[{\text{CrO}}_4^{2-}\right]$

Where,

• $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

Value of $\left[{\text{CrO}}_4^{2-}\right]$ is $1.00\times {10}^{-5}$ .

Substitute the values in above equation.

  $\begin{array}{c}Q=\left[{\text{CrO}}_4^{2-}\right]\\ =1.00\times {10}^{-5}\end{array}$

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

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

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 $0.242\text{ V}$ .

Value of $n$ is 2.

Value of $Q$ is $1.00\times {10}^{-5}$ .

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log Q\\ =0.242-\frac{0.0591}{2}\log \left[1.00\times {10}^{-5}\right]\\ =0.352\end{array}$

Hence, $E_{\text{cell}}$ of electrochemical cell is $0.352\text{ V}$ .

(d)

Interpretation Introduction

Interpretation:Concentration of ${\text{CrO}}_4^{2-}$ for unknown solution 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.

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 62E

Value of $\left[{\text{CrO}}_4^{2-}\right]$ is $1.35\times {10}^{-9}\text{ mol/L}$ .

Explanation of Solution

Given equilibrium reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+2{\text{e}}^{-}\to 2\text{Ag}+{\text{CrO}}_4^{2-}$

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

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left[{\text{CrO}}_4^{2-}\right]$

Rearrange above equation for $\log \left[{\text{CrO}}_4^{2-}\right]$ .

  $\log \left[{\text{CrO}}_4^{2-}\right]=\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}$

Where,

• $E_{\text{cell}}$is the cell potential.
• $E_{\text{cell}}^0$is the standard cell potential.
• $n$is total electrons transferred.
• $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

Value of $E_{\text{cell}}$ is $0.504\text{ V}$ .

Value of $E_{\text{cell}}^0$ is $0.242\text{ V}$ .

Value of $n$ is 2.

Substitute the value in above equation.

  $\begin{array}{c}\log \left[{\text{CrO}}_4^{2-}\right]=\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}\\ =\frac{2\left(0.242\text{ V}-0.504\text{ V}\right)}{0.0591}\\ =-8.87\end{array}$

Value of $\left[{\text{CrO}}_4^{2-}\right]$ is as follows:

  $\left[{\text{CrO}}_4^{2-}\right]=1.35\times {10}^{-9}$

Hence, value of $\left[{\text{CrO}}_4^{2-}\right]$ is $1.35\times {10}^{-9}\text{ mol/L}$ .

(d)

Interpretation Introduction

Interpretation: Solubility product of 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.

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 62E

Value of $K_{\text{sp}}$ is $1.55\times {10}^8$ .

Explanation of Solution

Given reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+2{\text{e}}^{-}\to 2\text{Ag}+{\text{CrO}}_4^{2-}$

The equilibrium of above reaction is as follows:

  ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\rightleftharpoons 2{\text{Ag}}^{+}\left(aq\right)+{\text{CrO}}_4^{2-}\left(aq\right)$

At equilibrium cell potential is zero. Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2$

Rearrange above equation for $\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2$ .

  $\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2={10}^{\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}}$

Where,

• $E_{\text{cell}}$is the cell potential.
• $E_{\text{cell}}^0$is the standard cell potential.
• $n$is total electrons transferred.
• $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

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

Value of $E_{\text{cell}}^0$ is $0.242\text{ V}$ .

Value of $n$ is 2.

Substitute the value in above equation.

  $\begin{array}{c}\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2={10}^{\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}}\\ ={10}^{\frac{2\left(0.242-0\right)}{0.0591}}\\ =1.55\times {10}^8\end{array}$

  $K_{\text{sp}}$ of ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is as follows:

  $K_{\text{sp}}=\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2$

Where,

• $K_{\text{sp}}$ is solubility product constant.

    $\left[{\text{CrO}}_4^{2-}\right]$ is concentration of ${\text{CrO}}_4^{2-}$ .

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

  Value of $\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2$ is $1.55\times {10}^8$ .

  Substitute value in above equation.

    $\begin{array}{c}{K}_{\text{sp}}=\left[{\text{CrO}}_4^{2-}\right]{\left[{\text{Ag}}^{+}\right]}^2\\ =1.55\times {10}^8\end{array}$

  Hence, value of $K_{\text{sp}}$ is $1.55\times {10}^8$ .