Chapter 8, Problem 8.35E

Interpretation Introduction

(a)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have $E$ equal to $0.050\text{V}$ is to be calculated.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Answer to Problem 8.35E

The ratio of $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Explanation of Solution

The given concentration cell is,

${\text{Fe}}^{2+}\left[\text{react}\right]\to {\text{Fe}}^{2+}\left[\text{prod}\right]$

For a concentration cell, standard electrode potential $E°$ is zero. The given electrode potential is $0.050\text{V}$.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

$E=E°-\frac{RT}{nF}\ln \text{Q}$

Where,

• $n$ is the number of electrons

• $T$ is the temperature

• $\text{Q}$ is the reaction quotient

• $F$ is the Faraday constant.

• $E°$ is the standard electrode potential.

• $E$ is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

$\begin{array}{rll}0.050& =& 0-\frac{\left(8.314\frac{\text{J}}{\text{K}\text{mol}}\right)\left(298\text{K}\right)}{\left(2\text{mol}{\text{e}}^{-}\right)\left(96485\frac{\text{C}}{\text{mol}{\text{e}}^{-}}\right)}\ln \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}\\ 0.050& =& -\frac{2477.572}{192970}\ln \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}\\ 0.050& =& -0.0128\ln \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}\\ -\frac{0.050}{0.0128}& =& \ln \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}\end{array}$

Solve for the value of ratio of $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ as shown below.

$\begin{array}{rll}\ln \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}& =& -3.91\\ \frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}& =& 0.020\end{array}$

Thus, the ratio of $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Conclusion

The ratio of $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Interpretation Introduction

(b)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have $E$ equal to $0.050\text{V}$ is to be determined.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Answer to Problem 8.35E

The ratio of $\frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.0029$.

Explanation of Solution

The given concentration cell is,

${\text{Fe}}^{3+}\left[\text{reacts}\right]\to {\text{Fe}}^{3+}\left[\text{prods}\right]$

For a concentration cell standard electrode potential $E°$ is zero. The given electrode potential is $0.050\text{V}$.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

$E=E°-\frac{RT}{nF}\ln \text{Q}$

Where,

• $n$ is the number of electrons

• $T$ is the temperature

• $\text{Q}$ is the reaction quotient

• $F$ is the Faraday constant.

• $E°$ is the standard electrode potential.

• $E$ is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

$\begin{array}{rll}0.050& =& 0-\frac{\left(8.314\frac{\text{J}}{\text{K}\text{mol}}\right)\left(298\text{K}\right)}{\left(3\text{mol}{\text{e}}^{-}\right)\left(96485\frac{\text{C}}{\text{mol}{\text{e}}^{-}}\right)}\ln \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}\\ 0.050& =& -\frac{2477.572}{289455}\ln \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}\\ 0.050& =& -0.00856\ln \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}\\ -\frac{0.050}{0.00856}& =& \ln \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}\end{array}$

Solve for the value of ratio of $\frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}$ as shown below.

$\begin{array}{rll}\ln \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}& =& -5.84\\ \frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}& =& 0.0029\end{array}$

Thus, the ratio of $\frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.0029$.

Conclusion

The ratio of $\frac{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{3+}}\right]}_{\text{react}}}$ for the given concentration cell is $0.0029$.

Interpretation Introduction

(c)

Interpretation:

The ratio of molarity of the ions in the given concentration cells required to have $E$ equal to $0.050\text{V}$ is to be determined.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentration cell.

Answer to Problem 8.35E

The ratio of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Explanation of Solution

The given concentration cell is,

${\text{Co}}^{2+}\left[\text{reacts}\right]\to {\text{Co}}^{2+}\left[\text{prods}\right]$

For a concentration cell standard electrode potential $E°$ is zero. The given electrode potential is $0.050\text{V}$.

The electrode potential at the given concentration can be calculated by using the Nernst equation as shown below.

$E=E°-\frac{RT}{nF}\ln \text{Q}$

Where,

• $n$ is the number of electrons

• $T$ is the temperature

• $\text{Q}$ is the reaction quotient

• $F$ is the Faraday constant.

• $E°$ is the standard electrode potential.

• $E$ is the electrode potential at given conditions.

Substitute the values of given potential, standard electrode potential and concentration of the species in the formula.

$\begin{array}{rll}0.050& =& 0-\frac{\left(8.314\frac{\text{J}}{\text{K}\text{mol}}\right)\left(298\text{K}\right)}{\left(2\text{mol}{\text{e}}^{-}\right)\left(96485\frac{\text{C}}{\text{mol}{\text{e}}^{-}}\right)}\ln \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}\\ 0.050& =& -\frac{2477.572}{192970}\ln \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}\\ 0.050& =& -0.0128\ln \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}\\ -\frac{0.050}{0.0128}& =& \ln \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}\end{array}$

Solve for the value of ratio of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ as shown below.

$\begin{array}{rll}\ln \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}& =& -3.91\\ \frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}& =& 0.020\end{array}$

Thus, the ratio of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Conclusion

The ratio of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ for the given concentration cell is $0.020$.

Interpretation Introduction

(d)

Interpretation:

The answers are to be compared and similarities or differences are to be stated.

Concept introduction:

The type of electrochemical cells which is comprised of two same electrodes but different concentrations or pressures is known as concentrations cells.

Answer to Problem 8.35E

The ratio of molarities is equal in case of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ and $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ as the number of electron change involved in these cells are same.

Explanation of Solution

The ratio of molarities is equal in case of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ and $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ as the number of electron change involved in these cells are the same. The concentration cells do not depend on the chemical identities of the half-cells involved.

Conclusion

The ratio of molarities is equal in case of $\frac{{\left[{\text{Co}}^{2+}\right]}_{\text{prod}}}{{\left[{\text{Co}}^{2+}\right]}_{\text{react}}}$ and $\frac{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{prod}}}{{\left[{\text{Fe}}^{\text{2+}}\right]}_{\text{react}}}$ as the number of electron change involved in these cells are same.