Chapter 10, Problem 10.6P

(a)

To determine

The reactions that occur at the two electrode.

Explanation of Solution

Introduction:

The voltage developed in an electrochemical cell is directly related to the Gibbs free energy change of the reaction. The standard-electrode potentials are determined under equilibrium conditions at room temperature $25°\text{C}$ and 1 atmosphere of pressure. The reduction reactions occurs at the cathode and the oxidation reaction at the anode.

The copper reduction reaction is more electropositive than the iron reduction reaction.

Copper deposits on the copper electrode in accordance with the reduction reaction.

The copper reduction reaction or the reaction at cathode is given by,

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

The oxidation reaction at the anode is given by,

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

Conclusion:

Thus, the reaction at the cathode is ${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu}$ and at anode is $\text{Fe}\to {\text{Fe}}^{2+}+2{\text{e}}^{-}$.

(b)

To determine

The voltage developed in the cell.

Answer to Problem 10.6P

The voltage developed in the cell is $0.87\text{V}$.

Explanation of Solution

Given:

The mass of copper in the solution is $635.4\text{g}$.

The volume of solution is $1\times {10}^{-3}{\text{m}}^3$.

The mass of iron is $0.5585\text{g}$.

Formula Used:

Write the expression for the concentration of copper.

  $C_{\text{Cu}}=\frac{m_{\text{Cu}}}{63.54}$....... (I)

Here, $C_{\text{Cu}}$ is the concentration of copper and $m_{\text{Cu}}$ is mass of copper ion in the solution.

Write the expression for the concentration of iron.

  $C_{\text{Fe}}=\frac{m_{\text{Fe}}}{55.85}$...... (II)

Here, $C_{\text{Fe}}$ is the concentration of iron and $m_{\text{Fe}}$ is mass of iron ion in the solution.

Write the expression for the half-cell voltage at the copper electrode.

  $V_{\text{Cu}}={\left(V_0\right)}_{\text{Cu}}+\frac{0.0592}{Z}\log C_{\text{Cu}}$...... (III)

Here, $V_{\text{Cu}}$ is the half-cell voltage at the copper electrode and ${\left(V_0\right)}_{\text{Cu}}$ is the voltage at copper electrode with $1\text{-molar}\text{solution}$.

Write the expression for the half-cell voltage at the iron electrode.

  $V_{\text{Fe}}={\left(V_0\right)}_{\text{Fe}}+\frac{0.0592}{Z}\log C_{\text{Fe}}$...... (IV)

Here, $V_{\text{Fe}}$ is the half-cell voltage at the iron electrode and ${\left(V_0\right)}_{\text{Fe}}$ is the voltage at iron electrode with $1\text{-molar}\text{solution}$.

Write the expression for the voltage developed in the cell.

  $V=V_{\text{Cu}}+V_{\text{Fe}}$...... (V)

Here, $V$ is the voltage developed in the cell.

Calculation:

Substitute $635.4\text{g}$ for $m_{\text{Cu}}$ in equation (I).

  $\begin{array}{c}{C}_{\text{Cu}}=\frac{635.4}{63.54}\\ =10\end{array}$

Substitute $0.5585\text{g}$ for $m_{\text{Fe}}$ in equation (II).

  $\begin{array}{c}{C}_{\text{Fe}}=\frac{0.5585}{55.85}\\ =0.01\end{array}$

Substitute $0.340\text{V}$ for ${\left(V_0\right)}_{\text{Cu}}$ , $2$ for $Z$ and $10$ for $C_{\text{Cu}}$ in equation (III).

  $\begin{array}{c}{V}_{\text{Cu}}=\left(0.340\text{V}\right)+\frac{0.0592}{2}\log \left(10\right)\\ =\left(0.340+0.0296\right)\text{V}\\ =\text{0}\text{.3696}\text{V}\end{array}$

Substitute $-0.44\text{V}$ for ${\left(V_0\right)}_{\text{Fe}}$ , $2$ for $Z$ and $0.01$ for $C_{\text{Cu}}$ in equation (IV).

  $\begin{array}{c}{V}_{\text{Fe}}=\left(-0.44\text{V}\right)+\frac{0.0592}{2}\log \left(0.01\right)\\ =\left(-0.44-0.0592\right)\text{V}\\ =-0.4992\text{V}\end{array}$

Here, negative sign indicate the voltage at anode.

Substitute $0.3696\text{V}$ for $V_{\text{Cu}}$ and $0.4992\text{V}$ for $V_{\text{Fe}}$ in equation (V).

  $\begin{array}{c}V=0.3696\text{V}+0.4992\text{V}\\ =0.8688\text{V}\\ =\text{0}\text{.87}\text{V}\end{array}$

Conclusion:

Therefore, the voltage developed in the cell is $0.87\text{V}$.