Chapter 15, Problem 15.1P

(a)

Interpretation Introduction

Interpretation:

Half-reaction for the $\text{Cu}$ electrode has to be written.

Concept Introduction: Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{oxidation}}^{°}+E_{\text{reduction}}^{°}$

Expression to compute $E_{\text{cell}}$ as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$

Here,

• $E_{\text{cell}}$ denotes overall cell potential.
• $E_{\text{cell}}^{°}$ denotes standard cell potential.
• $n$ is the moles of electrons transferred in each reaction.
• $\left[P\right]$ denotes concentration of products.
• $\left[R\right]$ denotes concentration of reactant.

Explanation of Solution

Since $\text{Cu}$ electrode is attached to postive terminal of potentiometer it serves as cathode and therefore reduction of $\text{Cu}$ occurs at this electrode.

Reduction half reaction for the $\text{Cu}$ electrode is therefore written as follows:

  ${\text{Cu}}^{2+}+2{\text{e}}^{-}\rightleftharpoons \text{Cu}\left(s\right)$

(b)

Interpretation Introduction

Interpretation:

Nernst equation corresponding to $\text{Cu}$ electrode has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression to compute $E_{\text{cell}}$ as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$        (1)

Since $\text{Cu}\left(s\right)$ is formed as product, value of $\left[\text{Cu}\right]$ is taken as unity.

Substitute 2 for $n$, $\left[{\text{Cu}}^{2+}\right]$ for $\left[R\right]$ and 1 for $\left[P\right]$ in equation (1).

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{2}\log \frac{1}{\left[{\text{Cu}}^{2+}\right]}$        (2)

(c)

Interpretation Introduction

Interpretation:

Cell voltage has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Substitute $0.339\text{ V}$ for $E_{\text{cell}}^{°}$ and $0.10\text{ M}$ for $\left[{\text{Cu}}^{2+}\right]$ in equation (2).

  $\begin{array}{c}{E}_{\text{cell}}=0.339\text{ V}-\frac{0.0592}{2}\log \left(\frac{1}{0.10\text{ M}}\right)\\ =0.309\text{ V}\end{array}$

Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{oxidation}}^{°}+E_{\text{reduction}}^{°}$        (3)

Substitute $0.339\text{ V}$ for $E_{\text{reduction}}^{°}$ and $-0.197\text{ V}$ for $E_{\text{oxidation}}^{°}$ in equation (3).

  $\begin{array}{c}{E}_{\text{cell}}^{°}=-0.197\text{ V}+0.339\text{ V}\\ =0.112\text{ V}\end{array}$

Thus, cell potential is $0.112\text{ V}$.