Chapter 14, Problem 14.13P

(a)

Interpretation Introduction

Interpretation:

Reduction recations corresponding to each half cell along with potential and overall cell voltage have to be calculated.

Concept Introduction:

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

Reduction half cell reaction in anodic half cell is as follows:

  ${\text{Br}}_{\text{2}}\left(l\right)+2{\text{e}}^{-}\rightleftharpoons 2{\text{Br}}^{-}$

Reduction half cell in cathodic-half cell is as follows:

  ${\text{Al}}^{3+}+3{\text{e}}^{-}\rightleftharpoons \text{Al}\left(s\right)$

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

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

Substitute $\text{1}\text{.078 V}$ for $E_{\text{cell}}^{°}$, 2 for $n$, ${\left[{\text{Br}}^{-}\right]}^2$ for $\left[P\right]$ and 1 for $\left[R\right]$ in equation (1) to obtain $E_{\text{cell}}$ for anodic half cell.

  $E_{\text{cell}}=\text{1}\text{.078 V}-\frac{0.0592}{2}\log {\left[{\text{Br}}^{-}\right]}^2$        (2)

Substitute $0.10\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in equation (2).

  $\begin{array}{c}{E}_{\text{cell}}=\text{1}\text{.078 V}-\frac{0.0592}{2}\log {\left(0.10\text{ M}\right)}^2\\ =1.1372\text{ V}\end{array}$

Substitute $-\text{1}\text{.677 V}$ for $E_{\text{cell}}^{°}$, 3 for $n$, $\left[{\text{Al}}^{3+}\right]$ for $\left[R\right]$ and 1 for $\left[P\right]$ in equation (1) to obtain $E_{\text{cell}}$ for cathodic half cell.

  $E_{\text{cell}}=-\text{1}\text{.677 V}-\frac{0.0592}{3}\log \frac{1}{\left[{\text{Al}}^{3+}\right]}$        (3)

Substitute $0.010\text{ M}$ for $\left[{\text{Al}}^{3+}\right]$ in equation (3).

  $\begin{array}{c}{E}_{\text{cell}}=-\text{1}\text{.677 V}-\frac{0.0592}{3}\log \left(\frac{1}{0.010\text{ M}}\right)\\ =-1.716\text{ V}\end{array}$

Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{right}}-E_{\text{left}}$        (4)

Substitute $-1.716\text{ V}$ for $E_{\text{right}}$ and $1.1372\text{ V}$ for $E_{\text{left}}$ in equation (4).

  $\begin{array}{c}{E}_{\text{cell}}^{°}=-1.716\text{ V}-\left(1.1372\text{ V}\right)\\ =-2.854\text{ V}\end{array}$

Hence net cell voltage is $-2.854\text{ V}$.

(b)

Interpretation Introduction

Interpretation:

Cell-diagram and net cell reaction in spontaneous direction has to be given.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Cell-diagram is illustrated as follows:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode. Thus electrons move spontaneously from $\text{Al}$ towards $\text{Pt}$ electrode.

Net cell reaction in spontaneous direction is as follows:

  $\frac{3}{2}{\text{Br}}_{\text{2}}\left(l\right)+\text{Al}\left(s\right)\rightleftharpoons 3{\text{Br}}^{-}+{\text{Al}}^{3+}$

(c)

Interpretation Introduction

Interpretation:

Limiting reagent has to be identified.

Concept Introduction:

Reagent present in less quantity governs the overall reaction as it is consumed first. Such reagent is termed Limiting reagent.

Explanation of Solution

Moles of ${\text{Br}}_2$ in $\text{14}\text{.3 mL}$ are calculated as follows:

  $\begin{array}{c}{\text{Moles of Br}}_2=\left(\frac{3.12\text{ g}}{1\text{ mL}}\right)\left(\text{14}\text{.3 mL}\right)\left(\frac{1\text{ mol}}{159.808\text{ g}}\right)\\ =0.2791\text{ mol}\end{array}$

Moles of $\text{Al}$ are calculated as follows:

  $\begin{array}{c}\text{Moles of Al}=\left(\frac{12\text{ g}}{1\text{ s}}\right)\left(\frac{1\text{ mol}}{\text{26}\text{.98 g}}\right)\\ =0.4447\text{ mol}\end{array}$

Since ${\text{Br}}_2$ is present in lesser amount it acts as limiting reagent as it would be consumed rapidly.

(d)

Interpretation Introduction

Interpretation:

Electrical work due to $0.231\text{ mL}$ of ${\text{Br}}_2$ has to be calculated.

Concept Introduction:

Work done is evaluated by expression as follows:

  $W=V\cdot Q$

Here,

$Q$ denotes charge .

$V$ denotes voltage.

$W$ denotes electrical work done.

Explanation of Solution

Moles of ${\text{Br}}_2$ in $0.231\text{ mL}$ are calculated as follows:

  $\begin{array}{c}{\text{Moles of Br}}_2=\left(\frac{3.12\text{ g}}{1\text{ mL}}\right)\left(0.231\text{ mL}\right)\left(\frac{1\text{ mol}}{159.808\text{ g}}\right)\\ =0.004507\text{ mol}\end{array}$

Work done in terms of Faradays is evaluated by expression as follows:

  $W=V\cdot n\cdot Z\cdot F$        (5)

Substitute $0.004507\text{ mol}$ for $n$, 3 for $Z$, $9.649\times {10}^4\text{ C/mol}$ for $F$ and $\text{1}\text{.50 V}$ for $V$ in equation (5).

  $\begin{array}{c}W=\left(\text{1}\text{.50 V}\right)\left(0.004507\text{ mol}\right)\left(3\right)\left(9.649\times {10}^4\text{ C/mol}\right)\\ =1.149\times {10}^3\text{ J}\end{array}$

Hence, electrical work due to $0.231\text{ mL}$ of ${\text{Br}}_2$ is $1.149\times {10}^3\text{ J}$.

(e)

Interpretation Introduction

Interpretation:

Rate of dissolution of $\text{Al}\left(s\right)$ has to be calculated.

Concept Introduction:

On faraday is equivalent to flow of $9.649\times {10}^4\text{ C}$ charge per mol.

Explanation of Solution

Since $2.89\times {10}^{-4}\text{C/s}$ flows and one faraday constitutes $9.649\times {10}^4\text{ C/mol}$ thus amount of electron that flow per hour is calculated as follows:

  $\begin{array}{c}{\text{Flow rate for 3e}}^{-}=\frac{1}{3}\left(\frac{2.89\times {10}^{-4}\text{C}}{1\text{ s}}\right)\left(\frac{1\text{ mol}}{9.649\times {10}^4\text{ C}}\right)\\ =9.98\times {10}^{-10}\text{ mol/s}\end{array}$

Since $\text{1 mol Al}$ has mass of $\text{26}\text{.98 g/mol}$ so rate in terms of $\text{g/s}$ is calculated as follows:

  $\begin{array}{c}\text{Flow rate}\left(\text{g/s}\right)=\left(\frac{9.98\times {10}^{-10}\text{ mol}}{1\text{ s}}\right)\left(\frac{\text{26}\text{.98 g}}{1\text{ mol}}\right)\\ =2.69\times {10}^{-8}\text{ g/s}\end{array}$

Hence, rate of dissolution of $\text{Al}\left(s\right)$ is $2.69\times {10}^{-8}\text{ g/s}$.