Chapter 16, Problem 16.3P

(a)

Interpretation Introduction

Interpretation:

Balanced titration reaction between ${\text{Cu}}^{\text{+}}$ and ${\text{Ce}}^{\text{4+}}$ has to be written.

Concept Introduction:

Titration is an analytical method of quantitative chemical analysis which determines the concentration of the analyte. A reagent called titrant is prepared as a standard solution. This titrant reacts with titrand (analyte) solution to determine the concentration of the analyte. The volume of titrant required for titration is termed as the titration volume.

Explanation of Solution

In this reaction, ${\text{Ce}}^{\text{4+}}$, acts as an oxidizing agent. It oxidizes ${\text{Cu}}^{\text{+}}$ to ${\text{Cu}}^{\text{2+}}$ and itself reduced to ${\text{Ce}}^{\text{3+}}$. Balanced chemical reaction between ${\text{Cu}}^{\text{+}}$ and ${\text{Ce}}^{\text{4+}}$ can be written as follows:

    ${\text{Ce}}^{4+}+{\text{Cu}}^{+}\to {\text{Ce}}^{3+}+{\text{Cu}}^{2+}$

(b)

Interpretation Introduction

Interpretation:

Two half-reactions for the indicator electrode have to be written.

Concept Introduction:

A redox reaction is a type of reaction in which oxidation state of all of the species changes. This type of reaction is actually characterized by the transfer of electron from one species to another.

Explanation of Solution

Complete cell reaction can be classified into oxidation half cell reaction and reduction half cell reaction. In oxidation half, cell oxidation occurs and in reduction half cell reduction occurs. For this case, two half-reactions for the indicator electrode can be written as follows:

Oxidation half cell reaction is as follows:

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

Reduction half cell reaction is as follows:

  ${\text{Ce}}^{4+}+{\text{e}}^{-}\to {\text{Ce}}^{3+}$

(c)

Interpretation Introduction

Interpretation:

Two Nernst equations for the cell voltage have to be written.

Concept Introduction:

Nernst equation in electrochemistry is the equation that shows relation between the reduction potential of a reaction to the standard reduction potential value of a particular species.

Consider a hypothetical reaction as follows:

    $\text{Ox}+{\text{ne}}^{-}\to \text{Red}$

For this reaction Nernst equation can be written as follows:

    ${\text{E}}_{\text{cell}}={\text{E}}^0-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{a}}_{\text{Red}}}{{\text{a}}_{\text{Ox}}}$

Here, ${\text{E}}_{\text{cell}}$ is the cell potential or electromotive force at that temperature of interest.

${\text{E}}^{\text{0}}$ is standard reduction potential.

$\text{R}$ is universal gas constant.

$\text{T}$ is temperature of interest in Kelvin scale.

$\text{n}$ is number of electron transfer.

$\text{F}$ is Faraday constant.

${\text{a}}_{\text{red}}$ is activity of the reduced form.

${\text{a}}_{\text{ox}}$ is activity of the oxidized form.

Explanation of Solution

Here oxidation half cell reaction is written as follows:

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

Hence Nernst equation for oxidation half cell at $\text{25}{}^{\text{o}}\text{C}$ can be written as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Cu}}^{\text{+}}{\text{/Cu}}^{\text{2+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =\text{0}\text{.161}\text{V}-0.05916{\log }_{10}\frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}-0.197\text{V}\\ =-0.03\text{6}\text{V}-0.05916{\log }_{10}\frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\end{array}$

Here, $\left[{\text{Cu}}^{\text{+}}\right]$ is the concentration of ${\text{Cu}}^{\text{+}}$.

$\left[{\text{Cu}}^{\text{2+}}\right]$ is concentration of ${\text{Cu}}^{\text{2+}}$.

Reduction half cell reaction is written as follows:

    ${\text{Ce}}^{4+}+{\text{e}}^{-}\to {\text{Ce}}^{3+}$

Hence Nernst equation for reduction half cell at $\text{25}{}^{\text{o}}\text{C}$ can be written as follows:

    $\begin{array}{c} {\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Ce}}^{\text{3+}}{\text{/Ce}}^{\text{4+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =1.70\text{V}-0.05916{\log }_{10}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}-0.197\text{V}\\ =1.503\text{V}-0.05916{\log }_{10}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\end{array}$

Here, $\left[{\text{Ce}}^{\text{3+}}\right]$ is the concentration of ${\text{Ce}}^{\text{3+}}$.

$\left[{\text{Ce}}^{\text{4+}}\right]$ is concentration of ${\text{Ce}}^{\text{4+}}$.

(d)

Interpretation Introduction

Interpretation:

Electrode potential at $\text{1}\text{.0}\text{mL}$, $\text{12}\text{.5}\text{mL}$, $\text{24}\text{.5}\text{mL}$, $\text{25}\text{.0}\text{mL}$, $\text{25}\text{.5}\text{mL}$, $\text{30}\text{.0}\text{mL}$, and $\text{50}\text{.0}\text{mL}$ volumes for ${\text{Cu}}^{\text{+}}$ have to be calculated. Titration curve has to be sketched.

Concept Introduction:

Refer to part (c).

Explanation of Solution

Here titration value of titrant at equivalence point is calculated as follows:

    $\begin{array}{c}\text{Volume}\text{of}{\text{Cu}}^{\text{+}}=\frac{\left(\text{Volume}\text{of}{\text{Ce}}^{4+}\right)\left(\text{Concentration}\text{of}{\text{Ce}}^{\text{4+}}\right)}{\left(\text{Concentration}\text{of}{\text{Cu}}^{\text{+}}\right)}\\ =\frac{\left(\text{100}\text{.0}\text{mL}\right)\left(\text{0}\text{.010}\text{M}\right)}{\left(\text{0}\text{.040}\text{M}\right)}\\ =25\text{mL}\end{array}$

Here titration value of titrant at equivalence point is $\text{25}\text{.0}\text{mL}$. Hence when $\text{10}\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Ce}}^{\text{3+}}{\text{/Ce}}^{\text{4+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =1.70\text{V}-0.05916{\log }_{10}\frac{\text{1}\text{.0}\text{mL}}{\left(25.0-1.00\right)\text{mL}}-0.197\text{V}\\ =1.58\text{V}\end{array}$

When $12.5\text{mL}{\text{Cu}}^{\text{+}}$  is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Ce}}^{\text{3+}}{\text{/Ce}}^{\text{4+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =1.70\text{V}-0.05916{\log }_{10}\frac{\text{12}\text{.5}\text{mL}}{\left(25.0-12.5\right)\text{mL}}-0.197\text{V}\\ =1.50\text{V}\end{array}$

When $24.5\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Ce}}^{\text{3+}}{\text{/Ce}}^{\text{4+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =1.70\text{V}-0.05916{\log }_{10}\frac{\text{24}\text{.5}\text{mL}}{\left(25.0-24.5\right)\text{mL}}-0.197\text{V}\\ =1.40\text{V}\end{array}$

When $25\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, then concentration of all the species will be equal. Hence cell potential is calculated as follows:

    $\begin{array}{c}\text{E}=\frac{{\text{E}}_{{\text{Ce}}^{\text{3+}}{\text{/Ce}}^{\text{4+}}}^{\text{0}}+{\text{ E}}_{{\text{Cu}}^{\text{+}}{\text{/Cu}}^{\text{2+}}}^{\text{0}}}{2}-{\text{E}}_{\text{Calomel}}\\ =\frac{1.70\text{V}+0.161\text{V}}{2}-0.197\text{V}\\ =\text{0}\text{.7335}\text{V}\end{array}$

When $25.5\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Cu}}^{\text{+}}{\text{/Cu}}^{\text{2+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =0.161\text{V}-0.05916{\log }_{10}\frac{\left(25.5-25.0\right)\text{mL}}{\left(25.0\right)\text{mL}}-0.197\text{V}\\ =0.065\text{V}\end{array}$

When $30\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Cu}}^{\text{+}}{\text{/Cu}}^{\text{2+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =0.161\text{V}-0.05916{\log }_{10}\frac{\left(30-25.0\right)\text{mL}}{\left(25.0\right)\text{mL}}-0.197\text{V}\\ =0.005\text{V}\end{array}$

When $50\text{mL}$ ${\text{Cu}}^{\text{+}}$ is added, the cell potential is calculated as follows:

    $\begin{array}{c}{\text{E}}_{\text{Cell}}={\text{E}}_{{\text{Cu}}^{\text{+}}{\text{/Cu}}^{\text{2+}}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\ln \frac{\left[{\text{Cu}}^{\text{+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}-{\text{E}}_{\text{Ag}|\text{AgCl}}\\ =0.161\text{V}-0.05916{\log }_{10}\frac{\left(50.0-25.0\right)\text{mL}}{\left(25.0\right)\text{mL}}-0.197\text{V}\\ =-0.036\text{V}\end{array}$

The titration curve is as follows:

(e)

Interpretation Introduction

Interpretation:

Suitable indicator for the redox titration has to be chosen.

Concept Introduction:

Titration is an analytical method of quantitative chemical analysis which determines the concentration of the analyte. A reagent called titrant is prepared as a standard solution. This titrant reacts with titrand (analyte) solution to determine the concentration of the analyte. The volume of titrant required for titration is termed as the titration volume.

A redox reaction is a type of reaction in which oxidation state of all of the species changes. This type of reaction is actually characterized by the transfer of electron from one species to another.

A redox indicator is a chemical species that undergoes sharp color change at a specific electrode potential. Mostly redox indicators are organic compounds.

Explanation of Solution

Complete cell reaction can be classified into oxidation half cell reaction and reduction half cell reaction. In oxidation half cell oxidation occurs and in reduction half cell reduction occurs. For this case, two half-reactions for the indicator electrode can be written as follows:

Oxidation half cell reaction is as follows:

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

Reduction half cell reaction is as follows:

  ${\text{Ce}}^{4+}+{\text{e}}^{-}\to {\text{Ce}}^{3+}$

Equivalence point potential for this reaction is $\text{0}\text{.7335}\text{V}$, hence suitable redox indicator for this reaction should be diphenylbenzidine sulfonic acid (violet to colorless) or diphenylamine sulfonic acid (red-violet to colorless).