Chapter 16, Problem 16.1P

(a)

Interpretation Introduction

Interpretation:

Consider the titration given in the figure, balanced titration reaction has to be written.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 16.1P

The balanced titration reaction is,

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

Explanation of Solution

Figure 1 shows the titration of iron (II) with cerium (IV).   $\text{100}\text{.0 mL}$ of $\text{0}\text{.0500M}{\text{Fe}}^{\text{2+}}$ with $\text{0}\text{.100M}{\text{Ce}}^{\text{4+}}$ in $\text{1M}{\text{HClO}}_{\text{4}}$ .  In this reaction, each Ceric ion ( ${\text{Ce}}^{\text{4+}}$) reduces one mole of ferrous ion frequently.  Thus balanced titration reaction is,

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

(b)

Interpretation Introduction

Interpretation:

Consider the titration given in figure, two different half-reaction for the indicator electrode has to be written.

Concept introduction:

Half reaction:

The reaction component in which either the oxidation or reduction reaction (redox reaction) takes place.  A half-reaction is formed by change in oxidation state of specific substance involved in redox reaction.

Answer to Problem 16.1P

The half-reaction for this indicator is,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}{\text{E}}^{\text{o}}\text{=0}\text{.767}\text{V}\\ \\ {\text{Ce}}^{\text{4+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Ce}}^{\text{3+}}{\text{E}}^{\text{o}}\text{= 1}\text{.70}\text{V}\end{array}$

Explanation of Solution

In the Pt indicator electrode, there are two reactions come to the equilibrium,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}{\text{E}}^{\text{o}}\text{=0}\text{.767}\text{V}\\ \\ {\text{Ce}}^{\text{4+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Ce}}^{\text{3+}}{\text{E}}^{\text{o}}\text{= 1}\text{.70}\text{V}\end{array}$

Ferrous reduced to ferric and cerium reduced to ceric.

(c)

Interpretation Introduction

Interpretation:

Consider the titration given in the figure, two different Nernst equation for cell voltage has to be written.

Concept introduction:

Nernst equation is,

${\text{E = E}}_{\text{o}}\text{-}\frac{\text{RT}}{\text{nF}}\text{ln}\text{Q}$

Where,

$\begin{array}{l}{\text{E}}_{\text{o}}\text{-Standard potential of the cell}\\ \\ \text{R - Gas constant (8}\text{.314J/K}\text{.Mol)}\\ \\ \text{T - Absolute temperature (Kelvin)}\\ \\ \text{n - Number of electrons}\\ \\ \text{F- Faraday's constant (96485 C/mole)}\\ \\ \text{Q - Reactant quotient}\end{array}$

Answer to Problem 16.1P

$\text{E=}\left\{\text{0}\text{.767-0}\text{.05916log}\frac{\left[{\text{Fe}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]}\right\}-\left\{0.241\right\}$

$\text{E=}\left\{\text{1}\text{.70-0}\text{.05916log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\right\}-\left\{0.241\right\}$

Explanation of Solution

Nernst equation for the cell voltage

The two indicator reactions are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}{\text{E}}^{\text{o}}\text{=0}\text{.767}\text{V}\\ \\ {\text{Ce}}^{\text{4+}}{\text{+ e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Ce}}^{\text{3+}}{\text{E}}^{\text{o}}\text{= 1}\text{.70}\text{V}\end{array}$

We should write Nernst equation for these reactions.

${\text{ln (oxidized) + ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{ln (reduced)}$

${\text{E = E}}_{\text{o}}\text{-}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\left(\frac{\left[\text{ln (reduced)}\right]}{\left[\text{ln (oxidized)}\right]}\right)$

So,

$\text{E=}\left\{\text{0}\text{.767-0}\text{.05916log}\frac{\left[{\text{Fe}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]}\right\}-\left\{0.241\right\}$

Formal potential for reduction of ${\text{Fe}}^{\text{3+}}$ in ${\text{1M HClO}}_{\text{4}}$ is $0.767$V.

Potential o saturated calomel electrode is $0.241V$.

$\text{E=}\left\{\text{1}\text{.70-0}\text{.05916log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\right\}-\left\{0.241\right\}$

(d)

Interpretation Introduction

Interpretation:

Consider the titration given in figure, value of E has to be calculated for given volume.

Concept introduction:

Electrode potential (E): The electromotive force between two electrodes called electrode potential.  Cell consists of two electrode, one is standard electrode (such as calomel electrode and standard hydrogen electrode) and another one is given electrode.

${\text{E = E}}_{\text{+}}{\text{+ E}}_{\text{-}}$

Where,

${\text{E}}_{\text{+}}$ is the potential of electrode connected to the positive terminal

${\text{E}}_{\text{-}}$ is the potential of electrode connected to the negative terminal

Answer to Problem 16.1P

$\begin{array}{l}\text{10}\text{.0 mL:}\text{E= 0}\text{.490 V}\\ \\ \text{25}\text{.0 mL:E= 0}\text{.526 V}\\ \\ \text{49}\text{.0 mL: E= 0}\text{.626V}\\ \\ \text{50}\text{.0 mL: E= 0}\text{.99V}\\ \\ \text{51}\text{.0 mL: E= 1}\text{.36V}\\ \\ \text{60}\text{.0 mL: E= 1}\text{.42V}\\ \\ \text{100}\text{.0 mL: E= 0}\text{.46V}\end{array}$

Explanation of Solution

Titration:

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

As each part of ${\text{Ce}}^{4+}$ is added it consumes ${\text{Ce}}^{4+}$ and forms an equal number of moles of ${\text{Ce}}^{3+}$ and ${\text{Fe}}^{3+}$.  Earlier to the equivalence point, extra unreacted ${\text{Fe}}^{2+}$ rests in the solution.

The equivalence point comes at $\text{50}\text{.0 mL}$.

Before the equivalence point,

$\begin{array}{l}{\text{E = E}}_{\text{+}}{\text{+E}}_{\text{-}}\\ \\ ={\text{E}}_{\text{+}}\text{+}E\left(calomal\right)\\ \\ \text{=}\left(\text{0}\text{.767 - 0}\text{.05916 log}\frac{\left[{\text{Fe}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]}\right)-0.241\\ \end{array}$

$\left(\text{0}\text{.526 - 0}\text{.05916 log}\frac{\left[{\text{Fe}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]}\right)\mathrm{......}\left(1\right)$

At $\text{10}\text{.0 mL}$:

This is the method to equivalence point. So, $10.0/50.0$ of the iron in the form of ${\text{Fe}}^{\text{3+}}$ and $40.0/50.0$ is the form of ${\text{Fe}}^{\text{2+}}$. Substituting this value in equation (1),

$\begin{array}{l}\text{=}\left(\text{0}\text{.526 - 0}\text{.05916 log}\frac{\left(\text{40/50}\right)}{\left(\text{10/50}\right)}\right)\\ \\ \text{=}\text{0}\text{.526 - 0}\text{.05916 log}\left(4.0\right)\\ \\ \text{=}\text{0}\text{.526 - 0}\text{.05916}\left(0.6020\right)\\ \\ \text{=}\text{0}\text{.526 - (0}\text{.0356)}\\ \\ \text{=}0.490V\\ \end{array}$

At $\text{25}\text{.0 mL}$:

$25.0/50.0$ of the iron in the form of ${\text{Fe}}^{\text{3+}}$ and $25.0/50.0$ is the form of ${\text{Fe}}^{\text{2+}}$.  Substituting this value in equation (1),

$\begin{array}{l}\text{=}\left(\text{0}\text{.526 - 0}\text{.05916 log}\frac{\left(\text{25/50}\right)}{\left(\text{25/50}\right)}\right)\\ \\ \text{=}\text{0}\text{.526 - 0}\text{.05916 log}\left(1\right)\\ \\ \text{=}\text{0}\text{.526 - 0}\text{.05916}\left(0\right)\\ \\ \text{=}0.526V\\ \end{array}$

At $\text{49}\text{.0 mL}$:

$49.0/50.0$ of the iron in the form of ${\text{Fe}}^{\text{3+}}$ and $1.0/50.0$ is the form of ${\text{Fe}}^{\text{2+}}$. Substituting this value in equation (1),

$\begin{array}{l}\text{=}\left(\text{0}\text{.767 - 0}\text{.05916 log}\frac{\left(\text{1}\text{.0/50}\right)}{\left(\text{49}\text{.0/50}\right)}\right)\text{-0}\text{.241}\\ \\ \text{=}\left(\text{0}\text{.767 - 0}\text{.05916 log}\left(\text{0}\text{.0204}\right)\right)\text{-0}\text{.241}\\ \\ \text{=}\left(\text{0}\text{.767 - 0}\text{.05916}\left(\text{-1}\text{.6903}\right)\right)\text{-0}\text{.241}\\ \\ \text{=}\left(\text{0}\text{.767 -(-0}\text{.1000)}\right)\text{-0}\text{.241}\\ \\ \text{= 0}\text{.867-0}\text{.241}\\ \\ \text{= }0.626V\end{array}$

At $\text{50}\text{.0 mL}$:

This is equivalence point $\left({\text{V}}_{\text{e}}\right)$, where $\left[{\text{Ce}}^{\text{3+}}\right]$ is equal to $\left[{\text{Fe}}^{\text{3+}}\right]$ and $\left[{\text{Ce}}^{\text{4+}}\right]$ is equal to $\left[{\text{Fe}}^{\text{2+}}\right]$.  For this reaction the Nernst equation is,

$\begin{array}{l}{\text{E}}_{+}\text{ = 0}\text{.767-0}\text{.05916log}\frac{\left[{\text{Fe}}^{\text{2+}}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]}\mathrm{......}\left(1\right)\\ \end{array}$

${\text{E}}_{+}\text{=1}\text{.70-0}\text{.05916log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\mathrm{......}(2)$

Adding these two voltage,

${\text{2E}}_{\text{+}}\text{= 2}\text{.467-0}\text{.05916}\text{log}\left(\frac{\left[{\text{Fe}}^{\text{2+}}\right]\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Fe}}^{\text{2+}}\right]\left[{\text{Ce}}^{\text{4+}}\right]}\right)$

$\left[{\text{Ce}}^{\text{3+}}\right]$= $\left[{\text{Fe}}^{\text{3+}}\right]$ and $\left[{\text{Ce}}^{\text{4+}}\right]$= $\left[{\text{Fe}}^{\text{2+}}\right]$, at the equivalence point, the ratio of concentration is unity. So, the logarithm is zero and

$\begin{array}{l}{\text{2E}}_{\text{+}}\text{= 2}\text{.467 V}\\ \\ {\text{E}}_{\text{+}}\text{=1}\text{.23}\text{V}\end{array}$

The cell voltage is,

$\begin{array}{l}{\text{E = E}}_{\text{+}}\text{-E}\left(\text{Calomel}\right)\\ \\ \text{=1}\text{.23-0}\text{.241= 0}\text{.99V}\end{array}$

At $\text{51}\text{.0 mL}$:

After the equivalence point, the formula is used to calculate voltage of the cell is,

$\begin{array}{l}\text{E=}\left[\text{1}\text{.70-0}\text{.05916 log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\right]\text{-0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70-0}\text{.05916 log}\frac{\text{50}\text{.0}}{\text{1}\text{.0}}\right]\text{- 0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70 - 0}\text{.05916×}\left(\text{1}\text{.6989}\right)\right]\text{- 0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70 - 0}\text{.1005}\right]\text{- 0}\text{.241}\\ \\ \text{= -1}\text{.5995 - 0}\text{.241}\\ \\ \text{=}1.36V\end{array}$

At $\text{60}\text{.0 mL}$:

$\begin{array}{l}\text{E=}\left[\text{1}\text{.70-0}\text{.05916 log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\right]\text{-0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70-0}\text{.05916 log}\frac{\text{50}\text{.0}}{\text{10}\text{.0}}\right]\text{- 0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70 - 0}\text{.05916×}\left(\text{0}\text{.6989}\right)\right]\text{- 0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70 - 0}\text{.0413}\right]\text{- 0}\text{.241}\\ \\ \text{=1}\text{.6587 - 0}\text{.241}\\ \\ \text{= }1.42V\\ \\ \text{=1}\text{.36}\text{V}\end{array}$

At $\text{100}\text{.0 mL}$:

$\begin{array}{l}\text{E=}\left[\text{1}\text{.70 - 0}\text{.05916 log}\frac{\left[{\text{Ce}}^{\text{3+}}\right]}{\left[{\text{Ce}}^{\text{4+}}\right]}\right]\text{-0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70-0}\text{.05916 log}\frac{\text{50}\text{.0}}{\text{50}\text{.0}}\right]\text{- 0}\text{.241}\\ \\ \text{=}\left[\text{1}\text{.70 -}\left(\text{0}\right)\right]\text{- 0}\text{.241}\\ \\ \text{= }1.46V\end{array}$