Chapter 19, Problem 22PS

a)

Interpretation Introduction

Interpretation:

Consider the following half reactions.

$\begin{array}{cc}\text{Half reaction}& {\text{E}}^{\text{o}}\text{(V)}\\ \begin{array}{l}{\text{MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+ 8H}}^{\text{+}}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{BrO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\to {\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq)+ 14H}}^{\text{+}}{\text{ +6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{\text{(aq)+7H}}_{\text{2}}\text{O(l)}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{ +3e}}^{\text{-}}\to {\text{NO(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{SO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}& \begin{array}{l}\text{+1}\text{.51}\\ \text{+1}\text{.47}\\ \text{+1}\text{.33}\\ \text{+0}\text{.96}\\ \text{+0}\text{.20}\end{array}\end{array}$

The strongest and weakest oxidizing agent has to be determined.

Concept introduction:

Electrochemical cells:

Therese are chemical energy is converted into electrical energy.

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

${\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Answer to Problem 22PS

Strongest oxidizing agent -${\text{MnO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Weakest oxidizing agent - ${\text{SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Explanation of Solution

The given table:

$\begin{array}{cc}\text{Half reaction}& {\text{E}}^{\text{o}}\text{(V)}\\ \begin{array}{l}{\text{MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+ 8H}}^{\text{+}}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{BrO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\to {\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq)+ 14H}}^{\text{+}}{\text{ +6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{\text{(aq)+7H}}_{\text{2}}\text{O(l)}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{ +3e}}^{\text{-}}\to {\text{NO(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{SO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}& \begin{array}{l}\text{+1}\text{.51}\\ \text{+1}\text{.47}\\ \text{+1}\text{.33}\\ \text{+0}\text{.96}\\ \text{+0}\text{.20}\end{array}\end{array}$

Highest ${\text{E}}^{\text{o}}$ value elements are strongest oxidizing agents and lowest ${\text{E}}^{\text{o}}$ value elements are weakest oxidizing agents.

From the given table,

Strongest oxidizing agent -${\text{MnO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Weakest oxidizing agent - ${\text{SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

(b)

Interpretation Introduction

Interpretation:

Consider the following half reactions.

$\begin{array}{cc}\text{Half reaction}& {\text{E}}^{\text{o}}\text{(V)}\\ \begin{array}{l}{\text{MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+ 8H}}^{\text{+}}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{BrO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\to {\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq)+ 14H}}^{\text{+}}{\text{ +6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{\text{(aq)+7H}}_{\text{2}}\text{O(l)}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{ +3e}}^{\text{-}}\to {\text{NO(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{SO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}& \begin{array}{l}\text{+1}\text{.51}\\ \text{+1}\text{.47}\\ \text{+1}\text{.33}\\ \text{+0}\text{.96}\\ \text{+0}\text{.20}\end{array}\end{array}$

The oxidizing agents which are capable of oxidizing ${\text{Br}}^{\text{-}}{\text{(aq) to BrO}}_{\text{3}}^{\text{-}}\text{(aq)}$ in acid solution have to be determined.

Concept introduction:

Electrochemical cells:

Therese are chemical energy is converted into electrical energy.

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

${\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Answer to Problem 22PS

Oxidizing agents are capable of oxidizing ${\text{Br}}^{\text{-}}{\text{(aq) to BrO}}_{\text{3}}^{\text{-}}\text{(aq)}$ in acid solution - ${\text{MnO}}_{\text{4}}^{\text{-}}$

Explanation of Solution

Higher than ${\text{Br}}^{\text{-}}$ ${\text{E}}^{\text{o}}$ value elements are capable of oxidizing ${\text{Br}}^{\text{-}}{\text{(aq) to BrO}}_{\text{3}}^{\text{-}}\text{(aq)}$ in acid solution

From the given table, ${\text{MnO}}_{\text{4}}^{\text{-}}$ have higher ${\text{E}}^{\text{o}}$ values than ${\text{Br}}^{\text{-}}$.

Therefore, ${\text{MnO}}_{\text{4}}^{\text{-}}$ Oxidizing agents are capable of oxidizing ${\text{Br}}^{\text{-}}{\text{(aq) to BrO}}_{\text{3}}^{\text{-}}\text{(aq)}$ in acid solution.

(c)

Interpretation Introduction

Interpretation:

Consider the following half reactions.

$\begin{array}{cc}\text{Half reaction}& {\text{E}}^{\text{o}}\text{(V)}\\ \begin{array}{l}{\text{MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+ 8H}}^{\text{+}}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{BrO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\to {\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq)+ 14H}}^{\text{+}}{\text{ +6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{\text{(aq)+7H}}_{\text{2}}\text{O(l)}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{ +3e}}^{\text{-}}\to {\text{NO(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{SO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}& \begin{array}{l}\text{+1}\text{.51}\\ \text{+1}\text{.47}\\ \text{+1}\text{.33}\\ \text{+0}\text{.96}\\ \text{+0}\text{.20}\end{array}\end{array}$

A balanced chemical equation for the reaction of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq) with SO}}_{\text{2}}\text{(g)}$ has to be given and to find whether this reaction is product – favoured or reactant – favoured at equilibrium.

Concept introduction:

Electrochemical cells:

Therese are chemical energy is converted into electrical energy.

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

${\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Answer to Problem 22PS

${\text{E}}^{\text{0}}\text{= 1}\text{.33 V}$ The value is positive it is a product –favoured.

Explanation of Solution

A balanced chemical equation for the reaction of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq) with SO}}_{\text{2}}\text{(g)}$ is as follows.

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq) + 3SO}}_{\text{2}}{\text{(g) + 2H}}^{\text{+}}\text{(aq) }\to {\text{2Cr}}^{\text{3+}}{\text{(aq) + 3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

Let’s calculate the ${\text{E}}^{\text{o}}$ value for the above reaction:

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= 1}\text{.33 V - (+0}\text{.20 V)}\\ \text{= 1}\text{.13 V}\end{array}$

The value is negative, it is reactant –favoured.

(d)

Interpretation Introduction

Interpretation:

Consider the following half reactions.

$\begin{array}{cc}\text{Half reaction}& {\text{E}}^{\text{o}}\text{(V)}\\ \begin{array}{l}{\text{MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+ 8H}}^{\text{+}}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\\ {\text{BrO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\to {\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\\ {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq)+ 14H}}^{\text{+}}{\text{ +6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{\text{(aq)+7H}}_{\text{2}}\text{O(l)}\\ {\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{ +3e}}^{\text{-}}\to {\text{NO(g)+2H}}_{\text{2}}\text{O(l)}\\ {\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq)+ 4H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{SO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\end{array}& \begin{array}{l}\text{+1}\text{.51}\\ \text{+1}\text{.47}\\ \text{+1}\text{.33}\\ \text{+0}\text{.96}\\ \text{+0}\text{.20}\end{array}\end{array}$

A balanced chemical equation for the reaction of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq) with Mn}}^{2+}\text{(aq)}$ has to be given and to find whether this reaction is product –favoured or reactant –favoured at equilibrium.

Concept introduction:

Electrochemical cells:

Therese are chemical energy is converted into electrical energy.

In all electrochemical cells, oxidation occurs at anode and reduction occurs at cathode.

An anode is indicated by negative sign and cathode is indicated by the positive sign.

Electrons flow in the external circuit from the anode to the cathode.

In the electrochemical cells two half cells are connected with salt bridge. It allows the cations and anions to move between the two half cells.

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

${\text{E}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Answer to Problem 22PS

${\text{E}}^{\text{0}}\text{= 0}\text{.14 V}$ The value is positive, it is product–favoured.

Explanation of Solution

A balanced chemical equation for the reaction of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{(aq) with Mn}}^{2+}\text{(aq)}$ is as follows.

${\text{5Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}{\text{ (aq) + 6Mn}}^{\text{2+}}{\text{(aq) + 22H}}^{\text{+}}\text{(aq) }\to {\text{10Cr}}^{\text{3+}}{\text{(aq) + 6MnO}}_{\text{4}}^{\text{-}}{\text{(aq) + 11H}}_{\text{2}}\text{O(l)}$

Let’s calculate the ${\text{E}}^{\text{o}}$ value for the above reaction:

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= 1}\text{.47 V - 1}\text{.33 V}\\ \text{= 0}\text{.14 V}\end{array}$

The value is positive, it is product –favoured.