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.

(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.

(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.

(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.