Chapter 19, Problem 19PS

Balance each of the following unbalanced equations; then calculate the standard potential, E°, and decide whether each is product favored at equilibrium as written. (All reactions are carried out in acid solution.)

1. (a) Sn²⁺(aq) + Ag(s) → Sn(s) + Ag⁺(aq)
2. (b) Al(s) + Sn⁴⁺(aq) → Sn²⁺(aq) +Al³⁺(aq)
3. (c) ClO₃⁻(aq) + Ce³⁺(aq) → Cl₂(g) + Ce⁴⁺(aq)
4. (d) Cu(s) + NO₃⁻ (aq) → Cu²⁺(aq) + NO(g)

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined and has to decide whether it is a product favoured at equilibrium in the direction.

$\begin{array}{l}\text{a) }\text{}{\text{Sn}}^{\text{2+}}\text{(aq) + Ag(s) }\to {\text{Sn(s)+Ag}}^{\text{+}}\text{(aq)}\\ {\text{b) Al(s) + Sn}}^{\text{4+}}\text{(aq) }\to {\text{Sn}}^{\text{2+}}{\text{(aq)+Al}}^{\text{3+}}\text{(aq)}\\ {\text{c) ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + Ce}}^{\text{3+}}\to {\text{ Cl}}_{\text{2}}{\text{(g)+Ce}}^{\text{4+}}\text{(aq)}\\ {\text{d) Cu(s) + NO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{Cu}}^{\text{2+}}\text{(aq)+NO(g)}\end{array}$

Concept introduction:

Electrochemical cells:

In this type of cells, 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.

Answer to Problem 19PS

${\text{E}}_{\text{cell}}^{\text{o}}\text{= -0}\text{.94 V}$

The value is negative, therefore it is not product favoured.

Explanation of Solution

The given reaction is as follows.

$\text{}{\text{Sn}}^{\text{2+}}\text{(aq) + Ag(s) }\to {\text{Sn(s)+Ag}}^{\text{+}}\text{(aq)}$

Balance the each half reaction:

$\begin{array}{l}{\text{Sn}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to \text{Sn(s)}\\ \text{Ag(s) }\to {\text{Ag}}^{\text{+}}{\text{(aq) + e}}^{\text{-}}\end{array}$

Find the overall reaction.

$\begin{array}{l}{\text{Sn}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\to \text{Sn(s)}\\ \text{2[Ag(s) }\to {\text{Ag}}^{\text{+}}{\text{(aq) + e}}^{\text{-}}\text{]}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{Sn}}^{\text{2+}}\text{(aq)+2Ag(s) }\to {\text{Sn(s)+2Ag}}^{\text{+}}\text{(Aq)}\end{array}$

Let’s write the half reactions:

$\begin{array}{l}\text{At anode:}\\ \text{Oxidation: Ag(s)}\to {\text{ Ag}}^{+}{\text{(aq) +e}}^{\text{-}}\\ \text{At cathode:}\\ {\text{Reduction: Sn}}^{2+}{\text{(aq) + e}}^{\text{-}}\to \text{ Sn(s)}\end{array}$

Let’s calculate the ${\text{E}}_{\text{cell}}^{\text{o}}$ value.

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= (-0}\text{.14 V) + (- 0}\text{.80 V)}\\ \text{= - 0}\text{.94 V}\end{array}$

The value is negative, therefore it is not product favoured.

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined and has to decide whether it is a product favoured at equilibrium in the direction.

$\begin{array}{l}\text{a) }\text{}{\text{Sn}}^{\text{2+}}\text{(aq) + Ag(s) }\to {\text{Sn(s)+Ag}}^{\text{+}}\text{(aq)}\\ {\text{b) Al(s) + Sn}}^{\text{4+}}\text{(aq) }\to {\text{Sn}}^{\text{2+}}{\text{(aq)+Al}}^{\text{3+}}\text{(aq)}\\ {\text{c) ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + Ce}}^{\text{3+}}\to {\text{ Cl}}_{\text{2}}{\text{(g)+Ce}}^{\text{4+}}\text{(aq)}\\ {\text{d) Cu(s) + NO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{Cu}}^{\text{2+}}\text{(aq)+NO(g)}\end{array}$

Concept introduction:

Electrochemical cells:

In this type of cells, 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.

Answer to Problem 19PS

${\text{E}}_{\text{cell}}^{\text{o}}\text{= +1}\text{.81 V}$

The value is positive, therefore it is a product favoured.

Explanation of Solution

The given reaction is as follows.

${\text{Al(s) + Sn}}^{\text{4+}}\text{(aq) }\to {\text{Sn}}^{\text{2+}}{\text{(aq)+Al}}^{\text{3+}}\text{(aq)}$

Balance the each half reaction:

$\begin{array}{l}{\text{Sn}}^{\text{4+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Sn}}^{2+}\text{(s)}\\ \text{Al(s) }\to {\text{Al}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}\end{array}$

Find the overall reaction.

$\begin{array}{l}{\text{3[Sn}}^{\text{4+}}{\text{(aq) + 2e}}^{\text{-}}\to {\text{Sn}}^{2+}\text{(s)}\\ \text{2[Al(s) }\to {\text{Al}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}\text{]}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2Al(s) + 3Sn}}^{4+}\text{(s) }\to 3{\text{Sn}}^{2+}{\text{(aq) + 2Al}}^{\text{3+}}\text{(aq)}\end{array}$

Let’s write the half reactions:

$\begin{array}{l}\text{At anode:}\\ \text{Oxidation: Al(s) }\to {\text{ Al}}^{3+}{\text{(aq)+ 3e}}^{\text{-}}\\ \text{At cathode:}\\ {\text{Reduction: Sn}}^{\text{4+}}{\text{(aq)+2e}}^{\text{-}}\to {\text{ Sn}}^{2+}\text{(aq)}\end{array}$

Let’s calculate the ${\text{E}}_{\text{cell}}^{\text{o}}$ value.

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= (+1}\text{.66 V) + (+0}\text{.15 V)}\\ \text{= +1}\text{.81 V}\end{array}$

The value is positive, therefore it is a product favoured.

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined and has to decide whether it is a product favoured at equilibrium in the direction.

$\begin{array}{l}\text{a) }\text{}{\text{Sn}}^{\text{2+}}\text{(aq) + Ag(s) }\to {\text{Sn(s)+Ag}}^{\text{+}}\text{(aq)}\\ {\text{b) Al(s) + Sn}}^{\text{4+}}\text{(aq) }\to {\text{Sn}}^{\text{2+}}{\text{(aq)+Al}}^{\text{3+}}\text{(aq)}\\ {\text{c) ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + Ce}}^{\text{3+}}\to {\text{ Cl}}_{\text{2}}{\text{(g)+Ce}}^{\text{4+}}\text{(aq)}\\ {\text{d) Cu(s) + NO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{Cu}}^{\text{2+}}\text{(aq)+NO(g)}\end{array}$

Concept introduction:

Electrochemical cells:

In this type of cells,  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.

Answer to Problem 19PS

${\text{E}}_{\text{cell}}^{\text{o}}\text{= -0}\text{.14 V}$

The value is negative, therefore it is  not a product favoured

Explanation of Solution

The given reaction is as follows.

${\text{ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + Ce}}^{\text{3+}}\to {\text{ Cl}}_{\text{2}}{\text{(g)+Ce}}^{\text{4+}}\text{(aq)}$

Balance the each half reaction:

$\begin{array}{l}{\text{2ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{+}{\text{+10e}}^{\text{-}}\to {\text{Cl}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\\ {\text{Ce}}^{\text{3+}}\text{(aq) }\to {\text{Ce}}^{\text{4+}}{\text{(aq) + e}}^{\text{-}}\end{array}$

The second equation is multiply with 10 .we get equal number of electrons.

$\begin{array}{l}{\text{2ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{+}{\text{+10e}}^{\text{-}}\to {\text{Cl}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\\ {\text{10Ce}}^{\text{3+}}\text{(aq) }\to 10{\text{Ce}}^{\text{4+}}{\text{(aq) +10 e}}^{\text{-}}\end{array}$

Find the overall reaction.

$\begin{array}{l}{\text{2ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + 12H}}^{\text{+}}{\text{+10e}}^{\text{-}}\to {\text{Cl}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\\ {\text{10Ce}}^{\text{3+}}\text{(aq) }\to {\text{10Ce}}^{\text{4+}}{\text{(aq) +10 e}}^{\text{-}}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{2ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + 10Ce}}^{\text{3+}}{\text{(aq) + 12H}}^{+}\to {\text{Cl}}_{\text{2}}{\text{(g) + 10Ce}}^{\text{4+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}\end{array}$

Let’s write the half reactions:

$\begin{array}{l}\text{At anode:}\\ {\text{Oxidation: Ce}}^{3+}\text{(aq) }\to {\text{ Ce}}^{4+}{\text{(aq) + e}}^{\text{-}}\\ \text{At cathode:}\\ {\text{Reduction: 2ClO}}_3^{-}{\text{(aq)+12H}}^{+}{\text{+10e}}^{-}\to {\text{ Cl}}_2{\text{(g) + 6H}}_2\text{O(l)}\end{array}$

Let’s calculate the ${\text{E}}_{\text{cell}}^{\text{o}}$ value.

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= 1}\text{.47 V+(- 1}\text{.61 V)}\\ \text{= -0}\text{.14 V }\end{array}$

The value is negative, therefore it is not a product favoured.

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}$ value of the given reaction has to be determined and has to decide whether it is a product favoured at equilibrium in the direction.

$\begin{array}{l}\text{a) }\text{}{\text{Sn}}^{\text{2+}}\text{(aq) + Ag(s) }\to {\text{Sn(s)+Ag}}^{\text{+}}\text{(aq)}\\ {\text{b) Al(s) + Sn}}^{\text{4+}}\text{(aq) }\to {\text{Sn}}^{\text{2+}}{\text{(aq)+Al}}^{\text{3+}}\text{(aq)}\\ {\text{c) ClO}}_{\text{3}}^{\text{-}}{\text{(aq) + Ce}}^{\text{3+}}\to {\text{ Cl}}_{\text{2}}{\text{(g)+Ce}}^{\text{4+}}\text{(aq)}\\ {\text{d) Cu(s) + NO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{Cu}}^{\text{2+}}\text{(aq)+NO(g)}\end{array}$

Concept introduction:

Electrochemical cells:

In this type of cells, 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.

Answer to Problem 19PS

${\text{E}}_{\text{cell}}^{\text{o}}\text{= +0}\text{.62 V}$

The value is positive, therefore it is a product favoured.

Explanation of Solution

The given reaction is as follows.

${\text{Cu(s) + NO}}_{\text{3}}^{\text{-}}\text{(aq) }\to {\text{Cu}}^{\text{2+}}\text{(aq)+NO(g)}$

Balance the each half reaction:

$\begin{array}{l}{\text{NO}}_{\text{3}}^{\text{-}}{\text{(aq) + 4H}}^{\text{+}}{\text{+3e}}^{\text{-}}\to {\text{NO(g) + 2H}}_{\text{2}}\text{O(l)}\\ \text{Cu(s) }\to {\text{Cu}}^{\text{2+}}{\text{(aq) + 2e}}^{\text{-}}\end{array}$

The first equation is multiplied with 3 and second equation is multiply with 2.we get equal number of electrons.

$\begin{array}{l}{\text{2NO}}_{\text{3}}^{\text{-}}{\text{(aq) + 8H}}^{\text{+}}{\text{+6e}}^{\text{-}}\to 2{\text{NO(g) + 4H}}_{\text{2}}\text{O(l)}\\ \text{3Cu(s) }\to 3{\text{Cu}}^{\text{2+}}{\text{(aq) + 6e}}^{\text{-}}\end{array}$

Find the overall reaction.

$\begin{array}{l}{\text{2NO}}_{\text{3}}^{\text{-}}{\text{(aq) + 8H}}^{\text{+}}{\text{+6e}}^{\text{-}}\to {\text{2NO(g) + 4H}}_{\text{2}}\text{O(l)}\\ \text{3Cu(s) }\to {\text{3Cu}}^{\text{2+}}{\text{(aq) + 6e}}^{\text{-}}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{3Cu(s) + 2NO}}_{\text{3}}^{\text{-}}{\text{(aq) + 8H}}^{\text{+}}\to {\text{3Cu}}^{\text{2+}}{\text{(aq) + 2NO(g) + 4H}}_{\text{2}}\text{O(l)}\end{array}$

Let’s write the half reactions:

$\begin{array}{l}\text{At anode:}\\ \text{Oxidation: Cu(s) }\to {\text{ Cu}}^{\text{2+}}{\text{(aq)+ 2e}}^{\text{-}}\\ \text{At cathode:}\\ {\text{Reduction: NO}}_{\text{3}}^{\text{-}}{\text{(aq) +4H}}^{\text{+}}{\text{(g)+ 3e}}^{\text{-}}\to {\text{ NO(g) +2H}}_{\text{2}}\text{O(l)}\end{array}$

Let’s calculate the ${\text{E}}_{\text{cell}}^{\text{o}}$ value.

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}\\ \text{= (-0}\text{.734 V) + 0}\text{.96 V}\\ \text{= -0}\text{.62 V}\end{array}$

The value is negative, therefore it is not  product favoured.