Chapter 17, Problem 104QRT

(a)

Interpretation Introduction

Interpretation:

The given redox reaction are to be balanced.

Concept Introduction:

A redox reaction is a kind of reaction in which one reacting species gets oxidize and another reacting species gets reduced simultaneously.  In the redox titration, either an oxidizing agent is titrated with reducing agent or a reducing agent is titrated with an oxidizing agent.  The standard reduction potential of a galvanic cell may be calculated in terms of standard reduction potential of cathode and anode as the relation mentioned below:

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

The relation between standard cell potential $\left(E_{\text{cell}}^{°}\right)$ and standard free energy change $\left({\Delta }_{\text{r}}{G}^{\circ }\right)$ is shown below.

    ${\Delta }_{\text{r}}{G}^{\circ }\text{=}-nF{E}_{\text{cell}}^{\circ }$

If the value of Gibbs free energy change $\left({\Delta }_{\text{r}}{G}^{\circ }\right)$ is negative for a chemical reaction, then that chemical reaction is product-favored whereas the positive value indicates that the reaction is reactant-favored.

Answer to Problem 104QRT

The first and second balanced chemical reactions are shown below.

  ${\text{2NO}}_3{}^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\left(\text{aq}\right)+6\text{Hg}\left(l\right)\to 3{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

  ${\text{2Hg}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

Explanation of Solution

The given first unbalanced redox reaction is shown below.

    ${\text{NO}}_3{}^{-}\left(\text{aq}\right)+{\text{H}}^{+}\left(\text{aq}\right)+\text{Hg}\left(l\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+\text{NO}\left(\text{g}\right)+{\text{H}}_2\text{O}\left(l\right)$

The oxidation number of mercury atom in $\text{Hg}$ and ${\text{Hg}}_2{}^{2+}$ is $0$ and $+1$ respectively.  Since there is increase in oxidation number, therefore, $\text{Hg}$ is oxidized to ${\text{Hg}}_2{}^{2+}$ and $\text{Hg}$ act as reducing agent.

The oxidation number of nitrogen atom in ${\text{NO}}_3{}^{-}$ and $\text{NO}$ is $+5$ and $+2$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{NO}}_3{}^{-}$ is reduced to $\text{NO}$ and ${\text{NO}}_3{}^{-}$ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: $\text{Hg}\to {\text{Hg}}_2{}^{2+}$

Reduction: ${\text{NO}}_3{}^{-}\to \text{NO}$

The given reaction is to be balanced in acidic medium.  Therefore, oxygen atoms are balanced by adding water molecules to the deficient side.

Oxidation: $\text{2Hg}\to {\text{Hg}}_2{}^{2+}$

Reduction: ${\text{NO}}_3{}^{-}\to \text{NO}+2{\text{H}}_2\text{O}$

After balancing oxygen atoms, now hydrogen atoms are balanced by adding protons $\left({\text{H}}^{+}\right)$ to the deficient side.

Oxidation: $\text{2Hg}\to {\text{Hg}}_2{}^{2+}$

Reduction: ${\text{NO}}_3{}^{-}+4{\text{H}}^{+}\to \text{NO}+2{\text{H}}_2\text{O}$

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: $\text{2Hg}\to {\text{Hg}}_2{}^{2+}+2{\text{e}}^{-}$

Reduction: ${\text{NO}}_3{}^{-}+4{\text{H}}^{+}+3{\text{e}}^{-}\to \text{NO}+2{\text{H}}_2\text{O}$

Since the electron gain is not equivalent to electron lost.  Thus, multiply the oxidation half reaction by $3$ and reduction half reaction by $2$.

Oxidation: $\text{6Hg}\to 3{\text{Hg}}_2{}^{2+}+6{\text{e}}^{-}$

Reduction: ${\text{2NO}}_3{}^{-}+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}$

Now, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    $\text{6Hg}+{\text{2NO}}_3{}^{-}+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}+3{\text{Hg}}_2{}^{2+}+6{\text{e}}^{-}$

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  ${\text{2NO}}_3{}^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\left(\text{aq}\right)+6\text{Hg}\left(l\right)\to 3{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

Thus, the first balanced chemical reaction is shown below.

  $\begin{array}{l}{\text{2NO}}_3{}^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\left(\text{aq}\right)+6\text{Hg}\left(l\right)\to \\ 3{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)\end{array}$

The second unbalanced redox reaction is shown below.

    ${\text{Hg}}^{2+}\left(\text{aq}\right)+{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

The oxidation number of mercury atom in ${\text{Br}}^{-}$ and ${\text{Br}}_2$ is $-1$ and $0$ respectively.  Since there is increase in oxidation number, therefore, ${\text{Br}}^{-}$ is oxidized to ${\text{Br}}_2$ and ${\text{Br}}^{-}$ act as reducing agent.

The oxidation number of nitrogen atom in ${\text{Hg}}^{2+}$ and ${\text{Hg}}_2{}^{2+}$ is $+2$ and $+1$ respectively.  Since there is decrease in oxidation number, therefore, ${\text{Hg}}^{2+}$ is reduced to ${\text{Hg}}_2{}^{2+}$ and ${\text{Hg}}^{2+}$ act as oxidizing agent.

Therefore, the unbalanced half reactions are shown below.

Oxidation: ${\text{Br}}^{-}\to {\text{Br}}_2$

Reduction: $\text{Hg}\to {\text{Hg}}_2{}^{2+}$

The given reaction is to be balanced in acidic medium.  Therefore, the atoms other than oxygen and hydrogen are balanced first.

Oxidation: ${\text{2Br}}^{-}\to {\text{Br}}_2$

Reduction: $\text{2Hg}\to {\text{Hg}}_2{}^{2+}$

The charge is on both sides is balanced by adding electrons to the more positive side of the half- reaction to equal the less positive side of the half- reaction.

Oxidation: ${\text{2Br}}^{-}\to {\text{Br}}_2+2{\text{e}}^{-}$

Reduction: ${\text{2Hg}}^{2+}+2{\text{e}}^{-}\to {\text{Hg}}_2{}^{2+}$

Since, there is the equal loss and gain of electron in the above two half-cell reactions.  Therefore, the overall cell reaction is get by addition the above two half-cell reactions.

    ${\text{2Hg}}^{2+}+2{\text{e}}^{-}+{\text{2Br}}^{-}\to {\text{Hg}}_2{}^{2+}+{\text{Br}}_2+2{\text{e}}^{-}$

The simplified chemical equation after removing the chemical species of the similar kind is shown below.

  ${\text{2Hg}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

Thus, the second balanced chemical reaction is shown below.

  ${\text{2Hg}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

(b)

Interpretation Introduction

Interpretation:

The cell diagram for the given redox reactions are to be drawn.

Concept Introduction:

Same as part (a).

Answer to Problem 104QRT

The cell representation for thefirst and second redox reaction is shown below.

  $\text{Hg}\left(l\right)|{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)\Vert {\text{NO}}_3{}^{-}\left(\text{aq}\right)|\text{NO}\left(\text{g}\right)|\text{Pt}$

  $\text{Pt}|{\text{Br}}^{-}\left(\text{aq}\right)|{\text{Br}}_2\left(l\right)\Vert {\text{Hg}}^{2+}\left(\text{aq}\right)|{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)|\text{Pt}$

Explanation of Solution

The first balanced redox reactions are shown below.

  ${\text{2NO}}_3{}^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\left(\text{aq}\right)+6\text{Hg}\left(l\right)\to 3{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

The half-cell reactions which take place at cathode and anode in the first redox reaction are shown below.

  $\begin{array}{cc}\text{Cathode}:& {\text{2NO}}_3{}^{-}+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}\\ \text{Anode}:& \text{6Hg}\to 3{\text{Hg}}_2{}^{2+}+6{\text{e}}^{-}\end{array}$

Thus, the cell representation for thefirst redox reaction is shown below.

  $\text{Hg}\left(l\right)|{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)\Vert {\text{NO}}_3{}^{-}\left(\text{aq}\right)|\text{NO}\left(\text{g}\right)|\text{Pt}$

The second balanced redox reactions are shown below.

  ${\text{2Hg}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

The half-cell reactions which take place at cathode and anode in the second redox reaction are shown below.

  $\begin{array}{cc}\text{Cathode}:& {\text{2Hg}}^{2+}+2{\text{e}}^{-}\to {\text{Hg}}_2{}^{2+}\\ \text{Anode}:& {\text{2Br}}^{-}\to {\text{Br}}_2+2{\text{e}}^{-}\end{array}$

Thus, the cell representation for thesecond redox reaction is shown below.

  $\text{Pt}|{\text{Br}}^{-}\left(\text{aq}\right)|{\text{Br}}_2\left(l\right)\Vert {\text{Hg}}^{2+}\left(\text{aq}\right)|{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)|\text{Pt}$

(c)

Interpretation Introduction

Interpretation:

The standard cell potential for the given cell reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 104QRT

The standard cell potential of first and second redox reaction is $\underset{\_}{0.164V}$ and $\underset{\_}{-1.0085V}$ respectively.

Explanation of Solution

The first balanced chemical reaction is shown below.

  ${\text{2NO}}_3{}^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\left(\text{aq}\right)+6\text{Hg}\left(l\right)\to 3{\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_2\text{O}\left(l\right)$

The half-cell reactions of $\text{Hg}$ electrode and ${\text{NO}}_3{}^{-}$ along with their standard reduction potentials are shown below.

Oxidation half reaction (Anode): $\text{6Hg}\to 3{\text{Hg}}_2{}^{2+}+6{\text{e}}^{-}$    $E^{\text{o}}=0.7960\text{V}$

Reduction half reaction (Cathode): ${\text{2NO}}_3{}^{-}+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}$    $E^{\text{o}}=0.96\text{V}$

The value of $E_{\text{anode}}^{\text{o}}$ is same as that of $E_{{\text{Hg/Hg}}_2{}^{2+}}^{\text{o}}$ which is $0.7960\text{V}$.

The value of $E_{\text{cathode}}^{\text{o}}$ is same as that of $E_{{\text{NO}}_3{}^{-}\text{/NO}}^{\text{o}}$ which is $0.96\text{V}$.

The standard cell potential is calculated as shown below.

    $E_{\text{cell}}^{\text{o}}=E_{\text{cathode}}^{\text{o}}-E_{\text{anode}}^{\text{o}}$        (1)

Substitute the value of $E_{\text{cathode}}^{\text{o}}$ and $E_{\text{anode}}^{\text{o}}$ in equation (1).

    $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=E_{{\text{NO}}_3{}^{-}\text{/NO}}^{\text{o}}-E_{{\text{Hg/Hg}}_2{}^{2+}}^{\text{o}}\\ =0.96\text{V}-0.7960\text{V}\\ =0.164\text{V}\end{array}$

The second balanced redox reaction is shown below.

  ${\text{2Hg}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Hg}}_2{}^{2+}\left(\text{aq}\right)+{\text{Br}}_2\left(l\right)$

The half-cell reactions of ${\text{Hg}}^{2+}$ electrode and ${\text{Br}}^{-}$ along with their standard reduction potentials are shown below.

Oxidation half reaction (Anode): ${\text{2Br}}^{-}\to {\text{Br}}_2+2{\text{e}}^{-}$    $E^{\text{o}}=1.066\text{V}$

Reduction half reaction (Cathode): ${\text{2Hg}}^{2+}+2{\text{e}}^{-}\to {\text{Hg}}_2{}^{2+}$    $E^{\text{o}}=0.0575\text{V}$

The value of $E_{\text{anode}}^{\text{o}}$ is same as that of $E_{{\text{Br}}_2/{\text{Br}}^{-}}^{\text{o}}$ which is $1.066\text{V}$.

The value of $E_{\text{cathode}}^{\text{o}}$ is same as that of $E_{{\text{Hg}}^{2+}{\text{/Hg}}_2{}^{2+}}^{\text{o}}$ which is $0.0575\text{V}$.

Substitute the value of $E_{\text{cathode}}^{\text{o}}$ and $E_{\text{anode}}^{\text{o}}$ in equation (1).

    $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=E_{{\text{Hg}}^{2+}{\text{/Hg}}_2{}^{2+}}^{\text{o}}-E_{{\text{Br}}_2/{\text{Br}}^{-}}^{\text{o}}\\ =0.0575\text{V}-1.066\text{V}\\ =-1.0085\text{V}\end{array}$

Thus, the standard cell potential of first and second redox reaction is $\underset{\_}{0.164V}$ and $\underset{\_}{-1.0085V}$ respectively.

(d)

Interpretation Introduction

Interpretation:

The standard change in Gibbs free energy for the given redox reactions has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 104QRT

The ${\Delta }_{\text{r}}{G}^{\circ }$ for the first and second redox reaction is $\underset{\_}{94.94124kJ/mol}$ and $\underset{\_}{194.610245kJ/mol}$ respectively.

Explanation of Solution

The half-cell reactions which take place at cathode and anode in the first redox reaction are shown below.

  $\begin{array}{cc}\text{Cathode}:& {\text{2NO}}_3{}^{-}+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}+4{\text{H}}_2\text{O}\\ \text{Anode}:& \text{6Hg}\to 3{\text{Hg}}_2{}^{2+}+6{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of first redox reaction is $6$. 

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the first redox reaction is $0.164\text{V}$.

The relation between standard cell potential and standard free energy change is shown below.

    ${\Delta }_{\text{r}}{G}^{\circ }\text{=}-nF{E}_{\text{cell}}^{\circ }$        (1)

Where,

• ${\Delta }_{\text{r}}{G}^{\circ }$ is the standard free energy change.
• $E_{\text{cell}}^{\circ }$ is the standard cell potential.
• $F$ is the Faradays constant.
• $n$ is the number of electron change in the half cell reaction.

The value of $n$ is $6$.

The value of $E_{\text{cell}}^{\circ }$ is $0.164\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$ and $F$ for the first redox reaction in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\circ }\text{=}-6{\text{e}}^{-}\times \frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\times \left(0.164\text{V}\right)\\ =-\frac{94941.24\text{J}}{1\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =-94.94124\text{kJ/mol}\end{array}$

Thus, the ${\Delta }_{\text{r}}{G}^{\circ }$ for the first redox reaction is $\underset{\_}{-94.94124kJ/mol}$.

The half-cell reactions which take place at cathode and anode in the second redox reaction are shown below.

  $\begin{array}{cc}\text{Cathode}:& {\text{2Hg}}^{2+}+2{\text{e}}^{-}\to {\text{Hg}}_2{}^{2+}\\ \text{Anode}:& {\text{2Br}}^{-}\to {\text{Br}}_2+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of second redox reaction is $2$. 

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the second redox reaction is $-1.0085\text{V}$.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $-1.0085\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$ and $F$ for the second redox reaction in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\circ }\text{=}-2{\text{e}}^{-}\times \frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\times \left(-1.0085\text{V}\right)\\ =\frac{194610.245\text{J}}{1\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =194.610245\text{kJ/mol}\end{array}$

Thus, the ${\Delta }_{\text{r}}{G}^{\circ }$ for the second redox reaction is $\underset{\_}{194.610245kJ/mol}$.

(e)

Interpretation Introduction

Interpretation:

The given redox reaction are product favoured has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 104QRT

The first redox reactions is the product favored reaction.

Explanation of Solution

The calculated values of the standard cell potential $\left(E_{\text{cell}}^{°}\right)$ in the part (c), for the first and second redox reaction is $0.164\text{V}$ and $-1.0085\text{V}$ respectively.

The calculated values of the standard Gibbs free energy change $\left({\Delta }_{\text{r}}{G}^{\circ }\right)$ in the part (d), for the first and second redox reaction is $-94.94124\text{kJ/mol}$ and $194.610245\text{kJ/mol}$ respectively.

The value of Gibbs free energy change $\left({\Delta }_{\text{r}}{G}^{\circ }\right)$ for the first redox reaction is negative.  It indicates that the overall reaction is spontaneous and product favored.

The value of Gibbs free energy change $\left({\Delta }_{\text{r}}{G}^{\circ }\right)$ for the second redox reaction is positive.  It indicates that the overall reaction is non-spontaneous and reactant favored.