Chapter 19, Problem 19.57QP

(a)

Interpretation Introduction

To determine: The standard cell potential $\left(E^{\text{ο}}\right)$ for the given overall reaction.

Interpretation: The standard cell potential $\left(E^{\text{ο}}\right)$ for the given overall reaction is to be calculated.

Concept introduction: The potential in an electrochemical cell is defined as a measure of energy per unit charge that is available from the oxidation and reductions reactions to carry out the reaction. The values standard electrode potential is obtained with reference to a standard hydrogen electrode.

Answer to Problem 19.57QP

The value of standard cell potential $\left(E^{\text{ο}}\right)$ for the overall reaction is $\underset{\_}{-0.27V}$.

Explanation of Solution

The given reaction is,

${\text{NH}}_4^{+}\left(aq\right)+2{\text{O}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{NO}}_3^{-}\left(aq\right)+{\text{2H}}^{+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The reduction half cell reaction for the given overall reaction is,

$2{\text{O}}_{\text{2}}\left(g\right)+{\text{8H}}^{+}\left(aq\right)+8{\text{e}}^{-}\stackrel{}{\to }{\text{4H}}_{\text{2}}\text{O}\left(l\right)E_{\text{cathode}}^{\text{ο}}=1.23\text{V}$ (1)

The oxidation half cell reaction for the given overall reaction is,

${\text{NH}}_4^{+}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{NO}}_3^{-}\left(aq\right)+10{\text{H}}^{+}\left(aq\right)+8{\text{e}}^{-}{E}_{\text{anode}}^{\text{ο}}=1.50\text{V}$ (2)

The standard cell potential $\left(E^{\text{ο}}\right)$ for the given overall reaction is calculated by the formula,

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

Substitute the values of $E_{\text{cathode}}^{\text{ο}}$ and $E_{\text{anode}}^{\text{ο}}$ from equations (1) and (2) in the above formula to calculate the value of standard cell potential $\left(E^{\text{ο}}\right)$ for the overall reaction.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=1.23\text{V}-1.50\text{V}\\ =\underset{\_}{-0.27V}\end{array}$

Hence, the value of standard cell potential $\left(E^{\text{ο}}\right)$ for the overall reaction is $\underset{\_}{-0.27V}$.

(b)

Interpretation Introduction

To determine: The ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ when the reaction is at equilibrium at $298\text{K}$.

Interpretation: The ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ when the reaction is at equilibrium at $298\text{K}$ is to be calculated.

Concept introduction: Nernst equation is defined as an equation that shows the relationship between the standard electrode potential of an electrochemical cell to the reduction potential, temperature and the concentrations of the chemical substances that undergoes oxidation and reduction.

The Nernst equation in electrochemistry is,

$E=E^{\text{ο}}-\frac{RT}{nF}\ln \frac{\left[\text{Red}\right]}{\left[\text{Ox}\right]}$

Answer to Problem 19.57QP

The ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ is $\underset{\_}{1.27\times 10^{-24}:1}$.

Explanation of Solution

The $\text{pH}$ of the reaction is $7.00$.

The partial pressure of oxygen gas is $0.21\text{atm}$.

The temperature of the reaction is $298\text{K}$.

The relation between $E_{\text{cell}}^{\text{ο}}$ and equilibrium constant is,

$nF{E}_{\text{cell}}^{\text{ο}}=RT\ln K$

Where,

• $n$ is the number of electrons involved in a chemical reaction.
• $F$ is the faradays constant $\left(96485\text{C}{\text{mol}}^{-1}\right)$.
• $R$ is the universal gas constant.
• $T$ is the temperature.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.
• $K$ is the equilibrium constant.

The above equation is rearranged to calculate the value of equilibrium constant at $298\text{K}$ as,

$\log K=\frac{n\times E_{\text{cell}}^{\text{ο}}}{0.0591}$

Substitute the values $n$ and $E_{\text{cell}}^{\text{ο}}$ in the above equation to calculate the value of equilibrium constant.

$\begin{array}{c}\log K=\frac{8\times -0.27\text{V}}{0.0591}\\ \log K=\frac{2.16}{0.0591}\\ \log K=-36.54\\ K={10}^{-36.54}\end{array}$

Simplify the above equation,

$K=2.88\times {10}^{-37}$

The concentration of ${\text{H}}^{+}$ ions is calculated by the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\text{pH}$ in the above formula to calculate the concentration of ${\text{H}}^{+}$ ions.

$\begin{array}{c}7.00=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]={10}^{-7.00}\\ =1\times {10}^{-7}\text{M}\end{array}$

The ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ is calculated by the formula,

$K=\frac{\left[{\text{NO}}_3^{-}\right]{\left[{\text{H}}^{+}\right]}^2}{\left[{\text{NH}}_4^{+}\right]{\left[{\text{P}}_{{\text{O}}_{\text{2}}}\right]}^2}$

Substitute the values of $K$, concentration of ${\text{H}}^{+}$ ions and partial pressure of oxygen gas in the above formula to calculate the ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$.

$\begin{array}{c}2.88\times {10}^{-37}=\frac{\left[{\text{NO}}_3^{-}\right]{\left(1\times {10}^{-7}\text{M}\right)}^2}{\left[{\text{NH}}_4^{+}\right]{\left(0.21\text{atm}\right)}^2}\\ \frac{\left[{\text{NO}}_3^{-}\right]}{\left[{\text{NH}}_4^{+}\right]}=\frac{2.88\times {10}^{-37}\times {\left(0.21\text{atm}\right)}^2}{{\left(1\times {10}^{-7}\text{M}\right)}^2}\\ \frac{\left[{\text{NO}}_3^{-}\right]}{\left[{\text{NH}}_4^{+}\right]}=\frac{1.270\times {10}^{-38}}{1\times {10}^{-14}}\\ \frac{\left[{\text{NO}}_3^{-}\right]}{\left[{\text{NH}}_4^{+}\right]}=\underset{\_}{1.27\times 10^{-24}:1}\end{array}$

Hence, the ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ is $\underset{\_}{1.27\times 10^{-24}:1}$.

Conclusion

1. a. The value of standard cell potential $\left(E^{\text{ο}}\right)$ for the overall reaction is $\underset{\_}{-0.27V}$.
2. b. The ratio of $\left[{\text{NO}}_3^{-}\right]$ to $\left[{\text{NH}}_4^{+}\right]$ is $\underset{\_}{1.27\times 10^{-24}:1}$.