Chapter 18, Problem 153CP

Consider the following galvanic cell:

A 15 0-mole sample of NH  is added to the Ag compartment (assume 1.00 L of total solution after the addition). The silver ion reacts with ammonia to form complex ions as shown:

$\begin{array}{cc}{\text{Ag}}^{+}\left(aq\right)+{\text{NH}}_3\left(aq\right)\rightleftharpoons {\text{AgNH}}_3{}^{+}\left(aq\right)& K_1=2.1\times {10}^3\\ {\text{AgNH}}_{\text{3}}{}^{+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\rightleftharpoons \text{Ag}{\left({\text{NH}}_3\right)}_2{}^{+}\left(aq\right)& K_2=8.2\times {10}^3\end{array}$

Calculate the cell potential after the addition of 15.0 moles of NH_3.

Interpretation Introduction

Interpretation: The cell potential after the addition of $15.0$ moles of ${\text{NH}}_{\text{3}}$ to the given cell compartment is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge that is the potential difference in an electrochemical cell containing two half cells. The cell potential in an electrochemical cell is measured by using Nernst equation.

To determine: The cell potential after the addition of $15.0$ moles of ${\text{NH}}_{\text{3}}$ to the $\text{Ag}$ compartment.

Answer to Problem 153CP

The cell potential after the addition of $15.0$ moles of ${\text{NH}}_{\text{3}}$ to the $\text{Ag}$ compartment is $\underset{\_}{0.64V}$.

Explanation of Solution

Explanation

The standard cell potential of the given cell is $\underset{\_}{1.20V}$.

The half cell reactions of the given cell are,

$\text{ Anode: Cd}\left(s\right)\stackrel{}{\to }{\text{Cd}}^{2+}+2{\text{e}}^{-}{E}_1{}^{\text{ο}}=-0.4\text{0V}$ (1)

$\text{Cathode: }\left({\text{Ag}}^{+}\left(aq\right)+{\text{e}}^{-}\stackrel{}{\to }\text{Ag}\left(s\right)\right)\times \text{2 }{E}_2{}^{\text{ο}}=0.80\text{V}$ (2)

Where,

• $E_1{}^{\text{ο}}$ is the standard electrode potential of equation (1).
• $E_2{}^{\text{ο}}$ is the standard electrode potential of equation (2).

The resultant cell reaction obtained by adding equation $\left(1\right)$ and $\left(2\right)$ is,

${\text{Cell: 2Ag}}^{+}\left(aq\right)+\text{Cd}\left(s\right)\stackrel{}{\to }{\text{Cd}}^{2+}\left(aq\right)+2\text{Ag}\left(s\right)$ (3)

The standard electrode potential of equation (3) is calculated by the formula,

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

0r, $E_3{}^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}$

Where,

• $E_3{}^{\text{ο}}$ is the standard electrode potential of equation (3).

Substitute the values of $E_1{}^{\text{ο}}$ and $E_2{}^{\text{ο}}$ in the above formula.

$\begin{array}{c}{E}_3{}^{\text{ο}}=0.8\text{0 V}-\left(-0.4\text{0 V}\right)\\ =1.20\text{V}\end{array}$

Hence, the standard cell potential of the given cell is $\underset{\_}{1.20V}$.

The formation constant $\left(K\right)$ for the reaction of formation of complex ion is $\underset{\_}{1.7\times 10^7}$.

The reactions for the formation of complex ion are given as,

${\text{Ag}}^{+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\stackrel{}{\to }\text{Ag}\left({\text{NH}}_{\text{3}}\right)\left(aq\right)K_1=2.1\times {10}^3$ (4)

$\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}^{+}\left(aq\right)+{\text{NH}}_{\text{3}}\left(aq\right)\stackrel{}{\to }\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{+}\left(aq\right)K_{\text{2}}=8.2\times {10}^3$ (5)

Where,

• $K_1$ is the equilibrium constant of equation (4).
• $K_2$ is the equilibrium constant of equation (5).

The resultant cell reaction obtained by adding equation $\left(1\right)$ and $\left(2\right)$ is,

${\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{\text{3}}\left(aq\right)\stackrel{}{\to }\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\left(aq\right)$ (6)

The formation constant of equation (6) is calculated by the formula,

$K=K_1\times K_2$

Where,

• $K$ is formation constant of equation (6).

Substitute the values of $K_1$ and $K_2$ in the above expression to calculate the formation constant.

$\begin{array}{c}K=K_1\times K_2\\ =\left(2.1\times {10}^3\right)\left(8.2\times {10}^3\right)\\ =\underset{\_}{1.7\times 10^7}\end{array}$

Hence, the equilibrium constant $\left(K\right)$ for the reaction of formation of complex ion is $\underset{\_}{1.7\times 10^7}$.

As the value of $\left(K\right)$ is large, the reaction goes to completion and backward reaction take place.

The equilibrium concentration of silver ion $\left[{\text{Ag}}^{\text{+}}\right]$ is $\underset{\_}{3.5\times 10^{-10}}$.

Given

Moles of ${\text{NH}}_{\text{3}}$ is $15.0$ moles

The reaction for the formation of complex ion is,

${\text{Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{\text{3}}\left(aq\right)\stackrel{}{\to }\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\left(aq\right)$

The concentration data for the complex formation reaction is,

$\begin{array}{l}{\text{ Ag}}^{\text{+}}\left(aq\right)+2{\text{NH}}_{\text{3}}\left(aq\right)\stackrel{}{\to }\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{\text{+}}\left(\text{aq}\right)\\ \text{Initial concentration}\left(\text{M}\right)\text{: }0.0\text{0 13}\text{.0 1}\text{.00}\\ \text{Change in concentration}\left(\text{M}\right):+x+x-x\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }x\text{ 13}.0+x\text{ 1}.00-x\end{array}$

The formation constant of complex ion is given by the expression,

$K=\frac{\left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_{\text{3}}\right]}^2}$

Substitute the value of equilibrium concentrations of $\left[{\text{NH}}_{\text{3}}\right]$ and $\left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{+}\right]$ in the above expression, the equilibrium concentration of silver ion $\left[{\text{Ag}}^{\text{+}}\right]$ is calculated.

$\begin{array}{l}K=\frac{\left[\text{Ag}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}^{+}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{NH}}_{\text{3}}\right]}^2}\\ 1.7\times {10}^7=\frac{1.00-x}{x{\left(13.0+2x\right)}^2}\end{array}$

Neglecting the change in concentration for ammonia and complex ion the expression becomes,

$\begin{array}{c}1.7\times {10}^7=\frac{1.00}{x{\left(13.0\right)}^2}\\ x=\underset{\_}{3.5\times 10^{-10}M}\end{array}$

So the equilibrium concentration of silver ion $\left[{\text{Ag}}^{\text{+}}\right]$ is $\underset{\_}{3.5\times 10^{-10}}$.

The cell potential is $\underset{\_}{0.64V}$.

Finally, the cell potential $\left(E\right)$ of the given cell is calculated by using the Nernst equation.

$E=E^{\text{ο}}-\frac{0.0591}{\text{n}}\log \frac{\left[{\text{Cd}}^{2+}\right]}{\left[{\text{Ag}}^{+}\right]}$.

Where,

• $E$ is the cell potential.
• $E^{\text{ο}}$ is the standard cell potential.
• n is the number of electrons in the given reaction.
• $\left[{\text{Cd}}^{\text{2}+}\right]$ is the equilibrium concentration of Cadmium ion.
• $\left[{\text{Ag}}^{+}\right]$ is the equilibrium concentration of silver ions.

Substitute the values of $E^{\text{ο}}$, n, $\left[{\text{Cd}}^{\text{2}+}\right]$ and $\left[{\text{Ag}}^{+}\right]$ in the above expression.

$\begin{array}{c}E=E^{\text{ο}}-\frac{0.0591}{\text{n}}\log \frac{\left[{\text{Cd}}^{2+}\right]}{\left[{\text{Ag}}^{+}\right]}\\ =1.2\text{0 V}-\frac{0.0591}{2}\log \frac{1.0}{\left(3.5\times {10}^{-10}\right)}\\ =\underset{\_}{0.64V}\end{array}$

Hence, the cell potential $\left(E\right)$ of the given cell is $\underset{\_}{0.64V}$.

Conclusion

Conclusion

The cell potential after the addition of $15.0$ moles of ${\text{NH}}_{\text{3}}$ to the $\text{Ag}$ compartment is $\underset{\_}{0.64V}$.