Chapter 6, Problem 6C.3AE

Interpretation Introduction

Interpretation: The potential of the given cell has to be estimated.

Concept introduction: To determine the activity coefficient of an ion in a dilute solution of known ionic strength, the Debye-Hückel limiting law is used.  The expression for the Debye-Hückel limiting law is,

    $\log {\gamma }_{\pm }=-A\left|z_{+}{z}_{-}\right|I^{1/2}$

The value of $A$ is $0.509$.

Where,

• $I$ is the ionic strength.
• $z_{+}$ and $z_{-}$ are the charges for the cation and anion, respectively.
• $A$ is the constant.
• ${\gamma }_{\pm }$ is the mean ionic coefficient.

Answer to Problem 6C.3AE

The potential of the given cell is $\underset{\_}{-0.6195V}$.

Explanation of Solution

The given cell representation is,

    $\text{Ag}\left(\text{s}\right)\text{|AgBr}\left(\text{s}\right)\text{|KBr}\left(\text{aq,}\text{0}\text{.050}\text{mol}{\text{kg}}^{-1}\right)\left|\right|\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq,0}\text{.010}\text{mol}{\text{kg}}^{-1}\right)\text{|Cd}\left(\text{s}\right)$

The half cell reactions are,

    $\begin{array}{l}\text{Left}\text{Hand}\text{Reaction:}\text{Ag(s)}\text{+}{\text{Br}}^{-}\left(\text{aq}\right)\to \text{AgBr}\left(\text{s}\right)+{\text{e}}^{-}\\ \text{Right}\text{Hand}\text{Reaction:}{\text{Cd}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cd}\left(\text{s}\right)\end{array}$

The cathode is the right-hand electrode and the anode is the left hand electrode.

The formula to calculate the standard potential of the cell is,

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

The value of $E_{\text{cathode}}^{\text{ο}}$ is $-0.40\text{V}$.

The value of $E_{\text{anode}}^{\text{ο}}$ is $0.0713\text{V}$.

Substitute the value of $E_{\text{cathode}}^{\text{ο}}$ and $E_{\text{anode}}^{\text{ο}}$ in the above expression,

    $\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=-0.40\text{V}-\left(0.0713\text{V}\right)\\ =-\text{0}\text{.4713}\text{V}\end{array}$

The formula to calculate the ionic strength for $\text{KBr}$ is,

    $I=\frac{1}{2}\left(b_{+}{z}_{+}^2+b_{-}{z}_{-}^2\right)$                                                                                        (1)

Where,

• $I$ is the ionic strength.
• $b_{+}$ and $b_{-}$ is the molality of the cation and the anion.
• $z_{+}$ and $z_{-}$ is the charge of the cation and anion.

The dissociation reaction of $\text{KBr}$ is shown below.

    $\text{KBr}\to {\text{K}}^{\text{+}}+{\text{Br}}^{-}$

The value of $b_{+}$ and $b_{-}$ is $0.050\text{mol}{\text{kg}}^{-1}$ and $0.050\text{mol}{\text{kg}}^{-1}$, respectively.

The value of $z_{+}$ is $+1$ and the value of $z_{-}$ is $-1$.

Substitute the value of $b_{+}$, $b_{-}$, $z_{+}$ and $z_{-}$ in equation (1).

    $\begin{array}{c}I=\frac{1}{2}\left(0.050\text{mol}{\text{kg}}^{-1}{\left(1\right)}^2+0.050\text{mol}{\text{kg}}^{-1}{\left(-1\right)}^2\right)\\ =0.050\text{mol}{\text{kg}}^{-1}\end{array}$

The formula to calculate the mean ionic activity coefficient for $\text{KBr}$ is,

    $\log {\gamma }_{\pm }=-A\left|z_{+}{z}_{-}\right|I^{1/2}$                                                                                   (2)

The value of $A$ is $0.509$.

Where,

• $I$ is the ionic strength.
• $z_{+}$ and $z_{-}$ are the charges for the cation and anion, respectively.
• $A$ is the constant.
• ${\gamma }_{\pm }$ is the mean ionic coefficient.

The value of $z_{+}$ is $+1$ and the value of $z_{-}$ is $-1$.

The value of $I$ is $0.050\text{mol}{\text{kg}}^{-1}$.

Substitute the value of $z_{+}$, $z_{-}$, $I$ and $A$ in equation (2).

    $\begin{array}{c}\log {\gamma }_{\pm }=-0.509|1\times \left(-1\right)|{\left(0.050\text{mol}{\text{kg}}^{-1}\right)}^{1/2}\\ =-0.1138\\ {\gamma }_{\pm }=e^{-0.1138}\\ =0.76\end{array}$

The formula to calculate the ionic strength for $\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is,

    $I=\frac{1}{2}\left(b_{+}{z}_{+}^2+b_{-}{z}_{-}^2\right)$                                                                                        (1)

Where,

• $I$ is the ionic strength.
• $b_{+}$ and $b_{-}$ is the molality of the cation and the anion.
• $z_{+}$ and $z_{-}$ is the charge of the cation and anion.

The dissociation reaction of $\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is shown below.

    $\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{Cd}}^{\text{2+}}+2{\text{NO}}_3^{-}$

The value of $b_{+}$ is $0.010\text{mol}{\text{kg}}^{-1}$.

The value of $b_{-}$ is $2\times 0.010\text{mol}{\text{kg}}^{-1}=0.020\text{mol}{\text{kg}}^{-1}$.

The value of $z_{+}$ is $+2$ and the value of $z_{-}$ is $-1$.

Substitute the value of $b_{+}$, $b_{-}$, $z_{+}$ and $z_{-}$ in equation (1).

    $\begin{array}{c}I=\frac{1}{2}\left(0.010\text{mol}{\text{kg}}^{-1}{\left(2\right)}^2+0.020\text{mol}{\text{kg}}^{-1}{\left(-1\right)}^2\right)\\ =0.030\text{mol}{\text{kg}}^{-1}\end{array}$

The formula to calculate the mean ionic activity coefficient for $\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is,

    $\log {\gamma }_{\pm }=-A\left|z_{+}{z}_{-}\right|I^{1/2}$                                                                                   (2)

The value of $A$ is $0.509$.

Where,

• $I$ is the ionic strength.
• $z_{+}$ and $z_{-}$ are the charges for the cation and anion, respectively.
• $A$ is the constant.
• ${\gamma }_{\pm }$ is the mean ionic coefficient.

The value of $z_{+}$ is $+2$ and the value of $z_{-}$ is $-1$.

The value of $I$ is $0.03\text{mol}{\text{kg}}^{-1}$.

Substitute the value of $z_{+}$, $z_{-}$, $I$ and $A$ in equation (2).

    $\begin{array}{c}\log {\gamma }_{\pm }=-0.509|2\times \left(-1\right)|{\left(0.03\text{mol}{\text{kg}}^{-1}\right)}^{1/2}\\ =-0.176\\ {\gamma }_{\pm }=e^{-0.176}\\ =0.66\end{array}$

The net reaction is,

    $\begin{array}{l}\text{2Ag(s)}\text{+}2{\text{Br}}^{-}\left(\text{aq}\right)\to 2\text{AgBr}\left(\text{s}\right)+2{\text{e}}^{-}\\ {\text{Cd}}^{2+}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cd}\left(\text{s}\right)\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \text{2Ag(s)}+{\text{Cd}}^{2+}\left(\text{aq}\right)+2{\text{Br}}^{-}\left(\text{aq}\right)\to \text{Cd}\left(\text{s}\right)+2\text{AgBr}\left(\text{s}\right)\end{array}$

The formula to calculate reaction quotient is,

    $Q=\frac{{\left(a_{\text{AgBr}\left(\text{s}\right)}\right)}^2\left(a_{\text{Cd}\left(\text{s}\right)}\right)}{{\left(a_{{\text{Br}}^{-}\left(\text{aq}\right)}\right)}^2{\left(a_{\text{Ag}\left(\text{s}\right)}\right)}^2\left(a_{{\text{Cd}}^{2+}\left(\text{aq}\right)}\right)}$

Where,

• $a_{\text{AgBr}\left(\text{s}\right)}$ is the activity of $\text{AgBr}\left(\text{s}\right)$.
• $a_{\text{Cd}\left(\text{s}\right)}$ is the activity of $\text{Cd}\left(\text{s}\right)$.
• $a_{{\text{Br}}^{-}\left(\text{aq}\right)}$ is the activity of ${\text{Br}}^{-}\left(\text{aq}\right)$.
• $a_{\text{Ag}\left(\text{s}\right)}$ is the activity of $\text{Ag(s)}$.
• $a_{{\text{Cd}}^{2+}\left(\text{aq}\right)}$ is the activity of ${\text{Cd}}^{2+}\left(\text{aq}\right)$.

The activities of solids are considered to be one.

Hence, the above expression becomes,

    $Q=\frac{1}{{\left(a_{{\text{Br}}^{-}\left(\text{aq}\right)}\right)}^2\left(a_{{\text{Cd}}^{2+}\left(\text{aq}\right)}\right)}$                                                                                (3)

The general formula for the activity is,

    $a_{\pm }={\gamma }_{\pm }b$

Where,

• $a_{\pm }$ is activity of the ion.
• ${\gamma }_{\pm }$ is the mean activity coefficient of the salt.
• $b$ is the molality of the ion.

Thus,

  $a_{{\text{Br}}^{-}\left(\text{aq}\right)}={\gamma }_{\text{KBr}}{b}_{-}$

The value of ${\gamma }_{\text{KBr}}$ is $0.76$.

The value of $b_{-}$ is $0.050\text{mol}{\text{kg}}^{-1}$.

Substitute the value ${\gamma }_{\text{KBr}}$ and $b_{-}$ in the above expression.

    $\begin{array}{c}{a}_{{\text{Br}}^{-}\left(\text{aq}\right)}=0.76\times 0.050\text{mol}{\text{kg}}^{-1}\\ =0.038\end{array}$

The activity coefficient of ${\text{Cd}}^{2+}\left(\text{aq}\right)$ is,

    $a_{{\text{Cd}}^{2+}\left(\text{aq}\right)}={\gamma }_{\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{b}_{+}$

The value of ${\gamma }_{\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ is $0.66$.

The value of $b_{+}$ is $0.010\text{mol}{\text{kg}}^{-1}$.

Substitute the value of ${\gamma }_{\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$ and $b_{+}$ in the above expression.

    $\begin{array}{c}{a}_{{\text{Cd}}^{2+}\left(\text{aq}\right)}=0.66\times 0.010\text{mol}{\text{kg}}^{-1}\\ =6.6\times {10}^{-3}\end{array}$

Substitute the value of $a_{{\text{Cd}}^{2+}\left(\text{aq}\right)}$ and $a_{{\text{Br}}^{-}\left(\text{aq}\right)}$ in equation (3).

    $\begin{array}{c}Q=\frac{1}{{\left(0.038\right)}^2\left(6.6\times {10}^{-3}\right)}\\ =0.105\times {10}^6\end{array}$

The cell potential at the given concentration can be calculated by using the Nernst equation as shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{RT}{nF}\ln \text{Q}$

Where,

• $n$ is the number of electrons
• $T$ is the temperature
• $\text{Q}$ is the reaction quotient
• $F$ is the Faraday constant.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.
• $E$ is the cell potential at given conditions.

The value of $R$ is $8.314{\text{JK}}^{-1}{\text{mol}}^{-1}$, $T$ is $25°\text{C}$, $E_{\text{cell}}^{\text{ο}}$ is $-0.4713\text{V}$, $F$ is $96485\text{C}{\text{mol}}^{-1}$ and $\text{Q}$ is $0.105\times {10}^6$.

The conversion of temperature from degress to Kelvin is shown below.

    $\begin{array}{c}\text{K}=°\text{C+273}\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

Substitute the values of $n$, $T$, $\text{Q}$, $F$, $E_{\text{cell}}°$, $R$ in the above formula.

    $\begin{array}{c}E=-\text{0}\text{.4713}\text{V}-\frac{\left(8.314\frac{\text{J}}{\text{K}\text{mol}}\right)\left(298\text{K}\right)}{\left(2\right)\left(96485\frac{\text{C}}{\text{mol}}\right)}\ln 0.105\times {10}^6\\ =-\text{0}\text{.4713}\text{V}-\frac{2477.572}{192970}\ln 0.105\times {10}^6\\ =-\text{0}\text{.4713}\text{V}-0.1484\text{V}\\ =\underset{\_}{-0.6195V}\end{array}$

Hence, the cell potential is $\underset{\_}{-0.6195V}$.