Chapter 22, Problem 22.5QAP

Interpretation Introduction

(a)

Interpretation:

The potential of a silver electrode in contact with solution, which is $0.0255\text{M}$ in $I_2$ and saturated with $\text{AgI}$ should be determined.

Concept introduction:

The electrode potential of the cell is the difference of potential difference of two electrodes. Here, the electrode potential of the cell is potential difference of electrode and solution.

Answer to Problem 22.5QAP

The potential of a silver electrode in contact with solution of $0.0255\text{M}$ in $I_2$ and saturated with $\text{AgI}$ is $-0.\text{0567}\text{V}$.

Explanation of Solution

The concentration of $I_2$ is $0.0255\text{M}$.

The he half cell reaction for $\text{AgI}$ is:

${\text{AgI}}_{\left(s\right)}+e^{-}\to {\text{Ag}}_{\left(s\right)}+{\text{I}}^{-}$

The initial electrode potential is $-0.151\text{V}$.

The expression for the Nernst equation for the above reaction is:

$E=E_0-\frac{0.0592}{n}\log \frac{a_{\text{P}}}{a_{\text{R}}}$ ...... (I)

Here, the initial electrode potential is $E_0$, the number of electrons is $n$, the activity of the product is $a_{\text{P}}$, and the activity of the reactant is $a_{\text{R}}$.

The number of electrons for half cell reaction is $1$ and the activity of the product is $1\text{M}$.

Substitute $1\text{M}$ for $a_{\text{P}}$, $1$ for $n$, $0.0255\text{M}$ for $a_{\text{R}}$, and $-0.151\text{V}$ for $E_0$ in Equation (I).

$\begin{array}{c}E=\left(-0.151\text{V}\right)-\left(\frac{0.0592\text{V}}{1}\right)\log \left(\frac{0.0255\text{M}}{1\text{M}}\right)\\ =\left(-0.151\text{V}\right)-\left(0.0592\text{V}\right)\left(-1.593\right)\\ =\left(-0.151\text{V}\right)+0.0943\text{V}\\ =-0.\text{0567}\text{V}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The potential of a silver electrode in contact with solution, which is $0.0035\text{M}$ in ${\text{CN}}^{-}$ and $0.0550\text{M}$ in $\text{Ag}{\left(\text{CN}\right)}_2^{-}$ should be determined.

Concept introduction:

The electrode potential of the cell is the difference of potential difference of two electrodes. Here, the electrode potential of the cell is potential difference of electrode and solution.

Answer to Problem 22.5QAP

The potential of a silver electrode in contact with solution, which is $0.0035\text{M}$ in ${\text{CN}}^{-}$ and $0.0550\text{M}$ in $\text{Ag}{\left(\text{CN}\right)}_2^{-}$ is $-0.\text{093}\text{V}$.

Explanation of Solution

The concentration in ${\text{CN}}^{-}$ is $0.0035\text{M}$ and concentration in $\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}$ is $0.0550\text{M}$.

The half cell reaction for $\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}$ is:

$\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}{}_{\left(aq\right)}+e^{-}\to {\text{Ag}}_{\left(s\right)}+2{\text{CN}}_{\left(aq\right)}^{-}$

The initial electrode potential is $-0.31\text{V}$.

Substitute $\left[\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}\right]$ for $a_{\text{R}}$ and ${\left[{\text{CN}}^{-}\right]}^2$ for $a_{\text{P}}$ in Equation (I).

$E=E_0-\frac{0.0592}{n}\log \frac{{\left[{\text{CN}}^{-}\right]}^2}{\left[\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}\right]}$ ...... (III)

The number of electrons for half cell reaction is $1$ and the activity of the product is $1\text{M}$.

Substitute $0.0035\text{M}$ for $\left[{\text{CN}}^{-}\right]$, $0.0550\text{M}$ for $\left[\text{Ag}{\left(\text{CN}\right)}_{\text{2}}^{-}\right]$, $1$ for $n$, and $-0.31\text{V}$ for $E_0$ in Equation (III).

$\begin{array}{c}E=\left(-0.31\text{V}\right)-\left(\frac{0.0592\text{V}}{1}\right)\log \left(\frac{{\left(0.0035\text{M}\right)}^2}{0.0550\text{M}}\right)\\ =\left(-0.31\text{V}\right)-\left(0.0592\text{V}\right)\left(-3.652\right)\\ =\left(-0.31\text{V}\right)+0.2162\text{V}\\ =-0.\text{093}\text{V}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The potential of a silver electrode in contact with solution, which is mixture of $25.0\text{mL}$ of $0.0500\text{M}$$\text{KBr}$ and $20.0\text{mL}$ of $0.100\text{M}$${\text{Ag}}^{+}$ should be determined.

Concept introduction:

The electrode potential of the cell is the difference of potential difference of two electrodes. Here, the electrode potential of the cell is potential difference of electrode and solution.

Answer to Problem 22.5QAP

The potential of a silver electrode in contact with solution, which is mixture of $25.0\text{mL}$ of $0.0500\text{M}$$\text{KBr}$ and $20.0\text{mL}$ of $0.100\text{M}$${\text{Ag}}^{+}$ is $0.\text{695}\text{V}$.

Explanation of Solution

The concentration in $\text{KBr}$ is $0.0500\text{M}$, the volume of $\text{KBr}$ is $25.0\text{mL}$, the volume of ${\text{Ag}}^{+}$ is $20.0\text{mL}$, and concentration in ${\text{Ag}}^{+}$ is $0.100\text{M}$.

The expression for the concentration in $\text{KBr}$ is:

$M_{\text{KBr}}=\frac{n_{{\text{Br}}^{-}}}{V_{\text{KBr}}}$ ...... (II)

Here, the solution volume of $\text{KBr}$ is $V_{\text{KBr}}$, and the number of moles of ${\text{Br}}^{-}$ is $n_{{\text{Br}}^{-}}$.

The expression for the concentration in ${\text{Ag}}^{+}$ is:

$M_{{\text{Ag}}^{+}}=\frac{n_{{\text{Ag}}^{+}}}{V_{{\text{Ag}}^{+}}}$ ...... (III)

Here, the solution volume of ${\text{Ag}}^{+}$ is $V_{{\text{Ag}}^{+}}$, and the number of moles of ${\text{Ag}}^{+}$ is $n_{{\text{Ag}}^{+}}$.

The expression for the total number of moles in after the reaction is:

$n_{\text{T}}=n_{{\text{Ag}}^{+}}-n_{{\text{Br}}^{-}}$ ...... (IV)

The expression for the total volume is:

$V_{\text{T}}=V_{\text{KBr}}+V_{{\text{Ag}}^{\text{+}}}$ ...... (V)

The expression for the concentration of $\left[{\text{Ag}}^{+}\right]$ is:

$M_{{\text{Ag}}^{+}}=\frac{n_{\text{T}}}{V_{\text{T}}}$ ...... (VI)

Substitute $\left[{\text{Ag}}^{+}\right]$ for $a_{\text{R}}$ and $1\text{M}$ for $a_{\text{P}}$ in Equation (I).

$E=E_0-\frac{0.0592}{n}\log \frac{1\text{M}}{\left[{\text{Ag}}^{+}\right]}$ ...... (VII)

The number of electrons for half cell reaction is $1$ and the activity of the product is $1\text{M}$.

The initial electrode potential of this reaction is $0.799\text{V}$.

Substitute $0.0500\text{M}$ for $M_{\text{KBr}}$ and $25.0\text{mL}$ for $V_{\text{KBr}}$ in Equation (II).

$\begin{array}{l}0.0500\text{M}=\frac{n_{{\text{Br}}^{-}}}{25.0\text{mL}}\\ n_{{\text{Br}}^{-}}=\left(0.0500\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(25.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ n_{{\text{Br}}^{-}}=\left(0.0500\text{mol}/\text{L}\right)\left(25.0\times {10}^{-3}\text{L}\right)\\ n_{{\text{Br}}^{-}}=1.25\times {10}^{-3}\text{mol}\end{array}$

Substitute $20.0\text{mL}$ for $V_{{\text{Ag}}^{+}}$ and $0.100\text{M}$ for $M_{{\text{Ag}}^{+}}$ in Equation (III).

$\begin{array}{l}0.100\text{M}=\frac{n_{{\text{Ag}}^{+}}}{20.0\text{mL}}\\ n_{{\text{Ag}}^{+}}=\left(0.100\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(20.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ n_{{\text{Ag}}^{+}}=2\times {10}^{-3}\text{mol}\end{array}$

Substitute $2\times {10}^{-3}\text{mol}$ for $n_{{\text{Ag}}^{+}}$ and $1.25\times {10}^{-3}\text{mol}$ for $n_{{\text{Br}}^{-}}$ in Equation (IV).

$\begin{array}{c}{n}_{\text{T}}=2\times {10}^{-3}\text{mol}-1.25\times {10}^{-3}\text{mol}\\ =0.75\times {10}^{-3}\text{mol}\end{array}$

Substitute $25.0\text{mL}$ for $V_{\text{KBr}}$ and $20.0\text{mL}$ for $V_{{\text{Ag}}^{+}}$ in Equation (V).

$\begin{array}{c}{V}_{\text{T}}=25.0\text{mL}+20.0\text{mL}\\ =\text{4}5.0\text{mL}\end{array}$

Substitute $0.65\times {10}^{-3}\text{mol}$ for $n_{\text{T}}$ for $\text{4}5.0\text{mL}$ for $V_{\text{T}}$ in Equation (VI).

$\begin{array}{c}{M}_{{\text{Ag}}^{+}}=\frac{0.75\times {10}^{-3}\text{mol}}{\left(\text{4}5.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)}\\ =\left(0.017\text{mol}/\text{L}\right)\left(\frac{1\text{M}}{1\text{mol}/\text{L}}\right)\\ =0.017\text{M}\end{array}$

Substitute $0.017\text{M}$ for $\left[{\text{Ag}}^{+}\right]$, $1\text{M}$ for $$

$1$ for $n$, and $0.799\text{V}$ for $E_0$ in Equation (VII).

$\begin{array}{c}E=\left(0.799\text{V}\right)-\left(\frac{0.0592\text{V}}{1}\right)\log \left(\frac{1\text{M}}{0.017\text{M}}\right)\\ =\left(0.799\text{V}\right)-\left(0.0592\text{V}\right)\left(1.77\right)\\ =\left(0.799\text{V}\right)-0.104\text{V}\\ =0.\text{695}\text{V}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The potential of a silver electrode in contact with solution, which is mixture of $25.0\text{mL}$ of $0.0500\text{M}$${\text{Ag}}^{+}$ and $20.0\text{mL}$ of $0.100\text{M}$$\text{KBr}$ should be determined.

Concept introduction:

The electrode potential of the cell is the difference of potential difference of two electrodes. Here, the electrode potential of the cell is potential difference of electrode and solution.

Answer to Problem 22.5QAP

The potential of a silver electrode in contact with solution, which is mixture of $25.0\text{mL}$ of $0.0500\text{M}$${\text{Ag}}^{+}$ and $20.0\text{mL}$ of $0.100\text{M}$$\text{KBr}$ is $0.\text{182}\text{V}$.

Explanation of Solution

The concentration in ${\text{Ag}}^{+}$ is $0.0500\text{M}$, the volume of ${\text{Ag}}^{+}$ is $25.0\text{mL}$, the volume of $\text{KBr}$ is $20.0\text{mL}$, and concentration in $\text{KBr}$ is $0.100\text{M}$.

The expression for the total number of moles in after the reaction is:

$n_{\text{T}}=n_{{\text{Br}}^{-}}-n_{{\text{Ag}}^{+}}$ ...... (VIII)

The expression for the concentration of $\left[{\text{Br}}^{-}\right]$ is:

$M_{{\text{Br}}^{-}}=\frac{n_{\text{T}}}{V_{\text{T}}}$ ...... (IX)

Substitute $\left[{\text{Br}}^{-}\right]$ for $a_{\text{R}}$ and $1\text{M}$ for $a_{\text{P}}$ in Equation (I).

$E=E_0-\frac{0.0592}{n}\log \frac{1\text{M}}{\left[{\text{Br}}^{-}\right]}$ ...... (VII)

The number of electrons for half cell reaction is $1$ and the activity of the product is $1\text{M}$.

The initial electrode potential of this reaction is $0.799\text{V}$.

Substitute $0.100\text{M}$ for $M_{\text{KBr}}$ and $20.0\text{mL}$ for $V_{\text{KBr}}$ in Equation (II).

$\begin{array}{l}0.100\text{M}=\frac{n_{{\text{Br}}^{-}}}{20.0\text{mL}}\\ n_{{\text{Br}}^{-}}=\left(0.100\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(20.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ n_{{\text{Br}}^{-}}=\left(0.100\text{mol}/\text{L}\right)\left(20.0\times {10}^{-3}\text{L}\right)\\ n_{{\text{Br}}^{-}}=2\times {10}^{-3}\text{mol}\end{array}$

Substitute $25.0\text{mL}$ for $V_{{\text{Ag}}^{+}}$ and $0.0500\text{M}$ for $M_{{\text{Ag}}^{+}}$ in Equation (III).

$\begin{array}{l}0.0500\text{M}=\frac{n_{{\text{Ag}}^{+}}}{25.0\text{mL}}\\ n_{{\text{Ag}}^{+}}=\left(0.0500\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(25.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ n_{{\text{Ag}}^{+}}=1.25\times {10}^{-3}\text{mol}\end{array}$

Substitute $1.25\times {10}^{-3}\text{mol}$ for $n_{{\text{Ag}}^{+}}$ and $2\times {10}^{-3}\text{mol}$ for $n_{{\text{Br}}^{-}}$ in Equation (VIII).

$\begin{array}{c}{n}_{\text{T}}=1.25\times {10}^{-3}\text{mol}-2\times {10}^{-3}\text{mol}\\ =0.75\times {10}^{-3}\text{mol}\end{array}$

Substitute $20.0\text{mL}$ for $V_{\text{KBr}}$ and $25.0\text{mL}$ for $V_{{\text{Ag}}^{+}}$ in Equation (V).

$\begin{array}{c}{V}_{\text{T}}=20.0\text{mL}+25.0\text{mL}\\ =\text{4}5.0\text{mL}\end{array}$

Substitute $0.65\times {10}^{-3}\text{mol}$ for $n_{\text{T}}$ for $\text{4}5.0\text{mL}$ for $V_{\text{T}}$ in Equation (VI).

$\begin{array}{c}{M}_{{\text{Br}}^{-}}=\frac{0.65\times {10}^{-3}\text{mol}}{\left(\text{4}5.0\text{mL}\right)}\\ =\frac{0.65\times {10}^{-3}\text{mol}}{\left(\text{4}5.0\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)}\\ =\left(0.014\text{mol}/\text{L}\right)\left(\frac{1\text{M}}{1\text{mol}/\text{L}}\right)\\ =0.014\text{M}\end{array}$

Substitute $0.014\text{M}$ for $\left[{\text{Br}}^{-}\right]$, $1$ for $n$, and $0.073\text{V}$ for $E_0$ in Equation (VII).

$\begin{array}{c}E=\left(0.073\text{V}\right)-\left(\frac{0.0592\text{V}}{1}\right)\log \left(\frac{0.014\text{M}}{1\text{M}}\right)\\ =\left(0.073\text{V}\right)-\left(0.0592\text{V}\right)\left(-1.853\right)\\ =\left(0.073\text{V}\right)+0.1097\text{V}\\ =0.\text{182}\text{V}\end{array}$