Chapter 15, Problem 15.6P

(a)

Interpretation Introduction

Interpretation:

Titration reaction has to be written and equivalence volume has to be calculated.

Concept Introduction:

Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{oxidation}}^{°}+E_{\text{reduction}}^{°}$

Expression to compute $E_{\text{cell}}$ as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$

Here,

• $E_{\text{cell}}$ denotes overall cell potential.
• $E_{\text{cell}}^{°}$ denotes standard cell potential.
• $n$ is the moles of electrons transferred in each reaction.
• $\left[P\right]$ denotes concentration of products.
• $\left[R\right]$ denotes concentration of reactant.

Expression to compute $E$ for silver ion electrode is given as follows:

  $E_{\text{cell}}=0.558+0.05916\log \left[{\text{Ag}}^{+}\right]$

Explanation of Solution

Titration reaction is written as follows:

  ${\text{Hg}}_2^{2+}+2{\text{Cl}}^{-}\rightleftharpoons {\text{Hg}}_{\text{2}}{\text{Cl}}_2\left(s\right)$        (1)

Dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$        (2)

Substitute  $\text{0}\text{.100 M}$ for $M_1$, $\text{0}\text{.100 M}$ for $M_2$ , $\text{50}\text{.0 mL}$ for $V_2$, in equation (2).

  $\left(\text{0}\text{.100 M}\right)V_1=\left(\text{0}\text{.100 M}\right)\left(\text{50}\text{.0 mL}\right)$        (3)

Rearrange equation (3) to calculate value of $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(\text{0}\text{.100 M}\right)\left(\text{50}\text{.0 mL}\right)}{\left(\text{0}\text{.100 M}\right)}\\ =50\text{ mL}\end{array}$

Since volume of ${\text{Hg}}_2^{2+}$ required to neutralize ${\text{Cl}}^{-}$ is half of its volume as per equation (1), so equivalence volume is calculated as follows:

  $\begin{array}{c}{V}_{\text{e}}=\frac{50\text{ mL}}{2}\\ =25\text{ mL}\end{array}$

Thus equivalence volume is $25\text{ mL}$.

(b)

Interpretation Introduction

Interpretation:

Equation to compute cell voltage has to be derived.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Electrochemical cell for mercury electrode is written as follows:

  ${\text{Hg}}_2^{2+}+2{\text{e}}^{-}\rightleftharpoons 2\text{Hg}\left(l\right)$

Expression to compute $E_{\text{cell}}$ for as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$        (4)

Since $\text{Hg}\left(l\right)$ is formed as product, value of $\left[\text{Hg}\left(l\right)\right]$ is taken as unity.

Substitute 2 for $n$, $\left[{\text{Hg}}_2^{2+}\right]$ for $\left[R\right]$ and 1 for $\left[P\right]$ in equation (4).

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{2}\log \frac{1}{\left[{\text{Hg}}_2^{2+}\right]}$        (5)

Rearrange equation (5).

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{2}\log {\left[{\text{Hg}}_2^{2+}\right]}^{-1}\\ =E_{\text{cell}}^{°}-\frac{0.0592}{2}\left(-1\right)\log \left[{\text{Hg}}_2^{2+}\right]\\ =E_{\text{cell}}^{°}+\frac{0.0592}{2}\log \left[{\text{Hg}}_2^{2+}\right]\end{array}$

Reduction potential correspoding to mercury electrode is $\text{0}\text{.796 V}$.

Cell potential is evaluated by expression as follows:

  $E_{\text{cell}}^{°}=E_{\text{right}}^{°}-E_{\text{left}}^{°}$        (6)

Substitute $\text{0}\text{.796 V}+\left(0.0592\text{/}2\right)\log \left[{\text{Hg}}_2^{2+}\right]$ for $E_{\text{right}}^{°}$ and $\text{0}\text{.241 V}$ for $E_{\text{left}}^{°}$ in equation (6).

  $\begin{array}{c}{E}_{\text{cell}}^{°}=\text{0}\text{.796 V}+\frac{0.0592}{2}\log \left[{\text{Hg}}_2^{2+}\right]-\text{0}\text{.241 V}\\ E_{\text{cell}}^{°}=0.555+\frac{0.0592}{2}\log \left[{\text{Hg}}_2^{2+}\right]\end{array}$

Hence cell voltage can be obtained from the equation as follows:

  $E_{\text{cell}}^{°}=0.555+\frac{0.0592}{2}\log \left[{\text{Hg}}_2^{2+}\right]$

(c)

Interpretation Introduction

Interpretation:

Value of $\left[{\text{Ag}}^{+}\right]$ and $E$ at indicated volumes has to be calculated and titration curve has to be sketched.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Titration reaction is written as follows:

  ${\text{Hg}}_2^{2+}+2{\text{Cl}}^{-}\rightleftharpoons {\text{Hg}}_{\text{2}}{\text{Cl}}_2\left(s\right)$

When $\text{0}\text{.1 mL}$ solution of ${\text{Hg}}_2^{2+}$ is added, concentration of unreacted ${\text{Cl}}^{-}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\frac{\left(5-0.01\right)\text{ mmol}}{\left(50+0.1\right)\text{ mL}}\\ =0.0996\text{ M}\end{array}$

Expression to compute metal ion concentration before equivalence volume is given as follows:

  $\left[{\text{Hg}}_2^{2+}\right]=\frac{K_{\text{sp}}}{{\left[{\text{Cl}}^{-}\right]}^2}$        (3)

Substitute $0.0996\text{ M}$ for $\left[{\text{Cl}}^{-}\right]$ and $1.2\times {10}^{-18}$ for $K_{\text{sp}}$ in equation (3).

  $\begin{array}{c}\left[{\text{Hg}}_2^{2+}\right]=\frac{1.2\times {10}^{-18}}{{\left(0.0996\text{ M}\right)}^2}\\ =1.209\times {10}^{-16}\text{ M}\end{array}$

Expression to compute $E$ for silver ion electrode is given as follows:

  $E_{\text{cell}}=0.555+\frac{0.05916}{2}\log \left[{\text{Hg}}_2^{2+}\right]$        (4)

Substitute $1.209\times {10}^{-16}\text{ M}$ for $\left[{\text{Hg}}_2^{2+}\right]$ in equation (4).

  $\begin{array}{c}{E}_{\text{cell}}=0.555+\frac{0.05916}{2}\log \left(1.209\times {10}^{-16}\text{ M}\right)\\ =0.084\text{ V}\end{array}$

When $\text{10 mL}$ solution is added, concentration of unreacted ${\text{Cl}}^{-}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}^{-}\right]=\frac{\left(5-1\right)\text{ mmol}}{\left(50+10\right)\text{ mL}}\\ =0.0667\text{ M}\end{array}$

Substitute $0.0667\text{ M}$ for $\left[{\text{Cl}}^{-}\right]$ and $1.2\times {10}^{-18}$ for $K_{\text{sp}}$ in equation (3).

  $\begin{array}{c}\left[{\text{Hg}}_2^{2+}\right]=\frac{1.2\times {10}^{-18}}{{\left(0.0667\text{ M}\right)}^2}\\ =2.697\times {10}^{-16}\text{ M}\end{array}$

Substitute $2.697\times {10}^{-16}\text{ M}$ for $\left[{\text{Hg}}_2^{2+}\right]$ in equation (4).

  $\begin{array}{c}{E}_{\text{cell}}=0.555+\frac{0.05916}{2}\log \left(2.697\times {10}^{-16}\text{ M}\right)\\ =0.102\text{ V}\end{array}$

Expression to compute $\left[{\text{Hg}}_2^{2+}\right]$ at equivalence volume is given as follows:

  $\left[{\text{Hg}}_2^{2+}\right]={\left(\frac{K_{\text{sp}}}{4}\right)}^{1/3}$        (5)

Substitute $1.2\times {10}^{-18}$ for $K_{\text{sp}}$ in equation (5).

  $\begin{array}{c}\left[{\text{Hg}}_2^{2+}\right]={\left(\frac{1.2\times {10}^{-18}}{4}\right)}^{1/3}\\ =6.7\times {10}^{-7}\text{ M}\end{array}$

Substitute $6.7\times {10}^{-7}\text{ M}$ for $\left[{\text{Hg}}_2^{2+}\right]$ in equation (4).

  $\begin{array}{c}{E}_{\text{cell}}=0.555+\frac{0.05916}{2}\log \left(6.7\times {10}^{-7}\text{ M}\right)\\ =0.372\text{ V}\end{array}$

Beyond equivalence volume when $\text{30 mL}$ solution is added excess $\left[{\text{Hg}}_2^{2+}\right]$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Hg}}_2^{2+}\right]=\frac{\left(5-0.1\right)\text{ mmol}}{\left(50+30\right)\text{ mL}}\\ =0.0063\text{ M}\end{array}$

Substitute $0.0063\text{ M}$ for $\left[{\text{Hg}}_2^{2+}\right]$ in equation (4).

  $\begin{array}{c}{E}_{\text{cell}}=0.555+\frac{0.05916}{2}\log \left(0.0063\right)\\ =0.490\text{ V}\end{array}$

Corresponding titration curve is drawn as follows: