Chapter 15, Problem 15.20P

a)

Interpretation Introduction

Interpretation:

The junction potential of $\text{0}\text{.1}\text{M}\text{HCl}|\text{0}\text{.1}\text{M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ has to be shown.

Concept Introduction:

Given:

The Henderson junction potential equation between the solutions $\text{α}\text{andβ}$ are given by

${\text{E}}_{\text{j}}\text{=}\frac{{\displaystyle \sum _{\text{i}}\frac{\left|{\text{z}}_{\text{i}}\right|{\text{u}}_{\text{i}}}{{\text{z}}_{\text{i}}}\left[{\text{C}}_{\text{i}}\left(\text{β}\right){\text{-C}}_{\text{i}}\left(\text{α}\right)\right]}}{{\displaystyle \sum _{\text{i}}\left|{\text{z}}_{\text{i}}\right|{\text{u}}_{\text{i}}\left[{\text{C}}_{\text{i}}\left(\text{β}\right){\text{-C}}_{\text{i}}\left(\text{α}\right)\right]}}\frac{\text{RT}}{\text{F}}\text{ln}\frac{{\displaystyle \sum _{\text{i}}\left|{\text{z}}_{\text{i}}\right|{\text{u}}_{\text{i}}\left[{\text{C}}_{\text{i}}\left(\text{α}\right)\right]}}{{\displaystyle \sum _{\text{i}}\left|{\text{z}}_{\text{i}}\right|{\text{u}}_{\text{i}}\left[{\text{C}}_{\text{i}}\left(\text{β}\right)\right]}}$

Explanation of Solution

In phase $\text{α}$,

$\text{0}\text{.1}\text{M}{\text{H}}^{\text{+}}$ Has $\text{u=36}{\text{.3×10}}^{\text{-8}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-1}}{\text{V}}^{\text{-1}}$

$\text{0}\text{.1}\text{M}{\text{Cl}}^{\text{-}}$ Has $\text{u=7}{\text{.91×10}}^{\text{-8}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-1}}{\text{V}}^{\text{-1}}$

In phase $\text{β}$

$\text{0}\text{.1}\text{M}{\text{K}}^{\text{+}}$ Has $\text{u=7}{\text{.62×10}}^{\text{-8}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-1}}{\text{V}}^{\text{-1}}$

$\text{0}\text{.1}\text{M}{\text{Cl}}^{\text{-}}$ Has $\text{u=7}{\text{.91×10}}^{\text{-8}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-1}}{\text{V}}^{\text{-1}}$

Using the Henderson equation substitute the values of phase $\text{α}\text{and}\text{β}$.

$\begin{array}{l}{\text{E}}_{\text{j}}\text{=}\frac{\left(\text{36}{\text{.3×10}}^{\text{-8}}\right)\left[\text{0-0}\text{.1}\right]\text{+7}{\text{.62×10}}^{\text{-8}}\left[\text{0-0}\text{.1}\right]\text{-}\left(\text{7}{\text{.91×10}}^{\text{-8}}\right)\left[\text{0}\text{.1-0}\text{.1}\right]}{\left(\text{36}{\text{.3×10}}^{\text{-8}}\right)\left[\text{0-0}\text{.1}\right]\text{+7}{\text{.62×10}}^{\text{-8}}\left[\text{0-0}\text{.1}\right]\text{-}\left(\text{7}{\text{.91×10}}^{\text{-8}}\right)\left[\text{0}\text{.1-0}\text{.1}\right]}\text{×0}\text{.05916log}\frac{\left(\text{36}{\text{.3×10}}^{\text{-8}}\right)\left[\text{0}\text{.1}\right]\text{-}\left(\text{7}{\text{.91×10}}^{\text{-8}}\right)\left[\text{0}\text{.1}\right]}{\left(\text{36}{\text{.3×10}}^{\text{-8}}\right)\left[\text{0}\text{.1}\right]\text{-}\left(\text{7}{\text{.91×10}}^{\text{-8}}\right)\left[\text{0}\text{.1}\right]}\\ \text{=26}\text{.9}\text{mV}\end{array}$

Conclusion

The junction potential of $\text{0}\text{.1}\text{M}\text{HCl}|\text{0}\text{.1}\text{M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ was showed.

b)

Interpretation Introduction

Interpretation:

The junction potential of $\text{0}\text{.1}\text{M}\text{HCl}|\text{x M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ where x varies from $\text{1mM}\text{to}\text{4M}$ has to be plotted.

Explanation of Solution

The junction potential of $\text{0}\text{.1}\text{M}\text{HCl}|\text{x M}\text{KCl}$ values are plotted, where x-axis has junction potential and $\text{KCl}$ concentration $\left(\text{M}\right)$

Figure 1

Conclusion

The junction potential of $\text{0}\text{.1}\text{M}\text{HCl}|\text{x M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ where x varies from $\text{1mM}\text{to}\text{4M}$ was plotted.

c)

Interpretation Introduction

Interpretation:

The behavior of the junction potential for $\text{y M}\text{HCl}|\text{x}\text{M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ has to be shown.

Explanation of Solution

Given,

The behavior of the junction potential for $\text{y M}\text{HCl}|\text{x}\text{M}\text{KCl}$

Where,

$\begin{array}{l}{\text{y=10}}^{\text{-4}}{\text{,10}}^{\text{-3}}{\text{,10}}^{\text{-2}}{\text{,10}}^{\text{-1}}\\ \text{x=1mM}\text{or}\text{4}\text{M}\end{array}$

And values are tabulated as,

$\left[\text{HCl}\right]$	$\text{y}\text{M}\text{HCl}|\text{1}\text{mM}\text{KCl}$	$\text{y}\text{M}\text{HCl}|\text{4mM}\text{KCl}$
${\text{10}}^{\text{-4}}\text{M}$	$\text{9}\text{.1 mV}$	$\text{4}\text{.6 mV}$
${\text{10}}^{\text{-3}}\text{M}$	$\text{26}\text{.9 mV}$	$\text{3}\text{.6 mV}$
${\text{10}}^{\text{-2}}\text{M}$	$\text{57}\text{.3 mV}$	$\text{3}\text{.0 mV}$
${\text{10}}^{\text{-1}}\text{M}$	$\text{93}\text{.6 mV}$	$\text{4}\text{.7 mV}$

Conclusion

The behavior of the junction potential for $\text{y M}\text{HCl}|\text{x}\text{M}\text{KCl}$ is $\text{26}\text{.9}\text{mV}$ at ${\text{25}}^{\text{0}}\text{C}$ was showed.