Chapter 19, Problem 19D.3P

(a)

Interpretation Introduction

Interpretation:

The zero-current potential of the ${\text{Fe}}^{\text{2+}}\text{/Fe}$ cathode and overpotential of each given potential have to be calculated.

Explanation of Solution

The zero-current potential of the ${\text{Fe}}^{\text{2+}}\text{/Fe}$ cathode and overpotential are calculated as,

    $\begin{array}{l}{\text{Fe}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\underset{}{\overset{}{\to }}\text{Fe}\text{ν}\text{=}\text{2}{\text{E}}^{\text{0}}\text{=}\text{-0}\text{.447}\\ \\ {\text{E}}_{\text{0}}\text{=}{\text{E}}^{\text{0}}\text{-}\frac{\text{RT}}{\text{νF}}\text{lnQ}\\ \\ \text{=}{\text{E}}^{\text{0}}\text{-}\frac{\text{RT}}{\text{νF}}\frac{\text{1}}{{\text{Fe}}^{\text{2+}}}{\text{γ}}_{{\text{Fe}}^{\text{2+}}}\text{=}\text{1}\\ \\ \text{=}\text{-0}\text{.447V}\text{-}\frac{\text{25}\text{.693}\text{×}{\text{10}}^{\text{-3}}\text{V}}{\text{2}}\text{ln}\left(\frac{{\text{moldm}}^{\text{-3}}}{\text{1}\text{.70}\text{×}{\text{10}}^{\text{-6}}\text{mol}{\text{dm}}^{\text{3}}}\right)\\ \text{=}\text{-0}\text{.618V}\text{(618mV)}\\ \\ \text{η}\text{=}{\text{E}}^{\text{'}}\text{-}{\text{E}}_{\text{0}}\text{η}\text{values}\text{are}\text{reported}\text{as,}\\ \\ \text{η}\text{=}\text{84,}\text{109,}\text{134,}\text{194}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cathodic current density from the rate of deposition of iron ion has to be calculated.

Explanation of Solution

The cathodic current density from the rate of deposition of iron ion is calculated as,

    $\begin{array}{l}\text{j}\text{=}\frac{\text{νF}}{\text{A}}\frac{{\text{dn}}_{\text{Fe}}}{\text{dt}}\text{=}\frac{\text{2(96485C/mol)}}{{\text{91cm}}^{\text{2}}}\frac{{\text{dn}}_{\text{Fe}}}{\text{dt}}\\ \\ \text{j}\text{=}{\text{j}}_{\text{0}}{\text{{e}}^{\text{(1-α)}}{}^{\text{fη}}\text{-}{\text{e}}^{\text{-αfη}}\text{}}\text{=}{\text{j}}_{\text{0}}{\text{e}}^{\text{-αfη}\left\{{\text{e}}^{\text{fη}\left\{\right\}}\text{-1}\right\}}\\ \\ \text{=}{\text{-j}}_{\text{0}}\left\{{\text{e}}^{\text{fη}}\text{-1}\right\}\\ \\ {\text{j}}_{\text{0}}\text{=}\frac{\text{-j}}{{\text{e}}^{\text{fη}}\text{-1}}.\end{array}$

${\text{j}}_{\text{0}}$ values are reported as,

$\frac{{\text{dn}}_{\text{Fe}}}{\text{dt}}{\text{(10}}^{\text{-12}}\text{mol/s)}$	-E’/mV	$\text{-η/mV}$	$\text{j/}{\text{(μAcm}}^{\text{-12}}\text{)}$	$\text{jc/}{\text{(μAcm}}^{\text{-12}}\text{)}$
1.47	702	84	0.0312	0.0324
2.18	727	109	0.0462	0.0469
3.11	752	134	0.0659	0.0663
7.26	812	194	0.154	0.154

(c)

Interpretation Introduction

Interpretation:

The exchange-current density has to be determined by analysing the given data.

Explanation of Solution

The exchange-current density using the given data is calculated as,

    $\begin{array}{l}\text{j}\text{=}{\text{j}}_{\text{0}}{\text{e}}^{\text{-αfη}}\\ \\ {\text{lnj}}_{\text{0}}\text{=}\text{ln}{\text{j}}_{\text{0}}\text{-α}\text{fη}\\ \\ \text{Performing}\text{linear}\text{regression}\text{analysis}\text{of}\text{the}\text{ln}{\text{j}}_{\text{c}}\text{versus}\text{η}\text{data,}\text{we}\text{find}\\ \\ {\text{lnj}}_{\text{0}}\text{=}\text{4}\text{.608}\text{standard}\text{deviation}\text{=}\text{0}\text{.015}\\ \text{αf}\text{=}\text{0}\text{.0413}\text{standard}\text{deviation}\text{=}\text{0}\text{.00011}\\ \\ \text{R}\text{=}\text{0}\text{.99994}\end{array}$

The correlation coefficient and the standard deviation indicate that the plot provides an excellent description of the data.

    $\begin{array}{l}{\text{j}}_{\text{0}}\text{=}{\text{e}}^{\text{4}\text{.608}}\text{or}{\text{j}}_{\text{0}}\text{=}\text{0}{\text{.00997μAcm}}^{\text{-2}}\\ \\ \text{α}\text{=}\frac{\text{0}\text{.01413}}{\text{f}}\text{=}\text{(0}{\text{.01413mV}}^{\text{-1}}\text{)(26}\text{.693mV)}\\ \\ \text{α}\text{=}\text{0}\text{.363}\text{.}\end{array}$