Chapter 25, Problem 25.15QAP

Interpretation Introduction

(a)

Interpretation:

The surface area should be determined in cm², neglecting the area of the nanotube attachment.

Concept introduction:

Surface area of a sphere = $4\pi r^2$

r = radius of the sphere.

Answer to Problem 25.15QAP

$1.92\times {10}^{-12}{\text{ cm}}^2$

Explanation of Solution

Using formula:

Surface area of a sphere = $4\pi r^2$

Substituting the values:

$\begin{array}{l}A=4\times \frac{22}{7}\times {\left(3.91\times {10}^{-7}\text{ cm}\right)}^2\\ A=1.92\times {10}^{-12}{\text{ cm}}^2\end{array}$

Interpretation Introduction

(b)

Interpretation:

The concentration gradient and the current for A at a concentration of 1.00 mM at different times should be calculated.

Concept introduction:

$\frac{\partial c_A}{{\partial }_x}=c_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

$\frac{\partial c_A}{{\partial }_x}$ - concentration gradient

$c_A^0$ - concentration of A

D − diffusion coefficient

t- time after the voltage is applied

r − radius of the sphere

$i=nFAD{c}_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

i − time dependent faradaic current

n − number of moles of electrode

F − Faraday constant

A − surface area

D − diffusion coefficient

$c_A^0$ - concentration of A

t- time after the voltage is applied

r − radius of the sphere

Answer to Problem 25.15QAP

t, s	Concentration gradient	I, A
1.00E-08	4.55E+05	6.75E-15
1.00E-07	3.19E+05	4.73E-15
1.00E-06	2.76E+05	4.09E-15
1.00E-05	2.62E+05	3.88E-15
1.00E-04	2.58E+05	3.82E-15
1.00E-03	2.56E+05	3.80E-15
1.00E-02	2.56E+05	3.79E-15
1.00E-01	2.56E+05	3.79E-15
1.00E+00	2.56E+05	3.79E-15
1.00E+01	2.56E+05	3.79E-15

Explanation of Solution

When the time is $1\times {10}^{-8}\text{s}$

$\begin{array}{l}\frac{\partial c_A}{{\partial }_x}=1.00\times {10}^{-3}\text{ M}\left(\frac{1}{\sqrt{3.14\times 8\times {10}^{-10}{\text{ m}}^2\text{/s}\times 1\times {10}^{-8}\text{ s}}}+\frac{1}{3.91\times {10}^{-9}\text{ m}}\right)\\ \frac{\partial c_A}{{\partial }_x}=1.00\times {10}^{-3}\text{ M}\left(\frac{1}{5.01\times {10}^{-9}\text{ m}}+\frac{1}{3.91\times {10}^{-9}\text{ m}}\right)\\ \frac{\partial c_A}{{\partial }_x}=1.00\times {10}^{-3}\text{ M}\times 4.554\times {10}^8{\text{ m}}^{-1}\\ \frac{\partial c_A}{{\partial }_x}=4.554\times {10}^5{\text{ Mm}}^{-1}\end{array}$

$\begin{array}{l}i=1\text{ mol}\times 96485\text{ C/mol}\times 1.92\times {10}^{-16}{\text{ m}}^2\times 8\times {10}^{-10}{\text{ m}}^2\text{/s }\times 4.554\times {10}^5\text{ M/m}\\ i=6.75\times {10}^{-15}\text{ A}\end{array}$

Likewise, the other concentration gradients and current can be calculated using a spreadsheet.

t, s	d, m	1/d + 1/r	Concentration gradient	I, A
1.00E-08	5.012E-09	4.55E+08	4.55E+05	6.75E-15
1.00E-07	1.5849E-08	3.19E+08	3.19E+05	4.73E-15
1.00E-06	5.012E-08	2.76E+08	2.76E+05	4.09E-15
1.00E-05	1.5849E-07	2.62E+08	2.62E+05	3.88E-15
1.00E-04	5.012E-07	2.58E+08	2.58E+05	3.82E-15
1.00E-03	1.5849E-06	2.56E+08	2.56E+05	3.80E-15
1.00E-02	5.012E-06	2.56E+08	2.56E+05	3.79E-15
1.00E-01	1.5849E-05	2.56E+08	2.56E+05	3.79E-15
1.00E+00	5.012E-05	2.56E+08	2.56E+05	3.79E-15
1.00E+01	0.00015849	2.56E+08	2.56E+05	3.79E-15

Interpretation Introduction

(c)

Interpretation:

The steady state current should be found

Concept introduction:

If r << d, which occurs at long times, the 1/r term predominates, the electron transfer process reaches a steady state. The steady state current only depends on the size of the electrode.

Answer to Problem 25.15QAP

$i=3.79\times {10}^{-15}\text{ A}$

Explanation of Solution

$\begin{array}{l}i=nFAD{c}_A^0\left(\frac{1}{r}\right)\\ i=1\text{ mol}\times 96485\text{ C/mol}\times \text{1}\text{.92}\times {\text{10}}^{-16}{\text{ m}}^2\times 8\times {10}^{-10}{\text{ m}}^2\text{/s}\times \text{1}\times {\text{10}}^{-3}\text{ M}\times \frac{1}{3.91\times {10}^{-9}\text{ m}}\\ i=3.79\times {10}^{-15}\text{ A}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The time required for the electrode to achieve steady state current following the application of the voltage step should be determined.

Concept introduction:

Current should be used for the calculation is 1.01 x steady state value.

$\frac{\partial c_A}{{\partial }_x}=c_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

$\frac{\partial c_A}{{\partial }_x}$ - concentration gradient

$c_A^0$ - concentration of A

D − diffusion coefficient

t- time after the voltage is applied

r − radius of the sphere

$i=nFAD{c}_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

i − time dependent faradaic current

n − number of moles of electrode

F − Faraday constant

A − surface area

D − diffusion coefficient

$c_A^0$ - concentration of A

t- time after the voltage is applied

r − radius of the sphere

Answer to Problem 25.15QAP

$t=7.9\times {10}^{-5}\text{ s}$

Explanation of Solution

$\begin{array}{l}i=1.01\times 3.79\times {10}^{-15}\text{ A}\\ i=3.83\times {10}^{-15}\text{A}\end{array}$

Using equation,

$\begin{array}{l}i=nAFD\times Gradient\\ Gradient=\frac{3.83\times {10}^{-15}\text{ A}}{1\text{ mol}\times \text{1}\text{.92}\times {\text{10}}^{-16}{\text{ m}}^2\times 96485\text{ C/mol}\times \text{8}\times {\text{10}}^{-10}{\text{ m}}^2/\text{s}}\\ Gradient=2.58\times {10}^5\text{ M/m}\end{array}$

$\begin{array}{l}2.58\times {10}^5\text{ M/m}=1\times {10}^{-3}\text{ M}\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{3.91\times {10}^{-9}\text{ m}}\right)\\ 2.58\times {10}^8=\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{3.91\times {10}^{-9}\text{ m}}\right)\\ \frac{1}{\sqrt{\pi Dt}}=2.246\times {10}^6\\ \frac{1}{\pi Dt}=5.042\times {10}^{12}\\ \frac{1}{3.14\times 8\times {10}^{-10}\times t}=5.042\times {10}^{12}\\ t=\frac{1}{3.14\times 8\times {10}^{-10}\times 5.042\times {10}^{12}}\\ t=7.9\times {10}^{-5}\text{ s}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The calculations should be repeated for a 3 µm spherical platinum electrode and for a spherical iridium electrode with a surface area of 0.785 mm²

Concept introduction:

The formula used:

$\frac{\partial c_A}{{\partial }_x}=c_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

$\frac{\partial c_A}{{\partial }_x}$ - concentration gradient

$c_A^0$ - concentration of A

D − diffusion coefficient

t- time after the voltage is applied

r − radius of the sphere

$i=nFAD{c}_A^0\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{r}\right)$

i − time dependent faradaic current

n − number of moles of electrode

F − Faraday constant

A − surface area

D − diffusion coefficient

$c_A^0$ - concentration of A

t- time after the voltage is applied

r − radius of the sphere

Answer to Problem 25.15QAP

For platinum electrode with radius of 3 3 µm, time required to achieve steady state is $35.77\text{ s}$ and for iridium electrode with radius of 0.00025 m, time required to achieve steady state is $2.5\times {10}^5\text{ s}$

Explanation of Solution

For the spherical platinum electrode area is calculated as:

$\begin{array}{l}A=4\pi r^2\\ A=4\times 3.14\times {(}^3\\ A=1.13\times {10}^{-10}{\text{ m}}^2\end{array}$

Steady state current is calculated as:

$\begin{array}{l}i=nFAD{c}_A^0\left(\frac{1}{r}\right)\\ i=1\text{ mol}\times 96485\text{ C/mol}\times \text{1}\text{.13}\times {\text{10}}^{-10}{\text{ m}}^2\times 8\times {10}^{-10}{\text{ m}}^2\text{/s}\times \text{1}\times {\text{10}}^{-3}\text{ M}\times \frac{1}{3\times {10}^{-6}\text{ m}}\\ i=2.91\times {10}^{-12}\text{ A}\end{array}$

$\begin{array}{l}i=1.01\times 2.91\times {10}^{-12}\text{ A}\\ i=2.94\times {10}^{-12}\text{A}\end{array}$

$\begin{array}{l}i=nAFD\times Gradient\\ Gradient=\frac{2.94\times {10}^{-12}\text{ A}}{1\text{ mol}\times \text{1}\text{.13}\times {\text{10}}^{-10}{\text{ m}}^2\times 96485\text{ C/mol}\times \text{8}\times {\text{10}}^{-10}{\text{ m}}^2/\text{s}}\\ Gradient=336.67\text{ M/m}\end{array}$

$\begin{array}{l}\text{336}\text{.67 M/m}=1\times {10}^{-3}\text{ M}\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{3\times {10}^{-6}\text{ m}}\right)\\ 336670=\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{3\times {10}^{-6}\text{ m}}\right)\\ \frac{1}{\sqrt{\pi Dt}}=3336.67\\ \frac{1}{\pi Dt}=1.113\times {10}^7\\ \frac{1}{3.14\times 8\times {10}^{-10}\times t}=1.113\times {10}^7\\ t=\frac{1}{3.14\times 8\times {10}^{-10}\times 1.113\times {10}^7}\\ t=35.77\text{ s}\end{array}$

For the spherical iridium electrode,

$\begin{array}{l}A=4\pi r^2\\ 0.785\times {10}^{-6}{\text{ m}}^2=4\times 3.14\times r^2\\ r=2.5\times {10}^{-4}{\text{ m}}^2\end{array}$

Steady state current

$\begin{array}{l}i=nFAD{c}_A^0\left(\frac{1}{r}\right)\\ i=1\text{ mol}\times 96485\text{ C/mol}\times 0.785\times {\text{10}}^{-6}{\text{ m}}^2\times 8\times {10}^{-10}{\text{ m}}^2\text{/s}\times \text{1}\times {\text{10}}^{-3}\text{ M}\times \frac{1}{2.5\times {10}^{-4}\text{ m}}\\ i=2.42\times {10}^{-10}\text{ A}\end{array}$

$\begin{array}{l}i=1.01\times 2.42\times {10}^{-10}\text{ A}\\ i=2.45\times {10}^{-10}\text{A}\end{array}$

$\begin{array}{l}i=nAFD\times Gradient\\ Gradient=\frac{2.45\times {10}^{-10}\text{ A}}{1\text{ mol}\times 0.785\times {10}^{-6}{\text{ m}}^2\times 96485\text{ C/mol}\times \text{8}\times {\text{10}}^{-10}{\text{ m}}^2/\text{s}}\\ Gradient=4.04\text{ M/m}\end{array}$

$\begin{array}{l}\text{4}\text{.04 M/m}=1\times {10}^{-3}\text{ M}\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{2.5\times {10}^{-4}\text{ m}}\right)\\ 4040=\left(\frac{1}{\sqrt{\pi Dt}}+\frac{1}{2.5\times {10}^{-4}\text{ m}}\right)\\ \frac{1}{\sqrt{\pi Dt}}=40\\ \frac{1}{\pi Dt}=1600\\ \frac{1}{3.14\times 8\times {10}^{-10}\times t}=1600\\ t=\frac{1}{3.14\times 8\times {10}^{-10}\times 1600}\\ t=2.5\times {10}^5\text{ s}\end{array}$

Interpretation Introduction

(f)

Interpretation:

The results for the three electrodes should be compared and the differences should be discussed.

Answer to Problem 25.15QAP

When the size of the electrode is higher than the thickness of the Nernst diffusion layer, it takes more time to achieve the steady state.

Explanation of Solution

For platinum electrode with radius of 3 3 µm, time required to achieve steady state is $35.77\text{ s}$ and for iridium electrode with radius of 0.00025 m, time required to achieve steady state is $2.5\times {10}^5\text{ s}$. For the nano onion with a radius of 3.91 nm, time required to achieve steady state is $t=7.9\times {10}^{-5}\text{ s}$ under the same conditions. This result indicates that, when the size of the electrode is higher than the thickness of the Nernst diffusion layer, it takes more time to achieve the steady state.

When r <<< d, the 1/r term predominates. The electron transport process reaches the steady state. At this time, steady state current depends only on the size of the electrode. If the size of the electrode is small than the thickness of the Nernst diffusion layer, the steady state is achieved very rapidly.