Surface Thermodynamics of Solid Electrode

Abstract

Surface thermodynamics is fundamentals for understanding of the electro-chemo-mechanical properties of solid electrode surfaces. The concept of surface phase (or interphase) is explained, and the general equations of surface thermodynamics are derived. Surface stress is the most important thermodynamic parameter to characterize a solid electrode, since the changes in surface stress are directly associated with the electro-chemo-mechanical properties. The relationship between surface stress and surface tension for a solid electrode is discussed to distinguish from surface tension which characterizes a liquid electrode such as mercury electrode. Furthermore, the electrocapillary curve (surface stress vs. potential or surface tension vs. potential) for the electrified interface of a solid electrode is derived and compared with that (surface tension vs. potential) derived for the electrified interface of a liquid electrode. The difference between electrocapillary curves derived for the electrified interfaces of solid and liquid electrodes is confirmed from the results obtained experimentally for the electrified interfaces of gold and mercury electrodes.

Appendices

Appendix 1: Derivation of the Tensor Equivalent of the Shuttleworth Equation

Let us follow the derivations made by Linford [11]. Consider the cube whose edges are unit length and parallel to x-, y-, and z-axes of a Cartesian reference frame. We perform two distinct operations on this cube as shown in Fig. 1.12 [10].

Fig. 1.12

Reprinted from [10] with permission of ASM International

Two distinct operations of stretching $(W_{0} )$ followed by separating $(W_{2} )$ or of separating $(W_{3} )$ followed by stretching $(W_{1} )$ on a cube [10]. The reversible works of $W_{0} + W_{2}$ are equivalent to those of $W_{1} + W_{3}$.

(1) Stretch the cube along the x-axis reversibly by an amount $dx$, maintaining the y edge at unit length, but allowing the z edge to vary its length. In this process, the flow of materials from the bulk by reducing the height of the crystal is permitted to form the extended surface. Let the work expended here denote by $W_{0}$. Next, let the stretched cube be separated (cleaved) along an $xy$ plane, which requires work $W_{2}$:

$$ W_{2} = 2\left( {\gamma + \Delta \gamma } \right)\left( {1 + dx} \right), $$

(1.135)

where $\Delta \gamma$ is the variation in surface tension γ arising from the stretch $dx$. The factor of two arises due to the formation of two surfaces, and the term of $\left( {1 + dx} \right)$ is the area of an $xy$ cross section of the stretched cube.

(2) Now let the cube be returned to its original dimensions and then be separated (cleaved) along an $xy$ plane into two halves. The work $W_{3}$ required to separate the cube into two halves is given by

$$ W_{3} = 2\gamma . $$

(1.136)

Furthermore, stretch each half in the x-direction by $dx$, keeping the y edge but the z edge not constant. Let this work denote by $W_{1}$. The work involved on the route (1), in which the cube is first stretched and then separated, is the same as that on the route (2), in which the cube is first separated and then stretched. As a result, the following relationship holds:

$$ W_{0} + W_{2} = W_{1} + W_{3} . $$

(1.137)

The substitution of Eqs. (1.135) and (1.136) into Eq. (1.137) leads to

$$ W_{0} + 2\left( {\gamma + \Delta \gamma } \right)\left( {1 + dx} \right) = W_{1} + 2\gamma . $$

(1.138)

The difference $(W_{1} - W_{0} )$ in the work required in the stretching stage of the two routes is two times as much as the product of the component $g_{xx}$ of a force (surface stress) in the newly formed surface and of the distance $dx$ through which the force acts, that is,

$$ W_{1} - W_{0} = 2g_{xx} dx. $$

(1.139)

It is noticed that the surface stresses of the unstretched two halves after the cleavage of the cube are balanced by elastic volume stresses in the interior of the two halves and thus are not included in Eq. (1.139) [10]. Since the strain $\Delta \varepsilon_{xx}$ is $\frac{dx}{1}$ in the unit cube, the substitution of Eqs. (1.139) into (1.138) leads to

$$ 2g_{xx} \Delta \varepsilon_{xx} = 2\gamma \Delta \varepsilon_{xx} + 2\Delta \gamma + 2\Delta \gamma \Delta \varepsilon_{xx} . $$

(1.140)

Because of $\Delta \gamma \Delta \varepsilon_{xx} \approx$ 0 and $\frac{\Delta \gamma }{{\Delta \varepsilon_{xx} }} \approx \frac{d\gamma }{{d\varepsilon_{xx} }}$, Eq. (1.140) can be rewritten as

$$ g_{xx} = \gamma + \frac{d\gamma }{{d\varepsilon_{xx} }}. $$

(1.141)

Consider the following two routes (3) and (4) analogous to those of (1) and (2) except that the stretching stages are replaced by shearing stages.

(3) The unit cube is first sheared, changing the shape, but not the area, of the $xy$ plane cross section, which requires work $W_{0}^{{\prime }}$, and then separated into two halves. The work $W_{2}^{{\prime }}$ required for the separation is given by

$$ W_{2}^{{\prime }} = 2\gamma + 2\Delta \gamma . $$

(1.142)

(4) The unit cube is first separated into two halves along an $xy$ plane, requiring work $W_{3}^{{\prime }} = W_{3} = 2\gamma$, and the subsequent shear of the two halves requires work $W_{1}^{{\prime }}$. Since $(W_{0}^{{\prime }} + W_{2}^{{\prime }} )$ is equal to $(W_{1}^{{\prime }} + W_{3}^{{\prime }} )$, the following relationship holds:

$$ W_{1}^{{\prime }} - W_{0}^{{\prime }} = 2\Delta \gamma . $$

(1.143)

The difference $W_{1}^{{\prime }} - W_{0}^{{\prime }}$ in the work required in the shearing stage of the two routes is two times as much as the product of the component $g_{xy}$ of a force in the newly formed surface and of the distance $dy$ through which the force acts. Since the strain $\Delta \varepsilon_{xy}$ is $\frac{dy}{1}$ in the unit cube, $W_{1}^{{\prime }} - W_{0}^{{\prime }}$ is given by

$$ W_{1}^{{\prime }} - W_{0}^{{\prime }} = 2g_{xy} \Delta \varepsilon_{xy} , $$

(1.144)

and

$$ g_{xy} = \frac{d\gamma }{{d\varepsilon_{xy} }}. $$

(1.145)

The combination of Eqs. (1.141) and (1.145) leads to the tensor equivalent of the Shuttleworth equation (see Eq. (1.71)): $g_{nm} = \gamma \delta_{nm} + \left( {\frac{\partial \gamma }{{\partial \epsilon_{nm} }}} \right)$. The $g_{nm}$ quantities are the components of the surface stress tensor g, where

$$ g = \left( {\begin{array}{*{20}c} {g_{xx} } & {g_{xy} } \\ {g_{yx} } & {g_{yy} } \\ \end{array} } \right), $$

(1.146)

and for an isotropic substance or for a crystal surface with a threefold or greater axis of symmetry,

$$ g = g\left( {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array} } \right), $$

(1.147)

in which $g_{xy}$ and $g_{yx}$ vanish, and the normal components of the tensor is equal to the tensor itself (see Eq. (1.69) or Eq. (1.80)).

Appendix 2: Calculation of the Magnitude of Surface Elastic Strain from the Curvature Change of Cantilever Bending

The following relationship between curvature radius of the cantilever bending R and surface stress g holds [21, 39] (see Eq. (2.14) in Sect. 2.3.1 of Chap. 2):

$$ \frac{1}{R} = \frac{{6\left( {1 - \nu_{\text{s}} } \right)}}{{E_{\text{s}} d_{\text{s}}^{2} }}g, $$

where $E_{\text{s}}$, $\nu_{\text{s}}$, and $d_{\text{s}}$ are Young’s modulus, Poisson’s ratio, and thickness of the cantilever substrate, respectively. The surface strain ε is given by

$$ \varepsilon = \frac{{d_{\text{s}} }}{2R}, $$

(1.148)

and a linear relationship between ε and g for small deformations is

$$ \varepsilon = \frac{{3\left( {1 - \nu_{\text{s}} } \right)g}}{{E_{\text{s}} d_{\text{s}} }}. $$

(1.149)

Furthermore, the term $\left( {\frac{\partial \varepsilon }{\partial E}} \right)_{{T,\mu_{i} }}$ can be derived from Eq. (1.149):

$$ \left( {\frac{\partial \varepsilon }{\partial E}} \right)_{{T,\mu_{i} }} = \frac{{3\left( {1 - \nu_{\text{s}} } \right)}}{{E_{\text{s}} d_{\text{s}} }}\left( {\frac{\partial g}{\partial E}} \right)_{{T,\mu_{i} }} . $$

(1.150)

In the experiments by Ibach et al. [22], the value of $\frac{{3\left( {1 - \nu_{\text{s}} } \right)}}{{E_{\text{s}} d_{\text{s}} }} =$ 5.23 × 10⁻⁸ m N⁻¹ was obtained from the values of $d_{\text{s}} =$ 0.3 mm, $E_{\text{s}} =$ 81.6 GPa, and $\nu_{\text{s}} =$ 0.573 for the Au (111) cantilever electrode. Very small value of $\varepsilon =$ 5.23 × 10⁻⁸ for $g =$ 1.0 J m⁻² is calculated from Eq. (1.149). The value of $\left( {\frac{\partial g}{\partial E}} \right)_{{T,\mu_{i} }} \approx 1 \text{N} \text{m}^{-1} \text{V}^{-1}$ was also obtained from the surface stress measurement of the Au (111) cantilever electrode [22]. Consequently, the value of $\left( {\frac{\partial \varepsilon }{\partial E}} \right)_{{T,\mu_{i} }} \approx$ 5 × 10⁻⁸ V⁻¹ is calculated from Eq. (1.150) [21].