convertin a docx into a structured dataframe¤
this feature engineering task transforms a docx to tidy data. it can eventually be exported as notebooks.
this work is important because we when open existing course content we want to avoid redundant work. spending <1 week to feature engineer existing content into structured notebooks is a reasonable time line. developers become familiar with the content through this data engineering activity.: typing.Sequence[docx.Type]
import docx, itertools, pandas, tempfile, shutil, typing, asyncio, subprocess
from IPython.display import Markdown, display; from toolz.curried import *
g = compose_left
PANDOC = shutil.which("pandoc")
document = docx.Document("/home/tbone/Downloads/2015-07-15-Textbook-Cambridge-U-P.docx")
the snippet discovers the style id's of each paragraph. this means we could remediate the the headings here and make them formal headings
paragraphs = pandas.Series(document.iter_inner_content())
df = paragraphs.to_frame("p").join(
paragraphs.apply(
g(operator.attrgetter("style"), operator.attrgetter("style_id"))
).rename("style_id")
)
end_of_toc = df.style_id.str.match("^TOC").pipe(lambda x: x[x]).index[-1] + 1
toc, df, refs = df.iloc[:end_of_toc], df.iloc[end_of_toc:], df.iloc[-1:]
def get_docx_subset(elements: typing.Sequence[docx.Type]) -> docx.document.Document:
"""create a new docx object with a subset of elements from an original"""
(new := docx.Document())._body._element.extend(x._element for x in elements)
return new
async def get_docx_subset_markdown(elements: typing.Sequence[docx.Type]) -> str:
"""shell out to pandoc to convert the docx elements to markdown"""
tmp = tempfile.NamedTemporaryFile(suffix=".docx")
get_docx_subset(elements).save(tmp.name)
return subprocess.check_output([PANDOC, "--to", "gfm", "--from", "docx", tmp.name]).decode()
define chapters by the 10 h1s we found and the nearest heading to identify notebooks.
notebooks = df.join(
df[df.style_id.str.match("^Heading1")].p.apply(operator.attrgetter("text")).rename("chapter")
).join(
df[df.style_id.str.match("^Heading")].p.apply(operator.attrgetter("text")).rename("notebook")
).ffill().fillna("empty").groupby(["chapter", "notebook"])
turned the grouped notebooks to markdown
(md := pandas.Series(
await asyncio.gather(*(get_docx_subset_markdown(notebooks.get_group(x).p) for x in notebooks.groups)),
notebooks.groups
).to_frame("markdown"))
below we see that most of this transformation renders pretty ok on notebooks
display(*md[md.markdown.str.contains("\$")].iloc[:10].markdown.apply(Markdown))
Ellingham diagram and buffered systems¤
One type of chemical reactions is between a pure element in liquid or solid states and its oxides involving one mole of oxygen gas, i.e.
Eq. 7‑7 $\frac{4}{v_{M}}M + O_{2}(gas) = M_{4/v_{M}}O_{2}$
with $v_{M}$ being the valence of the element $M$ in the oxide. Taking the pure $M$ and the gaseous $O_{2}$ at the reaction temperature and one atmospheric pressure as their respective reference states, both activities of the pure $M$ solid or liquid phase and its oxide are unity, and the activity of $O_{2}$ equals to its partial pressure in an ideal gas. becomes
Eq. 7‑8 $RTlnP_{O_{2}} = \mathrm{\Delta}{ }^{0}G = \mathrm{\Delta}S$}^{0}H - T\mathrm{\Delta}_{ }^{0
Based on , one can plot $\mathrm{\Delta}{ }^{0}G$ as a function of temperature for various oxidation reactions, which is called Ellingham diagram. The intercept at $T = 0K$ is given by $\mathrm{\Delta}H$, and the slope is represented by $- \mathrm{\Delta}_{ }^{0}S$, depicted by the following equation}^{0
Eq. 7‑9 $\mathrm{\Delta}{ }^{0}S = S$}}O_{2}} - S_{O_{2}} - \frac{4}{v_{M}}S_{M
Since the entropy of one mole of $O_{2}$ gas is significantly larger than those of the pure element and its oxide when they are in solid or liquid states and the entropy difference between the pure element and its oxide, the entropy of reaction is thus dominated by the reduction of the entropy by one mole of $O_{2}$ gas. Consequently, the entropies of reaction are approximately the same for most reactions when the pure elements and their oxides are solid as seen in where most lines have similar slopes.
Figure ‑: Ellingham Diagram for a number of metal-oxide systems.
For a chemical reaction on the Ellingham diagram at a given temperature, $\mathrm{\Delta}{ }^{0}G$ can be read from the y-axis of the diagram, and the equilibrium partial pressure of $O$ gas can be calculated using . Alternatively, one can plot the left part of for a given $P_{O_{2}}$ as a function of temperature on the Ellingham diagram, i.e.
Eq. 7‑10 $\mathrm{\Delta}{ }^{0}{G =}RTlnP$}
This results in straight lines, representing iso-partial-pressure lines of $O_{2}$, with the intercept being zero at $T = 0K$ and slopes of $RlnP_{O_{2}}$, which are negative for $P_{O_{2}}$ lower than one atmospheric pressure (1atm). The values of $P_{O_{2}}$ are marked on the secondary axis on the right of the Ellingham diagram. The intersection of the isoactivitiy line and the equilibrium line of the chemical reaction thus gives the relation of equilibrium temperature and equilibrium partial pressure of $O_{2}$. This is demonstrated in .
Figure ‑: Intersection of iso-partial-pressure lines of $O_{2}$ and equilibrium lines in the Ellingham diagram
For each chemical reaction in the Ellingham diagram, the three phases are in equilibrium on the line represented by , i.e. metal, metal oxides, and O2 gas. For conditions above the line, the value of $P_{O_{2}}$ is larger than its equilibrium value, and the metal will be oxidized. For conditions below the line, the value of $P_{O_{2}}$ is lower than its equilibrium value, and the metal oxide will be reduced. Therefore, the metal oxides in the upper part of the Ellingham diagram can be reduced by the metals in the lower part of the diagram, and vice versus, metals in the lower part of the diagram can be oxidized by the metal oxides in the upper part of the diagram. For example, in , Ca can reduce all oxides, and Cu2O is the least stable oxide.
From the above Ellingham diagram, it is noted that equilibrium partial pressures of $O_{2}$ are very low for most chemical reactions with many of them lower than $10^{- 12} atm$. One approach to obtain such a low pressure is to use auxiliary reactions containing $O_{2}$ that are easy to control and are independent of the equilibrium system of interest except the sharing of the oxygen partial pressure. Two common auxiliary reactions are the $H_{2}$/$H_{2}O$ and $CO$/$CO_{2}$ mixtures. For the $H_{2}$/$H_{2}O$ mixture, the chemical reaction is
Eq. 7‑11 $2H_{2}(gas) + O_{2}(gas) = 2H_{2}O(gas)$
The equilibrium oxygen partial pressure is obtained as
Eq. 7‑12 $- RTln\left{ {\frac{1}{P_{O_{2}}}\left( \frac{P_{H_{2}O}}{P_{H_{2}}} \right)}^{2} \right} = \mathrm{\Delta}{ }^{0}G = \mathrm{\Delta}S = - 498488 + 112.972T (J)$}^{0}H - T\mathrm{\Delta}_{ }^{0
where the $\mathrm{\Delta}{ }^{0}H$ and $\mathrm{\Delta}S$ are taken from the substance thermodynamic database (SSUB4) compiled by Scientific Group Thermodata Europe (SGTE) [59], which are slightly dependent on temperature, and the values in are evaluated at 1273K using Thermo-Calc [60]. At any given temperature, one has the following relation}^{0
Eq. 7‑13 $RTlnP_{O_{2}} = \mathrm{\Delta}{ }^{0}H - T\mathrm{\Delta} \right)T$}^{0}S - 2RTln\frac{P_{H_{2}}}{P_{H_{2}O}} = - 498488 + \left( 112.972 - 2Rln\frac{P_{H_{2}}}{P_{H_{2}O}
Its intercept at $T = 0K$ is given by $\mathrm{\Delta}{ }^{0}H = - 498,488 (J)$, and the slope by $- \mathrm{\Delta} \right)$. The values of the $\frac{P_{H_{2}}}{P_{H_{2}O}}$ ratio are marked on another secondary axis on the right of the Ellingham diagram. The intersection of the iso-partial-pressure-ratio line and the equilibrium line of the chemical reaction gives the relation of equilibrium temperature and equilibrium partial pressure ratio $\frac{P_{H_{2}}}{P_{H_{2}O}}$ for desired $P_{O_{2}}$ of the chemical equilibrium.}^{0}S - 2Rln\left( \frac{P_{H_{2}}}{P_{H_{2}O}} \right) = 112.972 - 2Rln\left( \frac{P_{H_{2}}}{P_{H_{2}O}
For the $CO$/$CO_{2}$mixture, the chemical reaction is shown below
Eq. 7‑14 $2CO(gas) + O_{2}(gas) = 2CO_{2}(gas)$
Similar, from the SSUB database, the following equation is obtained
Eq. 7‑15 $- RTln\left{ {\frac{1}{P_{O_{2}}}\left( \frac{P_{{CO}{2}}}{P}} \right)}^{2} \right} = \mathrm{\Delta{ }^{0}G = \mathrm{\Delta}S = - 562,927 - 172.020T (J)$}^{0}H - T\mathrm{\Delta}_{ }^{0
with $\mathrm{\Delta}{ }^{0}H$ and $\mathrm{\Delta}S$ calculated at 1273K, which are also slightly temperature-dependent as in the case of the $H_{2}$/$H_{2}O$ mixture. The iso-partial-pressure-ratio lines are written as}^{0
Eq. 7‑16 $RTlnP_{O_{2}} = \mathrm{\Delta}{ }^{0}H - T\mathrm{\Delta}}^{0}S - 2RTln\frac{P_{CO}}{P_{{CO{2}}} = - 562,927 + \left( 172.020 - 2Rln\frac{P \right)T $}}{P_{{CO}_{2}}
Its intercept at $T = 0K$ is given by $\mathrm{\Delta}{ }^{0}H = - 562,927 (J)$, and the slope by $172.020 - - 2Rln\left( \frac{P}}{P_{{CO{2}}} \right)$. The values of the $\frac{P}}{P_{{CO{2}}}$ ratio are marked on the third secondary axis on the right of the Ellingham diagram. The intersection of the iso-partial-pressure-ratio line and the equilibrium line of the chemical reaction gives the relation of equilibrium temperature and equilibrium partial pressure ratio $\frac{P}}{P_{{CO{2}}}$ for desired $P$ of the chemical equilibrium.}
Therefore, one can use a mixture of the $H_{2}$/$H_{2}O$ or $CO$/$CO_{2}$ to obtain the desired low $P_{O_{2}}$ values using and or calculating from the SSUB database. For example, $P_{O_{2}} = 10^{- 15} atm$ at 1273K, the calculated values from the SSUB database are $\frac{P_{H_{2}}}{P_{H_{2}O}} \approx 1.67$ and $\frac{P_{CO}}{P_{{CO}{2}}} \approx 2.78$, respectively, in which the temperature dependences of $\mathrm{\Delta}H$ and $\mathrm{\Delta}}^{0{ }^{0}S$ and the many more gaseous species in the gas phase are taken into account. On the other hand, the reading from the Ellingham diagram gives $\frac{P \approx 2.0$ and $\frac{P_{CO}}{P_{{CO}_{2}}} \approx 2.4$, and the agreement with the more accurate calculations above is remarkable keeping in mind the uncertainties in graphically drawing the lines and reading both values in the logarithmic scales from the diagram, indicating the robustness of the Ellingham diagram.}}}{P_{H_{2}O}
Internal process and differential and integrated driving forces¤
The driving force for an internal process can be defined as following using $U$, $H$, $F$, or $G$ as discussed in Chapter and depending on what system variables are kept constant
Eq. 7‑1 $- D = \left( \frac{\partial U}{\partial\xi} \right){S, V,N \right)} } = \left( \frac{\partial H}{\partial\xi{S, P,N \right)} } = \left( \frac{\partial F}{\partial\xi{T, V,N \right)} } = \left( \frac{\partial G}{\partial\xi{T,P,N$}
This can be termed as differential driving force as it relates the derivative of energy with respect to an internal process so the change is infinitesimally small and does not affect the properties of the system significantly. For a system under constant $T$ and $P$, let us consider an internal process for forming a new phase $\alpha$ with the composition $x_{i}^{\alpha}$ and Gibbs energy $G_{m}^{\alpha}\left( x_{i}^{\alpha} \right)$. The differential driving force for such an internal process can thus be defined as
Eq. 7‑2 $- D = G_{m}^{\alpha}\left( x_{i}^{\alpha} \right) - \sum_{i}^{}x_{i}^{\alpha}\mu_{i} = \sum_{i}^{}x_{i}^{\alpha}\left( \mu_{i}^{\alpha} - \mu_{i} \right)$
where $\mu_{i}$ is the chemical potential of component i in the system. This may also be called as nucleation driving force as if the $\alpha$ phase is nucleating in the system. As discussed in Chapter , chemical potentials are the intercepts on the Gibbs energy axis by the tangent plane of Gibbs energy. thus represents the distance between the tangent planes of the original system and the $\alpha$ phase at the composition of the $\alpha$ phase. Evidently, this distance is at its maximum when the two tangent planes are parallel to each other, i.e. the commonly called parallel tangent construction when evaluating nucleation driving force.
The situation is different for chemical reactions where the amount of each component in reactants is the same as that in products, i.e. there is a mass balance between reactants and products. The driving force for a chemical reaction is defined by the Gibbs energy difference between the reactants and products as if all the reactants are transferred to the products. This driving force may thus be called integrated driving force as it describes the energy difference of a system under two different states, i.e.
Eq. 7‑3 $- \int_{}^{}{Dd\xi} = \mathrm{\Delta}G = \sum_{p}^{}{n_{p}G_{p}} - \sum_{r}^{}{n_{r}G_{r}}$
where superscripts $p$ and $r$ denote products and reactants, $n_{p}$ and $n_{r}$ their corresponding moles, $G_{p}$and $G_{r}$ the Gibbs energies in per mole of formula of their respective stoichiometries as written in the chemical reaction. Conventionally, the products and reactants are represented by species or stoichiometric compounds rather than individual phases. This is particularly evident when various gaseous species are considered in a chemical reaction. Consequently, $G_{p}$and $G_{r}$ in represent the chemical potentials of product and reactant species. For species with a fixed composition, its chemical potential is the same as its Gibbs energy as shown by . For a species in a solution phase, its chemical potential is related to its activity as shown in . can thus be further written as
Eq. 7‑4 $\mathrm{\Delta}G = \sum_{p}^{}{n_{p}{ }^{0}G}} - \sum_{r}^{}{n_{r{ }^{0}G}} + RTln\frac{\prod_{p}^{}\left( a_{p} \right)^{n_{p}}}{\prod_{r}^{}\left( a_{r} \right)^{n_{r}}} = \mathrm{\Delta{ }^{0}G + RTln\frac{\prod$}^{}\left( a_{p} \right)^{n_{p}}}{\prod_{r}^{}\left( a_{r} \right)^{n_{r}}
It is evident from that for stoichiometric phases in a chemical reaction, their activities equal to one. At equilibrium, the integrated driving force becomes zero, i.e. $\mathrm{\Delta}G = 0$, and is re-arranged to become
Eq. 7‑5 $- RTln\frac{\prod_{p}^{}\left( a_{p} \right)^{n_{p}}}{\prod_{r}^{}\left( a_{r} \right)^{n_{r}}} = - RTlnK_{e} = \mathrm{\Delta}{ }^{0}G = \mathrm{\Delta}S$}^{0}H - T\mathrm{\Delta}_{ }^{0
where $K_{e}$ is often called reaction constant relating the activities of products and reactants at equilibrium. In a system with $\frac{\prod_{p}^{}\left( a_{p} \right)^{n_{p}}}{\prod_{r}^{}\left( a_{r} \right)^{n_{r}}} > K_{e}$, the chemical reaction goes to left, and the products are decomposed, while $\frac{\prod_{p}^{}\left( a_{p} \right)^{n_{p}}}{\prod_{r}^{}\left( a_{r} \right)^{n_{r}}} < K_{e}$, the chemical reaction goes to right, and the products are formed. is often recast into the following form by dividing $–RT$ on both sides of the equation
Eq. 7‑6 $\ln K_{e} = - \frac{\mathrm{\Delta}{ }^{0}H}{RT} + \frac{\mathrm{\Delta}$}^{0}S}{R
With $\ln K_{e}$ plotted with respect to $1/T$, indicates that the slope is $- \mathrm{\Delta}{ }^{0}H$/R and the intercept on the y axis is $\mathrm{\Delta}$ as shown in .}^{0}{S/R
Figure ‑: $\ln K_{e}$ plotted with respect to $1/T$ for several M-O systems at 1 bar and $K_{e}$ represented by the partial pressure ratio of CO2 and CO.
Maximum reaction rate and chemical transport reactions¤
Equilibrium thermodynamics can be used to calculate the maximum rates of reaction that are possible in dynamically reacting systems, such as when a corrosive gas passes over a heated sample. Other examples of such reactions include the reduction of a metal oxide in flowing hydrogen, the evaporation of a material in a vacuum, and the deposition of a thin film by a chemical vapour deposition process. The basic assumption used in calculating these maximum reaction rates is that local equilibrium exists at the location of the considered reaction.
This section examines several examples in which maximum reaction rates are calculated. The system can typically be divided into three regions: (1) the input region, which is usually near room temperature, (2) the high temperature region in which the primary reaction of interest is occurring, and (3) the exit region. Such a system is almost always at constant pressure throughout the system. Since the three regions have different temperatures, the key is to use the mass conservation of the carrier gas in all three regions. With the input gas at $T = 298K$ and $P = 1atm$, the volume of one mole of ideal gas is $0.0244\left( m^{3} \right) = 24.4(liter)$. The number of moles of input gas flowing through the system can be written as
Eq. 7‑17 $n_{gas}^{0} = \frac{Pf_{gas}}{RT} = \frac{f_{gas}}{24.4} = 0.0409f_{gas}$
where $f_{gas}$ is the input gas flow rate in liter per unit time at $T = 298K$ and $P = 1atm$, and $R$ the gas constant.
The first example is the evaporation of water in a flowing stream of dry hydrogen. A schematic diagram of the system being considered is given in . The goal of the calculation is to determine the maximum rate of $H_{2}O(liquid)$ loss, $n_{H_{2}O}^{ }$, through vaporization in a flowing stream of $H_{2}$, e.g. for generating a given $H_{2}O/H_{2}$ ratio for producing a given $O_{2}$ pressure in a high temperature system as related to the Ellingham diagram discussed in Chapter . The maximum rate is determined by saturating the $H_{2}$ with the equilibrium vapour pressure of $H_{2}O(liquid)$ with the number of moles of $H_{2}$ being $n_{H_{2}}^{0} = 0.0409f_{H_{2}}$ and $f_{H_{2}}$ being the input flow rate of $H_{2}$ in liter per unit time.
Figure ‑: Schematic diagram of vaporization of water in a flowing stream of dry hydrogen
In the first input region, the total number moles of gas is $N = n_{H_{2}}^{0}$. In the second high temperature region with the temperature and pressure being $T_{sys}$ and $P_{sys}$, the total number moles of gas is $N = n_{H_{2}}^{0} + n_{H_{2}O}^{ }$, which is the same for the third exit region with $T_{exit}$. To avoid condensation, one needs to maintain $T_{exit} > T_{sys}$. The vapour pressure of $H_{2}O$ at $T_{sys}$, $P_{H_{2}O }$, can be obtained from equilibrium thermodynamic calculations. Since the $H_{2}$ and $H_{2}O$ are occupying the same volume at the same temperature, the maximum number of moles $H_{2}O$ can be calculated from the following relation based on the ideal gas law
Eq. 7‑18 $n_{H_{2}O}^{ } = \frac{P_{H_{2}O}^{ }}{P_{H_{2}}^{ }}n_{H_{2}}^{0} = \frac{P_{H_{2}O}^{ }}{P_{sys}^{ } - P_{H_{2}O}^{ }}n_{H_{2}}^{0} = \frac{P_{H_{2}O}^{ }}{P_{sys}^{ } - P_{H_{2}O}^{ }}\frac{f_{H_{2}}}{24.4}$
For $P_{sys}^{ } = 101325Pa$, one can calculate $P_{H_{2}O}^{ }$ and plot $\frac{n_{H_{2}O}^{ }}{f_{H_{2}}}$ as a function of temperature from the SSUB database as shown in . The corresponding partial pressure ratio, $P_{H_{2}O}/P_{H_{2}}$ , is plotted in , which can be used to calculate $P_{O_{2}}^{ }$ at any given temperatures. An example is shown in with $T_{sys} = 348.15K$ and $P_{H_{2}O}/P_{H_{2}} = 0.607$.
Figure ‑: Ratio of maximum number of moles $H_{2}O$ with respect to hydrogen flow rate, $\frac{n_{H_{2}O}^{ }}{f_{H_{2}}}$, plotted as a function of temperature.
Figure ‑: Partial pressure ratio, $P_{H_{2}O}/P_{H_{2}}$, corresponding to .
Figure ‑: Partial pressure of oxygen, $P_{O_{2}}^{ }$, as a function of temperature with $T_{sys} = 348.15K$ and $P_{H_{2}O}/P_{H_{2}} = 0.607$.
The second example is the corrosion of $SiO_{2}$(s) by flowing $H_{2}$ gas at high temperatures. Considering $T_{sys}^{ } = 1700K$, $P_{sys}^{ } = 101325Pa$, and $f_{H_{2}} = 1 liter/min$, the system is thus defined with 0.0409 moles of $H_{2}$ and an equilibrium between the gas phase and the tridymite $SiO_{2}$. The equilibrium calculation gives 0.04092 moles of gas with its constitutions as $H_{2}:H_{2}O:SiO = 0.998887:4.833 \bullet 10^{- 4}:4.832 \bullet 10^{- 4}$. The corrosion rate of SiO2(s) is thus $0.04092x4.832 \bullet 10^{- 4} = 1.98 \bullet 10^{- 5}mol/min = 1.19gram/min.$
Another example is to consider that the flowing $CO$ gas of 298K and 1atm ($= 101325Pa$) at a rate of $1 liter/min$ passes through and equilibrates with single phase $C(s)$ at 1500K. The equilibrium system is defined by $T = 1500K$, $P = 1atm$, and 0.0409 moles of $CO$ with the equilibrium between the gas phase and C(s). The equilibrium calculation gives 0.040872 moles of gas phase with the mole fraction of $CO$ being 0.999327, resulting in the loss of CO or the deposition of C(s) at a rate of $0.0409 - 0.040872 \bullet 0.999327 = 5.55 \bullet 10^{- 5}mole/min$.
A chemical transport reaction is a reaction in which a condensed phase reacts with a gas phase to form vapour-phase products, which in turn undergo the reverse reformation of the condensed phase. Two well-known examples of such reactions are
Eq. 7‑19 $M(s) + \frac{n}{2}I_{2}(g) = MI_{n}(g)$
Eq. 7‑20 $Ni(s) + 4CO(g) = Ni{(CO)}_{4}(g)$
which are used in the purification of metals by the iodide process and in the purification of nickel by the Mond-Langer process. In both processes the forward reaction is favoured by lower temperatures and the reverse reaction by higher temperatures, resulting in the deposition of the metal. The most common technique for causing a chemical transport of a condensed substance makes use of the temperature dependence of the equilibrium constant. As was discussed previously, the enthalpy of reaction, $\mathrm{\Delta}{ }^{0}H$, determines the manner in which $K$ increases with increasing }$ changes with temperature (see ). The value of $K_{eT for $\mathrm{\Delta}{ }^{0}H > 0$, $K$ decreases with increasing T for $\mathrm{\Delta}{ }^{0}H < 0$, and $K}$ is independent of T for $\mathrm{\Delta{ }^{0}H = 0$. The $\mathrm{\Delta}S$ values for chemical transport reactions may be either positive or negative. For reactions by both are positive, and for reactions by both are negative.}^{0}H$ and $\mathrm{\Delta}_{ }^{0
In a typical experiment the starting solid is located at the point in a temperature gradient that corresponds to the largest $K_{e}$ value for the experimental condition. As the gaseous species migrate to other locations in the system with temperatures corresponding to lower $K_{e}$ values, the reverse reaction occurs to satisfy the new equilibrium requirements, and the solid phase is deposited. The dependence of $K_{e}$ on $\mathrm{\Delta}{ }^{0}H$ results in a material transport from hot to cold for $\mathrm{\Delta}H > 0$ (the same as for vaporization-condensation reactions), from cold to hot for $\mathrm{\Delta}}^{0{ }^{0}H < 0$, and no transport for $\mathrm{\Delta}H = 0$.}^{0
The success of a particular reaction in causing an appreciable transport of a condensed phase depends mainly upon the partial pressure gradients or concentration gradients of the gaseous species in the system. A reaction whose equilibrium is extreme toward either the reactant side or the product side will not give an appreciable transport of material. The concentration gradients are too small in such a system. Reactions with equilibrium constants near unity at the experimental temperatures usually give the largest transport since small changes in $K_{e}$ cause large changes in concentrations. The general condition required for obtaining a $K_{e}$ value near unity at a reasonable temperature is that $\mathrm{\Delta}{ }^{0}H$ and $\mathrm{\Delta}S$ both have the same sign, resulting from the equalities of .}^{0
Trends of entropies of reactions¤
The reaction entropy, $\mathrm{\Delta}_{ }^{0}S$ in , plays an important role in determining equilibria of high-temperature reactions. The most important single factor that determines the entropy of a reaction is the net change in the number of moles of gas as briefly mentioned in the discussion of the Ellingham diagram above. The reason this is true can be explained as follows.
The entropy of a substance can be thought of as being the sum of four parts: (i) translational, (ii) rotational, (iii) vibrational, and (iv) electronic. The translational entropy of a gas is the largest entropy term under most conditions. To the extent that the other contributions cancel between reactants and products, the entropy of reaction is determined by the change in the number of moles of gaseous molecules. Based on the literature data or calculations from the SSUB database, the net change in the number of moles of gas in a reaction results approximately in an entropy of reaction of about 175±45 J/K/mole-gas at 298K for many halides and oxides. The chemical reactions of and discussed above both reduce the gas by one mole, and their entropies of reaction are -113 and -172 J/K at 1273K, and -89 and -173J/K at 298K, respectively, indicating that the chemical reaction of is an exception of the empirical rule. For chemical reactions shown in the Ellingham diagram, their entropies of reaction follow this empirical rule pretty well with some of them shown in calculated from the SGTE database.
Table ‑: Entropies of reactions with gas at 298.15K, J/K
Reaction: Si+O2 =SiO2 -182
Reaction: Ti+O2=TiO2 -185
Reaction: 2Mg+O2=2MgO -217
Reaction: 2Ca+O2=2CaO -212
Reaction: 2Mn+O2=2MnO -150
Since the entropy of a reaction is primarily determined by the net change in the number of moles of gas, the entropies for reactions involving only condensed phases must be small. The entropies of fusion of monatomic solids are usually in the range 8-15 J/K/mole-atom as shown for some elements in . Most metals and many ionic salts have values that lie in this range when given in terms of per mole of atom of material. There are few exceptions such as silicon and boron shown in the table. For solid-state reactions, the average values can be approximated as 0±8J/K/mole-atom as also shown in the table.
Table ‑: Entropies of reactions of condensed phases at 298.15K, J/K
Reaction: Si(s)=Si(l) 29.762
Reaction: Ti(s2)=Ti(l) 7.288
Reaction: Mg(s2)=Mg(l) 9.184
Reaction: Ca(s2)=Ca(l) 7.659
Reaction: Mn(s2)=Mn(l) 11.443
Reaction: W(s)=W(l) 14.158
Reaction: B(s)=B(l) 21.380
Reaction: 3Fe+C=CFe3 17.060
Reaction: S+Mn=MnS 13.909
Reaction: NiO+Fe2O3=Fe2NiO4 0.464
Aqueous solution and Pourbaix diagram¤
The importance of aqueous solutions in all aspects of life is so well known and needs not be discussed further. Since many electrochemical processes involve electrolyte solutions in an aqueous solvent, electrochemical processes including water, hydrogen, and/or oxygen are discussed in more details. The hydrogen-oxygen cell can be described for both acidic electrolytes and alkaline electrolytes. With acidic electrolytes, H+ is in much higher concentrations than OH-, and thus half-cell reactions with H+ as an ionic transport species are more important than those involving OH-. The reverse is true for alkaline electrolytes that contain high OH-concentrations. Other than for nearly neutral acid-base systems, either H+ or OH- dominates the other by several orders of magnitude as can be seen from the value of the 298 K dissociation constant for H2O:
Eq. 8‑35 H2O(l) = H+(aq) + OH-(aq)
with the reaction constant being Ke = [H+][OH-] = 10-14 and $\mathrm{\Delta}_{ }^{0}G$= -RT ln Ke β= +79,908 J. By convention, one defines pH = - log [H+] and pOH = - log [OH-], and then pH + pOH = 14.
Under acidic electrolyte conditions of low pH (high [H+] concentrations) the anode reaction in a hydrogen-oxygen cell is:
Eq. 8‑36 ½ H2(g) = H+(aq) + e-
with ε1o = 0.0 V and $\mathrm{\Delta}{ }^{0}G$= 0 J. The corresponding cathode (reduction) reaction is:
Eq. 8‑37 2 H+(aq) + ½ O2(g) + 2 e- = H2O(l)
with ε2o = 1.229 V and $\mathrm{\Delta}{ }^{0}G$ = -2*1.229*96,485 J = -237,160 J. The net cell reaction for acidic electrolytes is:
Eq. 8‑38 H2(g) + ½ O2(g) = H2O(l)
with εocell = 1.229 V and $\mathrm{\Delta}{ }^{0}G$ = -2*1.229*96,485 J = -237,160 J
Under alkaline electrolyte conditions of high pH (high [OH-] concentrations) the anode reaction in a hydrogen-oxygen cell is:
Eq. 8‑39 2 OH-(aq) + H2(g) = 2 H2O(l) + 2 e-
with ε3o = 0.828 V and $\mathrm{\Delta}{ }^{0}G$= -2*0.828*96,485 J = -159,779 J. The corresponding cathode (reduction) reaction is:
Eq. 8‑40 H2O(l) + ½ O2(g) + 2 e- = 2 OH-(aq)
with ε4o = 0.401 V and $\mathrm{\Delta}{ }^{0}G$ = -2*0.401*96,485 J = -77,381 J. The net cell reaction for alkaline electrolytes is:
Eq. 8‑41 H2(g) + ½ O2(g) = H2O(l)
with εocell = 1.229 V and $\mathrm{\Delta}{ }^{0}G$ = -2*1.229*96,485 J = -237,160 J.
Plots of ε versus pH for a given chemical system have been typically used to exhibit the stability relationships of ionic species and solid phases in aqueous-based electrochemical systems. These graphs are often called Pourbaix diagrams after the inventor and are at constant temperature and constant pressure diagrams for a constant concentration, usually for one metallic element. By convention, the ε in a Pourbaix diagram corresponds to the potential for the cathode reduction reactions in the electrochemical half-cell with electrons as reactants. Pourbaix diagrams can be extended to multi-component materials when thermodynamic properties of the components are available in both the materials and the aqueous solution.
An example of an ε versus pH diagram is shown in for the Ni-H2O system at a 298K, 1 bar, and $c_{{Ni}^{2 +}} = 0.001$ molality. Three stability regions for Ni species are shown: Ni(s), NiO(s), and [Ni2+]. The two dashed lines on this diagram correspond to hydrogen reduction () and oxygen reduction () reactions, respectively.
Figure ‑: An ε versus pH, Pourbaix diagram for the Ni-H2O at 298K, 1 bar, and $c_{{Ni}^{2 +}} = 0.001$ molality.
For the ε and pH conditions within the boundaries of the Ni(s) region, no solid phase other than Ni(s) is stable, no ionic species with a concentration of 1 molarity is stable, and no gas species with a pressure of 1 bar is stable. Similar statements could be made about the NiO(s) and [Ni2+] areas on the diagram. In the [Ni2+] area, introduction of Ni(s) or NiO(s) into the system would result in the dissolution of these solid phases since they are not stable with respect to the [Ni2+] aqueous solution. The corresponding chemical reactions proceed spontaneously to the right as follows until the solid phases are consumed:
Eq. 8‑42 Ni(s) → Ni2+(1 molarity) + 2 e-
Eq. 8‑43 NiO(s) + 2 H+(aq) → Ni2+(1 molarity) + H2O(l)
No H+(aq) in involved in the first reaction, , so the boundary line separating Ni(s) and Ni2+ is independent of pH. No oxidation or reduction occurs in the second reaction, , i.e. no electrons are reactants or products in the reaction, the boundary line separating NiO(s) and Ni2+ is independent of ε.
Note the convention that the ε is the potential for a cathode reduction reaction, and boundary lines between two stability regions depict conditions under which partial equilibrium occurs for the two species for the ε and pH values at any point on these lines. For the boundary line separating Ni(s) and Ni2+ in an ideal aqueous solution, i.e. the reverse of , the following equation is obtained.
Eq. 8‑44 ε = εo = -0.268 V
For the NiO(s)-Ni2+ boundary line of an ideal solution, the reaction, , is a complete equilibrium, and thus the relationship is
Eq. 8‑45 $0 = \mathrm{\Delta}{ }^{0}G + RTln\frac{1}{\left( cG + 2 \cdot 2.303 \cdot RT \cdot pH$}} \right)^{2}} = \mathrm{\Delta}_{ }^{0
Eq. 8‑46 $pH = - \frac{\mathrm{\Delta}_{ }^{0}G}{2 \cdot 2.303 \cdot RT}$
where $\mathrm{\Delta}_{ }^{0}G$ is obtained as follows and can be calculated from the SSUB database and the standard potential of Ni,
Eq. 8‑47
At a specified temperature, only one standard free energy and only one equilibrium constant exists for this chemical reaction, and thus only one specific value of $pH = 6.631$ exists for the reaction represented by in this Pourbaix diagram.
The diagonal line in represents the equilibrium between Ni(s) and NiO(s) and is for a partial equilibrium reaction that is the sum of reactions of and
Eq. 8‑48 NiO(s) + 2 H+(aq) + 2 e- ═ Ni(s) + H2O(l)
The reduction of Ni from a divalent state in NiO to metallic Ni(s) occurs, but the reaction also depends on the H+ concentration, the pH. The corresponding Gibbs energy and Nernst equations are,
Eq. 8‑49 $\mathrm{\Delta}G = \mathrm{\Delta}{ }^{0}G + RTln\frac{1}{\left( c = - 23,939 + 2 \cdot 2.303 \cdot RT \cdot pH$}} \right)^{2}
Eq. 8‑50 $\varepsilon = \varepsilon^{0} - \frac{RT}{2f}\ln\frac{1}{\left( c_{H^{+}} \right)^{2}} = 0.124 – \frac{2.303 \cdot RT}{f}pH$
where $\mathrm{\Delta}_{ }^{0}G$ can be calculated as follows
Eq. 8‑51
The two additional lines in correspond to the reduction reactions related to H2 and O2 gases, i.e. the stability of H2O. The lower one is for the reverse of under εo = 0 and $P_{H_{2}} = 1$ with the Nernst equation being
Eq. 8‑52 $\varepsilon = \varepsilon^{0} - \frac{RT}{f}\ln\frac{\left( P_{H_{2}} \right)^{1/2}}{c_{H^{+}}} = - \frac{2.303 \cdot RT}{f}pH$
As the pH increases from 0, ε becomes more negative as is depicted. The top dashed line corresponds to the oxygen reduction reaction represented by under εo = 1.225 calculated from the aqueous solution database in Thermo-Calc [60] and $P_{O_{2}} = 1$ with the Nernst equation being
Eq. 8‑53 $\varepsilon = \varepsilon^{0} - \frac{RT}{2f}\ln\frac{\left( P_{O_{2}} \right)^{1/2}}{\left( c_{H^{+}} \right)^{2}} = 1.225 – \frac{2.303 \cdot RT}{f}pH$
The dependence of ε on pH is identical for both reduction reaction and , and their intercepts at $pH = 0$ differ by their difference in their εo values.
In this simple Pourbaix diagram of Ni in an ideal aqueous solution, all boundary lines are straight because there is only one ionic species of Ni in the aqueous solution, i.e. Ni2+. When there are more than one ionic species in the aqueous solution, the boundary lines may no longer be straight due to the competition between species. One example is Cu with two main ionic species of Cu+2 and CuOH+, and the reduction reaction between the metallic Cu and the aqueous solution involves both two species, i.e.
Eq. 8‑54 ${xCu}^{2 + } + (1 - x){CuOH}^{+} + (1 - x)H^{+} + 2 e - = Cu(s) + {(1 - x)H}_{2}O$
with
Eq. 8‑55 $\mathrm{\Delta}G = \mathrm{\Delta}{ }^{0}G + RTln\frac{1}{\left( c}^{2 +}} \right)^{x}\left( c_{{CuOH}^{+}}c_{H^{+}} \right)^{1 - x}} = \mathrm{\Delta{ }^{0}G + RTln\frac{1}{\left( c + 2.303(1 - x) \cdot RT \cdot pH$}^{2 +}} \right)^{x}\left( c_{{CuOH}^{+}} \right)^{1 - x}
Eq. 8‑56 $\varepsilon = \varepsilon^{0} - \frac{RT}{2f}\ln\frac{1}{\left( c_{{Cu}^{2 +}} \right)^{x}\left( c_{{CuOH}^{+}} \right)^{1 - x}} - \frac{2.303(1 - x) \cdot RT}{2f}pH$.
It is evident that both the slope and the intercept at $pH = 0$ are a function of the concentration of ${CuOH}^{+}$, which is a function of $pH$. Consequently, the boundary between the metallic Cu and the aqueous solution is no longer a straight line as shown in .
Figure ‑: An ε versus pH, Pourbaix diagram for the Cu-H2O system at 298K, 1 bar, and $c_{Cu} = 0.001$ molality.
The concentrations of various species in the aqueous solution, i.e. commonly called speciation, are plotted in , showing the change of dominant species as a function of pH value.
Figure ‑: Concentrations of ionic species in the aqueous solution at $\varepsilon = 0.3 V$ from .
In Pourbaix diagrams for alloys with two or more elements, activities of individual elements are to be used in calculating the potentials of reduction reactions. Considering a Fe-Ni alloy with Fe2+ and Ni2+ in the aqueous solution, the reduction reactions for Fe and Ni can be written separately as
Eq. 8‑57 Ni2+(cNi) + 2 e- → Ni (aNi in alloy)
Eq. 8‑58 Fe2+(cFe) + 2 e- → Fe (aFe in alloy)
with their potentials as
Eq. 8‑59 $\varepsilon_{Ni} = \varepsilon_{Ni}^{0} - \frac{2.303RT}{2f}\ln\frac{a_{Ni}}{c_{Ni}} = - 0.268 - \frac{2.303RT}{2f}\ln\frac{a_{Ni}}{c_{Ni}}$
Eq. 8‑60 $\varepsilon_{Fe} = \varepsilon_{Fe}^{0} - \frac{2.303RT}{2f}\ln\frac{a_{Fe}}{c_{Fe}} = - 0.441 - \frac{2.303RT}{2f}\ln\frac{a_{Fe}}{c_{Fe}}$
In principle, there are two scenarios for a given set of $a_{Ni}$ and $a_{Fe}$ of the alloy. The first scenario is at the limit of a dilute aqueous solution, i.e. $c_{Ni} = c_{Fe} = 0.001 $molarity, $\varepsilon_{Ni}$ and $\varepsilon_{Fe}$ can be calculated, and the element with the lower potential has the tendency to dissolve first, which can result in the so-called dialloying effect. The second scenario is for the equal potential, i.e. $\varepsilon_{Ni} = \varepsilon_{Fe}$ due to the externally imposed potential, and the equilibrium concentrations of Fe+2 and Ni+2 can be calculated from and .
Concentrations, activities, and reference states of electrolyte species¤
Thermodynamic descriptions of ionic species in solutions are different from those of neutral species, which leads to a need for defining concentration units, standard states, activities, and activity coefficients of ionic solutions. In most studies of electrochemical corrosion and electrodeposition, and in applied work of electrochemical engineers, ionic species concentrations are given in units of molarity, the number of moles of a species in a liter of solution (mol/l) symbolically represented in equations by either ci or [M+Z]. The other common concentration used for ionic species is molality, which is defined as the number of moles of a species in 1000g of solvent. For dilute aqueous solutions, molarity and molality values are very similar.
As discussed in Chapter , a practical definition of the activity of a species i is the thermodynamic reactivity, or tendency to react, of species i in the system of interest as compared to i in its reference state form. The reference state of a species is typically chosen as a specific chemical/physical state of the species at 1 atm external pressure and the temperature of interest. Similarly, a typical reference state for ionic species in aqueous solutions is the 1 molar ideal solution at 1 bar external pressure and the temperature of interest. If an electrolyte solution behaves ideally, then the activity of species i in solution is
Eq. 8‑6 $a_{i} = \frac{c_{i}\left( \frac{mol}{l} \right)}{c_{i}^{0}\left( \frac{mol}{l} \right)} = \frac{c_{i}\left( \frac{mol}{l} \right)}{1\left( \frac{mol}{l} \right)} = c_{i}(dimensionless)$
where $c_{i}$ is the molar concentration of i in the solution divided by $c_{i}^{0}$, the 1 molar reference state ideal solution concentration. Thus, in ideal solutions, the activity of an electrolyte species is numerically equal to its molar concentration. The above treatment of ionic species is equivalent to the common practice of depicting the activity of a gas by the value of its ideal gas partial pressure in units of bar.
The activity coefficient corrects for the nonideality of the species in solution as defined in . If the solution is ideal, $\gamma_{i} = 1$ for all concentrations of a species in solution. For all solutions, one expects $\gamma_{i} \rightarrow 1$ as $c_{i} \rightarrow 1$. It is not possible to measure $\gamma_{i^{+}}$ or $\gamma_{i^{-}}$ for individual charged ions, only a geometric mean of the positive and negative ion values. Consider the following ionic solution
Eq. 8‑7
Its chemical potential can be written as
Eq. 8‑8
Its geometric average or mean activity and activity coefficient are defined as
Eq. 8‑9
Eq. 8‑10
For example, one can define $\gamma_{\pm} = \left( \gamma_{{Na}^{+}}\gamma_{{Cl}^{-}} \right)^{1/2}$ and $\gamma_{\pm} = \left( \gamma_{{Al}^{3 +}}^{2}\gamma_{{{SO}{4}}^{2 -}}^{3} \right)^{1/5}$ for NaCl and Al2(SO4)3, respectively. For idea, weak electrolytes, $\gamma = 1$, and for non-ideal, strong electrolytes, $\gamma_{\pm} \neq 1$.
Electrical batteries¤
Batteries utilize electrochemical reactions to generate electricity for various devices. The theoretic voltage of a battery can be calculated from and , i.e.
Eq. 8‑73 $\varepsilon = - \frac{\mathrm{\Delta}G}{zf} = \varepsilon^{0} - \frac{RTlnQ}{zf}$
with $\mathrm{\Delta}G$ being the driving force of the net cell reaction and $Q$ being the reaction activity quotient. The actual voltage of a battery is lower than the theoretical one due to kinetic limitations of cell reactions and resistance to ion diffusion through the electrolyte. Based on whether the cell reactions are reversible or not, batteries typically categorized as either primary disposable or secondary rechargeable batteries. The net cell reactions in primary disposable batteries are not easily reversible, and electrode materials may not return to their original forms by applying a higher external potential of opposite sign. Consequently, primary batteries cannot be reliably recharged. On the other hand, the net cell reactions in secondary batteries are easily reversible. Furthermore, two half-cells in batteries may use different electrolytes with each half-cell enclosed in a container and a separator permeable to conducting ions but not the bulk of the electrolytes.
One common primary battery is zinc-carbon battery with a zinc anode cylinder and a carbon cathode central rod. The electrolytes are ammonium or zinc chloride next to the zinc anode and a mixture of ammonium chloride and manganese dioxide next to the carbon cathode. The half cell and net reactions with ammonium chloride are as follows
Eq. 8‑74 Zn + 2NH3 → Zn(NH3)22+ + 2 e-
Eq. 8‑75 2NH4Cl + 2MnO2 + 2 e- → 2NH3 + Mn2O3 + H2O+2Cl−
Eq. 8‑76 Zn + 2MnO2 + 2NH4Cl → Mn2O3 + Zn(NH3)2Cl2 + H2O.
The electric potential of the reaction is, treating all compounds as stoichiometric compounds
Eq. 8‑77 $\varepsilon = - \frac{\mathrm{\Delta}G}{2f} = - \frac{\mathrm{\Delta}^{0}G}{2f} = \frac{1}{2f}\left({ }^{0}G^{Zn} + 2}^{0}G^{{MnO{2}} + 2}^{0}G^{{NH{4}Cl} - -}^{0}G^{H_{2}O{ }^{0}G^{Zn\left( {NH} \right){2}{Cl} -}{ }^{0}G^{{{Mn} \right)$.}O}_{3}
The Gibbs energy of $Zn\left( {NH}{3} \right)$ is not available in current databases and has been recently estimated to be −505,375 J/mole-formula [62]. The value of at 298.15K is thus obtained as 1.67 V, which is pretty close to the actual operating voltage of the battery around 1.5 V.}{Cl}_{2
While with zinc chloride, the cell reactions and electric potential may be written as
Eq. 8‑78 Zn + ZnCl2 + 2OH− → 2ZnOHCl + 2 e-
Eq. 8‑79 MnO2 + H2O + e- → MnOOH + OH-
Eq. 8‑80 Zn + 2 MnO2 + ZnCl2 + 2 H2O → 2 MnOOH + 2 ZnOHCl
Eq. 8‑81 $\varepsilon = \frac{1}{2f}\left({ }^{0}G^{Zn} + 2}^{0}G^{{MnO{2}} +}^{0}G^{Zn{Cl{2}} + 2 \right)$}^{0}G^{H_{2}O} - 2_{ }^{0}G^{MnOOH} - 2_{ }^{0}G^{ZnOHCl
Secondary batteries can be recharged by applying an external electrical potential, which reverses the net cell reaction that occur during discharging. The oldest form of rechargeable battery is the lead-acid batteries used in automotive, and the latest development is the lithium-ion (Li-ion) batteries. A lead-acid battery typically uses Pb and PbO2 as the cathode and anode electrodes and a 35% sulfuric acid and 65% water solution as the electrolyte. Its anode and cathode reactions can be simplified as follows
Eq. 8‑82 $Pb + SO_{4}^{2 -} = PbSO_{4} + 2e^{-}$
Eq. 8‑83 $PbO_{2} + 4H^{+} + SO_{4}^{2 -} + 2e^{-} = PbSO_{4} + 2H_{2}O$
The net cell reaction is
Eq. 8‑84 $Pb + PbO_{2} + 2H_{2}SO_{4}^{ } = 2PbSO_{4} + 2H_{2}O$.
Its electric potential is represented by the following equation
Eq. 8‑85 $\varepsilon = - \frac{1}{2f}\left( 2_{ }^{0}G^{H_{2}O} + 2_{ }^{0}G^{{PbSO}{4}} - -}^{0}G^{Pb{ }^{0}G^{{PbO} \right)$}} - 2_{ }^{0}G^{H_{2}SO_{4}
with the value at 298.15K being 2.651 V calculated from Thermo-Calc [60]. During discharging, the reaction goes to right, and $PbSO_{4}$ is formed on both anode and cathode, while during charging, the reaction goes to the left, and $Pb$ and $PbO_{2}$ are restored. In practical applications, other ionic species such as ${H_{3}O}^{+}$ and $HSO_{4}^{-}$ may form in the electrolyte, complicating the reactions and affecting its potential.
In lithium ion batteries, lithium ions migrate in electrolytes between electrodes made of intercalated lithium compounds during charging and discharging. LiCoO2 and LiFePO4 are two of the several cathode materials used in lithium ion batteries, and the anode is typically made of carbon or metallic Li. The anode and cathode reactions for LiCoO2 batteries can be written in simple forms as follows
Eq. 8‑86 ${Li}{x}C + 6C$} = x{Li}^{+} + xe^{-
Eq. 8‑87 $x{Li}^{+} + xe^{-} + {Li}{1 - x}CoO$} = LiCoO_{2
with the net reaction and electric potential being
Eq. 8‑88 ${Li}{x}C} + {Li{1 - x}CoO + 6C$} = LiCoO_{2
Eq. 8‑89 $\varepsilon = - \frac{1}{xf}\left{ 6_{ }^{0}G^{C} +{ }^{0}G^{{LiCoO}}} - G^{{{Li{x}C}}} - G^{{{Li{1 - x}CoO}}} \right} = - \frac{1}{f}\left{ \left( \mu_{Li}^{{Li{1 - x}CoO}} - \mu_{Li}^{{Li{x}C} \right) - \frac{1}{x}\left( \mu}}^{{Li{1 - x}CoO -}{ }^{0}G^{{LiCoO} \right) \right}$}
The electric potential is a function of $x$. The value in the first parenthesis in the above equation denotes the chemical potential difference of Li between two electrodes, and the value in the second parenthesis represents the chemical potential difference of ${LiCoO}{2}$ between the states in the solution phase of ${Li}}CoO_{2}$ and by itself. Gibbs energies of ${{Li{x}C}$ and ${{Li}{1 - x}CoO}$ need to be obtained as a function $x$ in order to calculate the electric potential of the battery.
LiFePO4 uses metallic lithium as the anode with following half-cell and net cell reactions
Eq. 8‑90 $xLi = x{Li}^{+} + xe^{-}$
Eq. 8‑91 $x{Li}^{+} + xe^{-} + {Li}{1 - x}FePO$} = LiFePO_{4
Eq. 8‑92 $xLi + {Li}{1 - x}FePO$.} = LiFePO_{4
Its electric potential is also a function of $x$, i.e.
Eq. 8‑93 $\varepsilon = - \frac{1}{xf}\left{{ }^{0}G^{LiFePO}} - x ^{0}G^{Li} - G^{{Li{1 - x}FePO}} \right} = - \frac{1}{f}\left{ \left( \mu_{Li}^{{Li{1 - x}FePO}} - ^{0}\mu_{Li} \right) - \frac{1}{x}\left( \mu_{LiFePO_{4}}^{{Li{1 - x}FePO -}{ }^{0}G^{LiFePO \right) \right}$}
The value in the first parenthesis in the above equation denotes the chemical potential difference of Li between two electrodes, and the value in the second parenthesis represents the chemical potential difference of $LiFePO_{4}$ between the states in the solution phase of ${Li}{1 - x}FePO$ and by itself. Consequently, Gibbs energy of ${Li}{1 - x}FePO$ needs to be obtained as a function $x$ in order to calculate the electric potential of the battery. It is known that there are several miscibility gaps in the $FePO_{4}$ and $LiFePO_{4}$ psuedo-binary system, in which the chemical potentials are constants, so is the electric potential, resulting in stable battery output.
Electrolyte reactions and electrochemical reactions¤
Electrolytes that dissolve in a polar solvent such as water to produce ionic species do not necessarily exhibit valence changes. A simple example is the strong electrolyte NaCl(s) dissolving in water to produce solvated ions
Eq. 8‑1 $NaCl(s) = {Na}^{+}(aq) + {Cl}^{–}(aq)$
where the $(aq)$ indicates the ionic species in an aqueous solution. In this system, the ion concentrations must become quite large before the solution is saturated and can exist in equilibrium with $NaCl(s)$. Its reaction constant, defined by , is shown as $K_{e} = a_{{Na}^{+}}a_{{Cl}^{-}}$. If the product of the ion activities is less than $K_{e}$, the solution is not saturated, and more $NaCl(s) $ can be dissolved.
The precipitation of $AgCl(s)$, a weak electrolyte, occurs quite readily when ${Cl}^{–}$ ions are added to an aqueous solution containing ${Ag}^{+}(aq)$:
Eq. 8‑2 ${Ag}^{+}(aq) + {Cl}^{–}(aq) = AgCl(s)$
The equilibrium constant for this reaction, $K_{e} = \frac{1}{\left( a_{{Ag}^{+}}a_{{Cl}^{-}} \right)}$ is quite large, so the equilibrium product of the ion activities, proportional to their concentrations, is quite small. In the laboratory, the above reaction could occur as a result of adding hydrochloric acid to a silver nitrate solution. The accompanying $H^{+}(aq)$ [or ${H_{3}O}^{+}(aq)$] and ${NO_{3}}^{-}(aq)$ ions in solution are not directly involved in the silver chloride precipitation reaction so are not shown in reaction represented by .
The above ionic equilibria in the $AgCl(s) - H_{2}O$ system is not only important for understanding this electrolyte system, but also critical in electrochemical systems in which Ag(s) undergoes a valence change at one electrode and reacts with a ${Cl}^{–}(aq)$ ion to produce AgCl(s), and an electron that is externally transported finite distances to another electrode. The oxidation reaction occurs at the Ag/AgCl electrode (anode half-cell reaction where electrons are added into the system)
Eq. 8‑3 $Ag(s) + {Cl}^{-}(aq) = AgCl(s) + e^{-}$
A reduction reaction occurs at the other electrode (cathode half-cell reaction where electrons are consumed by the reaction)
Eq. 8‑4 $\frac{1}{2}{Cl}_{2}(g) + e^{-} = {Cl}^{-}(aq) $
The net cell reaction results in the formation AgCl(s) from its elements
Eq. 8‑5 $Ag(s) + \frac{1}{2}{Cl}_{2} (g) = AgCl(s)$
Without knowledge of the physical system under which the reaction is occurring, it would not be possible to know if reaction of was a result of chlorine gas reacting directly with Ag(s), or if the reaction was part of an electrochemical cell with a transport of electrons and ions over finite distances. The addition of the two half-cell reactions gives the net cell reaction, which does not show electrons as either reactant or product species and may or may not include ionic species in the reaction. A schematic of an electrochemical cell for the above system is shown in Figure 8‑1.
Figure ‑: Schematic diagram of an electrochemical cell consisting of a chlorine electrode and a silver-silver chloride electrode.
Oxidation and reduction can occur in electrolyte reactions without creating an electrochemical cell. This is the case when chlorine gas reacts directly with silver on a Ag(s) surface. Reaction of above is the net reaction for this process, but the electrons produced from the oxidation of Ag(s) are not transported over finite distances before combining with Cl2(g) in its reduction to Cl–(aq). No anode or cathode half-cell reactions exist in this system. The electrons and ions involved in the reaction move only over atomic-scale distances.
Fuel cells¤
Fuel cells are devices to convert chemical energy to electricity and heat through electrochemical reactions with the fuel and oxygen supplied to the anode and cathode, respectively. Typical ions migrating through the electrolyte are $H^{+}$, ${OH}^{-}$, ${CO}_{3}^{2 -}$, and $O^{2 -}$. In fuel cells with $H^{+}$ as migrating ions, H2 molecules are dissociated into $H^{+}$ on the anode, which are combined with O2 on the cathode to form H2O and release heat with the half cell and the net cell reactions in their simplest form as shown by for the anode, for the cathode, and for the net cell, respectively. Commonly used electrolytes are polymer and phosphoric acid, and both anode and cathode reactions are facilitated by catalyst, typically platinum. The thermodynamic limit of power which can be generated by the fuel cell is represented by
Eq. 8‑63 $w = - \mathrm{\Delta}G = - \mathrm{\Delta}{ }^{0}G \right)$} + RTln\left( P_{H_{2}}P_{O_{2}}^{1/2
For fuel cells with anions as migrating ions, the anions are generated on the cathode with H2O formed and heat generated on the anode. Their representative cathode reactions are
Eq. 8‑64 $\frac{1}{2}O_{2} + H_{2}O + 2e^{-}{= 2OH}^{-}$
Eq. 8‑65 ${\frac{1}{2}O}{2} + CO$} + 2e^{-} = {CO}_{3}^{2 -
Eq. 8‑66 $\frac{1}{2}O_{2} + 2e^{-} = O^{2 -}$.
The anode reaction for is the reaction represented by , operating at low temperatures and using catalyst for both electrodes. The anode reactions for and are
Eq. 8‑67 ${CO}{3}^{2 -} + H$} = H_{2}O + CO_{2} + 2e^{-
Eq. 8‑68 $O^{2 -} + H_{2} = H_{2}O + 2e^{-}$
respectively. To enable the diffusion of ${CO}_{3}^{2 -}$ and $O^{2 -}$ through the cathode and the electrolyte, both fuel cells are operated at relative high temperatures, with the former typically in molten carbonate solutions and the latter through solid oxides. Due to the high operating temperatures, fuels are converted to hydrogen within the fuel cell itself by a process called internal reforming, removing the need for precious-metal catalyst and enabling the use of a variety of fuels. The net cell reaction for all three fuel cells is the same as in the case of $H^{+}$, represented by .
Half cell potentials¤
When electron current flows between electrodes, reactions are occurring at the electrodes and concentration gradients causing polarization develop around the electrodes. These gradients result in extraneous potentials to occur at the electrodes. In such cases cell equilibrium is not established and measured cell potentials are not those for true partial equilibrium. If a cell is short-circuited with the electrodes connected by a conductor, current will flow until the external potential becomes zero, i.e. εext = 0, and equilibrium is established with same conditions as non-electrochemical systems. If an external potential, εext, is applied to the cell, chemical reactions occur until the cell potential balances to εext, and no current flows. This potential is called open-circuit voltage (OCV) in the literature. It is important to realize that OCV includes all reactions that occur on the electrode surface when the electrode is in contact with the electrolyte, such as passivation discussed in Chapter . Partial equilibrium in a cell is achieved when the cell potential is balanced by an applied external potential. In such partial equilibrium cases, equilibrium thermodynamic analyses can be used even though the cell potential is not zero, i.e. εcell ≠ 0. This differentiates electrochemical systems from other equilibrium systems discussed previously.
The number of electrons involved in a net cell reaction is important in relating cell potential and the Gibbs energy change for the cell reaction. As will be illustrated later in this section, this number denotes the number of electrons involved in the half-cell reactions that were added to yield the net cell reaction. The electrical work achieved by the transport of an electrical charge through a cell potential can be written as
Eq. 8‑20 $w = z f \varepsilon$
where z represents the moles of electrons in cell reaction, f the Faraday constant equal to 96,485 J/V/mole-electron, and ε the potential difference, often referred as electromotive force (emf) in the literature. For a system at constant temperature, pressure, and composition, this work is the same as the Gibbs energy difference between the two electrodes, i.e.
Eq. 8‑21 $- \Delta G = w = zf\varepsilon$
where the negative sign is because the system does work to the surrounding when the Gibbs energy of the system is decreased. When the applied external potential is larger than the cell potential, the surrounding does work to the system, and a common example is the charging of a battery. Thermodynamic relations discussed in previous chapters can thus be directly applied to electrochemical systems with some common equations shown in .
Table ‑: Thermodynamic Equations for Electrochemical Cells
∆G = -z f ε
∆S = - (∂∆G/∂T)P = + z f (∂ε/∂T)P
∆H = [∂(∆G/T)/ ∂(1/T)]P = - z f [∂(ε/T)/∂(1/T)]P = z f [T(∂ε/∂T)P – ε]
∆CP = (∂∆H/∂T)P = T z f (∂2ε/∂T2)P
A half-cell reaction potential cannot be measured directly, only its potential relative to another half-cell reaction. By convention, a standard half-cell potential is measured relative to the standard hydrogen half-cell reduction reaction at 25oC (298K) and 1 bar, which has a defined standard potential of zero volts,
Eq. 8‑22 H+(aq, a=1) + e- = 1/2 H2(g, 1 bar)
with εo (H+/H2,g) = 0.00 volts. The standard half-cell reduction reactions of metals at 25oC are for the general reaction
Eq. 8‑23 Mz+(aq, a=1) + z e- = M(s)
with εo (Mz+/M) volts. Half-cell reactions with the most positive standard electrode potentials have a tendency to spontaneously proceed toward reduction (cathode reactions). Half-cell reactions with the most negative standard electrode potentials have a tendency to spontaneously proceed toward oxidation (anode reactions).
Consider, for example, a cell made up of a standard hydrogen electrode and a standard zinc electrode with εo (H+/H2,g) = 0.00 volts and εo (Zn2+/Zn) = -0.762 volts. Thus, the H+ would tend to be reduced, and the zinc metal would tend to be oxidized, and the spontaneous reaction if all species had unit activities would be
Eq. 8‑24 2 H+(aq, a=1)+ Zn = H2(1 bar) + Zn2+(aq, a=1)
with εocell = 0.762 volts and $\mathrm{\Delta}{ }^{0}G = - 296485\varepsilon $. The cathode half-cell reaction would be the same as , while the anode half-cell reaction would be}^{0
Eq. 8‑25 Zn = Zn2+(aq, a=1) + 2 e-
When ion concentrations and H2 gas do not all have unit activities, the Gibbs energy and cell potential of the cell reaction, , becomes
Eq. 8‑26 $\mathrm{\Delta}G = \mathrm{\Delta}{ }^{0}G + RTln\frac{a$}^{2 +}}P_{H_{2}}}{\left( a_{H^{+}} \right)^{2}
Eq. 8‑27 $\varepsilon_{cell} = \varepsilon_{cell}^{0} - \frac{RT}{zf}\ln\frac{a_{{Zn}^{2 +}}P_{H_{2}}}{\left( a_{H^{+}} \right)^{2}}$
The standard reduction potentials of some common metals at 25oC are given in .
Table ‑: Standard reduction potentials of some common metals
A cell reaction can be established by different half-cell reactions. For example, the following reaction can be derived from two different cells
Eq. 8‑28 $3 {Fe}^{2 + } = 2 {Fe}^{3 +} + Fe(s)$
cell A
Eq. 8‑29 $3{Fe}^{2 +} + 6 e - = 3Fe(s)$ εo1 = -0.440 V
Eq. 8‑30 $2 Fe(s) = 2 {Fe}^{3 +} + 6 e -$ εo2 = +0.036 V
cell B
Eq. 8‑31 $2{Fe}^{2 +} = 2{Fe}^{3 + } + 2 e^{-}$ εo4 = -0.772 V
Eq. 8‑32 ${Fe}^{2 + } + 2 e - = Fe(s)$ εo5 = -0.440 V
Both give the same net reaction shown by , but with 6 and 2 electrons and standard cell potentials being εocell A = -0.404 V and εocell B = -1.212 V, respectively. However, the standard Gibbs energies of both cells are the same, i.e.
Eq. 8‑33 $\mathrm{\Delta}_{ }^{0}G$cell A = -6 f (-0.404) = + 2.424 f
Eq. 8‑34 $\mathrm{\Delta}_{ }^{0}G$cell B = -2 f (-1.212) = + 2.424 f
It is shown that $\mathrm{\Delta}_{ }^{0}G$ values are independent of half-cell reactions and depend only on the net reaction because the net reaction is neutral in electron and balanced in mass.
- extra toc. this should be generated from our content. if we do our job then the two should match.
- split the rest on headings
- split out the references which are the last element