Cambridge IGCSE IGCSE & A-Level Sheet preview

Thermodynamics and Electrochemistry

AP Chemistry · Topic 9

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9.1

Introduction to Entropy
Syllabus
Learning ObjectiveEssential Knowledge

9.1.A
Identify the sign and relative magnitude of the entropy change associated with chemical or physical processes.

  • 9.1.A.1 Entropy increases when matter becomes more dispersed. For example, the phase change from solid to liquid or from liquid to gas results in a dispersal of matter as the individual particles become freer to move and generally occupy a larger volume. Similarly, for a gas, the entropy increases when there is an increase in volume (at constant temperature), and the gas molecules are able to move within a larger space. For reactions involving gas-phase reactants or products, the entropy generally increases when the total number of moles of gas-phase products is greater than the total number of moles of gas-phase reactants.
  • 9.1.A.2 Entropy increases when energy is dispersed. According to kinetic molecular theory (KMT), the distribution of kinetic energy among the particles of a gas broadens as the temperature increases. As a result, the entropy of the system increases with an increase in temperature.

Source: College Board AP Course and Exam Description

Entropy 熵 $S$ measures the dispersal of energy and matter – loosely, the number of ways to arrange a system. Entropy increases when a substance goes solid → liquid → gas, when a solid dissolves, when gas moles increase, or when temperature rises. More disorder means higher entropy.

Entropy rises from solid to liquid to gas Entropy rises from solid to liquid to gas

Vocabulary Train
English Chinese Pinyin
Entropy 熵 shāng
Exercise sheet
9.2

Absolute Entropy and Entropy Change
Syllabus
Learning ObjectiveEssential Knowledge

9.2.A
Calculate the standard entropy change for a chemical or physical process based on the absolute entropies (standard molar entropies) of the species involved in the process.

  • 9.2.A.1 The entropy change for a process can be calculated from the absolute entropies of the species involved before and after the process occurs.
    • Equation: $\Delta S^{\circ}_{reaction} = \Sigma S^{\circ}_{products} - \Sigma S^{\circ}_{reactants}$

Source: College Board AP Course and Exam Description

Every substance has a positive absolute entropy $S^\circ$. For a reaction,

$$\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}).$$
Predict its sign from the change in gas moles: making more gas raises entropy ($\Delta S>0$).

Exercise sheet
9.3

Gibbs Free Energy and Thermodynamic Favorability
Syllabus
Learning ObjectiveEssential Knowledge

9.3.A
Explain whether a physical or chemical process is thermodynamically favored based on an evaluation of $\Delta G^{\circ}$.

  • 9.3.A.1 The Gibbs free energy change for a chemical process in which all the reactants and products are present in a standard state (as pure substances, as solutions of 1.0 M concentration, or as gases at a pressure of 1.0 atm (or 1.0 bar)) is given the symbol $\Delta G^{\circ}$.

  • 9.3.A.2 The standard Gibbs free energy change for a chemical or physical process is a measure of thermodynamic favorability. Historically, the term "spontaneous" has been used to describe processes for which $\Delta G^{\circ} < 0$. The phrase "thermodynamically favored" is preferred instead so that common misunderstandings (equating "spontaneous" with "suddenly" or "without cause") can be avoided. When $\Delta G^{\circ} < 0$ for the process, it is said to be thermodynamically favored.

  • 9.3.A.3 The standard Gibbs free energy change for a physical or chemical process may also be determined from the standard Gibbs free energy of formation of the reactants and products.

    • Equation: $\Delta G^{\circ}_{reaction} = \Sigma \Delta G^{\circ}_{f\ products} - \Sigma \Delta G^{\circ}_{f\ reactants}$

  • 9.3.A.4 In some cases, it is necessary to consider both enthalpy and entropy to determine if a process will be thermodynamically favored. The freezing of water and the dissolution of sodium nitrate are examples of such phenomena.

  • 9.3.A.5 Knowing the values of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for a process at a given temperature allows $\Delta G^{\circ}$ to be calculated directly.

    • Equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\,\Delta S^{\circ}$

  • 9.3.A.6 In general, the temperature conditions for a process to be thermodynamically favored ($\Delta G^{\circ} < 0$) can be predicted from the signs of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ as shown in the table below:

    $\Delta H^{\circ}$ $\Delta S^{\circ}$ Symbols $\Delta G^{\circ} < 0$, favored at:
    $< 0$ $> 0$ $<\ >$ all $T$
    $> 0$ $< 0$ $>\ <$ no $T$
    $> 0$ $> 0$ $>\ >$ high $T$
    $< 0$ $< 0$ $<\ <$ low $T$

    In cases where $\Delta H^{\circ} < 0$ and $\Delta S^{\circ} > 0$, no calculation of $\Delta G^{\circ}$ is necessary to determine that the process is thermodynamically favored ($\Delta G^{\circ} < 0$). In cases where $\Delta H^{\circ} > 0$ and $\Delta S^{\circ} < 0$, no calculation of $\Delta G^{\circ}$ is necessary to determine that the process is thermodynamically unfavored ($\Delta G^{\circ} > 0$).

Source: College Board AP Course and Exam Description

Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:

$$\Delta G = \Delta H - T\Delta S.$$
A process is thermodynamically favorable 热力学有利 when $\Delta G<0$. So exothermic ($\Delta H<0$) and entropy-increasing ($\Delta S>0$) reactions are always favorable; when the two oppose, temperature decides.

Whether a reaction is favorable, from the signs of the enthalpy and entropy change Whether a reaction is favorable, from the signs of the enthalpy and entropy change

Worked example. A reaction has $\Delta H=+40\ \text{kJ/mol}$ and $\Delta S=+120\ \text{J/(mol K)}$. It is endothermic (unfavorable enthalpy) but entropy-increasing, so it becomes favorable only when hot enough. Setting $\Delta G=\Delta H-T\Delta S<0$ and matching units ($\Delta S=0.120\ \text{kJ}$):

$$T>\frac{\Delta H}{\Delta S}=\frac{40}{0.120}=333\ \text{K}\;(60\,{}^{\circ}\text{C}).$$

Vocabulary Train
English Chinese Pinyin
Gibbs free energy 吉布斯自由能 jí bù sī zì yóu néng
thermodynamically favorable 热力学有利 rè lì xué yǒu lì
Exercise sheet
9.4

Thermodynamic and Kinetic Control
Syllabus
Learning ObjectiveEssential Knowledge

9.4.A
Explain, in terms of kinetics, why a thermodynamically favored reaction might not occur at a measurable rate.

  • 9.4.A.1 Many processes that are thermodynamically favored do not occur to any measurable extent, or they occur at extremely slow rates.
  • 9.4.A.2 Processes that are thermodynamically favored, but do not proceed at a measurable rate, are under "kinetic control." High activation energy is a common reason for a process to be under kinetic control. The fact that a process does not proceed at a noticeable rate does not mean that the chemical system is at equilibrium. If a process is known to be thermodynamically favored, and yet does not occur at a measurable rate, it is reasonable to conclude that the process is under kinetic control.

Source: College Board AP Course and Exam Description

$\Delta G<0$ says a reaction can happen, not that it will happen fast. A reaction can be thermodynamically favorable yet kinetically slow because of a high activation energy (diamond → graphite). Thermodynamics gives the direction; kinetics gives the speed.

Exercise sheet
9.5

Free Energy and Equilibrium
Syllabus
Learning ObjectiveEssential Knowledge

9.5.A
Explain whether a process is thermodynamically favored using the relationships between $K$, $\Delta G^{\circ}$, and $T$.

  • 9.5.A.1 The phrase "thermodynamically favored" ($\Delta G^{\circ} < 0$) means that the products are favored at equilibrium ($K > 1$) under standard conditions.
  • 9.5.A.2 The equilibrium constant is related to free energy by the equations
    • Equation: $K = e^{-\Delta G^{\circ}/RT}$
    • Equation: $\Delta G^{\circ} = -RT \ln K$
  • 9.5.A.3 Connections between $K$ and $\Delta G^{\circ}$ can be made qualitatively through estimation. When $\Delta G^{\circ}$ is near zero, the equilibrium constant will be close to 1. When $\Delta G^{\circ}$ is much larger or much smaller than $RT$, the value of $K$ deviates strongly from 1.
  • 9.5.A.4 Processes with $\Delta G^{\circ} < 0$ favor products (i.e., $K > 1$) and those with $\Delta G^{\circ} > 0$ favor reactants (i.e., $K < 1$).

Source: College Board AP Course and Exam Description

Free energy links to the equilibrium constant:

$$\Delta G^\circ = -RT\ln K.$$
So $\Delta G^\circ<0$ gives $K>1$ (products favored), and $\Delta G^\circ>0$ gives $K<1$. At equilibrium $\Delta G=0$.

Exercise sheet
9.6

Free Energy of Dissolution
Syllabus
Learning ObjectiveEssential Knowledge

9.6.A
Explain the relationship between the solubility of a salt and changes in the enthalpy and entropy that occur in the dissolution process.

  • 9.6.A.1 The free energy change ($\Delta G^{\circ}$) for dissolution of a substance reflects a number of factors: the breaking of the intermolecular interactions that hold the solid together, the reorganization of the solvent around the dissolved species, and the interaction of the dissolved species with the solvent. It is possible to estimate the sign and relative magnitude of the enthalpic and entropic contributions to each of these factors. However, making predictions for the total change in free energy of dissolution can be challenging due to the cancellations among the free energies associated with the three factors cited.

Source: College Board AP Course and Exam Description

Whether a salt dissolves depends on the free-energy change of dissolving. Dissolving often increases entropy (ordered solid → dispersed ions) but may cost enthalpy; the sign of $\Delta G$ (and thus $K_{sp}$) follows from $\Delta H - T\Delta S$.

Exercise sheet
9.7

Coupled Reactions
Syllabus
Learning ObjectiveEssential Knowledge

9.7.A
Explain the relationship between external sources of energy or coupled reactions and their ability to drive thermodynamically unfavorable processes.

  • 9.7.A.1 An external source of energy can be used to make a thermodynamically unfavorable process occur. Examples include:
    • 9.7.A.1.i Electrical energy to drive an electrolytic cell or charge a battery.
    • 9.7.A.1.ii Light to drive the overall conversion of carbon dioxide to glucose in photosynthesis.
  • 9.7.A.2 A desired product can be formed by coupling a thermodynamically unfavorable reaction that produces that product to a favorable reaction (e.g., the conversion of $ATP$ to $ADP$ in biological systems). In the coupled system, the individual reactions share one or more common intermediates. The sum of the individual reactions produces an overall reaction that achieves the desired outcome and has $\Delta G^{\circ} < 0$.

Source: College Board AP Course and Exam Description

An unfavorable reaction ($\Delta G>0$) can be driven by coupling it to a favorable one ($\Delta G<0$) that shares a common intermediate, as long as the sum has $\Delta G<0$. This is how cells use ATP to power otherwise unfavorable processes.

Exercise sheet
9.8

Galvanic and Electrolytic Cells
Syllabus
Learning ObjectiveEssential Knowledge

9.8.A
Explain the relationship between the physical components of an electrochemical cell and the overall operational principles of the cell.

  • 9.8.A.1 Each component of an electrochemical cell (electrodes, solutions in the half-cells, salt bridge, voltage/current measuring device) plays a specific role in the overall functioning of the cell. The operational characteristics of the cell (galvanic vs. electrolytic, direction of electron flow, reactions occurring in each half-cell, change in electrode mass, evolution of a gas at an electrode, ion flow through the salt bridge) can be described at both the macroscopic and particulate levels.
  • 9.8.A.2 Galvanic, sometimes called voltaic, cells involve a thermodynamically favored reaction, whereas electrolytic cells involve a thermodynamically unfavored reaction. Visual representations of galvanic and electrolytic cells are tools of analysis to identify where half-reactions occur and in what direction current flows.
  • 9.8.A.3 For all electrochemical cells, oxidation occurs at the anode and reduction occurs at the cathode.
    • Exclusion statement: Labeling an electrode as positive or negative will not be assessed on the AP Exam.

Source: College Board AP Course and Exam Description

The galvanic cell

Redox reactions can move electrons through a wire:

A galvanic cell with a salt bridge and voltmeter A galvanic cell with a salt bridge and voltmeter

  • A galvanic (voltaic) cell 原电池 uses a favorable reaction ($\Delta G<0$) to produce electricity – a battery.
  • An electrolytic cell 电解池 uses electricity to force an unfavorable reaction ($\Delta G>0$).

In both, oxidation happens at the anode 阳极 and reduction at the cathode 阴极.

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Transfer electrons in a cell

In a galvanic cell a spontaneous redox reaction drives electrons through a wire, doing electrical work; oxidation at one electrode, reduction at the other.

Vocabulary Train
English Chinese Pinyin
galvanic (voltaic) cell 原电池 yuán diàn chí
electrolytic cell 电解池 diàn jiě chí
anode 阳极 yáng jí
cathode 阴极 yīn jí
Exercise sheet
9.9

Cell Potential and Free Energy
Syllabus
Learning ObjectiveEssential Knowledge

9.9.A
Explain whether an electrochemical cell is thermodynamically favored, based on its standard cell potential and the constituent half-reactions within the cell.

  • 9.9.A.1 Electrochemistry encompasses the study of redox reactions that occur within electrochemical cells. The reactions are either thermodynamically favored (resulting in a positive voltage) or thermodynamically unfavored (resulting in a negative voltage and requiring an externally applied potential for the reaction to proceed).
  • 9.9.A.2 The standard cell potential of electrochemical cells can be calculated by identifying the oxidation and reduction half-reactions and their respective standard reduction potentials.
  • 9.9.A.3 $\Delta G^{\circ}$ (standard Gibbs free energy change) is proportional to the negative of the cell potential for the redox reaction from which it is constructed. Thus, a cell with a positive $E^{\circ}$ involves a thermodynamically favored reaction, and a cell with a negative $E^{\circ}$ involves a thermodynamically unfavored reaction.
    • Equation: $\Delta G^{\circ} = -nFE^{\circ}$

Source: College Board AP Course and Exam Description

The cell potential 电池电势 $E^\circ_{\text{cell}}$ measures the driving force in volts, found from standard reduction potentials ($E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$). It links to free energy by

$$\Delta G^\circ = -nFE^\circ_{\text{cell}},$$
where $n$ is moles of electrons and $F$ the Faraday constant. A positive $E^\circ_{\text{cell}}$ means a favorable (galvanic) reaction.

The electrochemical series of standard electrode potentials The electrochemical series of standard electrode potentials

Worked example. A cell pairs a copper cathode ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$, $E^\circ=+0.34\ \text{V}$) with a zinc anode ($\text{Zn}^{2+}+2e^-\rightarrow\text{Zn}$, $E^\circ=-0.76\ \text{V}$). The cell potential is

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=0.34-(-0.76)=1.10\ \text{V}.$$
It is positive, so the reaction is spontaneous and the cell (a Daniell cell) acts as a battery.

Vocabulary Train
English Chinese Pinyin
cell potential 电池电势 diàn chí diàn shì
Exercise sheet
9.10

Cell Potential Under Nonstandard Conditions
Syllabus
Learning ObjectiveEssential Knowledge

9.10.A
Explain the relationship between deviations from standard cell conditions and changes in the cell potential.

  • 9.10.A.1 In a real system under nonstandard conditions, the cell potential will vary depending on the concentrations of the active species. The cell potential is a driving force toward equilibrium; the farther the reaction is from equilibrium, the greater the magnitude of the cell potential.

  • 9.10.A.2 Equilibrium arguments such as Le Châtelier's principle do not apply to electrochemical systems, because the systems are not in equilibrium.

  • 9.10.A.3 The standard cell potential $E^{\circ}$ corresponds to the standard conditions of $Q = 1$. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when $Q = K$). Deviations from standard conditions that take the cell further from equilibrium than $Q = 1$ will increase the magnitude of the cell potential relative to $E^{\circ}$. Deviations from standard conditions that take the cell closer to equilibrium than $Q = 1$ will decrease the magnitude of the cell potential relative to $E^{\circ}$. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.

  • 9.10.A.4 Algorithmic calculations using the Nernst equation are insufficient to demonstrate an understanding of electrochemical cells under nonstandard conditions. However, students should qualitatively understand the effects of concentration on cell potential and use conceptual reasoning, including the qualitative use of the Nernst equation:

    • Equation: $E = E^{\circ} - (RT/nF) \ln Q$

    to solve problems.

Source: College Board AP Course and Exam Description

Away from standard conditions, the potential shifts with concentration – the Nernst equation 能斯特方程 (qualitatively): as reactants are consumed, $Q$ rises and $E_{\text{cell}}$ falls, reaching zero at equilibrium (a dead battery). Changing a concentration shifts $E_{\text{cell}}$ the way Le Chatelier predicts.

Vocabulary Train
English Chinese Pinyin
Nernst equation 能斯特方程 néng sī tè fāng chéng
Exercise sheet
9.11

Electrolysis and Faraday's Law
Syllabus
Learning ObjectiveEssential Knowledge

9.11.A
Calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.

  • 9.11.A.1 Faraday's laws can be used to determine the stoichiometry of the redox reaction occurring in an electrochemical cell with respect to the following:
    • 9.11.A.1.i Number of electrons transferred
    • 9.11.A.1.ii Mass of material deposited on or removed from an electrode (as in electroplating)
    • 9.11.A.1.iii Current
    • 9.11.A.1.iv Time elapsed
    • 9.11.A.1.v Charge of ionic species
    • Equation: $I = q/t$

Source: College Board AP Course and Exam Description

Electrolysis

In electrolysis 电解, the charge passed determines how much substance is deposited or produced – Faraday's law 法拉第定律. Convert current × time to charge, charge to moles of electrons ($F=96{,}485$ C/mol), then use the half-reaction's electron ratio to get moles (and mass) of product.

Electrolysis: ions move to the electrodes and are discharged Electrolysis: ions move to the electrodes and are discharged

Worked example. A current of $2.0\ \text{A}$ flows for $30\ \text{minutes}$ through copper(II) sulfate ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$). How much copper is deposited? The charge is $Q=It=2.0\times1800=3600\ \text{C}$, giving $3600/96485=0.0373\ \text{mol}$ of electrons. Since each Cu needs $2$ electrons, $0.0187\ \text{mol}$ of Cu forms, a mass of $0.0187\times63.5=1.2\ \text{g}$.

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Electrolyse a molten salt

Electrolysis uses current to force a non-spontaneous reaction: positive ions gain electrons at the cathode, negative ions lose them at the anode. Charge sets the amount deposited.

Vocabulary Train
English Chinese Pinyin
electrolysis 电解 diàn jiě
Faraday's law 法拉第定律 fǎ lā dì dìng lǜ
Exercise sheet
9.11

Exam tips
  • A process is thermodynamically favourable when $\Delta G<0$; combine enthalpy and entropy with $\Delta G=\Delta H-T\Delta S$ (match units — kJ vs J).
  • Entropy increases solid→liquid→gas and when more gas moles are produced.
  • Favourable does not mean fast — a high activation energy can make a $\Delta G<0$ reaction extremely slow (kinetic control).
  • In electrochemistry a positive $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$ means a spontaneous (galvanic) cell; oxidation is at the anode, reduction at the cathode.
  • In electrolysis the charge passed ($Q=It$) fixes the amount deposited (Faraday's law).

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