1 Unit 5: Chemistry for Applied Biologists Chemical reactions can be described as feasible if they occur without an external source of energy. In this topic you will learn about the use of Gibbs energy changes as a measure of feasibility for all reactions, as well as the use of electrode potential data in the case of redox reactions. In the case of biochemical systems, unfeasible reactions may still occur if they are coupled to reactions that are feasible. On successful completion of this topic you will: • be able to relate the feasibility of reactions to thermodynamic quantities (LO1). To achieve a Pass in this unit you need to show that you can: • use values of thermodynamic state functions to assess the feasibility of reactions (2.1) • write oxidation and reduction half-equations for given reactions (2.2) • categorise reactions as oxidation/reduction (redox) on the basis of oxidation numbers (2.3) • justify the feasibility of redox reactions in terms of standard reduction potentials (2.4). Studying the feasibility of reactions 5 . 2
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Unit 5: Chemistry for Applied Biologists
Chemical reactions can be described as feasible if they occur without an external source of energy. In this topic you will learn about the use of Gibbs energy changes as a measure of feasibility for all reactions, as well as the use of electrode potential data in the case of redox reactions. In the case of biochemical systems, unfeasible reactions may still occur if they are coupled to reactions that are feasible.
On successful completion of this topic you will: • be able to relate the feasibility of reactions to thermodynamic quantities
(LO1).
To achieve a Pass in this unit you need to show that you can: • use values of thermodynamic state functions to assess the feasibility of
reactions (2.1) • write oxidation and reduction half-equations for given reactions (2.2) • categorise reactions as oxidation/reduction (redox) on the basis of
oxidation numbers (2.3) • justify the feasibility of redox reactions in terms of standard reduction
potentials (2.4).
Studying the feasibility of reactions5.2
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1 Thermodynamic quantitiesIn Topic guide 5.1 you met the thermodynamic quantity enthalpy; it is obvious that enthalpy changes are not, by themselves, a measure of feasibility since both exothermic and endothermic reactions can be feasible depending on the conditions.
EntropyThe key to evaluating feasibility is the idea of entropy. Entropy (symbol S) is a thermodynamic quantity that measures the disorder of a chemical system. Molecules that are free to move (as in the case of gases) or that are moving rapidly can be thought of as possessing high levels of entropy (are highly disordered).
The second law of thermodynamics
The importance of entropy (or rather of entropy changes) to the feasibility of a reaction stems from the second law of thermodynamics, which states:
‘in any spontaneous process, the total entropy of the universe will increase’.
In any chemical process, the total entropy change will be composed of a contribution from the system (the reacting chemicals) and the surroundings (the rest of the universe which is not part of the system but which can exchange heat with the system).
The entropy change of the system takes into account the change in disorder of the particles involved in the reaction; the entropy change of the surroundings takes into account the change in disorder due to the heat which is exchanged with the system.
This can be written:
ΔStotal
= ΔSsystem
+ ΔSsurroundings
From the second law of thermodynamics it follows that for a reaction to be feasible then ΔS
total must be positive.
Calculating entropy
Calculation of ΔSsystem
Tables of data provide information about the standard molar entropy of a range of substances. This is the entropy of a mole of that substance measured under standard conditions (see Topic guide 5.1).
ΔSƟsystem
can be calculated from these data:
ΔSƟsystem
= Σ SƟproducts
– Σ SƟreactants
Calculation of ΔSsurroundings
As ΔSsurroundings
is connected to the heat exchanged with the surroundings, it can be calculated from the enthalpy change, ΔH, of a reaction.
ΔH can be measured experimentally, or calculated using values of standard enthalpy changes of formation of substances (see Topic guide 5.1).
Key termsFeasible: A reaction that is thermodynamically favourable. A feasible reaction will occur spontaneously.
Entropy: A measure of the disorder of a system.
Spontaneous: A reaction that can occur without an input of energy. The rate of a spontaneous reaction may, however, be too small to allow it to be observed.
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ΔSsurroundings
can then be calculated using the following equation, as long as the temperature remains constant during the process:
ΔSsurroundings
= –
ΔHT
(ΔH measured in joules, T measured in Kelvin)
The values of ΔSsurroundings
and ΔS system
can be combined to give a value of ΔStotal
and thus allow the feasibility of a reaction to be deduced.
Take it furtherThermodynamic data, such as molar entropy values or enthalpy change data, can be found in online databases such as the NIST databook http://webbook.nist.gov/chemistry/.
Activity1 The vaporisation of water is represented by the following equation:
H2O(l) ➝ H
2O(g)
a Use tables of thermodynamic data to find or calculate a value for ΔSƟsystem
. Comment on the sign of this quantity. Does it suggest that the system is becoming more or less disordered?
b Find a value of ΔHƟ for this process and use it to calculate ΔSƟsurroundings
at 298 K.c Combine your answers to (a) and (b) and calculate a value for ΔS
total at 298 K. What can you
say about the feasibility of the process?d If the temperature is raised, at what value would you expect the reaction to become
feasible? Test your prediction by calculating a value for ΔSsurroundings
at this temperature and hence ΔS
total at the same temperature (assume that the value of ΔS
system does not change
significantly).2 The oxidation of ethanol to ethanoic acid under aerobic conditions can be represented by the
equation:C
2H
5OH(l) + O
2(g) ➝ CH
3COOH(l) + H
2O(l)
a Use tables of thermodynamic data to find values of the enthalpy of formation for all of the compounds involved in the reaction. Hence calculate a value for the enthalpy change of reaction, ΔHƟ.
b Use the value of ΔHƟ to calculate a value for ΔSƟsurroundings
at 298 K (Hint: remember to convert ΔHƟ into joules).
c Use tables of thermodynamic data to find values of SƟ for all the substances in this equation. Hence calculate a value for ΔSƟ
system.
d Use your answers from (b) and (c) to calculate a value for ΔSƟtotal
at 298 K.e Comment on the feasibility of this process. What assumptions have you made about the
states of the reacting substances? Is this justified?
Gibbs energyA much more convenient approach to evaluating the feasibility of chemical reactions – and in particular biochemical reactions – is to use a thermodynamic quantity known as the Gibbs energy (sometimes called the Gibbs free energy, G).
This has the same units as ΔH (kJ mol–1) but also takes into account the entropy change of the system. It is defined by the following equation:
ΔGƟ = ΔHƟ – T ΔSƟsystem
It can also be shown that ΔG is then related to the value of ΔStotal
for a reaction:
ΔGƟ = −T ΔSƟtotal
So, from the second law of thermodynamics:
ΔGƟ must be negative in order for a reaction to be feasible.
LinkThe meaning of the Ɵ superscript was explained in Topic guide 5.1.
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Values of ΔGƟ
Values of ΔGƟ for biochemical reactions are widely quoted in reference books or websites. Alternatively, ΔGƟ can be calculated from thermodynamic data in the following ways:
• from values of ΔHƟ and ΔSƟsystem
using the equation: ΔGƟ = ΔHƟ – T ΔSƟsystem
• from values of ΔGƟformation
using the equation:
ΔGƟsystem
= Σ ΔGƟformation
(products) – Σ ΔGƟformation
(reactants)
ΔGƟ gives much more information than simply the feasibility of the reaction; its magnitude also defines the amount of ‘useful work’ that can be extracted from a chemical process (useful work in this case can mean electrical energy or energy used to drive other reactions – anything other than the work used to cause a system to expand).
These are very important data to know about a biochemical reaction which may be used to drive other vital cell processes, such as active transport or biosynthesis.
Of course ΔGƟ values only apply to reactions occurring under standard conditions; ΔG values may then be quoted if a reaction is occurring under non-standard conditions.
Take it furtherThermodynamic tables of data are widely found in chemical reference books or online at sites such as http://www2.ucdsb.on.ca/tiss/stretton/database/organic_thermo.htm.
The sign of ΔGƟ (or ΔG if the conditions are non-standard) is very important, as you will see in the following sections, and the terms exergonic and endergonic are used to describe the sign of the Gibbs energy change.
ActivityFind, or calculate from appropriate data, values of ΔGƟ for the following processes (remember that ΔGƟ can be calculated from the ΔGƟ
formation values for the products and reactants):1 The combustion of methane: CH
4 (g) + 2O
2(g) ➝ CO
2(g) + 2H
2O(g)
2 Complete oxidation of glucose: C6H
12O
6(s) + 6O
2(g) ➝ 6CO
2(g) + 6H
2O(l)
3 Phosphorylation of ADP to ATP: ADP + Pi + H+ ➝ ATP + H
2O.
ΔG and equilibrium
For a given equation, it is important to realise that the value of ΔGƟ refers to a reaction that goes to completion.
Clearly many reactions – including the majority of biochemical processes – are regarded as reversible processes that will reach equilibrium in a closed system.
How can a reaction be reversible if only one direction is feasible? Well, we know that the Gibbs energy of reactants and products can be calculated, but it is also possible to calculate Gibbs energies of equilibrium mixtures with various ratios of products to reactants. These are shown graphically in Figure 5.2.1.
Key termsExergonic: A reaction accompanied by a negative Gibbs energy change (and hence from which useful work may be produced).
Endergonic: A reaction accompanied by a positive Gibbs energy change and hence which are not, by themselves, feasible.
Equilibrium (chemical): A situation in which the proportion of reactants and products remains constant because the rates of forward and backward reactions are equal.
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G G G
100%reactant
100%product
100%reactant
100%product
100%reactant
100%product
Equilibriumfavours products
Equilibriumfavours reactants
Equilibrium favours neither reactantsnor products
∆G –ve ∆G +ve ∆G = 0
Notice that in each of these cases there is a fixed proportion of products to reactants which gives a minimum value of Gibbs energy. The formation of this equilibrium mixture from either reactants or products will be feasible, since in both cases the Gibbs energy decreases during the reaction.
The sign of ΔGƟ allows us to make predictions about the proportion of products which will be formed at equilibrium:
1 if ΔGƟ is negative, the proportion of products present will be large (the equilibrium position favours the right-hand side)
2 if ΔGƟ is positive, the proportion of products present will be small (the equilibrium position favours the left-hand side)
3 if ΔGƟ is zero, there will be equal proportions of products and reactants.
Technically, all reactions result in an equilibrium mixture being formed, but if ΔGƟ is more negative than about −40 kJ mol–1 then the reaction is regarded as having gone to completion. If ΔGƟ is more positive than about +40 kJ mol–1 the reaction is regarded as not having occurred at all.
Portfolio activity (2.1)Choose a simple biological process involving substances for which thermodynamic data is readily available. A suitable example could be the hydrolysis of urea:
CO(NH2)
2 + H
2O ➝ CO
2 + 2NH
3
Use tables of thermodynamic data to investigate the feasibility of this reaction. In your answer: • use suitable data to calculate values of ΔHƟ
formation and ΔSƟ
system
• calculate a value for ΔSƟtotal
at 298 K • calculate a value for ΔGƟ
• explain what you can deduce from your calculations about the feasibility of the process at 298 K.
Figure 5.2.1: The relative values of the Gibbs energies (G) of
products and reactants determines the position
of equilibrium which will be reached.
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2 Oxidation and reductionMany biochemical reactions are described as redox reactions.
These reactions generally involve the transfer of electrons between biochemical molecules.
Half-equationsIt is often helpful to split up the overall reaction into half-equations, which show the processes of reduction and oxidation separately. Unlike full balanced equations, half-equations clearly show the way in which electrons are transferred in each process.
To take a simple, non-biological example, zinc displaces copper from solutions of copper(II) salts.
Zn(s) + Cu2+(aq) ➝ Zn2+(aq) + Cu(s)
The two half-equations are:
Reduction: Cu2+(aq) + 2e− ➝ Cu(s)
Oxidation: Zn(s) ➝ Zn2+(aq) + 2e−
Biological examples of redox reactions are a little more complex. For example, during anaerobic respiration, pyruvate is reduced to lactate by NADH (using the enzyme lactate dehydrogenase).
pyruvate + NADH + H+ ⇌ lactate + NAD+
The two half-equations are:
Reduction: pyruvate + 2H+ + 2e− ➝ lactate
Oxidation: NADH ➝ NAD+ + H+ + 2e−
Reduction and oxidation in terms of electron transfer
From the equations above it is evident that:
1 reduction involves the gain of electrons
2 oxidation involves the loss of electrons
3 zinc and NADH act as reducing agents, and in the process they are oxidised.
Notice that the overall equation can be obtained from the half-equations by simply adding the two half-equations together, so long as the number of electrons is identical, which enables them to be cancelled out, as shown in the following example:
in the case of the Zn / Cu2+ system
Cu2+(aq) + 2e- ➝ Cu(s) Zn(s) ➝ Zn2+(aq) + 2e-
Cu2+(aq) + Zn(s) ➝ Cu(s) + Zn2+(aq)
Key termsRedox reaction: A reaction in which both oxidation and reduction occur simultaneously.
NADH and NAD+: These are the reduced and oxidised forms of the molecule nicotinamide adenine dinucleotide. They play an important role in enabling electron transfer to take place during biochemical reactions.
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Activity1 Copper metal can oxidise silver ions to metallic silver. The two half-equations are:
Cu(s) ➝ Cu2+(aq) + 2e−
Ag+(aq) + e− ➝ Ag(s)
Write an overall equation for this process. (Hint: in order to be able to combine the equations together, the number of electrons must balance. To achieve this you will need to multiply the numbers in one of the equations by two before combining them.)
2 Oxygen can oxidise iron metal to Fe2+ ions in the presence of water. The two half-equations for the reaction are:
Fe(s) ➝ Fe2+ + 2e−
½O2(g) + H
2O(l) + 2e− ➝ 2OH−
Combine these two half-equations together to produce an overall equation for this process.3 Oxygen can also oxidise NADH during the electron-transport chain that occurs in oxidative
phosphorylation. The two half-equations are:NADH ➝ NAD+ + H+ + 2e−
½O2 + 2H+ + 2e− ➝ H
2O
Combine these half-equations together to produce an overall equation for this process.
Portfolio activity (2.2)Choose some simple chemical redox reactions. Suitable examples could include:
• displacement reactions involving halogens, such as the reactions between chlorine and bromide ions
• reactions between reactive metals and acids (represented by H+ ions).
In your answer you should: • write the overall equation for the reaction • write down two half-equations which combine to produce the overall equation • classify each half-equation as reduction or oxidation, explaining your answer.
Reduction and oxidation in terms of oxidation number
If half-equations are not written out, it can be difficult to analyse an equation for a process to deduce what is being oxidised and what is being reduced.
Chemists use a strategy that involves assigning an oxidation number (sometimes called an oxidation state) to each atom involved in the reaction.
Oxidation numbers describe the number of electrons lost, gained or shared by an atom as a result of forming bonds.
The oxidation number of an uncombined element is always zero (even for elements that exist as molecules such as O
2).
Rules for assigning oxidation number
Oxidation numbers are assigned to individual atoms in compounds using a set of rules. Although these are easy to apply to simple molecules and ionic compounds, they can become very difficult to use with molecules based on chains and rings of carbon atoms.
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• The total oxidation number of all the atoms in a neutral compound is zero. • The oxidation number of an ion is equal to the charge on the ion.
– Examples: a sodium ion has the formula Na+; the oxidation number of sodium is +1. The nitrate ion has the formula NO
3−; the total oxidation number of all the
atoms in the ion adds up to −1. • Some atoms have oxidation numbers that are almost always constant when
they are present in compounds:– oxygen is −2 (except in peroxides)– hydrogen is +1 (except in metal hydrides)– fluorine is −1 – chlorine is −1 (except when bonded to oxygen or fluorine).
Take it furtherYou can read about the rules for assigning oxidation numbers and how to apply them to compounds in most level 3 chemistry textbooks, or online at sites such as: http://www.chemguide.co.uk/inorganic/redox/oxidnstates.html.
ActivityUse the rules for assigning oxidation numbers to write down the oxidation number of:a Mg in MgCl
2
b N in NO2
c N in N2O
5 (hint: the oxidation number applies to a single atom of N)
d P in PO4
3– (hint: remember that the total oxidation number equals the charge on the ion) e O in hydrogen peroxide, H
2O
2 (hint: look back at the exceptions in the list).
Identifying redox reactions using oxidation numbers1 An atom has been oxidised if its oxidation number becomes more positive (or
less negative).
2 An atom has been reduced if its oxidation number becomes less positive (or more negative).
If no changes in oxidation numbers occur during the reaction, then it cannot be classified as a redox process.
For example, the reaction for the decomposition of hydrogen peroxide in the presence of catalase is:
2H2O
2 ➝ O
2 + 2H
2O
oxidation numbers: +1 −1 0 +1 −2
So during the reaction, the oxygen atoms have been both oxidised (−1 to 0) and reduced (−1 to −2).
ActivityWrite down the oxidation numbers of the atoms in each of the reactants and products in these equations for redox reactions:1 Mg + 2HCl ➝ MgCl
2 + H
2
2 2H2O + 2F
2 ➝ 4HF + O
2
3 MnO4
− + 8H+ + 5Fe2+ ➝ Mn2+ + 5Fe3+ + 4H2O
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Portfolio activity (2.3)Nitrogen atoms display a wide range of oxidation numbers in compounds.
Use the following reactions to explain how oxidation numbers can be used to help decide if a given reaction is a redox process.1 C + 4HNO
3 ➝ CO
2 + 4NO
2 + 2H
2O
2 NH4
+ + CO3
2− ➝ NH3 + HCO
3−
3 5NO2
− + 2MnO4
− + 6H+ ➝ 5NO3
− + 2Mn2+ + 3H2O
In your answer you should: • write down the oxidation number of the different atoms in each reactant and product for the
three equations • explain whether a redox reaction has occurred • identify the reducing and oxidising agent in each reaction.
3 Standard reduction potentialsElectrochemical cellsChemists have long studied redox reactions by constructing electrochemical cells in which the two half-reactions occur in different parts of the electrochemical cell. Reduction occurs at one electrode and oxidation at the other and electrons flow between the two electrodes through an external electrical circuit.
A simple example (including the half-equations occurring) is shown in Figure 5.2.2.
(a)
Salt bridge
VCu electrode Zn electrode
Cu2+ Zn2+
Cu2+ + 2e– ➝ Cu Zn ➝ Zn2+ + 2e–
(b)
Salt bridge
VPlatinum electrode Platinum electrode
Solution containingFe2+ and Fe3+ ions
Solution containingI2 and I– ions
Fe3+ + e– ➝ Fe2+ 2I– ➝ I2 + 2e–
Reduction potentials
By setting up electrochemical cells, chemists have been able to measure the standard reduction potential, EƟ (also known as the standard electrode potential), for a wide range of half cells.
Take it furtherStandard reduction potentials are measured under specified conditions, including the use of 1 mol dm–3 solutions, 1 bar pressure and 298 K. For more information about how standard reduction potentials are measured see websites such as http://www.chemguide.co.uk/physical/redoxeqia/introduction.html.
Figure 5.2.2: Some electrochemical cells. Different reactions occur in each
half cell and the reactions can be combined to write an equation for the
overall process occurring in the cell.
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The sign and magnitude of the EƟ provides a measure of the reducing power of the reducing agent present in each half cell – the more negative the EƟ value, the more powerful the reducing agent.
Notice that because each half cell process also contains an oxidising agent (on the opposite side of the equation to the reducing agent), EƟ also provides a measure of its oxidising power – the more positive the EƟ value, the more powerful the oxidising agent.
Take it furtherTables of standard reduction potentials for standard chemical processes can be found online at sites such as http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html. For information on biochemical redox reactions it may be necessary to search using the relevant names of the biological molecules involved.
Biochemical reduction potentials
Table 5.2.1 shows some reduction potentials for biochemically important half-reactions.
Half-reaction Standard reduction potential / V
NAD+ + H+ + 2e− ➝ NADH −0.32
NADP+ + H+ + 2e− ➝ NADPH −0.32
FAD + 2H+ + 2e− ➝ FADH2
−0.22
fumarate + 2H+ + 2e− ➝ succinate +0.03
½O2 + 2H+ + 2e− ➝ H
2O +0.82
Notice that, although these processes are shown with one-way arrows, they are all reversible processes and the reduction potential is a function of the process, regardless of the direction of reaction.
Using reduction potentialsReduction potentials for half cells can be used to predict the feasibility of reactions. As in the previous sections, these predictions only apply to standard conditions.
Calculating EƟcell
The half cells shown in Table 5.2.1 can be combined to make an electrochemical cell. The maximum potential difference between the two electrodes (under standard conditions) is known as the standard cell emf (EƟ
cell).
The significance of the EƟcell
value for a particular cell is that it relates to the reaction that will occur if the half cells are connected by a conductor. Reduction occurs in one of the half cells and oxidation in the other.
EƟcell
= reduction potential of − reduction potential of reduction half cell oxidation half cell
Table 5.2.1: The reduction potentials of some half cells that occur in biological
value for a cell in which this overall reaction could occur is therefore 0.82 − (−0.32) = + 1.14 V.
Using EƟcell values
The sign and magnitude of EƟcell
can be used to give valuable information about reactions:
• a reaction will be feasible if EƟcell
is positive • the more positive an EƟ
cell value is, the more favourable it will be
• if EƟcell
is greater than about +0.4 V, then the reaction is likely to be feasible under all conditions, not just under standard conditions.
Take it furtherAs EƟ
cell and ΔGƟ are both indicators of the feasibility of a reaction, it will be no surprise to learn that
they are related. You can read more about the relationship at websites such as http://chemistry.about.com/od/workedchemistryproblems/a/Cell-Potential-And-Free−Energy-Example−Problem.htm.
ActivityDuring the process of aerobic respiration, succinate ions are oxidised to fumarate ions by the action of the oxidising agent FAD. The reaction can be represented by the following equation:
succinate + FAD ➝ fumarate + FADH2
The half-equations for this process are:
succinate ➝ fumarate + 2H+ + 2e−
FAD + 2H+ + 2e− ➝ FADH2
• Classify each of these half-equations as reduction or oxidation. • Use Table 5.2.1 to find the standard reduction potentials for each of the half cells represented
by these half-equations. • Calculate the EƟ
cell corresponding to a cell in which this reaction could occur.
• Comment on the sign and magnitude of this EƟcell
Portfolio activity (2.4)Find an equation for a biological redox reaction, similar to those that have been discussed in this section.
A suitable example could be the oxidation of malate to oxaloacetate by NAD+:
malate + NAD+ ➝ oxaloacetate + NADH+ + H+
Use ideas about reduction potentials to comment on the feasibility of this process. In your answer you should:
• write down the two half-equations that occur during this process • find reduction potentials for half cells in which these two half-equations can occur • use the reduction potential data to calculate the EƟ
cell value corresponding to a cell in which
this reaction could occur • comment on the feasibility of the reaction.
ChecklistAt the end of this section you should be familiar with the following ideas:
the feasibility of a chemical reaction can be predicted from a range of thermochemical data
ΔStotal
, which includes contributions from ΔSsystem
and the enthalpy change of the reaction, must be positive for a reaction to be feasible
ΔG is related to the amount of useful work which can be extracted from a reaction and must be negative for a reaction to be feasible
many biochemical reactions involve reduction and oxidation
reduction and oxidation can be understood in terms of transfer of electrons or change in oxidation numbers
standard reduction potentials, measured in electrochemical cells, can be used to calculate EƟ
cell
EƟcell
must be positive if a reaction is to be feasible.
Further readingActivities and assignments in this topic guide will require access to a database containing values of ΔHƟ, ΔGƟ and SƟ, as well as reduction potentials (often called electrode potentials). As noted in the Topic 5.1 Further Reading guide, the NIST Chemistry WebBook is very detailed and wide-ranging (http://webbook.nist.gov/chemistry/). Data on reduction potentials can be found at http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html.
Coverage of entropy, Gibbs energy, redox reactions and redox equilibria will be found in many general chemistry texts such as Advanced Chemistry (Clugston & Flemming, 2000). Chapters 13 and 14 cover this material.
Chemguide (http://www.chemguide.co.uk) covers much of the material from the section on oxidation and reduction in the Inorganic Chemistry section and the material on reduction potential in the Physical Chemistry section.
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AcknowledgementsThe publisher would like to thank the following for their kind permission to reproduce their photographs:
In some instances we have been unable to trace the owners of copyright material, and we would appreciate any information that would enable us to do so.
