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DoITPoMS Teaching & Learning Packages The Nernst Equation and Pourbaix Diagrams The Nernst Equation and Pourbaix Diagrams (all content)

The Nernst Equation and Pourbaix Diagrams (all content)

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Contents

Main pages

Additional pages

Aims

On completion of this TLP you should be able to:

  • Introduce and explain the basic ideas of electrochemistry, including electrochemical potentials, half cell reactions and equilibria.
  • Describe the mechanisms of aqueous corrosion
  • Derive the Nernst equation and show how it can be used to derive Pourbaix diagrams.
  • Explain the information contained in a Pourbaix diagram, and demonstrate how this can be used to predict corrosion behaviour.

Before you start

This TLP is largely self-explanatory, but a basic knowledge of logarithms and thermodynamics may be of some help. A general introduction to relevant thermodynamics can be found here.

Introduction

Corrosion is the wastage of material by the chemical action of its environment. It does not include mechanisms such as erosion or wear, which are mechanical. Aqueous corrosion is the oxidation of a metal via an electrochemical reaction within water and its dissolved compounds. Aqueous corrosion is dependent on the presence of water to act as an ion conducting electrolyte.

An understanding of aqueous corrosion is essential for all industries. The lifetime and safety of chemical plants, offshore platforms and ships are all dependent on controlling and predicting corrosion rates and products.

This TLP introduces the concepts of electrochemical equilibrium reactions, electrode potentials, construction of Pourbaix diagrams using the Nerst equation and their interpretation. A Pourbaix diagram is a plot of the equilibrium potential of electrochemical reactions against pH. It shows how corrosion mechanisms can be examined as a function of factors such as pH, temperature and the concentrations of reacting species.

Background

Electrode potentials

The electrode potential, E, of a metal refers to the potential difference measured (in volts) between a metal electrode and a reference electrode.

Ee is the equilibrium potential (or reversible potential), which describes the equilibrium between two different oxidation states of the same element, at whatever concentration (or pressure) they occur. Ee varies with concentration, pressure and temperature. It describes the electrode potential when the components of the reaction are in equilibrium. This does NOT mean that they are in equilibrium with the standard hydrogen electrode. It means only that the reaction components are in equilibrium with each other. In the reaction

Az+ + ze- = A,

a concentration, CAz+, of AZ+ is in equilibrium with solid A. The reaction moves away from equilibrium only if there is a source or sink for electrons. If this were the case, then the potential would move away from Ee.

E0, the standard equilibrium potential (or standard electrode potential), is defined as the equilibrium potential of an electrode reaction when all components are in their standard states, measured against the standard hydrogen electrode (SHE). It describes the equilibrium between two different oxidation states of the same element. E0 is a constant for a given reaction, defined at 298 K. Values of E0 for various electrochemical reactions can be found in data books.

At equilibrium, the chemical driving force for an electrochemical reaction, ΔG is equal to the electrical driving force, Ee. ΔG corresponds to a charge, zE, taken through the potential, Ee. The measured potential for an electrochemical reaction is therefore directly proportional to its free energy change.

ΔG = -zE Ee

where z is the number of moles of electrons exchanged in the reaction and E is Faraday’s constant, 96 485 coulombs per mole of electrons. Similarly, under standard conditions,

ΔG0 = -zE E0

Aqueous corrosion

Diagram describing aqueous corrosion

Oxidation of a metal in an aqueous environment is dependent on potential,E, and pH.

If oxidation does occur, the metal species is oxidised and loses electrons, forming metal cations; and a corresponding reduction reaction that consumes electrons at the cathode

In aqueous corrosion water is the electrolyte, an ion-conducting medium. This means that the sites of oxidation and reduction can be spatially separate. This is different from a gaseous environment, as a gas cannot conduct ions.

A metal oxidising to produce metal ions may dissolve into the water, resulting in corrosion. This is different from corrosion in a gas, where the oxidised metal stays where it is produced, as an oxide film on the metal.

Electrochemical half-cell reactions

A half-cell reaction is an electrochemical reaction that results in a net surplus or deficit of electrons. It is the smallest complete reaction step from one species to another. Although this reaction may proceed as a sequence of more simple reactions, these intermediate stages are not stable.

A half-cell reaction can either be a reduction, where electrons are gained, or an oxidation, where electrons are lost.

The following mnemonic is often helpful:

OILRIG: Oxidation Is Loss; Reduction Is Gain (of electrons).

The anode is the site of oxidation –where electrons are lost.

The cathode is the site of reduction –where electrons are gained.

Anions, such as O2-, are negatively charged ions, attracted to the anode.

Cations, such as Fe2+, are positively charged ions, attracted to the cathode.

Reduction half-cell reactions

Reduction reactions occur at the cathode and involve the consumption of electrons. In corrosion these normally correspond to reduction of oxygen or evolution of hydrogen, such as:

O2 + H2O + 4e- = 4OH-

O2 + 4H+ + 4e- = 2H2O

2H2O + 2e- = H2 +2OH-

2H+ + 2e- = H2

Oxidation half-cell reactions

Oxidation reactions occur at the anode and involve the production of electrons. For the corrosion of metals, these reactions normally correspond to the various metal dissolution or oxide formation reactions, such as:

Fe = Fe2+ + 2e-

Fe2+ = Fe3+ + e-

Fe + 2OH- = Fe(OH)2 + 2e-

2Fe + 3H2O = Fe2O3 + 6H+ + 6e-

In addition to causing corrosion, oxidation may result in the formation of a passive oxide. The passive oxide produced may protect the metal beneath from further corrosion –significantly slowing further corrosion. An example of such passivation is that of aluminium in water, where aluminium is oxidised to from a layer of Al2O3 that protects the metal beneath from further oxidation.

Reference electrodes

Since only differences in potential can be measured, a benchmark electrode is required, against which all other electrode potentials can be compared. The particular reference electrode used must be stated as part of the units.

The Standard Hydrogen Electrode (SHE)

The electrode reaction

2H+ + 2e- = H2

is defined as having an electrode potential, EH+/H2 of zero volts, when all reactants and products are in the standard state. The standard chemical potential of H+ at 1 molar (M) concentration is by definition equal to zero.

The standard state is defined as 298 K, 1 bar pressure for gases and a concentration 1 molar (1 mol dm-3) for ions in aqueous solution.

As a direct result of this, the standard hydrogen electrode (SHE) is commonly used as a reference electrode. When coupled with an electrode, the potential difference measured is the electrode potential of that electrode, as the SHE establishes by definition the zero point on the electrochemical scale.

The standard hydrogen electrode consists of a platinum electrode suspended in a sulphuric acid solution with a one molar concentration of H+. Purified hydrogen is bubbled through to equilibrate the 2H+ + 2e- = H2 electrode reaction.

[Pop-up for other reference electrodes]

Diagram of electrochemical cell

The diagram above shows how the standard potential,E0 of nickel can be determined. The nickel electrode contains Ni2+ ions in equilibrium with nickel metal.

The hydrogen electrode is linked via a salt bridge to the deaerated solution in which the nickel electrode is immersed. This permits charge transfer and potential measurement but not mass transfer of the acid solution in the electrode.

When Ee or E0 are measured relative to the SHE (or some other reference electrode), a voltmeter is used. The voltmeter is required to have a high impedance to resist any current flowing between the electrode and the SHE. If a current were allowed to flow, the electrodes would become polarised and would no longer be at equilibrium.

In practice, it is often difficult or impossible to determine experimentally the standard electrode potential for electrochemical systems. Many systems lie outside the water stability zone or are passive. For example, zinc will immediately begin to oxidise when immersed in water.

It is very simple to determine accurately the standard equilibrium potential from the equation linking chemical driving force with the electrical driving force,

ΔG0 = -zF E0

Now ΔG0, the standard free energy of formation can be expressed as

ΔG0 = μ0(products) − μ0(reactants)

where μ0 is the standard chemical potential. By combining these equations,

\[{E^0} = \frac{{\Delta {G^0}}}{{zF}} = \frac{{{\mu ^0}({\rm{products}}) - {\mu ^0}({\rm{reactants}})}}{{zF}}\]

To obtain a standard equilibrium potential, E0, for an electrochemical reaction, all that is required is to look up relevant values of standard chemical potential.

How corrosion of metal occurs

If a metal surface is immersed in an electrolyte such as water, metal ions tend to be lost from the metal into the electrolyte, leaving electrons behind on the metal. This will continue to occur until the metal reaches its equilibrium potential and the system comes to equilibrium, with a certain concentration of dissolved ions. The metal is at its equilibrium potential, Ee. If the electrolyte were to be continuously replaced (by water flowing in through a pipe for example), more and more metal ions would be lost, resulting in continuous corrosion of the metal.

A cathodic reaction may occur that uses up the electrons lost by the metal species. In the reaction,

2H+ + 2e- = H2

if the hydrogen gas evolved is lost from the system, the reaction is prevented from reaching equilibrium.

The cathodic reaction acts as a sink for electrons liberated in the oxidation reaction of the metal. As a result of this, the metal will not reach its equilibrium potential. The metal oxidation reaction is therefore not in equilibrium and carries a net reaction. The difference between the potential, E, of the metal and its equilibrium potential, Ee is called the overpotential and is given the symbol η.

η = E − Ee

As corrosion occurs, the mass of metal is reduced due to the conversion of atoms to ions, which are subsequently lost. The sites of oxidation (the anode) and reduction (the cathode) can both be situated on the same piece of metal – there is no need for an external electrode to be present for the process to occur.

Rules for balancing electrochemical equations

The aim of this procedure is to balance electrochemical equations in terms of electronic charge and moles of components, given the main reaction product and reactant.

By convention, electrochemical reactions are written as the REDUCTION of the species concerned, proceeding to the right. The species with the lower oxidation state is written on the right hand side.

The rules are as follows:

1. Write down the main reaction components, with the reduced form (the form with the lowest valency) on the right.

2. Add stoichiometric numbers to balance the number of metal atoms. (Don’t worry about charge or oxygen being balanced at this point).

3. Balance the number of oxygen atoms by adding H2O to the appropriate side.

4. Balance the number of hydrogen atoms by adding hydrogen ions (H+) to the appropriate side.

5. Balance the residual charge by adding electrons (e-) to the appropriate side.

Now each side of the equation has the same number of atoms of each element and the same overall charge.

Examples of balancing electrochemical reactions

Find the electrochemical reaction for an equilibrium between Cr2O3 and CrO42-

1. Write reduced species on right

CrO 42-    Cr2O3

2. Balance Cr metal atoms

2 CrO 42-    Cr2O3

3. Balance oxygen atoms with water

2 CrO 42-    Cr2O3 + 5 H2O

4. Balance hydrogen atoms with hydrogen ions

2 CrO 42- + 10 H+    Cr2O3 + 5 H2O

5. Balance charge with electrons

2 CrO 42- + 10 H+ + 6 e-    Cr2O3 + 5 H2O

Check: Each side of the equation has: Two Cr, eight O, ten H and zero residual charge – so it is balanced.

Click for more examples:

Example 1

Example 2

Example 3

The Nernst equation

The Nernst equation links the equilibrium potential of an electrode, Ee, to its standard potential, E0, and the concentrations or pressures of the reacting components at a given temperature. It describes the value of Ee for a given reaction as a function of the concentrations (or pressures) of all participating chemical species.

In its most fundamental forms, the Nernst equation for an electrode is written as:

\[{E_e} = {E^0} - \frac{{2.303RT}}{{zF}}\log \frac{{[{\rm{reduced}}]}}{{[{\rm{oxidised}}]}}\] or

\[{E_e} = {E^0} - \frac{{RT}}{{zF}}\ln \frac{{[{\rm{reduced}}]}}{{[{\rm{oxidised}}]}}\]  [Click here for a full derivation of Nernst equation – popup]

R is the universal gas constant (8.3145 J K-1 mol-1)
T is the absolute temperature
z is the number of moles of electrons involved in the reaction as written
F is the Faraday constant (96 485 C per mole of electrons)

The notation [reduced] represents the product of the concentrations (or pressures where gases are involved) of all the species that appear on the reduced side of the electrode reaction, raised to the power of their stoichiometric coefficients. The notation [oxidised] represents the same for the oxidised side of the electrode reaction.

Explanation of Activity

Example 1

In the reaction O2 + 4H+ + 4e- = 2H2O

water is the reduced species and the oxygen gas is the oxidised species. By convention, electrochemical half-equations are written as

Oxidised State  +  ne-    Reduced State

Taking into account the stoichiometric coefficients of the species, the log term of the Nernst equation for this reaction appears as

\[\log \frac{{{{[{H_2}0]}^2}}}{{{p_{{O_2}}}{{[{H^ + }]}^4}}}\]

Some of the species that take part in electrode reactions are pure solid compounds. The standard state for these compounds is unit mole fraction, and as they are pure, and are in their standard states. In dilute aqueous solutions, water has an overwhelming concentration, so it may be considered pure. The standard state for a gas is taken as 1 atm (or 1 bar) and the standard state for solutes (such as ions) is taken as 1 mol dm-3. The log term of the Nernst equation can now be reduced to

\[\log \frac{1}{{{p_{{O_2}}}{{[{H^ + }]}^4}}}\]

The Nernst Equation under standard conditions:

At 298.15 K (25 °C), the numeric values of the constants can be combined to give a simpler form of the Nernst equation for an electrode under standard conditions:

\[{E_e} = {E^0} - \frac{{0.0591}}{z}\log \frac{{[{\rm{reduced}}]}}{{{\rm{[oxidised]}}}}\]

This equation can be applied both to the potentials of individual electrodes and the potential differences across a pair of half-cells. However, it is generally more convenient to apply the Nernst equation to one electrode at a time.

Click here for 2 other examples

General expression of the Nernst Equation

Taking the general equation for a half-cell reaction as,

aA + mH+ + ze = bB + H2O

the Nernst equation becomes

\[{E_e} = {E^0} + \frac{{0.0591}}{z}\log \frac{{{{[A]}^a}}}{{{{[B]}^b}}} - \frac{m}{z}0.0591pH\]

Construction of a Pourbaix Diagram

A Pourbaix diagram plots the equilibrium potential (Ee) between a metal and its various oxidised species as a function of pH.

The extent of half-cell reactions that describe the dissolution of metal

M  =  Mz+  +  ze-

depend on various factors, including the potential, E, pH and the concentration of the oxidised species, Mz+. The Pourbaix diagram can be thought of as analogous to a phase diagram of an alloy, which plots the lines of equilibrium between different phases as temperature and composition are varied.

To plot a Pourbaix diagram the relevant Nernst equations are used. As the Nernst equation is derived entirely from thermodynamics, the Pourbaix diagram can be used to determine which species is thermodynamically stable at a given E and pH. It gives no information about the kinetics of the corrosion process.

Constructing a Pourbaix Diagram

The following animation illustrates how a Pourbaix diagram is constructed from first principles, using the example of Zinc.

Anatomy of a Pourbaix Diagram

The Pourbaix diagram provides much information on the behaviour of a system as the pH and potential vary.

The following animation explains how a Pourbaix diagram is built up from fundamentals.

Examples of a Pourbaix Diagram

Pourbaix diagrams for gold, zinc and aluminium

Gold’s Pourbaix diagram explains why it is the most immune substance known. It is immune in all regions in which cathodic reactions can take place. So gold never* corrodes in an aqueous environment.

Immunity of aluminium only occurs at lower potentials. Therefore, unless under conditions that cause it to passivate, it is much more susceptible to corrosion than gold or zinc.

* provided that the water is pure; that no ion complexes are present to provide a cathodic half cell reaction that occurs at a potential higher than +1.5 V(SHE).

Constructing a 3D Pourbaix Diagram

A Pourbaix Diagram does not have to be limited to two dimensions. Three (or higher) dimension diagrams can be constructed by varying other parameters such as concentration or temperature.

Constructing a 3D Pourbaix Diagram

Summary

In this package:

  • the concepts of equilibrium potential and their measurement are introduced.
  • electrochemical half-cells are defined and treatment of electrochemical equations demonstrated.
  • the physical and chemical processes which lead to aqueous corrosion are examined.
  • the Nernst Equation has been derived and the way in which it links measured potentials at various conditions with standard equilibrium potentials is discussed.
  • the way in which some electrochemical reactions have equilibrium potentials that vary as a function of pH is considered and the concept and derivation of a Pourbaix diagram introduced.
  • through the use of specific examples, the characteristics of Pourbaix diagrams and their uses are examined. The stability of water is demonstrated through the use of the Pourbaix diagram.
  • cathodic and anodic reaction lines on Pourbaix diagrams are discussed and the way in which a point on the diagram corresponds to physical corrosion examined.

Questions

Quick questions

You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

Going further

Books

  • J.M. West: "Basic Oxidation and Corrosion"
  • Ellis-Horwood L.L. Shreir, R.A. Jarman and G.T. Burstein: "Corrosion", third edition, Butterworth-Heinemann

Websites

 


Other reference electrodes

It is often impractical to use the standard hydrogen electrode owing to the clumsy nature of using hydrogen gas. In practice, a range of alternative, secondary electrodes are used. 

The potentials of these electrodes are precisely known with respect to the SHE, so a measured potential can be easily converted to an equivalent one relative to the SHE. Three of the most common secondary electrodes are:

  • The saturated calomel electrode (SCE),
  • The silver/silver chloride electrode
  • The copper-copper(II) sulphate electrode.

The saturated calomel electrode (SCE)

The reaction is based on the reaction between this elemental mercury (Hg) and mercury(I) chloride (Hg2Cl2, "calomel").

\[{\rm{H}}{{\rm{g}}_2}{\rm{C}}{{\rm{l}}_2} + 2{e^ - } \Leftrightarrow 2{\rm{Hg}} + 2{\rm{C}}{{\rm{l}}^ - }\]

A one molar solution of potassium chloride in water forms the aqueous phase in contact with the mercury and the mercury(I) chloride.

The Nernst equation for this electrode can be expressed as

\[{E_e} = {E^o} - \frac{{2.303RT}}{{2F}}\log {[C{l^ - }]^2} = {E^0} - 0.0591\log [C{l^ - }]\]

In cell notation the electrode is written as: Cl- (saturated) | Hg2Cl2(s) | Hg(l) | Pt

The measured potential, E, of the SCE is +0.241 V (SHE) for a saturated chloride ion solution at 298 K.

The silver/silver chloride electrode

This is based on the reaction is between the silver metal (Ag) and silver(I) chloride (AgCl). The half-cell reaction is

AgCl + e- = Ag + Cl-

which gives a Nernst equation of

\[{E_e} = {E^0} - \frac{{2.303RT}}{F}\log [C{l^ - }] = {E^0} - 0.0591\log [C{l^ - }]\]

Changing the electrolyte concentration with this electrode changes the equilibrium electrode potential, so fixed values of chloride concentration are required.

In cell notation, this is written as Ag | AgCl | KCl(1M).

The measured potential, E = +0.235 V (SHE) at 298 K.

The copper-copper(II) sulphate electrode

The copper-copper(II) sulphate electrode is based on the redox reaction between copper metal and its salt - copper(II) sulphate.
The corresponding equation can be presented as follows:

Cu2+ + 2e- = Cu

The Nernst equation below shows the dependence of the potential of the copper-copper(II) sulphate electrode on the concentration copper-ions:

\[E = {E^0} + \frac{{2.303RT}}{{2F}}\log [C{u^{2 + }}] = {E^0} + 0.0295\log [C{u^{2 + }}]\]

The equilibrium potential of a copper-copper sulphate electrode is -0.318 V with respect to the standard hydrogen electrode for a saturation concentration of copper ions at 298 K.

Example 1

Find the electrochemical reaction for an equilibrium between Zn and Zn(OH)42–

1. Write reduced species on right

 Zn(OH) 4 2–    Zn

2. Balance zinc atoms

 Zn(OH) 4 2–    Zn

3. Balance oxygen atoms with water

 Zn(OH) 4 2–    Zn +   4 H2O

4. Balance hydrogen atoms with hydrogen ions

 Zn(OH) 4 2–  +  4 H+   Zn   +   4 H2O

5. Balance charge with electrons

  Zn(OH) 4 2– + 4 H+ + 2 e-     Zn  + 4 H2O

Check: Each side of the equation has: One Zn, four O, four H and zero residual charge – so it is balanced.

Example 2

Find the electrochemical reaction for an equilibrium between NO3- and NO.

1. Write reduced species on right

NO3-     NO

2. Balance nitrogen atoms

NO3-     NO

3. Balance oxygen atoms with water

NO3-     NO + 2 H2O

4. Balance hydrogen atoms with hydrogen ions

NO3-  + 4 H +    NO + 2 H2O

5. Balance charge with electrons

NO3-  + 4 H + + 3e-    NO + 2 H2O

Check: Each side of the equation has: One N, three O, four H and zero residual charge – so it is balanced.

Example 3

Find the electrochemical reaction for an equilibrium between MnO4- and MnO2

1. Write reduced species on right

 MnO4-    MnO2

2. Balance Mn atoms

 MnO4-    MnO2

3. Balance oxygen atoms with water

 MnO4-    MnO2 + 2H2O

4. Balance hydrogen atoms with hydrogen ions

 MnO4-  + 4 H+   MnO2 + 2H2O

5. Balance charge with electrons

 MnO4-  + 4 H+ + 3 e-   MnO2 + 2H2O

Check: Each side of the equation has: One Mn, four O, four H and zero residual charge – so it is balanced.

Activity

When the concentration of a species is high or a complex is involved, concentration should be replaced by the term, activity. The system can be more or less active than its concentration suggests.

Detailed derivation of the Nernst equation

Consider the following reaction at equilibrium: \({{\rm{A}}^{z + }} + {\raise0.5ex\hbox{$\scriptstyle z$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{{\rm{H}}_2} \Leftrightarrow {\rm{A}} + z{{\rm{H}}^ + }\)

This can be expressed as two half equations:

 Az+ + ze- = A         and    \({\raise0.5ex\hbox{$\scriptstyle z$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{{\rm{H}}_2} = z{{\rm{H}}^ + } + z{e^ - }\)

The left hand reaction represents the equilibrium between atoms of A on a metal surface and Az+ ions in solution. The term ‘equilibrium’ refers to the fact that the rate of reaction in one direction equals the rate of the reverse reaction.

For the above reaction, the free energy change, ΔG, is given by

\[\Delta G = \Delta {G^0} + RT\ln \frac{{{{{X_A}} \mathord{\left/ {\vphantom {{{X_A}} {X_A^0}}} \right. \kern-0em} {X_A^0}}}}{{{{{C_{{A^{z + }}}}} \mathord{\left/ {\vphantom {{{C_{{A^{z + }}}}} {C_{{A^{z + }}}^0}}} \right. \kern-0em} {C_{{A^{z + }}}^0}}}}\]

where XA is the mole fraction of A in the metal and CA is the concentration of Az+ in solution. When the metal, A, is pure, XA = 1. Also, the standard states X0A and C0Az+ can be omitted, as the standard state for the metal phase is unit mole fraction, and for the dissolved ions is 1 mol dm-3. Thus, the free energy change can be expressed as

\[\Delta G = \Delta {G^0} + RT\ln \frac{1}{{{C_{{A^{z + }}}}}}\;\;\;\;\;\ (1)\]

At equilibrium, the chemical driving force, ΔG is always equal to the electrical driving force, Ee. As discussed previously, this can be expressed as

ΔG0 = -z F Ee

where z is the number of moles of electrons exchanged in the reaction and F is Faraday’s constant, 96 485 coulombs per mole of electrons. Under standard conditions,

ΔG0 = -z F E

From the fundamental thermodynamic equation

ΔG0 = -R T ln K

E0 can therefore be expressed as

\[{E^0} = \frac{{RT}}{{zF}}\ln K\]

where K is the equilibrium constant for the reaction.

However, only the standard equilibrium potential, E0, is related directly to K. The non-standard potential, Ee is not.

Equation (1) can now be expressed in terms of electrode potential by substituting for ΔG and ΔG0

\[ - zF{E_e} = - zF{E^0} + RT\ln \frac{1}{{C{}_{{A^{z + }}}}}\]

The equilibrium potential is therefore given by

\[{E_e} = {E^0} - \frac{{RT}}{{zF}}\ln \frac{1}{{C{}_{{A^{z + }}}}}\]

It is conventional to work in decadic logs rather than natural logs, since this is arithmetically more convenient and the pH scale is expressed in the decadic form: ln X = 2.303 log X so the equation may be written as:

\[{E_e} = {E^0} - \frac{{2.303RT}}{{zF}}\log \frac{1}{{C{}_{{A^{z + }}}}}\]

Note that for this simple reaction, the Nernst equation shows that the equilibrium potential, Ee is independent of the pH of the solution. Many half-cell reactions contain H+ ions and their Nernst equations therefore depend on the pH.

For the half-cell reaction

MnO4- + 4H+ + 3e- = MnO2 + 2H2O , E = 0.588 V(SHE)

the Nernst equation appears as

\[{E_e} = {E^0} - \frac{{2.303RT}}{{3F}}\log \frac{{[Mn{O_2}]{{[{H_2}O]}^2}}}{{[MnO_4^ - ]{{[{H^ + }]}^4}}}\]

\[ = {E^0} - 0.0197\log \frac{1}{{[MnO_4^ - ]{{[{H^ + }]}^4}}}\]

= 0.588 + 0.197 log [MnO4-] - 0.789 pH

The Nernst equation can therefore be generalised as

\[{E_e} = {E^0} - \frac{{2.303RT}}{{zF}}\log \frac{{[reduced]}}{{[oxidised]}}\]

The notation [reduced] represents the product of the concentrations (or pressures where gases are involved) of all of the species that appear on the reduced side of the electrode reaction, raised to the power of their stoichiometric coefficients. The notation [oxidised] represents the same for the oxidised side of the electrode reaction.

Solutions in which components, such as Cl- ions, can complex with the metal ions require a different treatment.

Further examples of the Nernst equation

Example 2

The reaction Al = Al3+ + 3e-  has a Nernst equation of

\[{E_e} = E_{Al/A{l^{3 + }}}^0 - \frac{{2.303RT}}{{3F}}\log \frac{{[Al]}}{{[A{l^{3 + }}]}}\]

= -1.66 + 0.0197 log[Al3+]

at 298 K, as E0 is –1.66 V(SHE) and the activity of pure aluminium is 1. In this simple reaction, the resulting equilibrium potential is independent of pH.

Example 3

Fe(OH)3 + 3H+ + e- = Fe2+ + 3H2O          E0 = 1.060 V(SHE), the Nernst equation is

\[{E_e} = {E^0} - \frac{{2.303RT}}{F}\log \frac{{[F{e^{2 + }}]{{[{H_2}O]}^3}}}{{[Fe{{(OH)}_3}]{{[{H^ + }]}^3}}}\]

\[ = 1.060 - 0.0591\log \frac{{[F{e^{2 + }}]}}{{{{[{H^ + }]}^3}}}]\]

= 1.060 - 0.0591 log[Fe2+] + (3 × 0.0591) log[H+]

at 298 K.

Since the pH scale is defined as  pH = -log[H+]   

Ee = 1.060 - 0.0591 log[Fe2+] + 0.177 log[H+]


Academic consultant: G. Tim Burstein (University of Cambridge)
Content development: Andy Bennett, Andy Collier, Carol Newby
Photography and video: Brian Barber and Carol Best
Web development: Lianne Sallows and David Brook

DoITPoMS TLP was funded by the UK Centre for Materials Education.

Additional support for the development of this TLP came from the Worshipful Company of Armourers and Brasiers'