Short Course _Thermodynmics

8/12/2019 Short Course _Thermodynmics

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To be a partner of choice in corro

research

orrosion Thermodynamics

5/27/2014 1Corrosion Electrochemistry :

Thermodynamics


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Thermodynamics deals with energy and itschanges in reactions.

Reactions are viewed in terms of changes in freeenergy.

According to the first law of thermodynamicsenergy can be neither created nor destroyed.

The second law states that the free energy isreleased from the system to surroundings in allspontaneous changes.

Corrosion reactions are spontaneous and aregoverned by the laws of thermodynamics.

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Tendency to corrode and rate of corrosion

Tendency to corrodeis determined by the free energy difference

between a metal and its corrosion product.

Consider the reaction:

Rate of corrosionis determined by the size of the energy barrier, the free energy of

activation,

G


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For the forward reaction the temperature dependence is given by theArrhenius equation.

A= undefined constant; R= gas constant, T= absolute temperature

The relationship between free energy and equilibrium constant for a

reaction:


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For a spontaneous transition from high energy(initial

state, Gi) to low energy (final state, Gf), the free energy

change

G = Gf- Gi< 0 (negative)

For a spontaneous reaction to occur, Gmust benegative !

.Mg + H2O + O2 = Mg(OH)2 (

G

o

= -596 kJ/mol)Cu + H2O + O2 = Cu(OH)2 (G o= -119 kJ/mol)

Au + 3/2H2O + O2 = Au(OH)3 (G o= +66 kJ/mol)


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Nernst equation

The free energy change is related to electrical potential by

Faraday's law:

G=

nFE and G=

nFE

Where, F=Faraday constant, 96,500 coulombs/mole; n

number of electrons transferred in the corrosion reaction and

E is the measured potential in volts.

Under standard conditions, G= -nFEo

From G = G+ (RT)ln{[C][D]/{[A][B]},

we have

- nFE= -nFEo+ (RT)ln{[C][D]}/{[A][B],the Nernst equation:

E= Eo- (RT/nF)ln{[[C][D]}/{[A][B]}

This is one of the most fundamental equations incorrosion science and engineering.


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Under standard conditions: T=298k, R=8.3143 J (mol k)-1,

ln X = 2.3 logX

Nernst equation can be written as

E= Eo- (0.059/z)log{[Products]/[Reactants]}

E is the non equilibrium potential generated by the corrosionreaction;

[reactants] = concentration of reactants and

[products] = concentration of products.


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Standard Electrode Potentials

The potential difference between the anode and cathode can be

measured by voltage measuring device. The absolute potential of the

anode and cathode cannot be measured directly.

To define a standard electrode, all other potential measurements are

made against the standard electrode. If the standard electrode

potential is arbitrarily set to zero, the potential difference measured

can be considered as the absolute potential.

Standard Hydrogen ElectrodeThe half-cell in which the hydrogen reaction takes place is called the

standard hydrogen electrode, often abbreviated SHE


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Electrode Standard Electrode Potential,EoV

(SHE)

Au3++ 3e-= Au + 1.50

Cl2 + 2e- = 2Cl- + 1.360

O2+ 2H++ 2e = H2O + 1.228Br2+ 2e = 2Br

_ + 1.065

Ag++ e = Ag + 0.799

Hg22++ 2e = 2Hg + 0.789

Fe2++ e_= Fe3+ + 0.771

I2+ 2e = 2I-

+ 0.536Cu++ e = Cu + 0.520

Cu2++ 2e = Cu + 0.337

2H++ 2e = H2 0.000 (by definition)

Pb2++ 2e = Pb 0.126

Sn2++ 2e = Sn 0.136

Ni2++ 2e = Ni 0.250

Fe2++ 2e = Fe 0.440

Cr3++ 3e = Cr 0.740

Zn2++ 2e = Zn 0.763

Al3++ 3e = Al 1.663

Mg2++ 2e = Mg

2.370+ + =

The logical choice

is for all reactants

in their standard

states and

potentials for this

condition are

described asstandard

electrode

potent ials, Eo.

Noble

or

Cathodic

Activeor

Anodic


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Standard Electrode Potential

The potential difference measured between metal, M, and the

hydrogen electrode, under well defined conditions.

Example:

Iron corrodes in a solution of H+

(a) iron dissolves: half-cell reaction: Fe = Fe2++ 2e-(b) hydrogen gas formed: half-cell reaction: 2H++ 2e- = H2

(c) overall reaction: corrosion reaction: Fe + 2H+= Fe2++ H2


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Substituting into Nernst equation:

E=Eo- (0.059/2)lg{[Fe2+][H2]/[Fe][H+]2}

The terms [H+] and [H2] have been made to 1, [Fe] can be

approximated as unity, so under standard conditions, [Fe2+] = 1 M,

E= Eo

The measured potential difference is the electrode potential of the

iron under standard conditions.

Points to note: if the measured potential is positive

dG=-nFE < 0, spontaneous reaction.

Eo= + 0.44 V, dG is negative, iron corrodes spontaneously in acid.


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The Basic Wet Corrosion Cell

Four essential components of a corrosion cell

The Anode

The Cathode

The Ionic Conductor (electrolyte)

The Metallic Conductor (electrical connection)


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(1) The anode

The anode corrodes by loss of electrons: M= Mz++ ze-

Anodic reaction

Oxidation reaction

Electron generation

(2) The cathode

The cathode does not corrode. Most important cathodic reactions:(i) pH < 7 2H++ 2e- =H2

(ii) pH > = 7 2H2O + O2+ 4e- = 4OH-

Other cathodic reactions are possible Depending on the environment.Cathode

Cathodic reaction

Reduction reaction

Electron consumption


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3) An electrolyte (ionic conductor)

a solution conducting electricity

(4) Electrical connection

the anode and cathode in a corrosion cell must be in electrical contact.Difference in free energies between the anode and the cathode

produces electrical potential which is the driving force for corrosion

reaction.

Current: flow of electrons; Corrosion current:corrosion rate

Points to note: all aqueous corrosion reactions can bethought of in

terms of the simple corrosion cell.

Separation of anode and cathode

permanent separation

differential aeration

random distribution


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Cell Potential

Dry batteries and Daniell cell

Symbolism of a corrosion cell: Zn | Zn2+|| Cu2+| Cu

Zn electrode on the left, immersed in Zn2+ions; Cu electrode on

the right, immersed in Cu2+

ions. Ionic species are separated by || .Zinc is the anode, copper is the cathode.

The two half-cell reactions are:

Zn = Zn2++ 2e-

Cu2++ 2e- = Cu

The overall reaction Zn + Cu2+= Zn2++ Cu


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To find the theoretical cell potential, using Nernst equation for each

of the half-cell reactions:

E(Zn/Zn2+)=Eo(Zn/Zn2+) - (0.059/2)lg[Zn2+], andE(Cu2+/Cu)=Eo(Cu2+/Cu) - (0.059/2)lg{1/[Cu2+]}.

So the cell potential

E(cell)=Eo(cell) - (0.059/2)lg{[Zn2+]/[Cu2+]}

Convention:

do not use oxidation potential

do use reduction potential

Eo(cell) = Eo(cathode) + Eo(anode)

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Short Course _Thermodynmics