electrochemical thermodynamics and concepts sensitivity of electrochemical measurements measurements...

Electrochemical Thermodynamics and Concepts

Sensitivity of electrochemical measurements

Measurements of electrochemical processes are made by measuring electrical currents and voltages. The currents results from the flow of charged ions (positive or negative) and or electrons. Current measurement can be extremely sensitive as small as 10-15 amps (coulombs per second). Let’s get “feeling” for the sensitivity of such a measurement.

monolayer: ≈1.5 x1015 atoms/cm2 . Suppose each atom on the surface dissolves as a single charged positive ion (i.e., a cation).

corresponding to 1.6 x 10-19 C/atom of charge. Today we can easily measure at the nanoamp scale. Suppose we measure a current density of 10-6 A cm-2 for 1 s or 10-6 C. A full monolayer of metal dissolving as a +1 cation results in a charge of 1.6 x 10-19 C/atom x 1.5 x 1015 atoms/cm2 or ≈ 2.4 x 10-4 C cm-2. Then 10-6 C corresponds to only of order 0.01 ML.

+1

1 cm1 cm

A

An electrochemical reaction is a chemical reaction involving electron transfer.

An ordinary chemical reaction does not involve electron transfer

Chemical reactions: Note the mass balance

N2 +3H2 = 2NH3 synthesis of ammonia

Electrochemical Reactions: Note the mass and charge balance

2H+ +2e- = H2 Hydrogen ion reduction

Zn = Zn2+ + 2e- Zn oxidation

Since electrochemical reactions involve electron transfer we can measurethe rate of these processes by designing an appropriate electrical circuit and measuring the flow of an electric current.

Faraday’s Law relates the flow of electric current to the mass of metalreacting.

Itam

nF

m = mass of metal reacting (gm)t = time (s, min, hours, years)I = current (Amps)a = atomic weight of metaln = number of electrons transferred in the reactionF = 96,500 C; Faraday’s constant which is equal to the

charge associated with 1 mole of electrons.

If , for example, we divide Faraday’s law by the time and area of A dissolving surface, we obtain a relation describing the rate of this process.

m iar

tA nF

A = surface area of corroding metal (cm2)i = I/A: the current density (A/cm2)r = Rate measured as a weight per unit area per unit time (gm/cm2-s).

Examples

Consider the case for steel or iron corrosion. Suppose we make a measurementof the current associated with the corrosion of 100 cm2 of a steel surface and find a current of 0.1A. Then dividing through by the area of the sample,

0.1 A/100 cm2 = 0.001 A/cm2 = 1 mA/cm2 current density.

m iar

tA nF

i = 1 mA/cm2 a = 55.85 gF = 96,500 Cn = 2

2 2Fe Fe e

27 20.001 / 55.85

2.89 10 /2 96,500

A cm ggm cm s

C

We can convert this to a penetration rate (cm/s) by dividing by the density of Fe.

27 20.001 / 55.85

2.89 10 /2 96,500

A cm gr gm cm s

C

37.88 /Fe g cm

7 28

3

2.89 10 /3.67 10 /

7.88 /Fe

r gm cm sp cm s

gm cm

Next consider:

H2 at 1Atm

porous barrier

{H+} =1{Zn2+} =1

Zn Pt

V

Electrochemical Cells

Pt is inert and acts as a catalyst: 2H+ + 2e- = H2

Zn is oxidized: Zn = Zn2+ +2e- Total reaction: Zn + 2H+ = Zn2+ + H2

We measure a potential difference of - 0.762 V

2H+ + 2e- = H2 E0 = 0.000 V (defined as zero by convention)

Zn2+ +2e- = Zn E0 = - 0.762 V (NHE) sign taken to be negative since Zn is oxidized.

H2 at 1Atm

porous barrier

{H+} =1{Cu2+} =1

Cu Pt

V

Pt is inert and acts as a catalyst: H2 = 2H+ + 2e- : hydrogen is oxidizedCu plates; Cu2+ +2e- = Cu Total reaction Cu2+ + H2 = Cu + 2H+

We measure a potential difference of 0.342 V

2H+ + 2e- = H2 E0 = 0.000 V (defined as zero by convention)

Cu2+ +2e- = Cu E0 = 0.342 V (NHE) This sign is positive since Cu is reduced.


1 M ZnSO4

1 M CuSO4

+1.104 V

If we construct this “cell” we observe the following:Zn dissolves; Zn = Zn2+ +2e-; oxidation occurs at the anodeCu plates; Cu2+ +2e- =Cu reduction occurs at the cathode

If we connect a voltmeter as shown we measure a voltage of 1.104 V.


Daniell Cell

Zn dissolves; Zn = Zn2+ +2e-; oxidation occurs at the anodeCu plates; Cu2+ +2e- =Cu reduction occurs at the cathode

For the Cu-Zn cell we measured a voltage difference of 1.104 V.

Zn2+ +2e- = Zn E0 = - 0.762 V (NHE)

Cu2+ +2e- = Cu E0 = +0.342 V (NHE)

+1.104 V

This difference is 1.104 V

Standard Reduction Potentials: EMF Series

3+ -

+ -2 2

+ -

3+ - 2+

2

Au +3e =Au +1.498

O +4H +4e =2H 0 (pH = 0) +1.229

Ag +1e =Ag +0.798

Fe +1e =Fe +0.771

Cu + -

+ -2

2+ -

2+ -

+2e =Cu +0.342

2H +2e =H +0.000

Pb +2e =Pb -0.126

Sn +2e =Sn -02+ -

2+ -

2+ -

2+ -

.138

Ni +2e =Ni -0.250

Co +2e =Co -0.277

Cd +2e =Cd -0.403

Fe +2e =Fe 2+ -

3+ -

-0.447

Zn +2e =Zn -0.762

Al +3e =Al -1.662

Noble

Active

Cathode

Anode

Volts (NHE): E0

Standard conditions

Standard states:

• For a solid; a =1: pure metal, metal oxide, etc.

• For a gas, 1 Atm pressure is taken as unit activity.

• For dilute solutes typically found in most instances of corrosion, activity is reasonably approximated by the concentration in M. The standard state is 1 M.

• Temperature is taken as 25ºC = 298 K

definition: pH = -log [H+]

Thermodynamics of Chemical Equations

A, B … reactantsX, Y … products

a, b, …., x, y,…. Stoichiometric Coefficients defining how many moles of A and B produce moles of X and Y

The chemical reaction can proceed as indicated if the energy change isnegative. That is if the energy of the products is less then that of thereactants. Mathematically we say

aA + bB xX + yY

0G OR 0 reactproducts

GG

0 lni i iG G RT a

Energy of the standard state

Activity or concentration “correction”-non standard state

Chemical Equilibrium

aA + bB + … xX + yY + …

A, B … reactantsX, Y … products

a, b, …., x, y,…. Stoichiometric Coefficients

Equilibrium is defined by the condition

Since

0 lni i iG G RT a ≣ standard free energy change /moleG

K is the equilibrium constant for the reaction.

Solving for G

The equilibrium constant is defined by

When components are not in standard state:

exp( )G

KRT

Chemical Equilibrium

Electrochemical Equilibrium

Free energy is measured in units of kcal or kJ.

1 calorie = 4.186 Joules1Joule = 6.24 x 1018 eV

Since 1 Volt = Joule/Coulomb

G Q E where Q is the total charge. The total charge transferredin an electrochemical reaction will be equal to the numberof moles of electrons participating in the reaction times thecharge/mole, Q = nF

By convention a minus sign relates ΔG and E.


If we flip the numerator and denominator in the argument of the ln function,


Nernst Equation

Formal Potential

Electrochemical Equilibrium: pH effects

We can write a general metal dissolution reaction in the following way:

2 0aA mH ne bB dH

Here A could correspond to some oxidized species such as Fe2+ and Bwould be its reduced form, metallic Fe. Alternatively A could be an oxide such asFeO and B its reduced form, also metallic Fe.

The Nernst equation for this general reaction is

It’s convenient at this stage to convert the natural log (Ln) to base 10.In general:

Ln (x) = 2.303 Log (x)

Also

2.303 8.31 / 2980.059 V

96,500 /

RT J moleK K

F C mole

Electrochemical Equilibrium and Corrosion

The Nernst Equation above can now be rewritten as:



Examples:

(1) Consider the reaction2 2Fe Fe e

First note that for this and all simple metal dissolution reactions there is noH+ in the reaction and no dependence of the reaction on pH, m = 0

The Fe2+ is equivalent to A. Also a =1. The quantity [Fe2+ ] is the concentration of Fe2+ in the electrolyte.

Fe is equivalent to B. Since Fe is a pure metal [Fe ] = 1

Then:

This type of reaction is favored at low values of pH.


ΔE(V)

[Fe2+]M

-0.477 1

-0.506 0.1

-0.534 0.01

-0.565 0.001

-0.595 10-4

-0.624 10-5

-0.654 10-6

For every decade change in ferrous cationsthere is ~ 30 mV decrease in the equilibriumpotential.

What is the difference between an open circuit potential (corrosion potential) and an Equilibrium Potential????? – Lab.


(2) Consider 2 3 26 6 2 3Al O H e Al H O

Both Al and Al2O3 are solids with unit activity, A=1, B=1. On a complete EMF series one could find thatAlso note that m = 6, n = 6 so,

The equilibrium potential decreases by ~ 60 mV per pH unit.

Examples:

This type of reaction occurs at some intermediate value of pH.

2 3/ 1.55 V.oAl Al OE


(3) One other type of corrosion reaction that can occur involves the formationof a soluble metal oxide anion at some high pH,

2 24 3 2AlO H e Al H O

The ΔE for this reaction is a function of both the dissolved ion contentand the pH of the electrolyte.

Examples:

The above reaction occurs in an acid solution.

An equivalent reaction in neutral or alkaline solutions is:

pHEEHHHH

059.00

// 22

pHEEHHHH

059.00

// 22

Cathode reactions Supporting Corrosion

Hydrogen reduction

2/0 0.059

H HE pH

The above reaction occurs in an acid solution.

An equivalent reaction in neutral or alkaline solutions is:

Cathode reactions Supporting Corrosion

Oxygen reduction

pHEE OHOOHO 059.00// 2222

2 22 4 4O H O e OH pHEE OHOOHO 059.00// 2222

2 2/ 1.229 0.059O H OE pH

Potential/pH (Pourbaix) Diagram

It turns out to be very convenient to represent the results of all these anodic metal oxidation processes and cathodic reduction process on a Map-like “phase diagram”.

The map considers the parameters of voltage and solution pH and shows the possible electrochemical/corrosion reactions that can occur for a particular metal such as Fe, Al, Cd,Zn, … and water.


2/0 0.059

H HE pH

oxygen evolutionand acidification

hydrogen evolutionand alkalization

water thermodynamically stable2 2/ 1.229 0.059O H OE pH

Water


Generic diagram for a metal

-2 0 2 4 6 8 10 12 14

1.2

pH

0.8

0.4

-0.4

0.0

-0.8

-1.2

-1.6

-2.0

-2.4

PO

TE

NT

IAL,

E(V

)

Corrosion

Corrosion

Passivation

Immunity

Corrosion-soluble ions of the metal are stable

Passivation- oxides are stable

Immunity-reduced form of the metal is stable


Reaction 1: Metal oxidizes to aqueous cations

Pourbaix diagram for Al

3

3

3

/

/

0 0.059log 0.059

0.0591.662 log

3

a

bAl Al

Al Al

A me e pH

n

e

n

l

B

A

Independent of pH since no H+ is involved.

Only depends on Al3+ activity


Reaction 2: Metal reacts to metal hydroxide or oxide


2 3/ 1.55 0.059Al Al OE pH

At higher pH Al2O3 is formed. At lower pH Al2O3 chemically dissolves to Al3+

Intersection depends depends on Al3+ activity (dashed lines are portions of the reactions with no significance)


The reaction rate constant for


is known

For (Al3+)=10-6,

3 6 11

3

.4

3

lo

log 6log 2log 6 2log

g 10 0.5log(10 ) 3

6 5.7 3

3.9

Al pH

K H Al pH Al

pH

pH

Independent of potential


Reaction 3: Metal reacts to form soluble aqueous anions


eHAlOOHAl 342 22

At higher pH, Al2O3 dissolves to AlO2-

2 3 2 22 2Al O H O AlO H

214.6 log[ ]pH AlO

2

2/1.262 0.020 log 0.079

Al AlOAlO pHE

2 2 29.22 10K H AlO

electrochemical thermodynamics and concepts sensitivity of electrochemical measurements measurements...

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