electrochemical thermodynamics and concepts sensitivity of electrochemical measurements measurements...
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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.
Electrochemical Cells
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.
Electrochemical Cells
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.
Electrochemical Equilibrium
If we flip the numerator and denominator in the argument of the ln function,
Electrochemical Equilibrium
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:
Electrochemical Equilibrium and Corrosion
Electrochemical Equilibrium and Corrosion
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.
Electrochemical Equilibrium and Corrosion
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.
Electrochemical Equilibrium and Corrosion
(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
Electrochemical Equilibrium and Corrosion
(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.
Potential/pH (Pourbaix) Diagram
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
Potential/pH (Pourbaix) Diagram
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
Potential/pH (Pourbaix) Diagram
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
Potential/pH (Pourbaix) Diagram
Reaction 2: Metal reacts to metal hydroxide or oxide
Pourbaix diagram for Al
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)
Potential/pH (Pourbaix) Diagram
The reaction rate constant for
Pourbaix diagram for Al
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
Potential/pH (Pourbaix) Diagram
Reaction 3: Metal reacts to form soluble aqueous anions
Pourbaix diagram for Al
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
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