Dieter Landolt
Corrosion and Surface Chemistry
of Metals
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© 2007, Dieter Landolt
1
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CHAPTER 1
1.1
Fe + H 2 SO 4 --> FeSO 4 + H 2
Cu + H 2 SO 4 + ½ O 2 --> CuSO 4 + H 2 O
2 Cu + ½ O 2 + 4 HCl --> 2 CuCl- + 3 H+ + H 2 O
Zn + 2 HCl --> ZnCl 2 + H 2
1.2
v cor = 30Pm/year
i = nF v cor
table 1.3: 1 mm/year = 3.06 nU/M
iron: U = 7.86 g/cm3, M = 55.8 g/mol, n = 2 (in deaerated solution)
1 mm/year = 3.06 (2) (7.86)/55.8 = 0.862 A/m2
30 mm/year = (0.862) (30x10-3) = 2.59 x10-2 A/m2 = 2.6 PA/cm2
1.3
2 Al + 6 H+ --> 2Al3+ + 3 H 2
2 mol Al yield 3 mol H 2 ,
M Al = 27.0 g/mol, 1 g Al = 0.0370 mol Al
0.0379 mol Al --> 5.56x10-2 mol H 2
V H2 = N H2 RT/P = (5.56x10-2)(8.3)(298)/105 = 1.35x10-3 m3 = 1.375 liter
1.4
dV H2 /dt = 0.5 ml/h
dN H2 /dt = P (dV H2 /dt)/RT
= (105N/m2)(0.5cm3/h)(10-6m3/cm3)/(8.3Nm/molK)(298K) = 2.02x10-5 mol H 2 /h
= 5.62x10-9 mol H 2 /s
Ni + 2H+ --> Ni2+ + H 2
v cor = (5.62x10-9 mol Ni/s)/(20 cm2) = 2.81x10-10 mol Ni/ cm2s
table 1.3: 1 mol/cm2s = 3.5x108 (M/U) mm/year
© 2007, Dieter Landolt
2
M Ni = 58.7 g/mol,U Ni = 8.9 g/cm3
v cor = (3.15x108)(58.7)(2.81x10-10)/8.9 = 0.58 mm/year
1.5
Convert weight percent of Fe-13Cr into mol percent:
M Fe = 55.8 g/mol , M Cr = 52.0 g/mol
0.87 g Fe = 0.87/55.8 = 1.559 x 10-2 mol Fe
1 g alloy contains
0.13 g Cr = 0.13/52.0 = 0.250 x 10-2 mol Cr
--------------------------1.809 x 10-2 mol
Total
%mol Cr: (0.250x10-2)(100)/1.809x10-2 ) = 13.8 mol %Cr ---> mol fraction Cr: X Cr = 0.138
mol% Fe:
Anodic current density:
86.2 mol% Fe ---> mol fraction Fe: X Fe = 0.862
i a = i a,Fe + i a,Cr = 2 mA/cm2
i a,Cr = X Cr i a = (0.138)(2) = 0.276 mA/cm2
i a,Fe = X Fe i a = (0.862)(2) = 1.724 mA/cm2
Table 1.3: 1 PA/cm2 = 8.95x10-2 mg/dm2day
Cr:
v cor,Cr = (0.276x103)(8.95x10-2)(52.0)/2 =
Fe:
v cor,Fe = (1.724x103)(8.95x10-2)(55.8)/2 = 4305 mg/dm2day
Alloy: Vcor = v cor,Cr + v cor,Fe =
642 mg/dm2day
4947 mg/dm2day
1.6
Fe + 2 O 2 --> Fe 3 O 4
All the oxygen present reacts:
(8mg/l) (10-3g/mg) (300 l) /(32g/mol) = 7.50x10-2 mol O 2
this corresponds to (3/2)(7.50x10-2 ) = 0.113 mol Fe
surface: 10 m2
Thickness corroded:
L = (0.133mol)(55.8g/mol)/10m2)(7.86g/cm3)(104cm2/m2) = 9.44x10-6 cm = 0.094Pm
CHAPTER 2
2.1
4 Cu + O 2 --> 2 Cu 2 O
© 2007, Dieter Landolt
3
From Fig.2.2: 'Go # -130 kJ/mol for 1200oC
'Go = - RT ln K = + RT ln P O2
P O2 = exp('Go/RT) = exp{(- 130x103J/mol)/ (8.3J/molK)(1473K)} = 2.4 x10-5 bar
2.2
Fe + CO 2 = FeO + CO
T = 1373 K
'Go = - RT ln K = - RT ln (P co /P CO2 )
ǻG o
¦ Ȟ ǻG
i
o
i
'Go FeO = -264 + 64.7x10-3 T kJ/mol
'Go FeO = - 175 kJ/mol
For T= 1373K:
'Go CO2 = -394 - 1.13x10-3 T kJ/mol
'Go CO2 = - 392 kJ/mol
'Go CO = -112 –– 87.8 x10-3 T kJ/mol
'Go CO2 = + 8.55 kJ/mol
'Go = = -175 + 8.55 ––(-392) ––(0) = 225.5 kJ/mol
P CO /P CO2 = exp(-'Go/RT) = exp{(-225.5x103)/(8.3)(1373)}= 19.8
P CO + P CO2 = 1 bar
19.2 P CO2 + P CO2 = 1 bar
P CO2 = 4.8x10-2 bar
------>
2.3
Co + H 2 O = CoO + H 2
'Go = 'Go CoO + 'Go H2 - 'Go Co - 'Go H2O
'Go H2 = 0, 'Go Co = 0
'Go CoO = -240 + 78.0 x10-3 T
kJ/mol
For 1223K : 'Go CoO = -144.6 kJ/mol
'Go H2O = -246 + 54.9 x10-3 T
kJ/mol
'Go H2O = -178.9 kJ/mol
'Go = -144.6 ––(-178.9) = 34.3 kJ/mol
K = exp(-'Go/RT) = exp{-(34.3x103)/(8.3)(1223)}= 3.38
K = P H2 /P H2O
------->
P H2 + P H2O = 1 bar
P H2 =
P H2 = 3.38 P H2O
---> 3.38P H20 + P H2O = 4.38 P H2O
1/4.38 = 0.23 bar
P H2O = 1- 0.23 = 0.77 bar
2.4
Guntelberg equation:
© 2007, Dieter Landolt
log fr
A DB z z J 1 /2
1 J
1/ 2
with J
1
z 2i c i
¦
2
4
a) 0.01 M FeCl 2 solution
J = (1/2)[22(0.01)+(-1)2(0.02) = 0.06/2 = 0.03 mol/l
log f r = - (0.51)(2x1)(0.03)1/2 / (1+(0.03)1/2) = - 0.150
f r = 0.708
b) Solution of 0.01 M FeCl 2 + 0.05M HCl
J = (1/2) [(22)(0.01)+(12)(0.02) + 0.05 + 0.05] = 0.08 mol/l
log f r = - [0.51)(2)(0.08)1/2 / 1.283] = - 0.225
f r = 0.596
2.5
E rev = 0 + (RT/nF) ln [a H+ 2 / P H2 ]
n = 2 for 2 H+ + 2 e = H 2 ; T = 60oC = 333K;
P H2 = 0.5 bar; pH = 8
E rev = [(8.3)(333) / (2)(96485)] ln [10-8)2 / 0.5] = - 0.518 V
2.6
The corrosion reaction Cu + 2 H+ --> Cu2+ + H 2 occurs spontaneously provided E rev,Cu < E rev,H
T = 25oC, P H2 = 1 bar, c Cu2+ = 10-6 mol/l :
E rev,Cu = 0.34 + (0.059/2)log a Cu2+
= 0.34 + (0.059/2)log 10-6 = 0.163 V
E rev,H
= 0.0 –– 0.059 pH
= - 0.059 (0.5) = - 0.030 V
It follows that E rev,Cu > E rev,H
-------> no corrosion
2.7
Fe3+
+ H 2 O = FeOH2+ + H+
K 1 # c H+ c FeOH2+ / c Fe3+
FeOH2+ + H 2 O = Fe(OH) 2 + + H+
Calculate K 1 , K 2 :
-RT ln K 1 = 'G 1
K 2 # c H+ c Fe(OH)2+ / c Fe3+
and
-RT ln K 2 = 'G 2
'Go 1 = Po H+ + Po FeOH2+ - Po Fe3+ - Po H2O
. = exp(-'Go 1 /RT) = exp[-(12.6x103)/(8.3)(298)] = 6.1x10-3 [mol/l]
© 2007, Dieter Landolt
5
'Go 2 = Po H+ + Po Fe(OH)2+ - Po FeOH2+ - Po H2O
. = exp(-'Go 2 /RT) = exp[-(28.5x103)/(8.3)(298)] = 9.91x10-6 [mol/l]
For pH = 4:
c FeOH2+ / c Fe3+ = K 1 /c H+ = 6.1x10-3 / 10-4 = 61
c Fe(OH)2+ / c FeOH2+ = K 2 / H + = 9.9x10-6 / 10-4 = 0.099
total concentration of Fe :
c tot = c Fe3+ + c FeOH2+ + c Fe(OH)2+ = 0.1 mol/l
= (1/61) c FeOH2+ + c FeOH2+ + 0.099 c FeOH2+ = 1.12 c FeOH2+
It follows:
c FeOH2+ = 0.1/1.12
c Fe3+
= 0.090 mol/l
= (1/61) c FeOH2+ = 0.001 mol/l
c FeOH2+ = 0.099 c FeOH2+ = 0.009 mol/l
2.8
Ag+ + 2 CN- = Ag(CN) 2 -
K=
a Ag(CN)-
2
a Ag + a
2
CN -
Ag+ + e = Ag
Eo = 0.799 V
Table 2.115: Ag(CN) 2 - + e = Ag + 2 CN-
Eo’’ = - 0.31 V
Table 2.8 :
Equilibrium if:
E rev = E o ´+
RT a Ag(CN)-2
RT
ln 2
= Eo +
ln a Ag +
F
a CNF
Eo’’ –– Eo = (RT/F) ln (a Ag+ a CN- 2 / a Ag(CN)2 - ) = (RT/F) ln (1 / K)
25oC :
Eo’’ –– Eo = -0.31 –– 0.799 = - 1.109 = 0.059 log 10 (1/K) = - 0.059 log K
log K = 18.8 Æ K = 6.3x1018
2.9
Cell voltage at equilibrium: U = )’’’’ –– )’’ = 0.342 V
Cell reaction: 1/2 H 2 + AgCl = H+ + Ag + + Cl- The cell voltage U is equal to the reverible potentialof the cell reaction E rev .
- The standard potential for the cell reaction is equal to that of the halfcell
© 2007, Dieter Landolt
6
AgCl + e = Ag + Cl- which according to equation (2.117) is Eo = 0.222 V.
Fre energy of reaction: 'G = - nF E rev
'G = 'Go + RT ln(a H+ a Cl- / P H2 1/2 ) -->
E rev = Eo –– (RT/nF) ln(a H+ a Cl- / P H2 1/2 )
For P H2 = 1 bar :
E rev = Eo –– (RT/nF) ln(a H+ a Cl- ) = Eo - (RT/nF) ln (f + c H+ f - c Cl- ) = Eo - (RT/nF) ln
For 25oC:
0.342 = 0.222 –– 0.059 log ( f r 2 (0.12)2 ) = 0.331 –– 0.059 log f r 2
0.059 log f r 2 = - 0.186 Æ f r = 0.807
2.10
Pt’’¦ O 2 (0.002bar) ¦ KOH(0.01M) ¦ O 2 (0.2bar) ¦ Pt’’’’
U = )’’’’ –– )’’ = E rev ’’’’ –– E rev ’’
= (RT/nF) ln P O2 ’’’’ / P O2 ’’ = (8.3)(353)/(4(96485)) ln (0.2/0.002)
= 3.50x10-2 V
Æ U = 0.035 V
In this cell Pt’’’’ is the positive electrode (cathode) and Pt’’ is the anode.
2.11
E prot = Eo + (RT/nF) ln 10-6
(mol/l)
Ni2+ + 2e = Ni
Eo = - 0.257 V
E prot,Ni = -0.257 + (RT/nF) ln 10-6
25oC : E prot,Ni = -0.257 + (0.059/2) log 10-6 = - 0.434 V
E calomel = 0.241V
Protection potential versus saturated calomel electrode :
E prot,Ni = - 0.434 V –– 0.241 V = - 0.675 V vs SCE
2.12
(mol/m2 s)
maximum corrosion rate : v cor = k L c s
surface concentration of Cu2+ ions : c s = c Cu2+,s
Cu2+ + 2 e = Cu
Eo = 0.340 V
E = Eo + (0.059/2) log c Cu2+,s
(25oC)
0.160 = 0.340 + (0.059/2) log c Cu2+,s
log c Cu2+,s = - 6.10
--> c s = 7.91x10-7 mol/l
v cor = (10-5 m/s)(7.91x10-7 mol/dm3)(103 dm3/m3) = 7.91x10-9 mol/cm2 s
© 2007, Dieter Landolt
7
2.13
Pt ¦ H 2 (1bar) ¦ Ni2+, OH-, H 2 O ¦ Ni(OH) 2 ¦ Ni
Ni(OH) 2 = Ni2+ + 2 OHSolubility constant : K s = a Ni2 + a OH- 2 = 1.6x10-16
(mol3l-3)
Water dissociation constant : K w = a H+ a OH- = 10-14
For pH 8 :
a H+ = 10-8,
(mol2l-2)
a OH- = K w / a H+ = 10-6 mol/l
a Ni2+ = Ks / a OH- 2 = 10-16 / 10-12 = 1.6x10-4 mol/l
U = E rev,Ni –– E rev,H
E rev,Ni = -0.257 + (0.059/2)log a Ni2+ = 0.257 + (0.059/2)log 1.6x10-4 = -0.369 V
E rev,H = 0 –– 0.059 pH = -0.472 V
U = (-0.369) –– (-0.472) = 0.103 V
2.14
a) Calculation of standard potential::
Cu(NH 3 ) 2 + + e = Cu + 2 NH 3
Eo 1 = ––0.10 V
(1)
Cu(NH 3 ) 4 2+ + e = Cu(NH 3 ) 2 + + 2 NH 3
Eo 2 = 0.10 V
-----------------------------------------------------------------------------------
(2)
Cu(NH 3 ) 4 2+ + 2e = Cu + 4 NH 3
(3)
Eo 3
(3) = (1) + (2) :
Eo 3 = (n 1 /n 3 ) Eo 1 + (n 2 /n 3 ) Eo 2 = ½ (-0.1) + ½ (0.1) = 0 V
b) Dominating species in equilibrium with Cu metal : Cu(NH 3 ) 2 + (because Eo 1 < Eo 3 )
For reaction 1:
E rev,1 = Eo 1 + 0.059 log a Cu(NH3)2+ / a NH3 2
Protection potential: E prot = Eo 1 + 0.059 log (10-6/(0.1)2 )
= -0.1 +(0.059)(-4) = -0.336 V
Note that in ths case the value of the protection potential depends on the ammonia concentration!
c) From table 2:
Cu+ +
e = Cu
Eo = 0.520 V
= Eo 4
(4)
Cu2+ + 2 e = Cu
Eo = 0.340 V
= Eo 5
(5)
Complexation reactions:
© 2007, Dieter Landolt
8
Cu+ + 2 NH 3
=
Cu2+ + 4 NH 3
Cu(NH 3 ) 2 +
=
(6)
Cu(NH 3 ) 4 2+
(7)
Monovalent copper:
(6) = (4) - (1) Æ 'Go 6 = 'Go 4 -'Go 1 = - F (Eo 4 -Eo 1 ) = - RT ln K 6
where K 6 = a Cu(NH3)2+ / a Cu+ a NH3 2
(RT/F) ln K 6 = 0.520 –– (-0.1) = 0.420 V
25oC: log K 6 = 0.420/0.059 = 7.119 Æ K = 1.3x107
Divalent copper:
(7) = (5) –– (3) Æ 'Go 7 = 'Go 5 -'Go 3 = - F (Eo 5 -Eo 3 ) = - RT ln K 7
where K 6 = a Cu(NH3)42+ / a Cu2+ a NH3 4
(RT/2F) lnK 7 = 0.340 –– (0.0) = 0.340 V
25oC: log K 7 = 0.34/0.0295 = 11.5 Æ K = 3.3x1011
2.15
Cr2+ + 2e = Cr
Cr3+ + 3e = Cr
E° = ––0.90 V
E° = ––0.74 V
a) Equilibrium with Cr metal:
(1)
(2)
E rev,1 = E rev,2
-0.90 + (RT/2F)ln a Cr2+ = -0.74 + RT/3F ln a Cr3+
(-0.90) –– (-0.74) = - 0.16 = (RT/F)ln a Cr3+ 1/3 - (RT/F) ln a Cr2+ 1/2 = (RT/F)ln (a Cr3+ 1/3 / a Cr2+ 1/2)
For 25oC and replacing activities by concentrations:
-0.16/0.059 = - 2.71 = log (c Cr3+ 1/3 / c Cr2+ 1/2)
Æ c Cr3+ 1/3 / c Cr2+ 1/2 = 1.94x10-3
c Cr3+ = (1.94x10-3 )3 c Cr2+ 3 /2 = 7.31x10-9 c Cr2+ 3 /2
from this we deduce that in contact with chromium metal c Cr3+ << c Cr2+ .
b) Mol fraction of Cr3+ : X Cr3+ = c Cr3+ /c
X Cr3+ = 7.31x10-9 c Cr2+ 3 /2 /c = 7.31x10-9 c1/2 X Cr2+ 3/2
with X Cr2+ = c Cr2+ /c
X Cr3+ + X Cr2+ = 1
Hence : X Cr3+ = 7.31x10-9 c1/2 (1 - X Cr3+ )3/2
c) Concentration ratio Cr3+concentration in contact with chromium metal at a total chromium
ion concentration of 0.01 M:
Set
X Cr3+ << 1 : X Cr3+ = 7.31x10-9 c1/2
© 2007, Dieter Landolt
9
c = 0.01:
X Cr3+ = 7.31x10-10
it follows:
c Cr3+ = X Cr3+ c = 7,31x10-8 mol/l
c Cr2+ | 0.01 mol/l
CHAPTER 3
3.1
RT ln (P'/P o ) = 2JM /rU
Kelvin:
P o = 0.023 bar, T = 25oC
J= 0.072 Nm-1
(table 3.4)
U +2 = 1g/cm3, M H2O = 18 g/mol
ln (P'/P o ) = 2JM / RT r U (2) (0.072) (18) / (88.3) (298) (1x10-6 (1)) = 1.048x10-3
P'/P o = 1.001 --> P' = 0.023 bar
The small drop size has a negligible effect
3.2
Pressure difference to be applied : 'P = 2 J / r pore
J= 0.0720 , r pore = 3x10-6 m
'P = 2 (0.072)/ (3x10-6) = 4.80x104 N/m2
1 bar = 105 N/m2 ---> 'P = 0.48 bar
3.3
(111) plane, r = atom radius, h = (4 r) sin 60o = 3.464 r
h
surface of triangle: A = (4 r)(h) / 2 = 2 r h = 6.928 r2
number of atoms per triangle:
(3) (1/6) + (3) (½) = 2
For copper: r = 0.1278x10-9 m
number of atoms per m2:
© 2007, Dieter Landolt
N = 2 /(6.928)(0.1278x10-9)2 = 1.76 x1019 atoms Cu / m2
10
(100) plane, r = atom radius, a = 4 r cos45o = 2.828 r
a
surface of square: A = a2 = 7.952 r2
number of atoms per square:
(4) (1/4) + 1 = 2
For copper: r = 0.1278x10-9 m
number of atoms per m2:
N = 2 / 7.952 r2 = 1.54 x1019 atoms Cu / m2
3.4
v ads = s CO P CO / (2S M CO RT)1/2
P CO = 10-5 Pa
s CO = 0.5
M CO = 28 g/lmol = 28x10 -3 kg/mol
T = 298 K
(2S M CO RT)1/2 = 20.85 Ns/mol
One gets:
v ads = (0.5)(10-5)/20.85 = 2.40 x 10-7 mol/m2s
The number of moles forming a monolayer is obtained by dividing the number of adsorption
divided by the Avogadro number: N mono = 10-19 / 6x1023 = 1.67x10-5 mol/m2
Time to form a monolayer:
-5
t mono = N mono / v ads
-2
t = (1.67x10 mol m ) / (2.40x10-7 mol m-2s) = 70 s
3.5
Q ads /R = - d lnP/ d(1/T) for T = const, here: T = 0.5
from Figure:
T (K)
425
448
453
493
-3
10 /T (K-1)
2.35
2.23
2.21
2.03
108 P CO (Torr)
1.2
9
11
25
ln pCO
-18.2
-16.2
-16
-15.2
© 2007, Dieter Landolt
11
Log PCO
-15
-16
-17
-18
2.0 2.1 2.2 2.3 2.4
10-3/T
From this: Q ads /R = 8.57x103 (J mol-1/J mol-1 K) ---> Q ads = 7.1x104 J/mol = 71 kJ/mol
3.6
carbon
z0
z1
Lc
oxide
Lox
metal
z2
Cr3+ signal from the oxide film without contamination:
Lox
Io
N ³ c Cr 3 exp(z / / )dz
0
integration yields: I o = - N c Cr3+ / [exp(-L ox //) –– 1] = N c Cr3+ / [1 - exp(- L ox //) ]
Cr3+ signal from the oxide film with carbon contamination:
z2
Ic
N ³ cCr 3 exp(z / / )dz
z1
integration yields:
I c = - N c Cr3+ / [exp(-z 2 //) - exp(-z 1 //)]
with z 1 = L c and z 2 = L c + L ox
From this:
I c /I o
= [exp(-z 1 //) - exp(-z 2 //)] / [1 - exp(- L ox //) ]
= [exp(-L c //) - exp((-L c +L ox )//)] / [1 - exp(- L ox //) ]
= exp(-L c //) [ 1 - exp((L ox )//)] / [1 - exp(- L ox //) ] = exp(-L c //)
© 2007, Dieter Landolt
12
L c = 0.18nm ( = 2 x atomic radius) , / = 1.9 nm :
--> I c /I o = exp(-0.18/) = 0.91
3.7
Energy of ISS signals (E P = energy of primary beam) :
Fe: E Fe /E P = (M Fe ––M He )/(M Fe + M He )
Ni: E Ni /E P = (M Ni ––M He )/(M Ni + M He )
M He = 4 g/mol
M Fe = 55.8 g/mol
M Ni = 58.7 g/mol
Relative energy resolution:
'E /E P = [(M Ni ––M He )/(M Ni + M He )] –– [(M Fe ––M He )/(M Fe + M He )]
= [(58.7––4)/(58.7+ 4)] –– [(55.8 ––4)/(55.8 + 4)] = 0.8724 –– 0.8662 = 0.0062
Absolute energy resolution for E P = 1KeV:
'E 'E /E P ) E P = (1000 eV) (0.0062) = 6.2 eV
3.8
1/ 2
§ 2z 2 F2HH o c b ·
¸¸
C ¨¨
RT
©
¹
§ zF') GC ·
cosh¨
¸
© 2RT ¹
z = 1, H = 78, H o = 8,95x10-12 C/V m, T = 298 K,
1/ 2
§ 2z 2 F2HH o ·
¨¨
¸¸
© RT ¹
= 7.15x10-2 As/Vm2
§ zF') GC ·
cosh¨
¸ = 1.51
© 2RT ¹
Double layer capacity:
C = (7.15x10-2 )(1.51) c1/2 = 0.110 c1/2
For c = 0.001 mol/l:
C = 0.110 As/Vm2= 0.110 F/m2 = 11.0 PF/cm2
3.9
1
Mott-Schottky: 2
C
2
§ 2lc · § F')SC
·
¸¸ ¨
¨¨
1¸
¹
© HH o ¹ © RT
Flatband potential, E (')sc=0) :
or
1
C2
2
§ 2lc · F
§ 2lc ·
¨¨ HH ¸¸ RT ') SC ¨¨ HH ¸¸
© o¹
© o¹
2
from the potential of intersect of (1/C2) vs E with x-axis:
1/C2 = 0 : E = E intersect
© 2007, Dieter Landolt
13
0
§ 2l c
¨¨
© HH o
2
· § F') SC
·
¸¸ ¨
1¸
¹
¹ © RT
E = E intersect : ') SC = RT/F
-->
For semiconductor electrodes :
') SC = E –– E (')sc=0)
Therefore :
E (')sc=0) = E intersect –– ') SC = E intersect –– RT/F
Charge carrier density: from slope of 1/C2 = f(()
2
2
a = d(1/C )/d') SC
§ 2l · F
= ¨¨ c ¸¸
© HH o ¹ RT
with l c 2 = HH o RT/2F2n o
a = [4HH o RT/2F2n o (HH o )2](F/RT) = 2/F n o HH o
Measured values:
E(V)
-0.1
0
0.1
0.2
0.3
0.5
0.7
103C
14.0
7.7
6.0
5.0
4.4
3.6
3.2
5.1
16.9
27.8
40.0
51.7
77.2
97.7
(F/m2)
10-3C-2
(m2/F2)
From these values we get:
intersect : E intersect = - 0.15 V
--> Flatband potential: E (')sc=0) = E intersect –– RT/F = (-0.15)-(0.026) = -0.176 V
slope: a = 1.19x105 m4/V F2
Charge carrier density:
n o = 2/aFHH o
H H o = 8.95x10-12 As/Vm
n o = 2/(1.119x105)(8)(8.95x10-12)(96485) = 2.43 mol/m3 = 2.43x10-6 molcm3
CHAPTER 4
4.1
BV equation: i = i o [ exp(K/E a ) –– exp(- K/E c ) ]
Applied potential: E = - 0.6V vs SCE = - 0.359 V (NHE)
equilibrium potential: E rev = 0 –– 0.059 pH = - 0.177 V
overvoltage: K= E –– E rev = - 0.359 –– (- 0.177) = -0.182 V
cathodic Tafel region: i = - i o exp(- K/E c ) = - 10-4 exp(- (-0.182)/0.055) = - 2.74x10-3 A/cm2
© 2007, Dieter Landolt
14
4.2
i o = F k a c Fe2+ exp(E rev /E a ) = F k c c Fe3+ exp(- E rev /E c )
solution I:
c Fe3+ = 10-3 mol/l, c Fe2+ = 10-3 mol/l
E rev,I = Eo + (RT/F) ln (10-3/10-3 ) = Eo
solution II:
c Fe3+ = 10-3 + 0.019 = 0.020 mol/l, c Fe2+ = 10-3 mol/l
E rev,I = Eo + (RT/F) ln (2x10-2/10-3 ) = Eo +0.077 V
i o,II /i o,I = F k c (2x10-2) exp(- (Eo+0.077)/E c ) / F k c (10-3) exp(- Eo/E c )
= 20 exp[(-Eo-0.077+Eo)/E c )] = 20 exp (-0.077/E c )
numerical values: i o,I = 5x10-3 A/cm2, E c = 0.05V
i o,II = i o,I 20 exp (-0.077/E c ) = (5x10-3)20 exp (-0.077/) = 0.021 A/cm2
4.3
Equilibrium potential for pH = 1 :
E rev = 0 –– 0.059 = - 0.059 V
Saturated calomel electrode (SCE):
E rev = 0.241 V
equilibrium potential versus SCE :
E rev (SCE) = -0.059 +0.241 = - 0.300 V
Surface area: 6 cm2 : i = I/A = I/6 A/cm2
Experimental data:
E(SCE)
K(V)
I (A)
i (A/cm2)
log «i «
-0.360
-0.060
-0.0003
-5.0x10-5
-4.30
-4
-0-480
-0.180
-0.0029
-4.83x10
-3.32
-0.600
-0.300
-0.0310
-5.17x10-3
-2.29
0.720
-0.420
-0.3010
-5.02x10-2
-1.30
-1
-1.5
-2
-2.5
-3
-3.5
-4
-4.5
-5
-0.5
-0.4
-0.3
-0.2
© 2007, Dieter Landolt
-0.1
0
15
log«i «vs K
From graph
intercept with K = 0 ---> log i o = -4.8
--> i o = 1.6x10-5 A/cm2
slope: ( d log «i )/dK E c ! E c = 0.052 V
4.4
10
5
0
-5
-10
-20 -15 -10 -5
0
5
10
15
20
di/dE = 1/r c # 0.58 mA/cm2mV = 0.58 :-1cm-2 ---> r c = 1.72 :cm2
slope:
i cor = (1/r c )[(1/b a ) + (1/b c ) ]-1
= 0.58 [ (1/0.03)+(1/0.05)]-1 = 0.58/53.3 = 0.011 Acm-2
i cor = 11 mA/cm2
4.5
100
10
1
0.1
0.01
-0.8
-0.75
-0.7
-0.65
-0.6
-0.55
-0.5
E (V vs SCE)
E cor = - 0.55 V vs SCE
from figure extrapolation: I cor # 0.07 mA/cm2 ---> i cor # 0.035 mA/cm2
© 2007, Dieter Landolt
16
Table 1.1:
1 PA/cm2 = 3.27x10-3 (M/nU) mm/year
Ni:
n = 2, M = 58.7 g/mol , U= 8.9 g/cm3
i cor = 35PA/cm2 = (35)( 3.27x10-3)(58.7) /(2)(8.9) = 0.377 mm/year
4.6
i cor = i o,Fe exp[(E-E rev,Fe )/E a,Fe ]
(1)
i cor = i o,H exp[- (E-E rev,H )/E c,H ]
(2)
(2 equations with 2 unknowns: E cor , i cor )
E rev,H = 0 –– 0.059 pH = -0.059 V
E rev,Fe = -0.44 + (0.059/2) log 10-2 = - 0.499 V
E a,Fe = 0.0174 V, E c,H = 0.052 V
Corrosion potential:
i o,Fe exp[(E-E rev,Fe )/E a,Fe ] = i o,H exp[- (E-E rev,H )/E c,H ]
ln i o,Fe + (E cor /E a,Fe ) –– (E rev,Fe /E a,Fe ) = ln i o,H - (E cor /E c,H ) + (E rev,H /E c,H )
ln 10-8 + (E cor /) –– (-0.499 / ) = ln 10-7 - (E cor /) + (-0.059 /)
--> E cor = - 0.359 V
Corrosion current density:
i cor = i o,Fe exp[(E-E rev,Fe )/E a,Fe ]
ln i cor = ln i o,Fe + (E cor /E a,Fe ) –– (E rev,Fe /E a,Fe ) = - 18.42 +(-20.61)+28.68 = - 10.36
--> i cor = 3.2 x 10-5 A/cm2
Note: The same result for i cor ccould also be obtained from equation (2)
4.7
i cor = i o,H exp[- (E-E rev,H )/E c,H ]
Unknowns: i cor , i o,H
E cor = - 0.520 V (SCE) = - 0.279 V (NHE)
b c,H = 0.12 V
pH 4 :
---> E c,H = 0.052 V
E rev,H = 0 –– 0.059 pH = - 0.236 V
Calculation of i o,H from cathodic polarization:
For 1 mA/cm2 E = - 0.90 v (SCE) = - 0.659 V (NHE)
i = i o,H exp[- (E-E rev,H )/E c,H ]
10-3 = i o,H exp[- (-0.659 +0.236)/] = i o,H exp[8.135] = i o,H (3.41x103)
---> i o,H = 2.9x10-7 A/cm2
© 2007, Dieter Landolt
17
Corrosion current density:
i cor = (2.9x10-7) exp[- (-0.279 +0.236)/]
---> i cor = 6.6x10-7 A/cm2
4.8
pH 3 : E cor = -0.50 V, i cor = 10-6 A/cm2
E rev = 0 –– 0.059 (3) = - 0.177 V
i cor = 10-5 A/cm2
E rev = 0 –– 0.059 (1) = - 0.059 V
pH 1 : E cor = ? ,
Corrosion current density: i cor = i o,H exp[- (E cor -E rev,H )/E c,H ]
pH 3 :
with E c,H = 0.053
10-6 = i o,H(pH3) exp[- ((-0.50) ––(-0.177))/]
Æ i o,H(pH3) = 10-6/(4.43x102) = 2.26x10-9 A/cm2
Exchange curent density: i o,H = n F k c c H+ exp[- E rev,H /E c,H ]
pH 3 : i o,H(pH3) = n F k c (10-3) exp[- (-0.177) /] = n F k c (28.2x10-3)
pH 1 : i o,H(pH1) = n F k c (10-1) exp[- (-0.059) /] = n F k c (3.04x10-1)
i o,H(pH1) / i o,H(pH3) = 0.304/(28.2x10-3) = 10.8
i o,H(pH1) = 2.44x10-8
Corrosion potential: E cor –– E rev = - E c,H ln i cor /i o,H
E cor = E rev - E c,H ln i cor /i o,H
For pH 1: E cor = -0.059 –– 0.053 ln (10-6/2.44x10-8) = - 0.059 –– 0.197 = - 0.256 V
4.9
i l = 0.62 n F D2/3 Q-1/6 c b Z1/2
n=4
D = 2.51x10-5 cm2/s Æ D2/3 = 8.57x10-4
(cm2/s)2/3
Q = 10-2 cm2/s Æ Q = 2.15 (cm2/s)-1/6
c b = 8 mg/l = (8x10-3/32 )10-3 = 0.25x10-6 mol/cm3
Z= 1200 rpm = 2S (1200/60) = 126 rad/s
Æ Z1/2 = 11.21 (rad/s)1/2
Limiting current density at 1200 RPM:
i l = (0.62)(4)(96585)( 8.57x10-4)( 2.15)( 0.25x10-6)( 11.21) = 1.23x10-3 A/cm2
Diffusion layer thickness:
i l = n F D c b /G
Æ G = n F D cb / il
G = 4 (96585)( 2.51x10-5 )( 0.25x10-6) /(1.23x10-3) = 2.05x10-3 cm = 20.5 Pm
© 2007, Dieter Landolt
18
4.10
Mass transport controlled corrosion: i cor = i l,O2
Mass transport rate for turbulent pipe flow: Sh = 0.0115 Re7/8 Sc1/3
Re = 42000, Pipe diameter L = 1.9 cm , D O2 = 2.5x10-5 cm2/s
c O2 = 7 mg/l = 2.19x10-4 mol/l = 2.19x10-7 mol/cm3
Sc = Q/D O2 = 0.01/(2.5x10-5) = 400
Sh = (0.0115)( 42x103)7/8(400)1/3 = 9.39x102
Æ i l,O2 = Sh 4F D O2 c O2 / L
Sh = i l,O2 L / 4F D O2 c O2
2
-5
i l,O2 = (9.39x10 )(4)(96485)( 2.5x10 )( 2.19x10-7) /1.9 = 1.05x10-3 A/cm2 = 1.05 mA/cm2
4.11
Sh = 0.079 Sc0.35 Re0.7
Table 4.27 :
Re = Zr L /Q
L = 2r = 0.04 m
Z = 2S(4000/60) = 419 rad/s
Sc = Q/D H
Q = 10-6 m2/s
D H = 9.32x10-9 m2/s
Sh = i l L /n F D H c b
c b = 0.001mol/l = 1 mol/m3 n = 1
Re = (419)(0.02)(0.04)/(10-6) = 3.35x105
Sc = (10-6)/9.32x10-9 = 1.072x102
Sh = (0.079)( 1.072x102)0.35(3.35x105)0.7 = 2.99x103
i l = Sh (n F D H c b )/L = (2.99x103)(1)(96485)( 9.32x10-9)(1)/(0.04) = 67.2 A/m2
Corrosion rate: 1 A/m2 = 0.327 (M/n U) mm/year (Table 1.1); M Mg = 24.3 g/mol, U Mg = 1.74
g/cm3 , n = 2 :
--> v cor = (67.2)(0.327)(24.3)/(2)(1.74) = 1.53 mm/year
4.12
H2O, O2
bentonite
O2
copper
2 Cu + O 2 + 2 H 2 O Æ 2 Cu(OH) 2
Oxygen flux across bentonite layer of thickness L = 2 m : N O2 = D O2 c O2 /L
D O2 = 8x10-11 m2/s,
c O2 = 5.4 ppm = 5.4 g/106 g H 2 O = 5.4g/m3H 2 O = (5.4/32) mol/m3 = 0.169 mol/m3
N O2 = (8x10-11)(0.169) /2 = 6.76x10-12 mol/m2s
© 2007, Dieter Landolt
19
Corrosion rate of copper:
Æ N Cu = 1.35x10-11 mol/m2s
N Cu = 2 N O2
1 mm/year = (3.15x104)(M/U) mol/m2s where M Cu = 63.5 g/mol, U Cu = 8.96 g/cm3
v cor = (1.35x10-11 ) (3.15x104)(63.5/) = 3.01 x10-6 mm/year
Thickness corroded in 100000 years: 0.3 mm
4.13
c
O2
cs,OH
OH-
G
O 2 + H 2 O + 4 e Æ 4 OH-
y
hydroxyl ion flux: N OH = 4 N O2
N O2 = i l,O2 /4F = 0.62 D O2 2/3 Q-1/6 c b,O2 Z1/2
N OH = 0.62 D OH 2/3 Q-1/6 c s,OH Z1/2 assuming c b,OH | 0
N OH / N O2 = D OH 2/3 c s,OH / D O2 2/3 c b,O2 = 4
Æ
c s,OH = 4 (D O2 /D OH )2/3 c b,O2
D O2 = 2.5x10-9 , D OH = 5.25x10-9 , c b,O2 = 8 mg/l = 2.5x10-4 mol/l
c s,OH = 4 (2.5/5.25)2/3 (2.5x10-4) = 6.1x10-4 mol/l
Water dissociation equilibrium: c OH c H+ | 10-14 (mol/l)2
c H+ | 1.6x10-11 mol/l or pH | 10.8
Note: The surface pH does not vary with the rotation rate.
4.14
Limiting current density at RDE in a binary electrolyte:
i l = (1-(z + /z - )) 0.62 n F D + 2/3 Q-1/6 (c +s –– c +b ) Z1/2
FeCl 2 : z + = 2, z - = 1
Æ
1-(z + /z - ) = 3
D + = 0.72x10-9 m2/s
c +s = 4.25 mol/l, c +b = 2.0 mol/l
Æ c +s –– c +b = 2.25 mol/l
Q = 10-6 m2s , Z = 200 RPM = (200)2S/60 = 20.9 rad/s
i l = (3)(0.62)(2)(96485)( 0.72x10-9)2/3(10-6)-1/6(2.25)(20.9)1/2 = 2.97x104 A/m2
i l = 2.97 A/cm2
© 2007, Dieter Landolt
20
4.15
N = (F2/RT) ¦ z i 2 D i c i
M NaCl = 58.5 g/mol, M Ca(HCO3)2 = 162.1 g/mol
c NaCl = 1 mg/l = (10-3/58.5) = 1.71x10-5 mol/l = 1.71x10-2 mol/m3 = c Na+ = c Clc Ca(HCO3)2 = 15 mg/l = (15x10-3/162.1) = 9.25x10-5 mol/l = 9.25x10-2 mol/m3 = c Ca+ = ½ c HCO325oC: F2/RT = (96485)2/(8.3)(298) = 3.76x106 As/Vmol
Na+
-
D Na+ = 1.33x10-9 m2/s
:
-9
D Cl- = 2.03x10 m /s
z Cl- = -1
Ca2+ :
D Ca2+ = 0.79x10-9 m2/s
z Ca2+ = 2
HCO 3 - :
D HCO3- = 1.18x10-9 m2/s
z HCO3- = -1
Cl
:
z Na+ = 1
2
N = (3.76x106)[ z Na+ 2D Na+ c Na+ + z Cl- 2D Cl- c Cl- + z Ca2+ 2D Ca2+ c Ca+ + z HCO3- 2D HCO3- c HCO3- ]
= (3.76x106)[(1.33x10-9 )(1.71x10-2 )+ (2.03x10-9) (1.71x10-2 )
+ ( 4 )(0.79x10-9 )( 9.25x10-2 )+ (1.18x10-9 )(18.5x10-2)] = 2.14x10-3 :-1m-1
Units: (As/molV)(m2/s)(mol/m3) = A/Vm = :-1m-1
4.16
steady motion:
z i q i (d)/dy) = 6S r i K i v i
Definition of electric mobility: u e,i = v i /(d)/dy) [(m/s)/(V/m)]; u e,i = z i F u i with u i = D i /RT
Fe2+ : D Fe2+ = 0.72x10-9m2/s Æ u e,Fe2+ = (2)(96485)(0.72x10-9)/(8.3)(298) = 5.61x10-8 m2/Vs
Fe3+ : D Fe3+ = 0.61x10-9m2/s Æ u e,Fe2+ = (3)(96485)(0.61x10-9)/(8.3)(298) = 7.13x10-8 m2/Vs
ionic radius: r i = z i q i /6S K u e,i
Fe2+ : r Fe2+ = (2)(1.6x10-19) /(6S)(10-3)( 5.61x10-8) = 3.03x10-10 m
Fe3+ : r Fe3+ = (3)(1.6x10-19) /(6S)(10-3)( 7.13x10-8) = 3.57x10-10 m
r Fe3+ > r Fe2+ because the higher charge leads to stronger hydration
CHAPTER 5
5.1
Fe3+ + 2 e = Fe2+
Eo = 0.771 V
c Fe3+ = c Fe2+ = 0.001 mol/l
E rev = 0.771 + (RT/F) ln (a Fe3+ /a Fe2+ ) | 0.771 + (RT/F) ln (c Fe3+ /c Fe2+ ) = 0.771 V
Applied potential E = 0.595 V
© 2007, Dieter Landolt
21
Overvoltage : K = E –– Erev = 0.595 –– 0.771 = - 0.176 V
Assuming that the rection is charge transfer controlled (Tafel region):
i c = - i o exp(-K/E c ) = - 5x10-3 exp(+0.176/0.05) = - 0.169 A/cm2
Assuming that the reaction is transport controlled (limiting current):
i l = - nF D Fe3+ c Fe3+ / G
i l = - (1)(96485)(0.61x10-5)(10-5)/10-3) = - 5.89x10-3 A/cm2
Comparison of i c and i l shows: il <<ic. Under the conditions of the experiment Fe3+ will be
reduced at the limiting current density under mass transport control.
5.2
I
Fe + OH-
Æ FeOH ads + e
II
FeOH ads
Æ FeOH ads + + e
III
FeOH ads + + H+ Æ Fe2+ + H 2 O
I:
k I,a c OH (1-T) exp(Df E) = k I,c T exp(-(1-D) f E)
where T = surface coverage of FeOH ads and f = F/RT
for T<< 1: T/(1 –– T) | T = (k I,a / k I,c ) c OH exp(Df E + (1-D) f E) = K I c OH exp( f E)
where K I = (k I,a / k I,c )
II:
v II = k II,a T exp(Df E) = k II,a K I c OH exp( f E) exp(Df E) = k II,a K I c OH exp [(1+ Df E)]
anodic current density:
i a = i I + i II = F v I + F v II = 2 F v II
i a = 2F k a c OH exp [(1+ Df E)]
where k a = k II,a K I
Charge transfer coefficient:
E a = d ln i a /dE = (1 + D) f Æ E a = 1/((1+D) f )
D = 0.5 ; 1/f = (RT/F) = (8.3)(298)/96485 = 0.0256 V : E a = 0.0256 /1.5 = 0.0171 V
Reaction order for OH- :
§ d ln i ·
¸
p OH = ¨
¨ d ln c –– ¸
OH ¹c
©
Fe2 =1
Variation of exchange current density with Fe2+ cocnetration:
i o = 2 F k a c OH exp [(1+ Df
E rev )]
d ln i o /d ln c Fe2+ = d[(1+D) f E rev ] / d ln c Fe2+
E rev | Eo + (RT/2F) ln c Fe2+
© 2007, Dieter Landolt
22
(1+ Df E rev ) = (1+ Df Eo)+ ((1+ D ln c Fe2+
d[(1+D) f E rev ] / d ln c Fe2+ = (1+ D § d ln i0 ·
¨
¸
© d ln cFe2 ¹cOH ––
Potential shift with c OH- at constant i and c Fe2+ :
ln i a = ln (2F k a c OH ) + (1 + D)f E
E = [1/(1 + D)f] [ln i a –– ln (2F k a –– ln c OH )]
§ dE
·
¨
¸
© d ln cOH–– ¹i ,cFe2 = - 1/(1 + D)f = - 0.0171 V
5.3
Volmer-Tafel mechanism:
H+ + e
Æ H ads
H ads + H ads Æ H 2
I at quasi equilibrium:
k Ic c H+ (1 –– T) exp (–– (1 –– D)f E) = k Ia T exp(D f E)
2x
I
II
for small coverage: T /(1 –– T) | T
II (RDS) :
T k Ic / k Ia ) c H+ exp[-(1-D) f E –– D f E) = k Ic / k Ia ) c H+ exp (–– f E)
v II = k IIc T2 = k IIc k Ic / k Ia )2c H+ 2 exp (-2f E)
i c = - 2 F k IIc k Ic / k Ia )2c H+ 2 exp (-2f E) = - 2 F k IIc k Ic / k Ia )2c H+ 2 exp (-E/E c )
E c = dE/dlni c Æ 1/Ec = d ln i /dE = -(-2f) = 2f
Æ
E c = 1/ (2f)
1/f = (RT/F) = (8.3)(298)/96485 = 0.0256 V
Æ
E c = 0.0128 V
b c = 2.3 E c = 0.0295 V | 30 mV
5.4
For a charge transfer controlled reaction:
Z Re
Rt / Z 2C 2
Rt2 1/ Z 2C 2
Z Im
Rt2 / ZC
Rt2 1/ Z 2C 2
A Nyquist plot of the tabulated data yields a hemi-circle:
© 2007, Dieter Landolt
23
8
6
4
2
0
0
2
4
6
8
ZRe
Z=’
Z=0
Charge transfer resistance from graph: R t = Z Re (Z=0) - Z Re (Z=’) = 8 :
Æ r t = R t A = (8)(2) = 16 :cm2
A = 2 cm2
Double layer capacity from the maximum: |Z Re | = |Z Im |
Rt2 / ZC
Rt / Z 2C 2
= 2
Rt2 1/ Z 2C 2
Rt 1/ Z 2C 2
Æ 1/ZC = R t
Z max = 1.25 kHz (see figure and table) ; R t = 8 : Æ C = 1/Z R t = 10-4 Farad
c dl = C/A = 0.5 x 10-4 F/cm2 = 50 PF/cm2
Exchange current density:
At the equilibrium potential : r t = RT/nFi o Æ i o = RT/nFr t
n = 2; RT/F = 0.026V ; r t = 16 :cm2
Æ
i o = 0.013/16 = 8.1x10-4 A/cm2
5.5
8
6
ZI
m
4
2
0
0
2
4
6
8
Z Re
From graph: R : = 2 :,
R t = 8 –– 2 = 6 :
© 2007, Dieter Landolt
24
A = 5 cm2 ; r t = R t A = 30 :cm2
Tafel region: r t = dE/di = (dE/d ln i)(d ln i/di) = E a / i
i = 1 mA/cm2 Æ E a = r t i = (30 :cm2)(10-3 A/cm2) = 0.030 V
5.6
(I)
(II)
(III)
Ni + H 2 O = NiOH ads + H+ + e
NiOH ads o NiOH+ + e
NiOH+ + H+ = Ni2+ + H 2 O
I:
k a,I (1-T) exp(DfE) = k c,I T c H+ -1 exp(-(1-D)fE)
T(1-T) | T k a,I /k c,I ) c H+ -1 exp(fE)
II:
v II = k a,II T exp(DfE) = k a,II >k a,I /k c,I ) c H+ -1 exp(fE)] exp(DfE)
= (k a,II k a,I /k c,I ) c H+ -1 exp ((1+DfE))
i a = 2 F v II = 2F k a c H+ -1 exp ((1+Df E))
where
k a = (k a,II k a,I /k c,I )
Exchange current density:
i o = 2F k a c H+ -1 exp ((1+Df E rev )
ln i o = ln (2F k a ) –– ln c H+ + (1+Df E rev
log i o = log (2F k a ) –– log c H+ + (1+Df E rev /2.3
d log i o / d pH | - d log i o / d log c H+ = 1
§ d log i0 ·
¨
¸
© d pH ¹cNi2 =1
d log i o / d log c Ni2+ = (d log i o / dE rev )(dE rev / d logc Ni2+ ) = ((1+Df /2.3)(2.3/2f) = (1+D
( setting E rev | Eo + (RT/2F) ln c Ni2+ = Eo + (2.3/2f) log c Ni2+ )
§ d log i0 ·
¨
¸ = (1 + 0.5)/2 = 0.75
© d log cNi2+ ¹pH
5.7
I
Cu
= Cu+ + e
II
Cu+
Æ Cu2+ + e
(a)
D I = D II = D f = F/RT
K = E - E rev
reaction step I:
i I = F k a,I exp(Df E) –– F k c,I c Cu+ exp(-(1-D)f E)
i o,I = F k a,I exp(Df E rev ) = F k c,I c Cu+(eq) exp(-(1-D)f E rev )
© 2007, Dieter Landolt
25
Æ
i I = i o,I exp(Df K) –– i o,I (c Cu+ / c Cu+(eq) ) exp(-(1-D)f K)
eliminate (c Cu+ / c Cu+(eq) ):
i I = i o,I exp(Df K) –– i o,I (c Cu+ / c Cu+(eq) ) exp(-(1-D)f K)
(c Cu+ / c Cu+(eq) ) = - i I + i o,I exp(Df K) / i o,I exp(-(1-D)f K)
= - (i I / i o,I ) exp((1-D)f K) + exp(Df K)/ exp(-(1-D)f K)
= - (i /2 i o,I ) exp((1-D)f K) + exp(f K) because at steady state i I = i II = i/2
reaction step II:
i II = i o,II (c Cu+ / c Cu+(eq) ) exp(Df K) –– i o,II exp(-(1-D)f K)
(using same reasoning as for step I)
= i o,II [- (i /2 i o,I ) exp((1-D)f K) + exp(f K)] exp(Df K) –– i o,II exp(-(1-D)f K)
= (-i o,II i /2 i o,I ) exp((1-D)f K) + i o,II exp(f K)] exp(Df K) –– i o,II exp(-(1-D)f K) = i/2
rearrange:
(i/2)[1 + (i o,II /i o,I ) exp(f K)] = i o,II [exp((1+Df K) - exp(-(1-D)f K)]
i = 2 i o,II [exp((1+Df K) - exp(-(1-D)f K)] / [1 + (i o,II /i o,I ) exp(f K)] q.e.d
(b) For i o,I >> i o,II and K >> 0 (anodic Tafel region) :
from above equation it follows (setting i o,I = ’) : i = i a = 2 i o,II exp ((1+D) f K)
E a = 1/(1+D)f
For i o,I >> i o,II and K << 0 (cathodic Tafel region) :
i = i c = - i o,II exp ( -(1-D) f K)
E c = 1/(1-D)f
CHAPTER 6
6.1
The standard potential of the halfcell reaction
Fe 2 O 3 + 6 H+ + 6 e = 2 Fe + 3 H 2 O
corresponds to the standard free energy change of the reaction (Chapter 2):
Fe 2 O 3 + 3H 2 = 2 Fe + 3 H 2 O
'Go = 3 'Go (H2O) - 'Go (Fe2O3) = 3 (-237.2) ––(-742.2) = + 30.6 kJ/mol
Eo = -'Go/nF = (-30.6x103) /(6)(96485) = - 0.053 V
Equilibrium potential in 0.5M H 2 SO 4 of pH | 0 :
E rev = Eo + RT/6F ln (1/a H+ 6) = - 0.053 –– 0.059 pH = - 0.053 V
© 2007, Dieter Landolt
26
Passivation potential at pH | 0 (from Fig. 6.9 ) :
Ep = 0.86 V
Æ E p >> E rev !
6.2
3 Fe 2 O 3 + 2 H+ 2 e = 2 Fe 3 O 4 + H 2 O
(1)
E 1 o =?
One finds in Table 6.9 :
Fe 3 O 4 + 8 H+ + 8 e = 3 Fe + 4 H 2 O
(2)
E 2 o = - 0.087 V
Fe 2 O 3 + 6 H+ + 6 e = 2 Fe + 3 H 2 O
(3)
E 3 o = - 0.053 V
-------------------------------------------------------------------------------reaction (1) = 3 x reaction (3) –– 2 x reaction (2)
'Go 1 = 3 'Go 3 –– 2 'Go 2
-n 1 F Eo 1 = - 3 n 3 F Eo 3 + 2 n 2 F Eo 2
Æ Eo 1 = 3 (n 3 /n 1 ) Eo 3 –– 2 (n 2 /n 1 ) Eo 2
n 1 = 2, n 2 = 8, n 3 = 6 :
Eo 1 = 3 (3) (- 0.053) –– 2 (4) (- 0.087) = 0.219 V
6.3
ln | i |
Ecor
Ep
E
i p = i cor exp[(E p –– E cor )/E a ]
Tafel region
Æ
E p = 0.15 –– 0.059 pH
for pH = 2: Ep = 0.138 V
E cor = - 0.05 V; i cor = 0.1 mA/cm2; E a = 30 mV
i p = (0.1x10-3) exp[(0.138 + 0.05) / 0.03)] = 0.053 A/cm2 = 53 mA/cm2
6.4
ln | i |
E
Cathodic partial reaction: Fe3+ + e Æ Fe2+ ; anodic partial reaction : M Æ Mz+ + z e
© 2007, Dieter Landolt
27
What value of c Fe3+,b is needed to passivate the electrode: Passivation if il,Fe3+ • i p
Rotating cylinder: Sh = 0.079 Re0.7 Sc0.35
Æ i l,Fe3+ = Sh nF c Fe3+,b D Fe3+ / L
Sh = i l,Fe3+ L / nFc Fe3+,b D Fe3+
with L = cylinder diameter: L = 2r ; Re =Zr L/Q = 2 Z r2/Q
Sc = Q/D Fe3+
i l,Fe3+ = (nFc Fe3+,b D Fe3+ /2r)(0.079) (2 Z r2/QQ/D Fe3+ )0.35
i l,Fe3+ = 0.0642 nF c Fe3+,b D Fe3+ 0.65Qr0.4 Z Æ
c Fe3+,b = i l,Fe3+ / 0.0642 nF D Fe3+ 0.65Qr0.4 Z 100 RPM = 1000(2S/60) = 104.7 rad/s
i p = 0.03 A/cm2 ; Q = 0.011 cm2/s; D Fe3+ = 0.5x10-5 cm2/s; n = 1; i l,Fe3+ = 3x10-2 A/cm2
Æ c Fe3+,b = 6.55x10-5 mol/cm3
6.5
From Fig. 6.16 we read:
pH
log |i p |
1.8500
-1.6500
3.0200
-1.9500
4.5500
-3.7000
7.4500
-4.8000
8.4200
-5.3000
9.3700
-5.9000
11.500
-7.2000
A plot of log |i p | versus pH of yields:
y = -0.58094 - 0.57265x R= 0.99181
-
0
2
4
6
p
8
1
1
By regression analysis: log |i p | = -0.08 –– 0.57 pH
To reach spontaneous passivation i l,O2 • i p
(cf. problem 6.4)
Limiting current density for oxygen reduction:
© 2007, Dieter Landolt
28
i l,O2 = 4 F D O2 c O2,b / G
with
c O2,b = 0.5 ppm = 0.5 mg/l = 0.5x10-3/32 = 1.56x10-5 mol /l = 1.56x10-8 mol O 2 /cm3
D O2 = 2.5x10-5 cm2/s (from Table 4.1) ; G = 10-2 cm
Æ i l,O2 = 7.53x10-6 A/cm2 or
log |i l,O2 | = - 5.12 (A/cm2)
log |i l,O2 | = log |i p |
- 5.12 = -0.08 –– 0.57 pH
Æ
pH = (-5.12+0.08)/0.57 = - 7.97 | 8.0
--> Spontaneous passivation for pH • 8
6.6
From Fig. 6.15 :
Z(rad/s)
261.80
167.60
41.900
Z1/2
16.180
12.940
6.4700
i p (mA/cm2)
1550.0
1300.0
600.00
1600
1400
y = -28.182 + 99.313x R= 0.99702
ip(mA/cm2)
1200
1000
800
600
400
200
0
0
5
10
Z1/2
15
20
regression analysis: i p | 99 Z1/2 mA/cm2 = 99x10-3 A/cm2
Levich equation: i l = 0.62 nF D2/3 c sat Q-1/6 Z1/2
i l = i p | 99 Z1/2
Saturation concentration at electrode surface:
c sat = i l / 0.62 nF D2/3 Q-1/6 Z1/2 = 99x10-3 Z1/2 / 0.62 nF D2/3 Q-1/6 Z1/2 = 99x10-3 / 0.62 nF D2/3 Q1/6
For D = 10-5cm2/s ; Q = 10-2 cm2/s ; n =2 :
Æ c sat = (99x10-3)/[(0.62)(2)(96485)(2.92x10-4)(2.15)] = 1.29 x10-3 mol/cm3 # 1.3 mol/l
© 2007, Dieter Landolt
29
6.7
2 CrO 3(s) + 6 H+ + 6 e = Cr 2 O 3(s) + 3 H 2 O
= 6 H+ + 6 e
3H 2
-------------------------------------------------2 CrO 3 + 3H 2
= Cr 2 O 3 + 3 H 2 O
'Go = 'Go Cr2O3 + 3 'Go H2O –– 2 'Go CrO3
For this reaction:
Using Tables 6.8 and 2.9 :
'Go = (-1058) + 3(-237.2)-2(-510) = - 749.6 kJ/mol Cr 2 O 3
Eo = - 'Go / nF = - (- 749.6x103)/(6)(96485) = 1.295 V
pH 5 : E rev = Eo + RT/nF ln(1/a H+ 2) = Eo –– 0-059 pH
E rev = 1.295 –– (0.059) (5) = 1.00V
6.8
Data points:
i(A/cm2)
0.26200
0.25000
0.22700
0.20000
0.17200
0.13400
0.10700
Z(rad/s)
600.00
400.00
200.00
100.00
50.000
20.000
10.000
1/i
3.8170
4.0000
4.4050
5.0000
5.8140
7.4630
9.3460
1/Z1/2
0.041000
0.050000
0.071000
0.10000
0.14100
0.22400
0.31600
10
y = 2.9883 + 20.071x R= 0.99998
8
6
4
2
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Regression analysis: (1/i) = 2.99 + 20.1 (1/Z1/2)
Z Æ ’: 1/i = 2.99 # 3.0 or i (Z = ’) = 0.33 A/cm2
1/2
1/Z
6.9
© 2007, Dieter Landolt
30
H 2 CrO 4 = HCrO 4 - + H+
K 1 = 6.3
HCrO 4 - == CrO 4 2- + H+
K 2 = 3.3x10-7
2HCrO 4 - = Cr 2 O 7 2- + H 2 O
K 3 = 33.9
To solve this problem we replace all activities by concentrations.
K 1 | c H+ c HCrO4- / c H2CrO4 = 6.3
(1)
K 2 | c H+ c CrO4-- / c HCrO4- = 3.3x10-7
(2)
K 3 | c Cr2O7 / c HCrO4- 2 = 33.9
(3)
a H+ | c H+ = 10-3
(pH = 3)
(4)
Total chromium concentration: c = c H2CrO4 + c HCrO4- + c CrO4-- + c Cr2O7
(5)
Æ 5 equations with 5 unknowns: c H+ , c HCrO4- , c CrO4-- , c H2CrO4 , c Cr2O7
Approximate solution:
(1): c HCrO4- / c H2CrO4 = 6.3/10-3 = 6.3x103
Æ c HCrO4- >> c H2CrO4
(2): c CrO4-- / c HCrO4- = 3.3x10-7 / 10-3 = 3.3x10-4
Æ c CrO4-- << c HCrO4-
(3): c Cr2O7 / c HCrO4- 2 = 33.9
Æ c Cr2O7 = 33.9 c HCrO4- 2
(5): c = 0.1 mol/l = c HCrO4- + c CrO4-- + c H2CrO4 + c Cr2O7 | c HCrO4- + c Cr2O7
solving for (5) c HCrO4- yields :
c HCrO4- = 0.042 mol/l
with (3): c Cr2O7 = 33.9 (0.042)2
c Cr2O7 = 0.058 mol/l
with (1) c H2CrO4 = 0.042/6.3x103
c H2CrO4 = 6.7x10-6 mol/l
with (2): c CrO4-- = (3.3x10-4)( 0.042)
c CrO4-- = 1.4x10-5 mol/l
=
c HCrO4- + 33.9 c HCrO4- 2
CHAPTER 7
7.1
E rev = Eo + (RT/F) ln a Cu2+
Eo = 0.340 V
0.1 ppm = (0.1x10-6g/g)(103g/l)/63.5g/mol = 1.57x10-6 mol/l
a Cu2+ # c Cu2+ = 1.57x10-6 mol/l
95oC = 368 K :
E rev,95 = 0.340 + (8.3)(368)/896485)ln(1.57x10-6) = 0.128 V
55oC = 338 K :
E rev,55 = 0.340 + (8.3)(338)/896485)ln(1.57x10-6) = 0.146 V
'E = E rev,95 - E rev,55 = 0.018 V
For corrosion cells: E anode < E cathode
© 2007, Dieter Landolt
--> Metal at 95oC is anodic and corrodes preferentially.
31
7.2
Cell voltage for I = 0 :
U (I=0) = 2.0 V
Cell voltage for I z 0 :
U (I) = U (I=0) - ] a - |] c |- I R int
with
] a = 0.18 + 0.12 log I
] c = - 0.20 –– 0.09 log |I|
R int = 0.14 :
R
I = 0.1A :
U (I=0.1) = U (I=0) - ] a - |] c |- I R int
U (I=0.1) = 2.0 –– (0.18 + 0.12 log 0.1) - |(-0.20-0.09 log 0.1) | - 0.1(0.14)
= 2.0 –– 0.06 –– 0.11 –– 0.014 = 1.816 V
I = 1A :
U (I=1) = U (I=0) - ] a - |] c |- I R int
U (I=1) = 2.0 –– (0.18 + 0.12 log 1) - |(-0.20-0.09 log 1) | - 1(0.14)
= 2.0 –– 0.18 –– 0.20 –– 0. 14 = 1.480 V
7.3
Fe + ½ O 2 + 2H+ --> Fe2+ + H 2 O
--> V cor = N Fe2+ = 2 N O2 (mol/cm2s)
A = A Fe + A Cu = 2500 = A Fe + 5(0.5)
cm2
--> A Cu = 2.5 cm2
A Fe = 2497.5 cm2 | 2500 cm2
N O2 = 4x10-11 mol/cm2s
Number of moles O 2 reacting per second:
J O2 = N O2 A = (4x10-11)(2500) = 1.0x10-7 mol O 2 /s
number of moles Fe reacting per second:
J Fe = 2 J O2 = 2.0 x10-7 mol Fe/s
(a) Rivets made of iron: A Fe = 5x0.5 cm2
v cor = N Fe = 2.0 x10-7/2.5 = 8.0x10-8 molFe/cm2s
Table 1.3:
1 mol/cm2s = 3.15x108 (M/U) mm/year
M Fe = 55.8 g/mol ; U Fe = 7.86 g/cm3
--> v cor = 179 mm/year
(b) Rivets made of copper: A Fe = 2500 cm2
v cor = N Fe = 2.0 x10-7/2500 = 8.0x10-11 molFe/cm2s --> v cor = 0.179 mm/year
7.4
Anodic partial reaction: i a , Zn = i cor,Zn exp(E Zn -E cor,Zn )/E a,Zn
A Zn = A Cu =10cm2; i l,O2 = 0.05 A/cm2;
E cor,Zn = - 0.765 V ; E cor,Cu = 0.01 V ; R int = 1 : ; E a,Zn = 0.04V
For I a,Cu | 0 :
I a,Zn = i l,O2 (A Cu + A Zn ) = (0.05)(20) = 1.0 A
© 2007, Dieter Landolt
32
(a) i a,Zn = I a,Zn / A Zn = 1.0/10 = 0.1 A/cm2
(b) E Zn - E cor,Zn = E a,Zn ln (i a , Zn / i cor,Zn )
With i a , Zn = 1.0 A/cm2 ; i cor,Zn = i l,O2 = 0.05 A/cm2 :
E Zn = 0.765 ln (0.1 / 0.05) = - 0.737 V
(c) I = I a,Zn + I c,Zn # I a,Zn + i l,O2 A Zn = 1.0 –– 0.05 (10) = 0.5 A
(d) E Cu –– EZn = I R int --> E Cu = E Zn + I R int = -0.737 + (0.5)(1) = - 0.273 V
7.5
(a) Noble electrode, equilibrium conditions for oxygen:
O 2 + 4H+ + 4e = 2 H 2 O
E rev,I = Eo + (RT/4F) ln(P O2,I a H+ )
E rev,II = Eo + (RT/4F) ln(P O2,II a H+ )
E rev,I - E rev,II = (RT/4F) ln(P O2,I /P O2,II ) = 0.0064 ln(0.01/1.0) = - 0.0295 V
(b)
brass
ln |i|
O2,II
O2,I
E
k H = 1.3 x10-3 mol/dm3 bar
P O2,I = 0.01 bar :
c O2,I = k H (0.01) = 1.3x10-5 mol/l
P O2,II = 1.0 bar :
c O2,II = k H (1.0) = 1.3x10-3 mol/l
i lO2 = - nFD O2 C O2 /G !i lO2,I / i lO2,II = c O2,I / c O2,II
Dissolution of brass: i a = i cor exp(E-E cor )/E a
It follows for electrodes I and II: E I –– E cor = E a ln(i a,I /i cor ) and E II –– E cor = E a ln(i a,II /i cor )
--> E I –– E II = E a ln(i a,I /i a,II ) = E a ln(i lO2,I /i lO2,II )
= 0.017 ln(1.3x10-5/1.3x10-3) = - 0.078 V
--> E II is more noble than E I
7.6
Steel pipe:
E cor,I = 0.1 V
Rebar :
E cor,II = 0.6 V
soil resistance: U e = 3000 :cm2 = 1/N
Ohmic resistance at coating defect: R int (1/2 d N ) = 3000 /(2)(1) = 1500 :
Ohmic control: E cor,II - E cor,I = I (R int + R ext ) # I R int
-->
I = (0.6 –– 0.1)/1500 = 3.33x10-4 A
© 2007, Dieter Landolt
33
i a,I = I/A I = 3.33x10-4/0.785 = 4.25x10-4 A/cm2
Table 1.3: 1PA/cm2 = 3.27x10-3(M/nU) mm/year
vcor = (425) (3.27x10-3)(55.8)/(2)(7.86) = 4.92 mm/year
Wall thickness: 6 mm
Time to perforation: t p = 6/4.92 = 1.22 years
7.7
Fe-18w%Cr-8w%Ni-0.4w%C:
C dissolved --> C 23 Cr 6
initial concentrations of Cr and C :
18gCr/100g steel = (18/52.0)/100g steel = 0.346 molCr/100g steel
M Cr = 53.0 g/mol
0.4gC/100g steel = (0.4/12.0)/100g steel = 0.0333 molC/100g steel
M C = 12.0 g/mol
Reaction of 0.0333 mol C with 23 Cr/ 6C :
(0.033/6) 23 = 0.128 mol Cr/100g steel = 0.128(53.0) g Cr/100g steel = 6.64 g Cr/100g steel
Remaining Cr concentration : 18g/100g steel –– 6.64 g/100 g steel = 11.36 g Cr/100g steel
Chromium concentration in w% : 11.4 w%
7.8
Pit growth rate: (dL/dt)= (i a /nF)(M/U) = [D (c sat -c b )/L] (M/r)
For c b = 0:
L = (2 D Fe2+ c sat M/U)1/2 t1/2
M Fe = 55.8 g/mol;
L = pit depth
U Fe = 7.86 g/cm3
c sat = 4.25 mol/l
D Fe2+ = 0.72x10-5 cm2/s
-->
(Table 4. )
L = [(2)(0.72x10-5)(4.25x10-3)(55.8)/(7.86)]1/2 t1/2 = 6.66x10-4 t1/2
Wall thickness 3 mm :
Time to perforation: tp1/2 = (0.3 cm) /(6.66x10-4 cm s1/2 )= 455 s-1/2
tp = 2.07x105 s = 2.07x105 / 3600 = 57.5 h
(b) Binary electrolyte , 0.1 M FeCl (dL/dt)= (i a /nF)(M/U) = [(1 –– z + /z - ) D Fe+ (c sat -c b )/L] (M/U)
with (1 –– z + /z - ) = 1 –– 2/(-1) = 3 ; (c sat -c b ) = 4.25 –– 0.1 = 4.15 mol/l
-->
L = [(2)(3)(0.72x10-5)(4.15x10-3)(55.8)/(7.86)]1/2 t1/2 = 1.13 x10-3 t1/2
tp = (0.3/(1.13x10-4)2 = 6.94x10-4 s = 19.2 h
© 2007, Dieter Landolt
34
7.9
Pitting criterion: E b < E rev,O2
E b = 0.60 V(SCE) = 0.86 V (NHE)
O 2 + 4 H+ + 4 e = 2 H 2 O
Eo = 1.23 V
E rev,O2 = 1.23 + (RT/4F) log(P O2 a H+ 4) = 1.23 - (2.3 RT/F) pH + (2.3 RT/4F) log P O2
T = 40oC = 313 K
pH = 8 ; P O2 = 0.2 bar (air)
E rev,O2 = 1.23 –– (0.062)(8) + (0.0155)(-0.699) = 1.23 - 0.495 - 0.011 = 0.724 V
--> E rev,O2 < E b , no pitting corrosion
CHAPTER 8
8.1
Plot L versus log t:
L (nm
t (min)
Log t
1.5
2.0
2.6
3.1
3.6
1.0
11.0
98
1050
9960
0
1.04
1.99
3.02
4.0
4
y = 1.493 + 0.53084x R= 0.99912
3.5
3
2.5
2
1.5
1
-
0
1
2
3
log t
4
5
Regression line : 1.50 + 0.53 log t
10 years = 5.25x105 min
Oxide thickness after 10 years : L = 1.5 + 0.53 log (5.25x105) = 4.5 nm
© 2007, Dieter Landolt
35
8.2
Kelvin equation: ln (P c /P sat ) = - 2JM/U RT r c Æ r c = - 2JM / U RT ln(P c /P sat )
P c /P sat = 0.85 Æ ln(P c /P sat ) = -0.163
J = 0.072 J/m2 (Table 3.1)
M = 18 g/mol = 18x10-3 kg/mol
r = 1 g/cm3 = 103 kg/m3
r c = - (2)(0.072)(18x10-3) /(103)(8.31)(298)(-0.163) = 6.42x10-9 m = 6.4 nm
8.3
T = 30oC ; relative humidity: RH = 53 %
From Fig. 8.10:
absolute humidity for these conditions AH | 16 g /m3
for T = 20oC and AH = 16 g/m3 Æ RH | 95 %
Saturation in presence of ZnSO4 : (Table 8.11) :
Æ RH at 20oC > RH sat, ZnSO4
RH sat, ZnSO4 = 90%
condensation is possible
8.4
months
mass (mg)
log t
log m
2
12
22
36
20
69
101
148
0.301
1.08
1.38
1.56
1.30
1.84
2.00
2.17
y = 1.097 + 0.67628x R= 0.99835
2.2
2
1.8
1.6
1.4
1.2
0.2 0.4 0.6 0.8
1
1.2
1.4 1.6
log (months)
Regression line: log m = 1.097 + 0.676 log t
20 years = 240 months
log m = 1.097 + 0.676 log (240) = 2.71 --> corroded mass : m = 508 mg
© 2007, Dieter Landolt
36
steel: U = 7.86 g/cm3 ; surface : A = 25 cm2
Corroded depth after 20 years:
L = (508mg)(10-3g/mg) / (7.86 g/cm3)(25cm2) = 2.6x10-3 cm = 26 Pm
8.5
Reaction
number
1
2
3
4
5
Oxidation
state of iron
Fe(II)
Fe(II)
Fe(III)
Fe(III)
Fe(III)
Equation
Fe(OH) 2
Fe(OH) 2
Fe(OH) 3
Fe(OH) 3
Fe(OH) 3
Equilibrium constant
= Fe2+
+ 2 OH+
= FeOH
+ OH3+
= Fe
+ 3OH2+
= FeOH
+ 2OH+
= Fe(OH) 2 + OH-
K 1 = 8x10-16
K 2 = 4x10-10
K 3 = 10-36
K 4 = 6.8x10-25
K 5 = 1.7x10-15
Water dissociation equilibrium:
a OH- a H+ = 10-14
Æ log a OH- = -14 + pH ;
pH = 6: log a OH- = - 8
a) Fe(II):
Reaction (1): K 1 = a Fe2+ a OH- 2
log K 1 = -15.1 = log a Fe2+ + 2 log a OH- = log a Fe2+ - 16
log a Fe2+ = -15.1 + 16 = + 0.9
Æ c Fe2+ | a Fe2+ = 7.9 mol/l
Reaction (2): K 2 = a FeOH+ a OHlog K 2 = - 9.4 = log a FeOH+ + log a OH- = log a FeOH+ - 8
log a Fe2+ = -9.4 + 8 = -1.4
Æ c FeOH+ | a FeOH+ = 3.6x10-2 mol/l
b) Fe(III):
Reaction (3): K 3 = a Fe3+ a OH- 3
log K 1 = log a Fe3+ + 3 log a OH- = log a Fe2+ - 24
log a Fe3+ = -36 + 24 = - 12
Æ c Fe3+ | a Fe3+ = 10-12 mol/l
Reaction (4): K 3 = a FeOH2+ a OH- 2
log K 4 = log a FeOH2+ + 2 log a OH- = log a FeOH2+ - 16
log a FeOH2+ = -24.2 + 16 = - 8.2
Æ c FeOH2+ | a FeOH2+ = 6.3x10-9 mol/l
Reaction (5): K 5 = a Fe(OH)2+ a OHlog K 5 = log a Fe(OH)2+ + log a OH- = log a Fe(OH)2+ - 8
log a Fe(OH)2+ = -14.8 + 8 = - 6.8
Æ c Fe(OH)2+ | a Fe(OH)2+ = 1.6x10-7 mol/l
c)
Dominant Fe(II) species:
© 2007, Dieter Landolt
Fe2+
(saturation concentration at pH 6 | 7.9 mol/l )
37
Dominant Fe(III) species:
Fe(OH) 2 +
(saturation concentration at pH 6 | 1.6x10-7
mol/l)
Æ
Because of the lower saturation concentration of Fe(III) species oxidation of Fe(II) favors
precipitation of Fe(OH) 3 .
8.6
4Fe + 3O 2 + 2 H 2 O Æ 4 FeOOH
Maximum rate = limiting current density for O 2
Oxygen flus to surface: N O2 = D O2 c O2 /G
D O2 = 2.5x10-9 m2/s (Table 4. x )
c O2 = c O2, saturation = 2.5x10-4 mol/l = 2.5x10-1 mol/m3
G = 1 Pm = 10-6 m
N O2 = (2.5x10-9)(2.5x10-9)/(10-6) = 6.25x10-4 mol/m2s
N Fe = (4/3) N O2
Corrosion rate: N Fe = 8.33x10-4 mol/m2s
Table 1.1 : 1 mol/m2s = (3.15x10-4)(M/U) mm/year
Vcor = (3.15x10-4) (8.33x10-4)(55.8)/7.86 = 186.3 mm/year
The actual values are usually lower because of the precipitation of corrosion products on the
surface
CHAPTER 9
9.1
k p = k p ' P O2 1/n
k p ' = k o exp(-'G#/RT)
--> ln k p ' = ln k o - 'G#/RT
Activation energy at P O2 = 1bar: d ln kp' / d(1/T) = -'G#/R
Estimated from Fig. 9.13 for 1 bar :
T
1073
1173
1273
1373
1473
(1/T)x104 k p '
9.32
8.53
7.86
7.28
6.79
6.1E-13
4.5E-12
3.1E-11
1.5E-10
5.8E-10
© 2007, Dieter Landolt
ln k p '
-28.128
-26.135
-24.209
-22.649
-21.269
38
-21
y = -2.8161 - 2.7227x R= 0.99976
-22
-23
-24
-25
-26
-27
-28
-29
6.
7
7.5
8
8.
9
9.5
4
10 /T
Slope: 'G#/R = 2.72x104 K
--> 'G# = (8.3)( 2.72x104) = 22.6x104 J/mol = 226 kJ/mol
(b) ln k p = ln k p ' + (1/n) lnP O2
d ln k p /d(1/T) = d ln k p '/d(1/T) + [d(1/n)/d(1/T)] ln P O2
--> Only if n is independent of T do we find the same activation energy at 1 bar and 0.01 bar
oxygen pressure.
9.2
n-type oxide M 2 O :
M 2 O = ½ O 2 + 2M i x +2 e
K = P O2 1/2 c Mix 2 c e 2
where M i = interstitial cation
K = P O2 1/2 c Mix 4
c Mix = c e -->
c Mix 4 = K P O2 -1/2
--> c Mix v P O2 -1/8
---------------------
p-type oxide M 2 O :
M 2 O + 2 V M ' + 2h = ½ O 2
K = P O2 1/2 c h -2 c VM' -2 = P O2 1/2 c VM' -4
c VM' 4 = P O2 1/2 K-1
(since c h = c VM' )
--> c VM' v P O2 1/8
----------------
9.3
Cu 2 O is a p-type semiconductor that conducts by movement of metal vacancies V M '
Ni2+ is an electron donor in a monovalent copper oxide:
c Dx = c VM'
Wagner Hauffe rules: oxidation rate of Cu 2 O should increase in presence of Ni
© 2007, Dieter Landolt
39
9.4
600oC = 873 K
From Table 2.1:
= NiO
'Go NiO = -245000+(98.5)(873) = -159000 J/mol
Ni + ½ S (g) = NiO
'Go NiS = -187000+(72.0)(873) = -124100 J/mol
½ S (g) + O 2 = SO 2
'Go SO2 = -362000+(73.1)(873) = -290200 J/mol
Ni + ½ O 2
The following equilibria will be considered:
(1) Ni + ½ O 2
= NiO
K 1 = P O2 -1/2
(2) Ni + SO = NiS + O 2
K 2 = P O2 P SO2 -1
(3) NiO + SO = NiS + (3/2) O 2
K 3 = P O2 3/2P SO2 -1
log K 1 = (-1/2) log P O2
log K 2 = log P O2 - log P SO2
log K 3 = (3/2)log P O2 - log P SO2
Reaction (1): ln K 1 = - 'Go 1 /RT
log K 1 = - (-159000)/(2.3)(8.3)(873) = 9.54
= (-1/2) log P O2
--> log P O2 = -19.8
Reaction (2): ln K 2 = - 'Go 2 /RT = - ('Go NiS - 'Go SO2 )/RT
log K 2 = (-159000)/(2.3)(8.3)(873) = -((-124100)-(-290200))/RT = -166100/RT
= (-166100)/ (2.3)(8.3)(873) = - 9.97
= log P O2 - log P SO2
--> log P SO2 = 9.97 + log P O2
Reaction (3): ln K 3 = - 'Go 3 /RT with 'Go 3 = 'Go NiS - 'Go NiO 'Go SO2
ln K 3 = -[(-124100)-( -159000)-( -290200)]/RT = -325100 /RT
log K 3 = - (-325100)/ (2.3)(8.3)(873)= -19.48
= (3/2)log P O2 - log P SO2
--> log P SO2 = 19.48 +(2/3) log P O2
Plot these equations:
© 2007, Dieter Landolt
40
1
3
2
NiS
-8
-12
-16
NiO
Ni
-20
-20 -16
-12
-8
P tot = 1 bar : 2vol% SO 2 -->
(b)
P SO2 = 2x10-2 bar
P O2 = 10-16 bar
from diagram: --> NiS is the stable product
9.5
From Fig. 9.1 (Ellingham diagram) for 900oC:
'Go TiC > 'Go Cr23C6 > 'Go MoC > 'Go Fe3C
Æ all Ti will react with C to TiC, the rest forms Cr 23 C 6 . (If there were still C left it would form
MoC and Fe 3 C).
Element
w-%
M (g/mol)
Mol /g alloy
Fe
75
55.8
1.34x10-2
Cr
23.4
52.0
4.50x10-3
Ti
0.05
47.9
1.04x10-5
Mo
0.015
95.9
1.56x10-5
C
0.05
12.0
4.17x10-5
Ti + C Æ TiC
:
1.04x10-5 mol TiC formed
M TiC = 47.9 + 12 = 59.9 g/mol
Æ 1.04x10-5 mol TiC = (1.04x10-5)(59.9) = 6.20x10-4 g TiC/g alloy
C available for reaction with Cr:
4.17x10-5 -1.04x10-5 mol = 3.13 x10-5 mol C
23 Cr + 6 C Æ Cr 23 C 6
3.13 x10-5 mol C / 6 = 0.522 x10-5 mol Cr 23 C 6
M Cr23C6 = (23)(52)+(6)(12) = 1268 g / mol
Æ 0.522 x10-5 mol Cr 23 C 6 = (0.522 x10-5)( 1268) = 6.62x10 -3 g Cr 23 C 6 / g alloy
Result:
TiC: 0.062 w-%
© 2007, Dieter Landolt
Cr 23 C 6 : 0.66 w-%
41
CHAPTER 10
10.1
(a)
Energy dissipation:
Q f = f F N v s / A (J/m2s) where v s = sliding velocity
-1
Q f = (0.1) (100N)(50s x 2mm)(10-3m/mm) / (3x10-6 m2 ) = 3.33x105 J/m2s
(b) Temperature incease:
T –– To = (T 1 l H /k 1 ) Q f
where T 1 = ½ ; l H = thermal conduction length; k 1 = thermal conductivity
T –– To = (0.01m) /0.46 Jcm-1sK)(102 cm/m)) (3.33x105 J/m2s) = 36.2 oC
10.2
Adhesive wear rate :
v w = K WA (F N l s / 3H)
wear resistance v (1/v W ) v (H/ K WA )
If K WA is independent of hardness H the adhesive wear should vary hardness in the same way as
abrasive wear.
10.3
Dimensionless force: F* = F N /A H
H = 250 kg/mm2 = 2450 MPa
F* = (50N)/(4x10-6m2)(2450x106 Ns/m2) = 5.10x10-3
Dimensionless sliding velocity: v* = v s r o /D T
v s = 2S f r o = 2S (3/60)(8) = 2.57 cm/s
Thermal conductivity D T = h T /c p U
with h T = 0.46 J/cm s K; c p = 490 J/kg K ; U Fe = 7.86 g/cm3
D T = (0.46 J/cm s K ) / (490 J/kg K ) (7.86 g/cm3) = 0.117 cm2/s
v* = (2.57 cm/s )(8 cm)/( 0.117 cm2/s)= 176
Zone III, Severe oxidation wear
10.4
Friction coefficient: f = (F f /A)/E k
where F f = friction force, A wall area, E k = kinetic energy of fluid
Wall area:
A = 2S r L where L = 10 m (length of pipe)
Shear force acting on pipe wall: F f /A = 'P (4Sr2)/ 2S r L = 'P 2r / L
© 2007, Dieter Landolt
42
with 'P = P inlet - P outlet = 0.15 bar
F f /A = (0.15x105 Pa)(2)(0.05)/10 = 150 Pa = 150 N/m2
Flow velocity: v = volume flow rate/cross section = v vol / 4S r2
= 90x10-3/4S(5.10-2)2 = 2.87 m/s
Kinetic energy: E k = Uv2 /2
U = density of fluid
E k = (1/2) (103 kg/m3)(2.87m/s)2 = 4.12x103 J/m3
Friction coefficient: f = (150 N/m2)/(4.12x103 J/m3) = 0.036
10.5
Critical flow velocity : v crit = (2 W crit / Uf)1/2
f = 0.32 Re-1/4
Friction coefficient:
with Re = v L / Q
W crit = 9.6 N/m2; U= 103 kg/m3; Q = 10-6 m2/s ; (a) L = 0.05 m ; (b) L = 0.20 m
v crit
= [2 W crit / (U0.32 Re-1/4)]1/2 = [2 W crit / (0.32 U v crit -1/4 L-1/4 Q )]1/2
[2 W crit 0.32-1 U v crit +1/4 L+1/4 Q )]1/2 = [2 W crit (3.125) UQ1/4]1/2 v crit 1/8 L1/8
= [2 (9.6) (3.125) (10-3)(106/4)]1/2 v crit 1/8 L1/8
= [2 (9.6) (3.125) (10-3)(31.62)]1/2 v crit 1/8 L1/8 = [1.90]1/2 v1/8 L1/8 = 1.377 v crit 1/8 L1/8
v crit 7/8 = 1.377 L1/8
Æ v crit = 1.44 L1/7
(a) L = 0.05 m :
v crit = 1.44 (0.05)1/7 = 0.93 m/s
(b) L = 0.20 m :
v crit = 1.44 (0.20)1/7 = 1.14 m/s
10.6
Tinst -Ts =
with
fȕ
F*1/2 v*
2N1/2
N | 1 + (4 u 10––3 ) F*(1––F*)
f = 0.5
b=1
T inst -T s = 400oC
F*1/2 = (400) (2/(0.5)(1) N1/2 /v* = (1600/v*) N1/2
v* = 1600 N1/2 / F*1/2 = 1600 (1 + (4 u 10––3 ) F*(1––F*)) / F*1/2
© 2007, Dieter Landolt
43
F*
log F*
v*
log v*
-4
5
10
-4.0000
1.6x10
0.001
-3.0000
50400
4.70
0.01
-2.0000
16000
4.20
0.10
-1.0000
5000.0
3.70
1.0
0.0000
1600.0
3.20
5.20
5.5
y = 3.2 - 0.5x R= 1
5
4.5
4
3.5
3
-
-4
-
-
-
0
1
Wear map showing isotherm for f = 0.5 and 'T = 400
log v*
v* = v sl r /D T
Discussion:
F* = F N /A H
For a given set up the heating increases with either sliding velocity or applied normal force.
CHAPTER 11
11.1
Cold worked copper:
E rev,1 = E o,1 + (RT/2F) ln a Cu2+
Annealed copper:
E rev,2 = E o,2 + (RT/2F) ln a Cu2+
Cu annealed --> Cu coldworked
'Go = 'Go 2 ––'Go 1
'Go 2 = -nF E o,2
'Go 1 = -nF E o,1
'Go = - nF (E o,2 –– E o,1 ) = - nF E cell
'Go = 12 J/g = (12) (63.5) = 762 J/mol
E cell = - 'Go/nF = E o,2 –– E o,1 = -(762)/(2)(96485)= - 3.9 x10-3 V = - 3.9 mV
It follws :
E o,2 < E o,1
The cold worked electrode is anodic and corrodes preferentially
note:
The effect of cold work on the equilibrium potential is small, however.
© 2007, Dieter Landolt
44
11.2
K ISCC = f(a/w) V o (Sa)1/2
f(a/w) = 1
K ISCC = 5 MN m-3/2 = 5 x106 N/m3/2
Vo
= 700 MPa = 700 x106 N/m2
S1/2 a1/2 = K ISCC /V o = 5x106/700x106 = 0.714 x10-2
a1/2 = (0.714x10-2)/1.772 = 0.403x10-2 m1/2
critical crack length: a = 0.162x10-4 m = 16.2 Pm
11.3
cross section:
A = Sd2/4 = 0.0052 S/4 = 1.96x10-5 m2
applied stress:
V appl = F/A = (1.5x104 N)/(1.96x10-5m2) = 7.64x108 N/m2
yield strength:
V e = 1100 MPa = 1.1x109 N/m2
V e > V appl
(a) inert environment:
no rupture
(b) dissolved hydrogen:
in Fig. 11.27 we find for a steel with V e = 1100 MPa : V rupture = 30 –– 80 N/mm2
or
In presence of dissolved H:
V rupture << V appl
V rupture = 30 x107 –– 80 x107 N/m2
rupture occurs due to hydrogen
embrittlement
11.4
'G = 'Go + RT ln (X H / P H2 1/2)
½ H 2 = H (m)
at equilibrium:
'G = 0
(setting a H = X H )
--> ln X H = - 'Go/RT + ln P H2 1/2
P H2 = 20 bar --> lnP H2 1/2 = 1.5
ln X H = - 'Go/RT + 1.5
Table 11.20 for Fe D :
'Ho = 26.3 kJ/mol
'So = -50.3 J/mol K for 300- 900 oC;
T = 200oC = 473 K
'Go = 'Ho - T 'So
= 26.3x103 –– (473)(-50.3) = 5.01x104 J/mol H (m) = 50.1 kJ/mol mol H (m)
ln X H = (- 5.01x104 )/(8.31)(473) + 1.5 = -12.74 + 1.5 = -11.24
© 2007, Dieter Landolt
45
--> X H = 1.31x10-5
[mol H/(mol H + mol Fe)] # [mol H/mol Fe]
Volume H 2 dissolved per cm3 Fe:
U Fe = 7.86 g/cm3 ; M Fe = 55.8 g/mol
set: X H
=
--> 1 cm3 Fe = 7.86/55.8 = 0.14 mol Fe
n H /n Fe
where n H , n Fe are the number of mol of hydrogen and fer per cm3 Fe
n H = X H n Fe = 1.31x10-5 / 0.14 = 1.85x10-6 mol H /cm3 Fe
this corresponds to n H /2 = 9.23x10-7 mol H 2 /cm3 Fe
25oC, 1bar : V H2 = n H2 RT / P H2 = (9.23x10-7)(8.31)(298) / (105)
(1 bar = 105 N/m2)
V H2 = 2.28x10-8 m3 H 2 / cm3 Fe = 2.28 x 10-2 cm3 H 2 / cm3 Fe
11.5
reaction at anode :
H (m) --> H+ + e
i a = 80PA/cm2 = F [D H(m) c Hsat(m) /L]
diffusion controlled anodic current density:
where L
= thickness of Fe sheet; L = 20 Pm
c Hsat(m) = hydrogen concentration in Fe on cathode side
D H(m) = diffusion coefficient of H in Fe
hydrogen solubility:
c Hsat(m) = 1.62x10-3 cm3 H 2 /g Fe (25oC, 1 bar)
= 1.62x10-3 / 7.86 = 1.27x10-2 cm3 H 2 /cm3Fe
( U Fe = 7.86 g/cm3)
number of moles H 2 :
n H2 = PV/R T = (105)(1.27x10-8)/8.31)(298) = 5.13x10-7 mol H 2 /cm3 Fe
n H = 2 n H2 --> n H = 10.26x10-7 mol H / cm3 Fe
Diffusion coefficient :
D H(m) = i a L / F c Hsat(m)
D H(m) = (80x10-6A/cm2)(20x10-4cm)/(96485As/mol)( 10.26x10-7mol/cm3) = 1.62x10-6 cm2/s
CHAPTER 12
12.1
overall reaction:
Ni2+ + H 2 PO 2 –– + H 2 O o Ni + H 2 PO 3 –– + 2 H+
anodic partial reaction:
H 2 PO 2 –– + H 2 O o H 2 PO 3 –– + 2 H+ + 2 e
E = 0.085 + 0.065 ln
i a,HP
cathodic partial reaction:
© 2007, Dieter Landolt
Ni2+ + 2 e
o Ni
E = ––0.542 –– 0.039 ln |i c,Ni |
46
at open circuit, E = E ocp :
0.085 + 0.065 ln i dep
i a,HP = - i c,Ni = i dep
(i dep = Ni deposition current density)
= ––0.542 –– 0.039 ln i dep
(0.065 +0.039) ln i dep = –– 0.542 - 0.085 = - 0.627
ln i dep = - 0.627/0.104 = - 6.026
Æ i dep = 2.41x10-3 A/cm2
Ni deposition rate: v dep = (i dep /2 F)(M /U) [cm/s]
with M Ni = 58.7 g/mol ; n = 2; U Ni = 8.9 g/cm3
v dep = (2.41x10-3) (58.7) / (2)(96485)( 8.9) = 8.23 x 10-8 cm/s
Time to form 12 Pm deposit:
t = (12x10-4 cm)/(8.23 x 10-8 cm/s) = 1.46x104 s = 4.05 h
12.2
French hardness = total concentration of Mg2+ and Ca2+ ions exprssed in mg CaCO3 /l :
10 mg CaCO 3 = 1oF
c Ca2+ = 10-4 mol/l
c Mg2+ = 4x10-5 mol/l
c tot = c Ca2+ + c Mg2+ = 14 x 10-5 mol/l
10 mg CaCO 3 = 10-3 g / M CaCO3 = 10-3/100.1 = 10-5 mol CaCO 3
(M CaCO3 = 100.1 g/mol)
Æ 10 mg CaCO 3 /l corresponds to c tot = 10-5 mol Ca2+/l
c tot = 14 x 10-5 mol/l Æ 14 oF
Although degree hardness is often used in practice this concentration unit should be avoided
whenever possible.
12.3
'G° ads = ––RT ln b L
(a)
From Fig. 12.32:
c/T # 0.1 + c [mM/l]
Langmuir:
c/T = 1/b L + c
By comparison: 1/b L = 0.1x10-3 mol/l = 104 mol/l
b L = 104 l/mol
'G° ads = ––RT ln b L = - (8.3)(298) ln 104 = 2.28x104 J/mol
(b)
ln b L = - 'G° ads /RT
'G° ads = 'H° ads –– T 'S° ads
R (dln b L /dT) = - d('G° ads /T)/dT
= -[ (d'G° ads /dT) (1/T) -'G° ads /T2] = -[-'S° ads /T –– ('H° ads -T'S° ads )/T2 ]
© 2007, Dieter Landolt
47
dln b L /dT = 'H° ads / R T2
exothermic: 'H° ads < 0 --> dln b L /dT < 0
by definition: b L = k ads / k des ; as T increases k ads / k des decreases. An increasing temperature
therefore favors desorption.
12.4
log |i|
Fe + O 2 + H 2 O --> Fe2+ + 2OH-
Eprot
E
Ecor
i cor = - i l,O2 = nF v cor
i prot = - i cor = i l,O2
2
2
v cor = 10 mg/cm day = 10x10 mdd = 1000 mdd
Table 1.3 : 1 mdd = 0.112 (n/M) A/m2
n = 2 ; M = 55.8 g/mol
v cor = (1000) (0.112)(2)/ (55.8) = 4.01 /Am2
A = 2 m2 --> I prot = 8.02 A
12.5
E prot = Eo + (RT/nF) ln 10-6
Eo = -0.44 V; T = 15oC = 288K
[mol/l]
E prot = (-0.44) + [(8.31)(288)/(2)(96485)] ln10-6 = -0.44 –– 0.171 = -0.611 V
E Cu/CuSO4 = 0.316 V
E prot (Cu/CuSO4) = -0.611- 0.316 = - 0.927 V
12.6
Protection current:
I prot = i prot A = - i cor A = i l,O2 A
Estimation of corrosion current density from polarization resistance:
r p = (dE/di) E=Ecor or (1/ r p ) = (di/dE) E=Ecor
For charge transfer controlled anodic reaction and mass transport controlled cathodic reaction:
i = i cor exp(]/E a ) –– i l,O2
where polarization ] = (E-E cor )
(1/ r p ) = (di/dE) E=Ecor = (di/d]) ]=0 = i cor (1/E a )
-->
i cor = E a / r p
E a = 20 mV = 0.02 V ; r p = 4x104 :cm2 ; A = 9 m2
i cor = 0.02/4x104 = 5x10-7 A/cm2 = 5x10-3 A/m2
© 2007, Dieter Landolt
48
I prot = - i cor A = - (5x10-3) (9) = - 0.045 A
Charge for 6 years protection:
Q = I t /T
t = (6x365x24x3600) = 1.89x108 s
T= 0.5
(efficiency)
Q = (0.045)(1.89x108)/0.5 = 1.70 x 107 Coulomb
Mass magnesium required:
m Mg = Q M Mg /nF
where M Mg = 24.3 g/mol ; n=2
m Mg = (1.70 x 107) (24.3) /(2)(96485) = 2.14 x 103 g = 2.14 kg
12.7
RE1
RE2
y
r1 = h
r2
The stray current creates a potential gradient in the soil. For cylindrical geometry in infinite
space :
d)/dr = - i/N = –– U e i = –– U e I / 2S r L
where N = conductivity; U e = resisitivity; L = length of pipe, r = radial coordinate
integration yields: )1 ) 0
U e I r1 dr
2SL ro³ r
)2 )0
U e I r2
ln
2SL r0
Ue I r1
ln
2SL r0
Here r o is the pipe radius.
For a finite space we suppose a symetrical image field. The potential difference between
reference electrodes RE1 and RE2 is then obtained from the sum of the two potential fields:
) 2 ' –– ) 1 ' = 2 () 2 –– ) ) = - (U e I/SL) ln r 2 /r 1
set r 1 2 = h2 ; r 2 2 = h2 + y2 :
) 2 ' –– ) 0 ' = - (U e I/SL) ln r 2 /r 1 = (1/2)(U e I/SL) ln r 2 2/r 1 2 = (1/2)(U e I/SL) ln[(h2+y2)/h2]
q.e.d.
numerical values:
© 2007, Dieter Landolt
49
L = 1 m ; h = 2 m ; y = 20 m; U e = 3000 :cm = 30 :m
) 2 ' –– ) 1 ' = - (1/2)(I)/S) ln[(4+400)/4]= - (4.78 I) (4.615) = - 22.1 I
Measured potential difference between RE2 and RE1: ) 2 ' –– ) 1 ' = 1.25 V
Stray current per meter length:
I = 1.25/22.1 = 5.7 x 10-2 A/m
12.8
Fe + 2 H+ --Fe2+ + H 2
anodic partial reaction:
E = 0.08 + 0.05 log i a,Fe
cathodic partial reaction:
E = ––0.36 –– 0.12 log |i c,H |
(a) Corrosion current density, no inhibitor : E = E cor
E cor = 0.08 + 0.05 log i cor = ––0.36 –– 0.12 log i cor
log i cor = (-0.36-0.08)/(0.05+0.12) = -0.44/0.17 = - 2.58 --> i cor = 2.58x10-3 A/cm2
Corrosion potential:
E cor = 0.08 + 0.05 log i cor = 0.08 + 0.05 (-2.58) = - 0.05 V
(b) Hydrogen exchange current density, no inhibitor
i c = i oH exp (-K/E c ) --> K = - E c ln |i c |/i R+ = E c ln i R+ - E c ln |i cH |
E rev,H = 0 –– 0.059 pH = - 0.059(0.2) = - 0.012 V
at the corrosion potential E = E cor ; : |i cH | = i cor
K (Ecor) = E cor –– E rev,H = -0.05 ––(-0.012) = - 0.038 V
K (Ecor) = E c ln i R+ - E c ln i cor
ln i R+ = (K (Ecor) /E c ) + ln i cor
E c = 0.12/2.3 V = 0.052 V
ln i R+ - 0.038/0.052) + ln (2.58x10-3) = -0.731 -5.96 = - 6.69
i oH = 1.24x10-3 A/cm2
(c) with inhibitor: i oH ' = 10-4 i oH = 1.24x10-7
cathodic partial reaction, E = E cor :
K (Ecor) = E c ln i R+ ' - E c ln i cor ’’ = E cor –– E rev,H = E cor –– (- 0.012)
E cor ’’ = -0.012 + E c ln i R+ ' - E c ln i cor
= -0.012 + 0.12log (1.24x10-7) - log i cor ’’ = -6.92 –– 0.12 log i cor
© 2007, Dieter Landolt
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anodic partial reaction:
E cor ’’ = 0.08 + 0.05 log i cor ’’
(same as without inhibitor)
Combine:
E cor ’’ = -0.83 –– 0.12 log i cor ’’ = 0.08 + 0.05 log i cor ’’
(0.05+0.12) log i cor ’’ = - 0.83 - 0.08
0.17 log i cor ’’ = - 0.91
log i cor ’’= -5.35 Æ i cor ’’ = 4.4 x10-6 A/cm2
E cor ’’ = 0.08 + 0.05 log i cor ’’ = 0.08 + 0.05 (-5.35) = -0.19 V
(d) Inhibition efficiency:
T = (i cor –– i cor ’’) / i cor
T= (1.24x10-3 –– 4.4x10-6) /1.24x10-3 = 0.997
Æ T = 99.7%
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© 2007, Dieter Landolt
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