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Exercise 5
Formation of oxide layer on metals at high and low temperatures.
The purpose of the exercise is to give a basic knowledge of the formation of oxides at high
temperatures (500 -1000 °C) and at low temperatures (20 °C).
A: In the first case oxides are formed because of oxidation at high temperatures. These oxides can,
depending on the surrounding environment, either protect the material or cause the material to start
breaking down. This phenomenon is known from power plants where high temperature corrosion is a
big problem. To reduce the problem steel alloys which can form more stable oxides, are being
developed
B1 & B2: In the other case, where the corrosion takes place at low temperature, electrolytic oxidation
(controlled corrosion) of aluminum or titanium takes place. The purpose here is either to form a
surface layer with either a decorative look and/or great wear resistance. In the exercise we become
familiar with two very different ways of forming oxides and to understand the basic mechanisms for
oxide formation including the speed at which the formation takes place.
Start by reading the chapter on the DVD
"Thermodynamic Aspects of Metal-Oxygen Reactions". by P. Kofstad
Afterwards explain the following:
1.
2.
3.
Ellingham/Richardson diagram (figure 1.2). What can you read of out the diagram? On the
diagram the partial pressure for oxygen PO2. What is the connection between the partial pressure
and the temperature for the formation of oxides of a given metal?
Focus on the 3 ways an oxide layer can growth.
Also focus on how you can calculate the speed of oxide formation.
Read the following info about electrolytic oxidation of aluminium. Then explain the following:
1.
2.
In which of the 3 ways describes in "Thermodynamic Aspects of Metal-Oxygen Reactions" is
the oxide layer on the aluminium surface growing in the electrolytic process?
How is the process carried out and which chemical reactions take place at the cathode and anode?
Anodizing of aluminium.
Several metals can be anodized. Practically aluminium, magnesium, zinc, lead and titanium can be
anodized. The mechanism for the formation of the oxide layer on the different metals is not the same.
The electrolytic anodizing of aluminium was developed in the beginning of the 1930s. The process
forms a hard, wear resistant, protective and sometimes decorative surface on aluminium. This surface
treatment of aluminium has many applications. Aluminium is a soft metal and the surface is easily
scratched. Anodizing forms a ceramic like surface consisting of very hard aluminum oxide with small
pores, which can be dyed in a number of colours. Anodizes aluminium has many applications.
Sometimes very thick oxide layers are formed (hard anodizing) which are particularly suited for the
aviation industry.
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Figure 1. Anodized B&O-equipment.
The anodizing can be carried out in many chemical solutions depending on the later application. Most
commonly a solution consisting of 20% sulphuric acid by weight.
Anodic reaction:
2Al + 3H2O  Al2O3 + 6H+ + 6e
Cathodic reaction:
6H+ + 6e  3H2
The above reaction equations describe the anodic reaction at the anodic oxidation of aluminium in an
acidic solution.
The anodizing process is carried out by dipping the aluminium components in the sulphuric acid and
carry out the electrolysis with the specimens coupled as anode (+) oxidation of the aluminum.
As cathode (-) lead and aluminium plates are often used and there is a strong formation of hydrogen.
During the anodizing process the solution has to be agitated mechanically.
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To study the anodizing process further for a constant voltage you can follow the development of the
current during the first second and minutes of an anodizing process in sulphuric acid. Typically you
will get a curve looking like figure 3.
1. Cathode bar
3. Air supply
4. Cooling water inlet
7. Anode bar
8. Cathodes
9. Rack
10. Springs
Figure 2. Principles of an anodizing bath.
Figur 3. Curve showing the current during a
constant voltage and the development of pores
during the anodizing process.
During period (a) in figure 3 the current will be high until the oxide layer starts forming on the whole
surface. Hereafter the current will decrease linearly with the increasing thickness of the oxide layer.
The layer is an insulator covering the whole surface. This continues in period (b). In the end of (b) is an
increase which is caused by the formation of a more uneven layer where the thickness is smaller
(causing higher current density)
The uneven layer is caused by the electric field being concentrated around the areas with lower
thickness.
Hereby both the formation and the break down of the oxide in the areas increase. The reason of the
increased break down of the oxide is not yet clear. The electric field plays probably a role. The break
down follows the following equation.
+
Al2O3 + 6 H  2 Al
3+
+ 3 H2O
The formation of the well knows hexagonal pores is initiated is this way (figure 4) which characterizes
the amorphous oxide layer formed in many acids. The anodizing in (c) shows an increase in the
anodizing current. This is when the hexagonal pores are formed. The (c) period is revealed by the
stable (d) period where the actual oxide layer is formed.
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The oxide layer consists of a porous wall and a barrier layer in the bottom (figure 4). The electro
chemical process takes from the bottom and up where the aluminium is change into aluminium oxide.
The pore gets longer and longer during the process. . The uppermost layer is thus formed first and the
bottom layer it the last to be formed.
Figure 4. Figure shows the formation of small cells. The hole in the middle has a diameter of 150 Ångstrøm.
The pore diameter is a function of the electrolyte and voltage. The number of pores pr area (N), pore
diameter (dp), the thickness of the barrier layer. (B) and the wall thickness of the pores (V) is
determined by the voltage.
High voltage decreases the number of pores pr area but increases the size of the pores. The barrier layer
thickness (B) in the bottom and the wall thickness of the pores increase when the voltage in the bath is
increased.
If the bath voltage is very high it will give rise to a local temperature increase which will cause local
break down of oxide layer. This situation is critical, because the current will increase in the damaged
area and thus accelerates a rapid etching of the oxide layer. This is called burning and can be observed
by unstable and decreasing voltage at constant current.
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Enheds-celle
d
Porer
Tykkelse af oxidlag
BarriereLag
Pores/cm2 (N) ........................
Pore diameter (dp) ..........................................
Barrier layer thickness (B)...........................
Pore wall thickness (V)..........................
Pore length (L)...
..............................
Anodizing current density...............
Anodizing voltage...........................
Process temperature (decorative anodizing).....
Bath conc. (decorative anodizing) .............
Process temperature (hard anodizing) .........
Bath conc. (hard anodizing)........
1-3  1011
15 nanometer
5-15 nanometers
15-20 nanometers
up to 30 m
2 amp/dm2
18-22 Volt
18-22 C
200 g/l H2SO4
0-5 C
100 g/l H2SO4
(10Å = 1 nanometer, 10.000 Å = 1m )
Table 1. Typical factors for anodizing in sulphuric acid (also called GS: Gleichstrom Schwefelsäure).
The electrolyte contains oxalic acid and allows larger temperature variations without influencing the
quality of the oxide layer.
The pores are very small. The pore length relative to the pore diameter is very large.
The anodic efficiency is about 65% (the part of the faraday current that participates in the oxide
formation) and the layer thickness can be determined by the following formula:
H = 0.4 · W · T · J/F
where H
F
J
T
W
=
=
=
=
=
layer thickness in 
2
Surface in dm
Total current in amps
Time in minutes
Anodic efficiency
2
When calculating the layer thickness a rule of thumb says that 1 amp hour/dm gives rise to a layer
thickness of 15-16 µm, depending on the alloy – So it can be concluded that the anodic efficiency is not
100%.
Anodizing of aluminium can have many purposes. The most important are:
1.
Decorative: Special structures and colours
2.
Protection against corrosion and wear (also on mirror like surfaces)
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3.
Basis for organic layers, basis for galvanic plated metal layers. Basis for solid lubrication
agents.
4.
Technical purposes: Hard wear resistant surfaces.
5.
Electrically insolating.
Commercial oxide layers thickness in the literature are normally stated to be about 5-25 µ, while the
anodizing in chrome acid solutions gives values only between 2.5 and 5 µm.
In the table below are the recommended layer thicknesses for the GS-process for different purposes.
Oxide layer
thickness
25 m
20 m
15 m
10 m
5 m
Purpose
The surface is exposed to excessive corrosion or wear – especially in an
outdoor corrosive environment.
Strong or normal exposure outdoors for example building materials,
vehicles and ships. Strong exposure indoor from chemicals, in moist air
for example for machinery in the food industry.
Medium strong exposure indoor for example handles and clean
decorative surfaces.
Normal exposure indoor and outdoor in clean dry atmosphere and for
reflectors, decorative lists for cars and sports equipment.
Normal exposure indoor.
Table 2. Thickness of aluminium oxide layer for different purposes.
Anodizing Titanium
Anodizing of titanium is exceptionally easy and is a totally different process from anodizing
aluminum. Simply, you prepare a mild ionic solution, then you apply a controlled DC voltage to the
titanium specimen (the titanium is coupled to the anode). A sheet of stainless steel is used as an
electrode in the bath, to which can be connected to the cathode of the power supply.
Batteries can be stacked in series with an adjustable voltage supply to achieve the necessary voltage for
the desired color. Measure the actual voltage using a DC volt meter.
Figure 5
In the image to the left (figure
5), each titanium wire specimen
was clamped in a stainless-steel
alligator clip during the
anodizing process, which is why
the top of each wire is not
anodized (without color). The
entire process of anodizing takes
only a few seconds, and is easily
visible.
The surface has a color due to the optical properties and thickness of the titanium oxide coating (see
appendix 1). The basic physics is similar to the color spectrum in an oil slick on water. Read appendix
1 and understand the color formation. In a lot of cases different surfaces for instance stainless steel
some times starts to be discolored in connection with a corrosion attack. Often that is caused because
thin and transparent oxide layers are formed.
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Instructions for high temperature oxidation.
A: The experiment takes place in the metallurgical laboratory.
Study how a piece of iron is exposed at a certain temperature develops an oxide layer over time. The
formation of a fairly thick layer can take several days or weeks depending on the temperature. The
exercise has thus been started several days ago.
A number of samples have been exposed and taken out of the oven after certain duration. The weight
before exposure is stated in the journal that you receive along with the weight gain after the exposure.
The results of the last sample you will receive on the exercise day.
In the journal you will find info about the heat treatment temperature. The test results you will receive
from Steffen Sonne Munch (The metallurgic laboratory). He will tell about the heat treatment. You are
not going to carry out any experimental work but you are going to look at the data and answer
questions. You must agree in the groups when to carry out the experiments.
Questions
1.
Image the weight gain as function of time in Excel and state whether the oxide layer grows
parabolic of linearly.
2.
Using "Thermodynamic Aspects of Metal-Oxygen Reactions” equation 1.10. Calculate the kp
(speed constant) for the given situation and state its dimensions. What physical-chemical material
properties influence the parabolic growth constant?
3.
Look at the state diagram for (Fe-O) and state what oxides are formed during the oxidation process
at the given temperature for which the experiments were carried out.
4.
Estimate the layer thickness/material loss for the formed oxide if the test sample were heat treated
for a year at the given temperature.
5.
Explain the cause that an iron thread is turned into a tube of iron oxides during long-time
oxidation. Especially look at the formation of wüstit and magnetite.
6.
In "Thermodynamic Aspects of Metal-Oxygen Reactions" on figure 1.2 is an
Ellingham/Richardson or just Ellingham diagram. What pressure is necessary to remove oxides
from aluminium at a temperature close to the melting temperature of aluminium?
7.
Look at the Ellingham/Richardson diagram 2. Explain the formation of oxides for the following
metals: Zn, Hg, Au and Ag.
8.
What alloy elements are used if one wishes to increase the oxidation resistance of steel? Look at:
http://www.haynesintl.com/ - Explain with background in the Ellingham/Richardson diagram why
these alloying elements have an significant effect. Try especially to look at HAYNES® HR-224®
alloy.
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Ellingham/Richardson diagram 2
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Procedure for electro-chemical oxidation:
B1: Experiments take place in the Galvan laboratory (remember coat and safety
glasses)
Carry out an anodizing in sulphuric acid in 200 g/l of 5 aluminium plates.
The anodizing time is increased with 8 minutes for every experiment so that the last plate is exposed
for 40 minutes in the bath.
The current density is set at 2 A/dm2 and is supervised during the experiment.
The corresponding voltage is about 15-20 volts.
The anodizing is carried out on at 6063 alloy
Before the anodizing the plate is pickled in a solution of 50 g/l NaOH to clean the surface. Afterwards
the plate is treated with a diluted saltpeter solution HNO3.
Rise thoroughly with water after every treatment.
Measure the layer thickness using an eddy current instrument.
Measure the layer thickness in the 4 corners and in the middle of the plate. Use the average for the
further calculations.
B2: Anodizing of titanium
The experiment is carried out under supervision of the teachers since the voltage in the experiment is
120 volts. The anodizing takes place in a sulfamic acid electrolyte 50 g/l.
Anodizing of 5 plates is carried out.
A3 30 volt – 50 volt – 70 volt – 90 volt and 120 volt.
The formed layer is relatively thin and it is not possible to determine the weight using a scale.
Therefore you must determine the thickness from the colour of the plates. This colour is due to
interference. See Appendix 1
Questions
1.
State the direction analysis for the 6063 aluminium alloy and give a short explanation tor its
characteristics.
2.
State the chemical reaction on the aluminium surface in connection with the NaOH treatment and
in the saltpeter acid solution HNO3 .
3.
Image the layer thickness in Excel as function of time and state whether the layer grows parabolic
or linearly.
4.
Contact the other team and explain why the two metals behave differently. Would you expect
aluminium at high temperate oxidation to grow following the same mechanisms as for electrolytic
oxidation?
5.
In connection with electrolytic oxidation of titanium (anodizing).
See the appendix about determination of thickness by interference colored surfaces
Calculate the thickness of the formed oxide layers using the assumption that we have first order
interferes.
Calculate furthermore the thickness of the thinnest oxide layer that can give rise to an interference
colour when we focus upon titanium oxide.
6.
State whether the growth of the oxide layer can be assumed to e parabolic or linear. Explain how
the mechanism for oxide formation can be assumed to take place and if t has any similarities with
the mechanism known from the electrolytic oxide formation of aluminium.
Please hand in one report for each group.
Please include the following: Course number, exercise number, group number and your names.
Opload your report at Campus in the group called TRAINING and under the right Exercise
number. Then it will be corrected and uploaded at Campus again
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Appendix 1.
Theory for determination of thickness by interference colored surfaces
of professor Per Møller
One of the best known examples of interference appears when light interacts with a thin transparent
film, as known from anodized titanium, which consists of TiO 2 forming a transparent film, located on
top of the titanium metal.
t
Figure 1 show how a plane wave hits a thin film and is partially reflecting at the top and at the interface between
the thin film and the material titanium. The thickness of the thin film is t
The waves reflected from the surface respectively at the titan oxide and the interface between metal and
oxide interfere with each other. Interference can cause either a constructive interference or a destructive
interference depending on the phase difference between the two reflected waves.
Interference is the phenomenon which occurs when, for instance, light waves are mixed and the
individual contributions are added to a new signal. This “mixing” is called interference and the result
may be "nothing" (the absence of resultant wave or signal called destructive interference), or
fluctuations are amplified and constructive interference is the result.
Constructive interference
Destructive interference
If the wavelength of the incoming wave named  a constructive incoming wave is considered between
the A and B wave, if the thickness of the oxide layer could be expressed wih the following relationship
2t = m. where m is the (0, 1, 2,...)
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Wave length interval
Color (measured in
vacuum)
~ 625-740 nm
red
Medie
Refraktions
indeks
Vacuum
1,000000
orange ~ 590-625 nm
yellow ~ 565-590 nm
Diamond
2,4
TiO2
2,6
green ~ 520-565 nm
cyan ~ 500-520 nm
Glass
1,5
~ 450-500 nm
Rubin
1,77
blue
indigo ~ 430-450 nm
violet ~ 380-430 nm
Figure 2
In the above figure 2 the wavelength of the different colours of light and refractive index of selected
transparent materials is given together with the value in vacuum.
Light moves at a speed depending on the medium, the waves move in. When a plane wave move from
an area with a given propagation rate to another area with a different propagation rate, the direction of
the wave propagation is changed, that phenomenon is called refraction.
Light refraction in glass, water and other transparent substances is an example of refraction. The light
moves, in glass with only 70% of the speed it has in the surrounding air and if the medium is titanium
dioxide, the speed is reduced to 40%. The light also changes direction when it hits the interface
between glass and air.
This lead to, that a simple relationship between the thickness of the oxide layer and the wavelength of
the incoming light cannot be established.
Two factors complicate that issue because one of the wavelengths of light is modified in the titanium
dioxide.
The second is that there is a phase shift of 0,5 when light is reflected on a surface, where light from
an area with a lower refractive index pass to an area of higher refractive index. This occurs when light
is reflected on the oxide surface See Figure 3
Transparant
film of TiO2
Figure 3
Titan
substrate
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The light changes the wavelength, when it passes TiO2 - The relationship between the wavelengt 0 in
air and the wavelengt n in material with refractive index n can be expressed with following
relationship n = 0/n.
In the figure above a light beam is directed from a medium with index n0 against a surface with an
index nf of 2.6. The incoming beam on the oxide surface is divided into a reflective and refractive part.
The refracted part of the beam is reflected at point B and leave the oxide film again in C with the same
direction as the reflected beam from A. Parts of the beam may be reflected again in C and be subject to
multiple reflections (m order reflection), resulting in loss of intensity.
The two parallel beam, leaving the surface in A and C can be combined together through a converging
lens as shown in the figure.
The two beams cross at the location P and interfere. As the two beams have moved along different
paths from point A onwards, one in air, the second part in a oxide film a phase difference will appear
and lead to constructive or destructive interference in P.
The optical difference can be expressed through the geometric length ABC which the refracted beam
has moved multiplied by the refractive index n of the film  = n(AB + BC).
If we assume that the beam is directed vertically against the surface the optical length difference can
be expressed as the thickness (t) of the oxide film.  = n(AB + BC) = n(2t).
If we consider a simple system without regarding the value of n of the oxide layer, and let this be
identical to air, then a new relationship can be established between the thickness and the wavelength
2nt = 0, where constructive interference of the two beams are fulfilled .
The issue is rather more complex, since as mentioned earlier, both a change in wavelength of the beam
in addition to the refracted and the reflected beam has gained a phase-shift on 0.5
Since we assume that at nf > n0 and nf > ns, we can conclude that a phase shift on 0,5 will be
obtained,when the beam is reflected in A. This implies, that we can add 0,50m to the simplified
equation 2nt = 0.
The equation = n(AB + BC) = n(2t) can be parted up in contribution p coming from the optical
length difference caused by the thickness of the film and r as the equivalent part derived from
reflection.
p + r = m, (m = 1, 2, 3, ...) where m correspond with the order of interference. When the present
equation is true constructive interference is possible.
For a thin film with thickness t and a refractive index nf placed in air the following equation can be
written  p = 2nf t according to the earlier equation  = n(AB + BC) = n(2t). As mentioned earlier
contributed expressed as follows. As mentioned earlier the contribution r comes from reflection and
can expressed as follows r = 0/2.
By combination of both contribution p + r the equation is obtained.
The equation is true for constructive interference.
With respect to destructive interference the following equation can be obtained.
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Anodized titanium has become very well known during the last years - used on Air-glasses
http://www.lindberg.com and for many other application. It is also possible without major problems to
produce similar effects on Niobium, tantalum and zirconium.
The present interference colouring of titanium is proposed to be a first order reflection. Estimate the
thickness of the oxide layer by using the established theory.
It is now possible also to colour stainless steel using the same principle by growing the oxide layer
thickness by a chemical oxidation.
Please hand in one report for each group.
Please include the following: Course number, exercise number, group number and your names The front page for the report are available at the DVD.
Upload you report to CampusNet , or send it directly to Per Møller pm@epc-info.dk.
You will receive the corrected exercises.
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