Tutorial 1  Corrosion in Action: Outdoor Activities

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• color photographs of the appearance of the corroded structure/component
• a brief description of the structure/component and its intended function
• roughly identify the materials of construction (e.g. stainless steel, aluminum, galvanized steel, copper, etc.)
• a detailed description of the observed corrosion phenomenon including its location, appearance, shape, size, distribution
• Try to explain why it happened, how it happened and what can be done to prevent it

Your write-up should be no more than two-A4 pages including color photographs. Submit your write-up by 20 August 2004. The best entry will receive one year membership with NACE (National Association of Corrosion Engineers, USA) sponsored by Dr. Qiu Jianhai.

Tutorial 2 

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2.1 Potassium chloride is an ionic substance, the aqueous solution of which is commonly used in corrosion experiments.

(a) What is its molecular weight? 

(b) How do you make a solution which is 0.1 M?

(c) What is meant by a 'saturated' solution? 

Answers:

(a) the molecular weight of KCl: 39.10 + 35.45 = 74.55

(b) A solution which is 1 M contains 74.55 g per litre of solution. A solution which is 0.1 M contains 0.1 x 74.55 = 7.46 g per litre.

(c) A saturated solution means that there is an excess of KCl solid crystals in the solution. Saturation represents an equilibrium state.

2.2 What is the pH of a solution containing 10^-3 M hydrogen ions?

Answer:

pH=-lg[H+] = -(-3) =3

2.3 Nitric acid is a liquid with the formula HNO3. When added to water it dissociates into a hydrogen ion and a nitrate ion, NO^3-.

(a) What is the molecular weight of nitric acid?

(b) If 6.3 g is contained in a litre of water, what is the concentration of hydrogen ions?

(c) What is the pH of this solution?
 

Answers:

(a) Molecular weight of HNO₃: 1 + 14.00 + 3x16.00 = 63.00

(b) 1 M of solution contains 63 g; 0.1 M of solution contains 6.3 g

HNO₃ = H+ + NO₃^-

1M        1M    1M

0.1M     0.1M  0.1M

So [H+] = 0.1 M

(c) pH = -lg[H+] = - (-1) =1

2.4 When iron rusts it forms a chemical compound with atoms of oxygen.
If iron can have a valency of either two or three, while oxygen only has a valency of two,

(a) Give two possible ratios of iron/oxygen atoms in rust.

(b) Suggest two possible formulae for rust.
 

Answers

(a) When Fe has 2 valence, Fe/O=1; When Fe has 3 valence, Fe/O=2/3

(b) FeO and Fe₂O₃

2.5 You are given three liquids:

1. A solution containing 0.1 M sodium chloride;
2. A solution containing 0.5 M hydrochloric acid;
3. Tap water.

Which solution will conduct electricity best?

Answer

0.5 M HCl > 0.1 M NaCl > Tap water

2.6

Express the standard electrode potential, E^o, of a metal in terms of the standard Gibbs free energy change, DG^o. Hence calculate the value of DG^o at standard temperature and pressure for the corrosion of iron, assuming a divalent reaction.

Answer

DG^o = -zE^oF
= -2 (+0.44) x 96,494
= -84.9 kJ mol^-1
 

Note
1. Negative DG indicates that corrosion occurs spontaneously.
2. The reduction potential (Table 4.1) is -0.44 V, so for the oxidation process, we use +0.44 V. See Section 4.5 for details of the sign convention.

2.7 A metal, M, of valency, z, atomic mass, W, and density, D (kg/m³), is corroding uniformly over its exposed surface area with a current density of i_corr A/m². Derive an expression for the number of millimetres of metal which will be lost during one year (mm/y), assuming that the build-up of corrosion product does not stop the corrosion reaction.

Answer:
Over 1 m² of exposed metal, the number of coulombs passed in one year will be:
icorr x 60 x 60 x 24 x 365
= 3.154 x 107 x i_corr
1 mol of metal of valency z converted into ions gives:
z x 96,494 coulombs
(Remember that a mole is expressed in grams.)
Thus the number of mol per square metre lost in a year
= (3.154 x 107 x icorr)/(z x 96,494)
= (326.8 x icorr)/z
Converting mol into kg, the number of kg lost per square metre per year
= (326.8 W x icorr)/(1000 x z)
The metal is of density D kg m^-3, thus if the mass lost were D kg m^-2 then the depth of penetration of the corrosion would be 1 metre (1000 mm). The actual penetration (in mm) is therefore:
= (326.8 W x 1000 x icorr)/(1000 x z x D)
= (326.8 W x i_corr)/(z x D)

Note: The expression derived above is commonly used as a measure of corrosion rate. It is known as the wastage rate, is measured in mm/y(mm per year), and is popular because it gives engineers a better 'feel' for the rate of corrosion. In the case of copper, a corrosion current density of 0.01 A/m² is commonly observed. Thus, with z = 2, W = 63.5, and D = 8960 kg m^-3, we calculate a wastage rate:
= (326.8 x 63.5 x 0.01)/(2 x 8960)
= 1.16 x 0.01 mm/y.
Thus we see that the wastage rate for copper is almost the same as the corrosion current density of copper in A/m². The units of icorr are important. It is both surprising and fortuitous that for many metals the quantity, W/(zD), is approximately constant. Hence the observation for copper is applicable to many other metals too:
icorr (A/m²) = mm/y

Tutorial 3

3.1 Briefly define each one of the following:
 

(a) Corrosion;
(b) Standard electrochemical potential;
(c) Exchange current;
(d) Polarisation;
(e) Double layer.

Answers:

(a) Corrosion is the degradation of a metal by electrochemical reactions with its environment.

(b) Standard electrode potential of metal M, is the potential difference measured between the metal, M, immersed in its own ions solution with unity concentration, and the standard hydrogen electrode at 298K and 1 atm.

(c) Exchange current is the anodic or cathodic current under equilibrium conditions where there is no net current flow.

(d) Polarisation is the potential shift away from the equilibrium/stabilized potential. Or polarisation is the potential disturbance to the equilibrium conditions.

(e) Double layer refers to the non-homogeneous distribution of ions when a metal is immersed in an electrolyte. The double layer consists of two parts: A compact Helmholtz layer in which potential changes linearly with the distance from the electrode surface and a more diffuse outer Gouy-Chapman layer in which the potential changes exponentially with distance.

3.2 Iron is connected to copper and then immersed in a solution containing both Fe2+ and Cu2+ ions.

(a) Which metal corrodes?
(b) Write equations to describe the reactions which occur at each electrode,
assuming each metal has a valency of 2.
(c) Calculate the maximum possible potential of the resulting corrosion cell.

Answers:

The metal with the most negative reduction potential will be the anode. From Table 4.1, E^o for iron is -0.44 V, while for copper it is +0.34 V. The iron will thus be the anode and will corrode. Convention requires that we write the cell as

Fe| Fe2+ || Cu2+ | Cu

and the cell potential as the reduction potential of the electrode on the right minus the reduction potential of the electrode on the left. Thus:
E(cell) = E(Cu redn) - E(Fe redn)
= (+ 0.34) - (-0.44)
= +0.78 V
The electrode on the left is the anode, and thus the electrode reactions are:
Fe  = Fe2+ + 2e-
Cu2+ + 2e-= Cu

Note:
A positive cell potential leads to a negative DG and tells us that the cell, as described, will corrode spontaneously.

3.3 Calculate the rest potential, versus the saturated calomel electrode, of a piece of copper in equilibrium with a solution containing 10^-6 M copper (II) ions.

Answer

Using the Nernst equation:
E = E^o + (0.059/2) lg(10^-6)
= +0.34 + (-0.177)
= +0.163 V SHE

To convert from SHE to SCE we subtract 0.242 V, thus:
E = 0.163 - 0.242
= -0.079 V SCE
 

3.4 For the electrochemical cell having nickel and cadmium electrodes in equilibrium with solutions of their ions:
(a) Write two equations to describe the reactions which occur.
(b) Determine the greatest potential which may be obtained from such a cell under standard conditions.
(c) State one other condition necessary to achieve this potential.

Answer

(a) Cd = Cd²⁺ + 2e E^o(Cd²⁺/Cd)=+0.40 V (oxidation potential)

Ni²⁺ + 2e = Ni E^o(Ni²⁺/Ni)= - 0.25 V

Total: Ni²⁺ + Cd = Ni + Cd²⁺

(b) E_cell = E^o(Cd²⁺/Cd) + Eo(Ni²⁺/Ni) = 0.15 V

(c) This potential is achieved under zero current flow.

3.5 (a) Explain clearly the meaning of the following terms when used in the context of E/pH diagrams: immunity; corrosion; passivation; pH.
(b) Use the following Figure to obtain a value for the free corrosion potential of zinc in water at pH 8.

(c) Calculate the same value by means of the Nernst equation and thus show how the equation is used in constructing part of the E/pH diagram.
(d) Describe, in principle, how it is possible to complete the remainder of the E/pH diagram.
(e) Explain how inspection of the E/pH diagram suggests two possible methods for controlling the corrosion of zinc in water.
(f) What factors, if any, might make these suggestions impractical?
(g) What other factors limit the application of E/pH diagrams to real situations?
(h) State the effect of an increase in temperature upon the potential you found in parts (b) and (c). Give a reason for your answer.

Answer

(a) (1) Immunity: The metal can be considered to be in a thermodynamically favorable condition for corrosion not to occur. Or When the concentration of metal ions in the environment is less than 10^-6M.

(2) Corrosion: thermodynamic conditions are favorable for corrosion to occur. Or the concentration of metal ions in the aqueous solution is greater than 10^-6M.

(3) Passivation: As region of the E - pH diagram in which an inert surface materials is thermodynamically favoured - usually an oxide or hydrated oxide.

(4) pH: Definition of acidity/alkalinity

pH = -log[H⁺]

pH=7 neutral

pH>7 alkaline

pH<7 acid

(b) Free corrosion potential of Zinc in water at pH=8 is -0.94 V (SHE)

(c) From Nernst Equation for Zn²⁺+2e = Zn

E=E_o+0.0295log[Zn²⁺]=-0.763 + 0.0295log[10^-6]=-0.94V

This oxidation of zinc is independent of pH and hence is a horizontal line on the E-pH chart. It is defined as occurring at the point where [Zn²⁺] = 10^-6M.

(d) The chart is constructed by considering all the possible species of metal ions which may exist in water, and their equilibria.

For zinc, only four species are possible: Zn, Zn²⁺, Zn(OH)₂ and ZnO₂^2-.

The equilibria which relate to each pair are written down and the Nernst equation solved in each case where current flows(ie an exchange of electrons). In those where no current flows, chemical data are used.

(e) Two methods for controlling corrosion of zinc:

(i) The application of a potential to take the zinc into the immunity region (Zn).

(ii) The alkalinization of the electrolyte so that it is taken into the passive region,

Zn(OH)₂, but not too far into ZnO₂^2- region.

(f) The Zn(OH)₂ which should be protective to passivate may be weak and easily broken by turbulence, or it may allow electrolyte through it by being porous.

(g) The diagrams relate to thermodynamic not kinetic reactions. They are calculated for simple electrolyte systems which often do not take account of other ions. They do not consider alloys although practical E-pH diagrams can be produced.

(h) An increase in temperature will result in E becoming more negative (anodic). This is because:

E=E^o+(RT/nF)ln[M+]

E^o is negative, R, n, F all positive. Thus as T increases, E becomes more negative.

Tutorial 4

4.1 A univalent metal, M, has ba =+0.2 V, bc = -0.2 V, and i_o = 20 mA/m². Calculate values for ia and ic when the metal is anodically polarised to +0.20 V. Hence determine the experimentally measured current density for the same polarisation.

Fig.4.1 plots the anodic and cathodic current densities for M. At +0.20 V:
ia = 200 mA m-2
ic = 2 mA m-2
and imeas = ia - ic
= 198 mA m-2

Alternatively, we can use the Tafel equation, thus:

h= lg(i/io)
Taking antilogs we get for the anodic reaction:

10 ^h/β = ia/io
But ^h/β= 1

Thus ia = 10io
= 200 mA m-2.

Similarly for the cathodic current,
^h/β = -1

Thus ic = 0.1io
= 2 mA m-2
Therefore imeas = 198 mA m-2, as before.

Note: The slope of the Tafel plot - the beta constant (+0.2 or -0.2 in the example) - is obtained by measuring the voltage change per logarithmic decade of current, e.g. the voltage change between 10-1 and 10-2 A m-2.

4.2 Draw and label a diagram to describe each of the following:
(a) The use of a polarisation curve to calculate a cathodic Tafel constant.
(b) An anodic polarisation curve exhibiting passivation.

Answer

The slope of the extrapolated straight line is the cathodic Tafel constant

4.3 (a) Give two useful functions of E/pH diagrams.
      (b) Give two limitations to their use.

Answer

(a)

• Identify Immune/Passivity/Corrosion
• Guideline for corrosion protection

• Thermodynamic calculation, no kinetic info
• Neglect of other ions (Cl-, S= etc)

Answer

(a) Fe/Ti has the greatest potential difference in the Galvanic series, so iron will corrode the fastest.

(b) Fe is cathodic to Zn and Cd. But Fe/Zn has the greatest potential difference. Thus in fresh water, Zn will provide best sacrificial protection to iron.

Tutorial 5

5.1 (a) Explain what is meant by the term 'passivity' in the context of corrosion.
      (b) Give two examples of natural passivity of metals.

Answer

(a) Passivity is the reduction in corrosion caused by the insoluble corrosion products, most often, oxide surface films.

(b) Natural passivity refer to the natural formation of a passive film on the metal surface. Examples are aluminium and its alloys, titanium and its alloys, stainless steels and high CrNi alloys

5.2 Imagine that you are asked to investigate the rate of corrosion of an Inconel 625 reactor in the particular environment of a chemical process.
(a) Describe the basic equipment you would need to carry out a laboratory investigation.
(b) Explain the role of each piece of equipment mentioned in (a).
(c) Describe one experiment which you would perform, clearly stating the conditions for the experiment.
(d) Explain how you would interpret data obtained from your experiment.
(e) List the other parameters which might usefully be varied and explain how further information relating to the corrosion performance in the reactor might be obtained as a result.
(f) Discuss the limitations of the experiment you have described in the context of other forms of corrosion to which the material may be susceptible.

Answer

(a) Refer to the drawings below:

(b) Specimen is the test electrode

Reference electrode to provide a stable potential against which other measurements of potential are made: must remain inert.

Auxiliary electrode is to act as current carrier in the circuit

Glass cell is to provide convenient holder for electrode and solution.

Potentiostat is to control the potential and measure the current.

(c) Tafel polarisation to determine the corrosion rate, or Cyclic polarisation to determine the pitting tendency.

Conditions for Tafel polarisation:

• - allow specimen to reach equilibrium in giiven electrolyte (~ 30 minute)
• - apply potential 300 mV (SCE) around the ccorrosion potential (open circuit potential) at 20 mV/min and measure the current.
• - Plot E vs log(current density)

E_corr provide data on corrosion tendency.

i_corr the corrosion current density indicates actual net corrosion rate under normal conditions. Assuming general corrosion of surface, icorr can be easily converted into a penetration rate expressed in mm per year. Slope of anodic Tafel curve indicates susceptibility to galvanic effects. Small anodic slope (Tafel constant) indicate corrosion will increase dramatically if even slight polarisation occurs. Any sign of passivity (reduction in corrosion rates at higher potential) can be helpful because of surface film formation.

Refer to the following figures for polarisation curves

For cyclic pitting scans, refer to the following figure

(e)

1. Vary pollutant levels in electrolyte. -Does pollutant really have a significant effect ?
2. Vary flow over specimen surface - compare stagnant with flowing solutions
3. Vary oxygen concentration- what is the effect of still waters where O₂ levels may be low eg in crevices or restricted areas.
4. Vary temperature - is there a critical pitting temperature?

5.3 Pure nickel is oxidized at 900^oC. It weight gain is monitored over a period of 1200 minutes:

Determine whether the data best follows parabolic or cubic oxidation kinetics.

Answer:

It is apparent that the square of weight gain plotted vs time closely approximates a straight line through the origin, while the similar plot for the cubic of weight gain does not. Therefore, the data seem to follow parabolic oxidation kinetics.

5.4 At high temperatures, copper reacts with air to form the oxide Cu₂0. Would this oxide be protective according to the Pilling-Bedworth criterion? Densities are: Cu = 8.92 g/cm³, Cu₂0 = 6.0 g/cm³.

Answer

The reaction is 2Cu + ½O₂ = Cu₂O. That is, two moles of copper will form one mole of oxide.
Volume of 1 mole of oxide = (63.55 g/mol Cu x 2 moles + 15.999 g/mol O) + 6.0 g/cm3 = 23.85 cm³.
Volume of metal reacted = 2 moles of Cu x 63.55 g/mol + 8.92 g/cm³ = 14.25 cm³

P-B ratio = 23.85 cm³ Cu₂O + 14.25 cm³ Cu = 1.67.
Since the Pilling-Bedworth ratio is greater than one, the oxide should be protective.

A further criterion might be added to Pilling and Bedworth's original concept: if the P-B ratio is much greater than two, the oxide may become unprotective in time as the oxide thickens, compressive stresses build up within it, and it blisters or cracks.

End of Tutorial