Electrochemistry and Corrosion Professor Brian Kinsella
Corrosion • Corrosion can be defined as the deterioration of a substance by reaction with its’ environment. • Strictly speaking this definition includes non-metals as well as metals, for example, plastics and ceramics. However, for this course corrosion will focus on the corrosion of metals which is mainly by electrochemical reactions.
Assignment 1 – Corrosion of Reinforcing Bar in Concrete • Often referred to as concrete cancer – this area of corrosion amounts to tremendous costs to government, industry and society. • It affects concrete structures of wharves, bridges, tall buildings and pathways. • The problem is caused by oxygen corrosion of the mild steel reinforcing bar. The iron oxide corrosion product occupies a larger volume then the iron and this results in high internal pressure causing the concrete to crack and spall (break in chips or flakes).
Assignment 1 – Corrosion of Reinforcing Bar in Concrete • Under normal circumstances the steel does not corrode due to the high pH of concrete and the formation of a passive (protective oxide) film which can form under this condition. • Destruction of the passive film can be attributed to ingression of chloride ions and carbon dioxide. • There are a number of techniques used to prevent corrosion taking place, including cathodic protection. • There are also a number of techniques used to treat the problem once it has occurred. • In your assignment you will need to cover the corrosion mechanism, the methods of prevention and the methods of treatment.
Assignment 1 – Corrosion of Reinforcing Bar in Concrete • Key topics to look for: • Pourbaix diagram • Cathodic protection methods • Inhibitors • Quality of concrete • Comment on the effectiveness of prevention and cure.
Basic Thermodynamics • Gas Laws
Gas Constant • The calorie is equal to 4.1840 joules • The equation is used to determine the number of moles of a gas if P, V and T are known or P, V or T if the other parameters are known. • It is important to remember that 1 mole of gas will occupy 22.4 liters at 1 atmosphere and 0oC.
Gas Constant • Note: The value of R = 8.314 joules deg-1 mole-1is used in the Nernst Equation
Daltons Law – The total pressure of a gas in a liquid is equal to the sum of the partial pressures of the gases in the mixture The total number of moles of a gas in a mixture is equal to the sum of the numbers of moles of the different gases Where lower case p is used to represent the partial pressure
Raoult’s Law • The mole fraction of a component in the vapor is equal to the pressure fraction in the vapor • E.g. for two components
Raoult’s Law – Ideal Solutions • Example: The vapor pressure of benzene and toluene at 60o are 385 and 139 mm. Calculate the partial pressure of benzene and toluene, the total vapor pressure of the solution, the mole fraction of toluene in the vapor above a solution with 0.60 mole fraction toluene. • pbenzene = 0.4 x 385 = 154 • ptoluene = 0.6 x 139 = 83.4 • Ptotal = 237.4 Note that the more volatile component is more concentrated in the vapor. In this case benzene is more volatile and the mole fraction of benzene in the vapor will be 154/237.4 = 0.649
Raoult’s Law – Ideal Solutions • The law enables the determination of volatile components such as pentane and hexane in the gas phase for top of the line corrosion mechanisms. • The law predicts that the vapor pressure of a pure component will decrease due to the presence of a non volatile soluble material, e.g. salt in water will decrease the mole fraction of water and its vapor pressure. • Does the corrosivity of CO2 gas decrease under super saline conditions?
Henry's Law - Non Ideal Solutions • For dilute solutions, the partial pressure of the component present at lower concentration is directly proportional to its mole X2 • For ideal solutions K2 = p2o and Henry’s law becomes identical to Raoult’s law
Henry's Law - Non Ideal Solutions • Calculate the solubility of CO2 in water at 25oC at a CO2 pp of 760 mm
Henry's Law - Non Ideal Solutions • X2 = moles CO2/moles CO2 + moles H2O • X2 = moles CO2/moles CO2 + 1000/18 H2O • X2 = moles CO2/moles CO2 + 55.556 H2O • X2 moles CO2/55.556 H2O • 1/X2 55.556/moles CO2 • 55.556nCO2
Consequence of Dalton’s Raoult’s and Henry’s Law • It is important to remember Dalton’s, Raoult’s and Henry’s laws when preparing corrosive gas mixtures. • E.g., If you need to determine the corrosive effect of 1 ppm dissolved oxygen. Knowing that seawater contains about 10 ppm dissolved oxygen at 20oC and 1 bar (760 mm), i.e., 80% N2 and 20% O2. A gas mixture of 98% N2 and 2% O2 should result in 1 ppm dissolved oxygen. • The same result can be obtained by bubbling nitrogen and oxygen into your test solution at the rates of 98 and 2 mL s-1 respectively (or 196 and 4 mL minute-1,respectively – etc).
Consequence of Dalton’s Raoult’s and Henry’s Law • Alternatively, to study the corrosivity of CO2 at 0.2 bar, you can achieve the required concentration by bubbling N2 and CO2 gas into your test solution at 20 and 80 mL minutes-1 respectively. • You are required to compare the corrosivity of acetic acid with CO2 at 0.2 bar to steel. Determine the molar concentration of acetic acid required since for comparison, both acids need to be at the same molar concentration. • CO2 + H2O↔ H2CO3
First Law of Thermodynamics • Energy can be transformed (changed from one form to another), but it can neither be created nor destroyed." • “ The increase in the internal energy of a system is equal to the amount of energy added by heating the system, minus the amount lost as a result of the work done by the system on its surroundings.
Reversible Expansion of a Gas The work performed during the expansion of a gas can be related to changes in pressure or volume.
Energy • For an ideal gas:
Enthalpy • Constant pressure process
Constant pressure and Relationship between CP and CV • The ratio is the heat capacity at constant pressure, Cp
Reversible Adiabatic Expansion of a Gas No loss or gain of heat to the environment – the system is thermally isolated so that q = 0
Reversible Adiabatic Expansion of a Gas • Calculate the temperature increase and final pressure of He if a mole is compressed adiabatically and reversibly from 44.8 liters at 0oC to 22.4 liters. The molar heat capacity of He = 3.00 cal deg-1 mole-1 . R must be expressed in the same units.
Heat Capacity of Solids • Debye showed that at sufficiently low temperature • The quantity , the characteristic temperature of a substance, is usually of the order of 100-400oK. The equation is suitable for extrapolating heat capacity data below 15oK
Standard State • The change in enthalpy ΔH or internal energy ΔE for a reaction depends on the states of the reactants and products. E.g. the heat of combustion of graphite is different from that of diamond. To facilitate the tabulation of thermodynamic data certain standard state are adopted and thermodynamic properties are tabulated to these standard states. The standard state of a gas is the ideal gas at 1 atm at the temperature concerned; for a liquid it is the pure liquid at 1 atm at the temperature concerned; for a solid it is a specified crystalline state at 1 atm at the temperature concerned. The standard form of carbon is graphite and the standard form of sulfur is rhombic sulfur. In writing equations, solids, liquids, and gases are designated by (s), (l), and (g) respectively, since the enthalpy change depends on the physical state of the reactants and products. The standard state refers to temperature at 25oC unless otherwise specified.
Calculation of the Heat of a Reaction at Constant Pressure, from the Heat of Reaction at Constant Volume The standard heat of combustion of n-heptane at constant volume and 25oC is 1148.93 kcal C7H16(l) + 11O2 (g) = 7CO2 (g) + 8 H2O(l) ΔEo = -1,148.93 kcal mole-1 Estimate the heat absorbed at constant pressure qp = qv + RTΔn = -1,148,930 – (1.987)(298.1)(4) = 1,151,300 cal mole-1 or1,151.30 kcal mole-1
Application of First Law to Thermodynamics • Calc. of ∆H when carbon burns to carbon monoxide • It is difficult to burn C only to CO Note the change in polarity for the reverse reaction
Enthalpy of Formation • Standard enthalpy change for a reaction is equal to the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants at the temperature concerned. • vi and vj are the stochiometric coefficients in the balanced chemical equations for reactants and products, respectively. • The molar enthalpy for each element in its standard state is given the value zero.
Enthalpy of Formation • For example the enthalpy of formation of CO2 is the enthalpy change for the following reaction. • Since the enthalpy of combustion of hydrogen to form liquid water at 298oK is –68,317.4 cal mole-1, the enthalpy of formation of H2O(l) is –68,317.4 cal mole-1. The enthalpy of formation of gaseous water is less negative by the molar heat of vaporisation of water at 25oC, which is 10,519.5 cal mole-1. Therefore the enthalpy of formation of H2O(g) @ 25oC is -57,797.9 cal mole-1
Enthalpy of Formation • When a substance cannot be formed directly in a rapid reaction from its elements, the enthalpy change for a series of suitable reactions may be utilized in calculating the enthalpy of formation. • Example: calculate the enthalpy change of H2SO4(l) from the enthalpy change for the combustion of sulfur to SO2, the oxidation of SO2 to SO3 using a platinum catalyst, and the heat of solution of SO3 in water to give H2SO4 at 25oC
Enthalpy of Formation • Consider the following
Enthalpy of Formation • The enthalpy of formation of many compounds, ions, and atoms are accurately know. From these enthalpies of formation the enthalpy change for many reactions may be calculated by use of the equation given. • The standard state of a solute in aqueous solution is taken as the hypothetical ideal state of unit molality, in which the enthalpy of the solute is the same as in the infinitely dilute solution.
Enthalpy of Formation Heat evolved solution heats up Heat absorbed solution becomes colder
Heat of Solutions • Substance dissolving in water • Calculate the integral heat of solution of 1 mole of HCl(g) in 200H2O(l) • The reaction involves HCl gas dissolving in two hundred moles of water to produce a solution of HCl (1 mole in 200 moles of water)
Heat of Dilution • Calculate the integral heat of dilution for the addition of 195 moles of water to 1 mole of HCl in 5 moles of water. The enthalpy of formation of HCl in 5 moles of H2O is -37.37 kcal and the enthalpy of formation in 200 moles of water is -39.798 kcal.
Enthalpies of Formation of Ions • Since for strong electrolytes in dilute solution the thermal properties of the ions are essentially independent of the accompanying ions, it is convenient to use relative enthalpies of formation of individual ions. The sum of the enthalpies of formation of H+ and OH- ions may be calculated.
Enthalpies of Formation of Ions • The separate enthalpies of formation of H+ and OH- cannot be calculated, but if the enthalpy of formation of H+(aq) is arbitrarily assigned the value zero, it is possible to calculate relative enthalpies of formation for other ions. Denoting the electron by e, the convention is that. • Therefore the enthalpy of formation of OH- is given by
Bond Energies • The bond energy E(A-B) is the contribution of the bond between a particular pair of atoms A-B in a molecule to the total binding energy of the molecule. The total binding energy is required to dissociate the molecule into atoms, each in its ground state. Bond energies are usually given for a reference temperature of 298oC. • For a diatomic molecule the bond energy is the heat of dissociation of the molecule. • In molecule of the type, A-Bn where all n bonds are identical, the bond energy of the A-B bond is 1/n times the heat of dissociation of into A + nB. • For polyatomic molecules having several types of bonds, it is not possible to make precise assignment of bond energies, but in a series of similar compounds bond energies for different bonds may be assigned in such a way as to reproduce as much as possible the heat of atomization of all compounds in the series.
Bond Energies • E(C-H) = 394/4 = 98 kcal. This may be compared with the energy of dissociation of successive hydrogen atoms of methane.