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Electrochemical Thermodynamics and Concepts Sensitivity of electrochemical measurements Measurements of electrochemical processes are made by measuring electrical currents and voltages. The currents results from the flow of charged ions (positive or negative) and or electrons. Current measurement can be extremely sensitive as small as 10-15 amps (coulombs per second). Let’s get “feeling” for the sensitivity of such a measurement. monolayer: ≈1.5 x1015 atoms/cm2 . Suppose each atom on the surface dissolves as a single charged positive ion (i.e., a cation). M Þ M + + e- A +1 1 cm 1 cm corresponding to 1.6 x 10-19 C/atom of charge. Today we can easily measure at the nanoamp scale. Suppose we measure a current density of 10-6 A cm-2 for 1 s or 10-6 C. A full monolayer of metal dissolving as a +1 cation results in a charge of 1.6 x 10-19 C/atom x 1.5 x 1015 atoms/cm2 or ≈ 2.4 x 10-4 C cm-2. Then 10-6 C corresponds to only of order 0.01 ML. An electrochemical reaction is a chemical reaction involving electron transfer. An ordinary chemical reaction does not involve electron transfer Chemical reactions: Note the mass balance N2 +3H2 = 2NH3 synthesis of ammonia Electrochemical Reactions: Note the mass and charge balance 2H+ +2e- = H2 Hydrogen ion reduction Zn = Zn2+ + 2e- Zn oxidation Since electrochemical reactions involve electron transfer we can measure the rate of these processes by designing an appropriate electrical circuit and measuring the flow of an electric current. Faraday’s Law relates the flow of electric current to the mass of metal reacting. Ita m nF m= t = I = a = n = F = mass of metal reacting (gm) time (s, min, hours, years) current (Amps) atomic weight of metal number of electrons transferred in the reaction 96,500 C; Faraday’s constant which is equal to the charge associated with 1 mole of electrons. If , for example, we divide Faraday’s law by the time and area of A dissolving surface, we obtain a relation describing the rate of this process. m ia r tA nF A = i = r= surface area of corroding metal (cm2) I/A: the current density (A/cm2) Rate measured as a weight per unit area per unit time (gm/cm2-s). Examples Consider the case for steel or iron corrosion. Suppose we make a measurement of the current associated with the corrosion of 100 cm2 of a steel surface and find a current of 0.1A. Then dividing through by the area of the sample, 0.1 A/100 cm2 = 0.001 A/cm2 = 1 mA/cm2 current density. m ia r tA nF a = 1 mA/cm2 55.85 g F = 96,500 C n = 2 i = Fe Fe2 2e 0.001A / cm 2 55.85 g 2.89 107 gm / cm 2 s 2 96,500C We can convert this to a penetration rate (cm/s) by dividing by the density of Fe. Fe 7.88 g / cm3 0.001A / cm 2 55.85 g r 2.89 107 gm / cm 2 s 2 96,500C p r Fe 2.89 107 gm / cm 2 s 8 3.67 10 cm / s 3 7.88 gm / cm Next consider: Electrochemical Cells V {Zn2+} =1 {H+} =1 Zn Pt We measure a potential difference of - 0.762 V H2 at 1Atm porous barrier Pt is inert and acts as a catalyst: Zn is oxidized: Total reaction: 2H+ + 2e- = H2 Zn = Zn2+ +2eZn + 2H+ = Zn2+ + H2 2H+ + 2e- = H2 E0 = Zn2+ +2e- = Zn E0 = - 0.762 V (NHE) sign taken to be negative since Zn is oxidized. 0.000 V (defined as zero by convention) Electrochemical Cells V {Cu2+} =1 {H+} =1 Cu Pt We measure a potential difference of 0.342 V H2 at 1Atm porous barrier Pt is inert and acts as a catalyst: Cu plates; Total reaction H2 = 2H+ + 2e- : hydrogen is oxidized Cu2+ +2e- = Cu Cu2+ + H2 = Cu + 2H+ 2H+ + 2e- = H2 E0 = Cu2+ +2e- = Cu E0 = 0.342 V (NHE) This sign is positive since Cu is reduced. 0.000 V (defined as zero by convention) Electrochemical Cells +1.104 V 1 M CuSO4 1 M ZnSO4 If we construct this “cell” we observe the following: Zn dissolves; Zn = Zn2+ +2e-; oxidation occurs at the anode Cu plates; Cu2+ +2e- =Cu reduction occurs at the cathode If we connect a voltmeter as shown we measure a voltage of 1.104 V. Daniell Cell +1.104 V Zn dissolves; Cu plates; Zn = Zn2+ +2e-; Cu2+ +2e- =Cu oxidation occurs at the anode reduction occurs at the cathode For the Cu-Zn cell we measured a voltage difference of 1.104 V. Zn2+ +2e- = Zn E0 = - 0.762 V (NHE) Cu2+ +2e- = Cu E0 = +0.342 V (NHE) This difference is 1.104 V Standard Reduction Potentials: EMF Series Volts (NHE): E0 Au 3+ +3e- =Au +1.498 O 2 +4H + +4e- =2H 2 0 (pH = 0) +1.229 Ag + +1e- =Ag +0.798 Fe3+ +1e- =Fe 2+ +0.771 Cu 2+ +2e- =Cu +0.342 2H + +2e- =H 2 +0.000 Pb 2+ +2e- =Pb -0.126 Sn 2+ +2e- =Sn -0.138 Ni 2+ +2e- =Ni -0.250 Co 2+ +2e- =Co -0.277 Cd 2+ +2e- =Cd -0.403 Fe 2+ +2e- =Fe -0.447 Zn 2+ +2e- =Zn -0.762 Al3+ +3e- =Al -1.662 Noble Cathode Anode Active Standard conditions Standard states: • For a solid; a =1: pure metal, metal oxide, etc. • For a gas, 1 Atm pressure is taken as unit activity. • For dilute solutes typically found in most instances of corrosion, activity is reasonably approximated by the concentration in M. The standard state is 1 M. • Temperature is taken as 25ºC = 298 K definition: pH = -log [H+] Thermodynamics of Chemical Equations aA + bB xX + yY A, B … reactants X, Y … products a, b, …., x, y,…. Stoichiometric Coefficients defining how many moles of A and B produce moles of X and Y The chemical reaction can proceed as indicated if the energy change is negative. That is if the energy of the products is less then that of the reactants. Mathematically we say Gproducts Greact 0 Gi Gi0 RT ln ai OR G 0 Activity or concentration “correction” -non standard state Energy of the standard state Chemical Equilibrium aA + bB + … ® xX + yY + … A, B … reactants X, Y … products a, b, …., x, y,…. Stoichiometric Coefficients Equilibrium is defined by the condition ( DG = xGx + yGy + Since Gi G RT ln ai 0 i ( DG = xGx0 + yG y0 + ) - ( aG A + bGB + )=0 G ≣ standard free energy change /mole ) - ( aG 0 A + bG B0 + ) +RT (x ln ax + y ln a y + ) - RT (aln aA + bln aB + ) = 0 Solving for G Chemical Equilibrium DG = -RT ln 0 x y X Y a b A B a a aa = -RT ln K K is the equilibrium constant for the reaction. The equilibrium constant is defined by G K exp( ) RT When components are not in standard state: DG = DG 0 + RT ln a Xx aYy aAaaBb Electrochemical Equilibrium DG = DG 0 + RT ln a Xx aYy aAaaBb Free energy is measured in units of kcal or kJ. 1 calorie 1Joule = = 4.186 Joules 6.24 x 1018 eV Since 1 Volt = Joule/Coulomb G QE DG = -nFE where Q is the total charge. The total charge transferred in an electrochemical reaction will be equal to the number of moles of electrons participating in the reaction times the charge/mole, Q = nF By convention a minus sign relates ΔG and E. Electrochemical Equilibrium If we flip the numerator and denominator in the argument of the ln function, Electrochemical Equilibrium Nernst Equation Formal Potential Electrochemical Equilibrium: pH effects We can write a general metal dissolution reaction in the following way: aA mH ne bB dH 2 0 Here A could correspond to some oxidized species such as Fe2+ and B would be its reduced form, metallic Fe. Alternatively A could be an oxide such as FeO and B its reduced form, also metallic Fe. The Nernst equation for this general reaction is a m éH ù é ù A RT ë û ë û E = E0 + ln nF é B ùb é H O ù d ë û ë 2 û + Electrochemical Equilibrium and Corrosion a m éH ù é ù A RT ë û ë û E = E0 + ln nF é B ùb é H O ù d ë û ë 2 û + It’s convenient at this stage to convert the natural log (Ln) to base 10. In general: Ln (x) = 2.303 Log (x) Also RT 2.303 8.31J / moleK 298K 0.059 V F 96,500C / mole Electrochemical Equilibrium and Corrosion The Nernst Equation above can now be rewritten as: a éë Aùû m 0.059 + é ù E=E + log + 0.059log H b ë û n n éë B ùû 0 a éë Aùû m 0.059 E=E + log - 0.059 pH b n n éë B ùû 0 Electrochemical Equilibrium and Corrosion a éë Aùû m 0.059 0 E=E + log - 0.059 pH b n n éë B ùû Examples: (1) Consider the reaction Fe Fe 2 2e This type of reaction is favored at low values of pH. First note that for this and all simple metal dissolution reactions there is no H+ in the reaction and no dependence of the reaction on pH, m = 0 The Fe2+ is equivalent to A. Also a =1. The quantity [Fe2+ ] is the concentration of Fe2+ in the electrolyte. Fe is equivalent to B. Since Fe is a pure metal [Fe ] = 1 Then: E = E 0 + 0.0295log[Fe 2+ ] = -0.447 + 0.0295log[Fe 2+ ]; (V) Electrochemical Equilibrium and Corrosion ΔE (V) [Fe2+] M -0.477 1 -0.506 0.1 -0.534 0.01 -0.565 0.001 -0.595 10-4 -0.624 10-5 -0.654 10-6 E = -0.447 + 0.0295log[Fe 2+ ]; (V) For every decade change in ferrous cations there is ~ 30 mV decrease in the equilibrium potential. What is the difference between an open circuit potential (corrosion potential) and an Equilibrium Potential????? – Lab. Electrochemical Equilibrium and Corrosion a éë Aùû m 0.059 E=E + log - 0.059 pH b n n éë B ùû 0 Examples: (2) Consider Al2 O3 6 H 6e 2 Al 3H 2 O This type of reaction occurs at some intermediate value of pH. Both Al and Al2O3 are solids with unit activity, A=1, B=1. o On a complete EMF series one could find that E Al / Al2 O3 1.55 V. Also note that m = 6, n = 6 so, E Al/ Al O = -1.55 - 0.059 pH 2 3 The equilibrium potential decreases by ~ 60 mV per pH unit. Electrochemical Equilibrium and Corrosion a éë Aùû m 0.059 E=E + log - 0.059 pH b n n éë B ùû Examples: 0 (3) One other type of corrosion reaction that can occur involves the formation of a soluble metal oxide anion at some high pH, AlO2 4 H 3e Al 2 H 2O o E Al/ AlO - = E Al/ + AlO 2 2 0.059 4 log éë AlO2- ùû - 0.059 pH 3 3 E Al/ AlO- = -1.262 + 0.020log éë AlO2- ùû - 0.079 pH 2 The ΔE for this reaction is a function of both the dissolved ion content and the pH of the electrolyte. Cathode reactions Supporting Corrosion Hydrogen reduction + - 2H + 2e = H2 EH / H2 EH0 / H2 0.059 pH The above reaction occurs in an acid solution. An equivalent reaction in neutral or alkaline solutions is: 2H2O + 2e- = H2 + 2OH - EH / H2 EH0 / H2 EH / H 0 0.059 pH 2 0.059 pH Cathode reactions Supporting Corrosion Oxygen reduction O2 + 4H + + 4e- ® 2H2O EO 2 / H 2O EO02 / H 2O 0.059 pH The above reaction occurs in an acid solution. An equivalent reaction in neutral or alkaline solutions is: O2 2 H 2 O 4e 4OH EO2 / H 2O EO02 / H 2O 0.059 pH EO2 / H 2O 1.229 0.059 pH Potential/pH (Pourbaix) Diagram It turns out to be very convenient to represent the results of all these anodic metal oxidation processes and cathodic reduction process on a Map-like “phase diagram”. The map considers the parameters of voltage and solution pH and shows the possible electrochemical/corrosion reactions that can occur for a particular metal such as Fe, Al, Cd, Zn, … and water. Potential/pH (Pourbaix) Diagram Water oxygen evolution and acidification EO2 / H2O 1.229 0.059 pH EH / H 0 0.059 pH 2 hydrogen evolution and alkalization Potential/pH (Pourbaix) Diagram Generic diagram for a metal Corrosion-soluble ions of the metal are stable 1.2 POTENTIAL, E(V) 0.8 0.4 Passivation- oxides are stable 0.0 -0.4 Corrosion Immunity-reduced form of the metal is stable -0.8 Passivation -1.2 Corrosion -1.6 -2.0 Immunity -2.4 -2 0 2 4 6 pH 8 10 12 14 Potential/pH (Pourbaix) Diagram Pourbaix diagram for Al Reaction 1: Metal oxidizes to aqueous cations M = M n + + neAl = Al 3+ + 3eeAl / Al 3 e0 A a 0.059 m log 0.059 pH b n B n eAl / Al 3 1.662 0.059 log Al 3 3 Independent of pH since no H+ is involved. Only depends on Al3+ activity Potential/pH (Pourbaix) Diagram Pourbaix diagram for Al Reaction 2: Metal reacts to metal hydroxide or oxide M + nH 2O = M (OH ) n + nH + + ne2Al + 3H 2O = Al2O3 + 6H + + 6e- EAl / Al2O3 1.55 0.059 pH At higher pH Al2O3 is formed. At lower pH Al2O3 chemically dissolves to Al3+ Intersection depends depends on Al3+ activity (dashed lines are portions of the reactions with no significance) Potential/pH (Pourbaix) Diagram Pourbaix diagram for Al The reaction rate constant for 2Al 3+ + 3H2O = Al2O3 + 6H + is known H ) ( K= ( Al ) + 6 3+ 2 = 10-11.4 For (Al3+)=10-6, log K 6log H 2log Al 3 6 pH 2log Al 3 log Al 3 106 0.5log(10 11.4 ) 3 pH 6 5.7 3 pH pH 3.9 Independent of potential Potential/pH (Pourbaix) Diagram Pourbaix diagram for Al Reaction 3: Metal reacts to form soluble aqueous anions Al 2H 2 O AlO2 4H 3e EAl / AlO 1.262 0.020log AlO2 0.079 pH 2 At higher pH, Al2O3 dissolves to AlO2 2 Al2O3 H 2O 2 AlO 2H pH 14.6 log[ AlO2 ] K H AlO2 1029.2 2 2