Thermodynamics and Electrode Potential ME Dr. Zuhair M. Gasem

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1 Thermodynamics and Electrode Potential ME Copyright Dr. Zuhair M. Gasem

2 Corrosion Science and Engineering 2 Corrosion Science Engineering: corrosion forms, and controlling methods Chpater2 Thermodynamics Chapter 3 Kinetics Predict corrosion tendency Answer whether corrosion is possible or not Predict speed of corrosion If corrosion is possible, how fast?

3 Objective 3 Given a metal in an electrolyte: Anodic reaction is M M +n + ne - Cathodic reaction is N n+ + n e - N Will the metal corrode? Corrosion is possible if the spontaneous direction of the reaction is M + N n+ M +n + N We need to determine the spontaneous direction of an electrochemical reaction.

4 Electrode Potential 4 An electrode is a metal surface allowing charge transfer in or out of an ionic solution. Metal conductors consist of free electrons and cations (+ve ions). Water molecules are polar. Water molecules polarize the metal surface and attract free electrons of the metal. Excess electrons exist at the surface of a metal in water. This creates the electrical double layer which prevents easy charge transfer. Electrode potential

5 Electrode Potential 5 The electrical double layer results in the formation of an electrical potential difference at the metal-solution interface. This potential is called the electrode potential or electrochemical potential. The electrode potential indicates the tendency of a metal towards electrochemical reactions: anodic tendency (M M +n + ne - ) cathodic tendency (M +n + ne - M) The electrode potential depends on the electrolyte, ions concentration, and temperature.

6 Reversible Electrode Potential 6 The presence of the electrical double layer drives the electrode to reach equilibrium condition where the net current crosses the double layer is zero. At equilibrium, there is still charge transfer between the metal and the electrolyte but the anodic current (M M +n + ne - ) is equal to the cathodic current (M +n + ne - M) and the net current is zero (i net = i anodic i cathodic =0). The potential of a reaction at equilibrium is called the reversible electrode potential. The reaction at equilibrium is written as M M +n + ne - or M = M +n + ne -

7 Standard Hydrogen Electrode (S.H.E.). 7 The absolute electrode potential can not be measured experimentally. The electrode potential is measured relative to a reference electrode such as the standard hydrogen electrode (S.H.E.). Assume that the reaction 2H + + 2e - H 2 has a reversible potential of 0 when [H+] = 1 M and P H2 = 1 atm, and the temperature is 25 C. This is called the standard condition 2H+ + 2e- H2

8 Emf Series 8 We will measure the reversible electrode potentials of all important reactions relative to the S.H.E. This produces the Electromotive Force Series (emf Series) which gives the reversible electrode potential of a reaction at standard conditions (ion concentration is 1 M, pressure of any gas at 1 atm, and a temperature of 25 C) [M +n ] = 1 M, P = 1 atm, T= 25 C.

9 9 Standard emf (Reduction Potentials)

10 Standard emf (Reduction Potentials) 10 Note the followings: The potential of a reaction is called the halfcell electrode potential The reactions in the emf series are all written as reduction reactions Noble metals such as Au and Pt have positive reduction potentials Active metals such as Mg and Na have negative reduction potentials

11 Example: half-cell potential for Zn 11 Measure the standard reversible half-cell reaction for Zn electrode relative to S.H.E. e o Zn/Zn+2 = V vs. S.H.E. Note that the +ve terminal of the voltmeter is connected to the SHE. Hence, SHE electrochemical potential is more +ve than that for Zn/Zn

12 Example: half-cell potential for Zn 12 Note that when the -ve terminal of the voltmeter is connected to the SHE, the voltmeter reads V vs. SHE. This indicates that Zn/Zn 2+ is ve towards SHE

13 Galvanic Cell Potentials 13 A galvanic cell consists of two electrodes: an anode and a cathode. Corrosion always involves two half-cell reactions. We will use the emf series to predict which half-cell reaction will act as the cathode and which will act as the anode when the two are connected.

14 Cell Potentials 14 Thermodynamics states that a spontaneous reaction is associated with a decrease in free energy. Free energy of an electrochemical reaction is given by: G cell = -n*f*e cell n = number of electrons exchanged F = Faraday s constant = Coulomb per mole of electrons E cell = (e c -e a ): the potential difference b/w the cathode and the anode in the cell.

15 Faraday s Constant 15 One mole of electrons, contains Avogadro s number (6.023x10 23 ) electrons Charge on one electron is Coulomb Hence the charge on one mole of electrons=6.023x10 23 *1.6x10-19 =96500 C This is known as Faraday s constant F=96500 C/(mole of e)

16 Example: Cu-Zn Galvanic Cell 16 A galvanic cell is at equilibrium (no current flow), consists of Zn and Cu electrodes at standard conditions. Which one will be the anode (the corroding metal) and which one is the cathode if we allow current to flow? Before current flows: the two reactions are at equilibrium and at standard conditions (we can use the emf series directly): Zn e - Zn e o = V Cu e - Cu e o = V

17 Cu-Zn Galvanic Cell 17 After current flows: Assume initially Zn will react in the cathodic direction (Zn e - Zn), and Cu will react in the anodic direction (Cu Cu e - ) (in essence we are assuming that Cu will corrode) Net reaction is Cu + Zn +2 Cu +2 + Zn e a = V e c = V E cell = e c -e a = = -1.1 V G = -n*f*e cell ( G is +ve) Hence Cu + Zn +2 Cu +2 + Zn is not the spontaneous direction of the reaction)

18 Cu-Zn Galvanic Cell 18 Assume Cu will react in the cathodic direction (Cu e - Cu), and Zn will react in the anodic direction (Zn Zn e - ) (in essence we are assuming that Zn will corrode) Net reaction is Cu +2 + Zn Cu + Zn +2 e a = V e c = V E cell = e c -e a = (-0.763) = 1.1 V G = -n*f*e cell ( G is ve) Hence Cu +2 + Zn Cu + Zn +2 is the spontaneous direction of the reaction) Conclusion: Zn will act as the anode (Zn will corrode) when coupled with Cu to form a galvanic cell (Daniel cell) at the standard condition.

19 Cu-Zn Galvanic Cell 19 C: (Cu e - Cu) A: (Zn Zn e - ) Overall reaction is the addition of A and C: Cu +2 + Zn Cu + Zn +2 The equilibrium cell potential is 1.1 V This can act as a battery supplying current. The maximum work available from the battery is ( G cell = -n*f*e cell ) G cell = -2*96500*1.1 = KJ

20 Example: Cu-Ag 20 A silver electrode is immersed in silver nitrate (AgNO 3 ) and a copper electrode is immersed in a copper nitrate Cu(NO 3 ) 2 to form a galvanic cell at the standard condition. Which electrode will act as the cathode and which is the anode? Cu e - Cu (e o = V vs. SHE) 2Ag + + 2e - 2Ag (e o = V vs. SHE) E cell = e c e a G = -n*f*e cell E cell = =0.462 V

21 Example: Galvanic cell Cu-Ag 21 In galvanic cell such as a battery: The anode is (-ve) where anodic reaction takes place The cathode is (+ve) where cathodic reaction takes place Cu Cu 2+ +2e - 2Ag + + 2e - 2Ag

22 Current circuit in galvanic cell 22 Salt bridge replaces a porous barrier Salt bridge allows ions flow Note the flow of negative current: Electrons in the wire Anions in solution

23 Determination of the Spontaneous Direction of a reaction 23 Example: Indicate the spontaneous direction of the reaction (assume standard conditions): 3Pb + 2Al 3+ 3Pb Al Assume the reaction goes from left to right A: 3Pb 3Pb e - e o a = V C: 2Al 3+ +6e - 2Al e o c = V E cell = e c -e a = V (note that e is used for halfcell potential while E for the overall cell) G cell = -n*f*e cell = +ve (hence the spontaneous direction of the reaction is from right to left or 3Pb Al 3Pb + 2Al 3+ and E cell = V

24 Concentration Effects on Reversible Standard Potential 24 The emf series lists the electrode potential at standard condition ([M +n ] = 1 M, P = 1 atm, T= 25 C). What happen if we change the ions concentration or the temperature or pressure? It would be impossible to measure the electrode potential for all conditions. Nernst s equation calculates the electrode potential at conditions different than the standard conditions.

25 Nernst s Equation 25 For a reaction written in the reduction direction: Nernst s Equation: e = e o RT/(nF)* Log ([Reactant]/[Product]) R : gas constant = J/(mol*K) T : temperature in K n: number of electron mole F: Farady s constant Log [solid] = 1 at room temperature 2.3RT/F=2.3*8.314*298/96500= V Nernst s equation at room temp is written as: e = e o /n* Log ([Reactant]/[Product])

26 Nernst s Equation 26 Example: what would be the half-cell electrode potential of Fe when [Fe 2+ ] = 1x10-6 M, at RT? Fe e - Fe e o = V vs. SHE e = /2* log(1x10-6 /1) = V Example: what would be the half-cell electrode potential for hydrogen reduction when [H + ] = 1x10-3, P H2 =0.1 atm; at 100 C? 2H + + 2e - H 2 eo = 0 V e = e o + 2.3*8.314*373/nF* Log ([H + ] 2 /P H2 ) e = /2*(2*log 1x10-3 -log 0.1) e = 0.074/2*(-6- (-1)) = V

27 Nernst s Equation 27 For a general reaction described by aa + mh + + ne - bb + dh 2 O The reversible potential for a given concentration different than the standard conditions are given by Nernst s equation as: e = e o + 2.3RT/nF log {[A] a [H + ] m /[B] b [H 2 O] d }

28 Derivation of Nernst s Equation 28 The free energy of a reaction in equilibrium as a function of composition is given by: G = - RT ln (K/Q) K is the equilibrium constant of the reaction Q is the reaction quotient (Q=[Product]/[Reactant]) G = -RT ln K + RT ln Q but G o = -n F E o, G = -n F E -n F E = -n F E o + RT ln Q E = E o RT/nF ln Q and (ln x = 2.3 log x) E = E o 2.3 RT/nF log Q Nernst s equation for a reduction reaction is written as: E = E o RT/nF log {[Reactant]/[Product]}

29 Example 29 Write the reaction (Zn +2 + Fe Zn + Fe +2 ) in the spontaneous direction and Calculate the cell potential : [Fe +2 ] = 0.1 M and [Zn +2 ]=1.5 M. Calculate the reduction potential for each half-cell reaction. Fe e - Fe ; e o = V e = /2*log(0.1/1)= V Zn e - Zn ; e o = V e= /2*log(1.5/1)= V Fe/Fe +2 Zn/Zn +2 (FeSO 4 ) (ZnSO 4 )

30 Example 30 Assume that the reaction has a tendency to proceed from L to R Cathode: Zn e - Zn e = V Anode : Fe Fe e - e = V E cell = = V Since G is +ve, the spontaneous direction is reversed (from R to L) The spontaneous direction Zn + Fe +2 Zn +2 + Fe (Zn is anode and Fe is cathode) Thus, Zn has the tendency to corrode in this cell. E cell = = 0.28 V

31 Predicting The Spontaneous Direction by Inspection 31 M +m + N M + N +n Calculate the potential for each half cell reaction as written in the reduction direction. Use Nernst equation if not in standard conditions. M +m + me - M e = e M N +n + ne - N e = e N Note that the electrode potential in the cathodic direction becomes: More ve than e o if the concentration is < 1 M More +ve than e o if the concentration is > 1 M Then The spontaneous cathode in the cell is the more +ve The spontaneous anode in the cell is the more ve The E cell = e c -e a

32 Example 32 Cu e - Cu; e o = v, when [Cu +2 ] =1M If [Cu +2 ] >1M, the electrode potential becomes more +ve. For [Cu +2 ] = 2M e = e o /n* Log ([Reactant]/[Product]) e = /2*log(2/1)= V If [Cu +2 ] <1M, the electrode potential becomes more ve. For [Cu +2 ] = 0.01M e = /2*log(0.01/1)= V

33 Concentration Cell 33 Use the inspection method to predict which electrode will corrode? Left electrode: Fe e - Fe e = /2*log(0.01/1) = V Right electrode e = /2*log(0.1/1) = V Since the more negative electrode acts as anode; then the left electrode is the anode and the right one is the cathode.

34 Concentration Cell 34 Use the inspection method to predict which electrode will corrode?

35 Example: 35 Use the inspection method to write the following reaction in the spontaneous direction. Cd + Fe +2 Cd +2 + Fe [Cd +2 ] = 1x10-6 M and [Fe +2 ] = 1.5 M Which electrode is the anode? Which electrode is the cathode? Draw a diagram for the galvanic cell and show the flow of electrons, anions, and cations? Which metal is corroding? Calculate E cell? Calculate the maximum energy available? Fe/Fe +2 Cd/Cd +2 Fe in a solution of FeSO 4 and Cd in a solution of CdSO 4

36 Example Problem 36 Determine the thermodynamic tendency for tin (Sn) to corrode in de-aerated sulfiric acid with [H + ]=10-2 M. Assume [Sn +2 ]=10-6 M and P H2 = 1 atm. 1. Identify the anodic and cathodic reactions. Sn +2 +2e - Sn e o = V 2H e - H 2 e o = 0 V 2. Calculate the half-cell potential for each reaction e Sn = /2*log(10-6 /1)= V e H2 = /2*log([10-2 ] 2 /1)=0.059*log10-2 = - 0/059*2 = V 3. By inspection, Sn is the anode and H 2 is the cathode 4. E cell = e c -e a = (-0.315) = V ( G is ve) 5. Spontaneous reaction is Sn + 2H +2 Sn +2 + H 2 6. Conclusion: Sn has a tendency to corrode (will corrode) in the given acidic solution.

37 Galvanic cell application1: Batteries 37 A battery is a galvanic cell which produces direct current from the electrochemical reactions of two electrodes (similar to corrosion or useful corrosion) Lead storage battery Anode reaction (-ve) Pb(s) +HSO 4- PbSO 4 + H + + 2e - (e o = V) Cathode reaction (+ve) PbO 2 + HSO H + + 2e - PbSO 4 + 2H 2 O (e o =1.69 V ) Pb+ PbO 2 + 2HSO 4- +2H + 2PbSO 4 (s) + 2H 2 O (E cell = = 2.04 V) Note that sulfuric acid (H 2 SO 4 ) is consumed during discharging and the battery must recharged (reactions are reversed by external current) to reproduce the acid.

38 Galvanic cell application2:fuel Cells 38 A fuel cell produces electricity from a continuous external supply of fuel (on the anode) and oxidant (on the cathode) in the presence of an electrolyte. A hydrogen fuel cell uses H 2 as fuel and O 2 as oxidant: Anode: 2H 2 + 4OH - 4H 2 O + 4e - Cathode: O 2 + 2H 2 O + 4e - 4OH - Net reaction: 2H 2 (g) + O 2 (g) 2H 2 O (l) Application Electrical power source (a plant in Tokyo) Electric battery to power electric motor in electric and hybrid vehicles.

39 Water Chemistry 39 Many metals corrode in water medium. Any water solution at 25 C, dissociate into [H + ] and [OH - ] ions such that: [H + ][OH - ] = 1x10-14 The solution is called neutral if [H + ] = [OH - ] The solution is called acidic if [H + ] > [OH - ] The solution is called basic if [H + ] < [OH - ] Examples of acidic solutions: HCl, HNO 3, H 2 SO 4 Basic Solutions: NaOH, KOH

40 Water Chemistry 40 The ph scale provides a convenient way to represent solution acidity: ph = - log [H + ] Thus, for a neutral solution where [H + ] = 1x10-7 ph = -log (1x10-7 ) = 7 Note that the ph decreases as the [H + ] increases. A solution of ph 3 has a [H + ] concentration 10 times higher than ph of 4 and 100 times that of ph 5.

41 Water Chemistry 41 The water ph and the level of dissolved O 2 determine the reduction reaction that takes place during corrosion of metals in water solution: I. Acid solutions with no dissolved O 2 2H + + 2e - H 2 (1) For [H + ] = 1M RT ; e o = 0 Apply Nernst s equation for reversible potential at different [H + ] RT and P H2 =1) e = /2 log {[H + ] 2 /P H2 } = 0.059/2 *2 log [H + ] The potential of the reaction at RT varies according to e = ph (a)

42 II. Water Chemistry Neutral or alkaline solutions Add [OH - ] to both sides of reaction (1) to neutralize it yields 2H + + 2OH - + 2e - H 2 + 2OH - 2H 2 O + 2e - H 2 + 2OH - (2) For neutral solution ; [H + ] = 1x10-7 RT e = *7 = V For alkaline solution ; [H + ] = 1x10-14 RT e = *14 = V Plotting Equation (a) as a function of ph yields the e-ph (Pourbaix) diagram 42

43 Pourbaix Diagram for Water 43

44 Pourbaix Diagram for Water 44 III. Acid solutions with dissolved O 2 Add [O 2 ] to both sides of reaction (1): 2H + + 2e - + O 2 H 2 + O 2 4H + + O 2 + 4e - 2H 2 O (3) For [H + ] = 1 M (ph=0), P O2 = 1 RT e o = V Apply Nernst s equation for reversible potential at different [H + ] (assume P O2 = 1 atm e = /4 log {[H + ] 4 P O2 } = ph The potential of the reaction at RT varies according to e = ph (b)

45 Pourbaix Diagram for Water 45 IV. Alkaline solutions with dissolved O 2 Add [OH - ] to both sides of reaction (3): 2OH - + 2H + + 2e - + O 2 H 2 + O 2 + 2OH - 2H 2 O+ O 2 + 4e - 4OH - (4) Apply Nernst s equation for reversible potential at different [H + ] (assume P O2 = 1 atm e = /4 log {P O2 /[OH - ]} = ph The potential of the reaction at RT varies according to e = ph (b) Plotting Equation (b) as a function of ph yields the e-ph (Pourbaix) diagram

46 Pourbaix Diagram for Water 46

47 Pourbaix Diagram for Water 47 Note the followings: Pourbaix diagram is a plot of reversible potential vs. ph for reactions in pure water. If we apply a more -ve potential than the eq value: 2H + + 2e - H 2 2H 2 O + 2e - H 2 + 2OH - Hydrogen is generated or evolved If we apply a more +ve potential than the eq value: 4H + + O 2 + 4e - 2H 2 O 2H 2 O+ O 2 + 4e - 4OH - Oxygen is evolved The region b/w lines a and b is the region where water is stable.

48 Pourbaix Diagram for Water 48 O 2 H 2

49 Pourbaix Diagram for Metals 49 If we consider all possible reactions of a metal in water gives three types of reactions: Reactions dependent only on potential such as Fe e - Fe. These reactions are represented as horizontal line in Pourbaix diagram (independent of ph) Reactions dependent only on ph such as Fe OH - Fe(OH) 2. These reactions are represented as vertical line in Pourbaix diagram (independent of potential) Reactions dependent on both ph and potential, i.e. Fe(OH) 2 + 2H + + 2e - Fe + 2H 2 O

50 Pourbaix Diagram for Water 50 For Al in water, the possible reactions: 1. Al e - Al e 0 = V Nernst s equation is written as e = /3*log([Al 3+ ]/1) For [Al 3+ ]=1x10-6 M, e = V Al +3 Al

51 Pourbaix Diagram for Water Al 2 O 3 + 6H + + 6e - Al + 3H 2 O; e 0 =-1.55V e = ph Al 2 O 3

52 Pourbaix Diagram for Water Al 2 O 3 + 6H + 2Al H 2 At equilibrium; K = (H + ) 6 /(Al +3 ) 2 = For [Al 3+ ]=1 M, ph=1.9 For [Al 3+ ]=1x10-6 M, ph=3.9

53 Pourbaix Diagram for Water AlO H + + 3e - Al + 2H 2 O e = log (AlO - 2 ) ph

54 Pourbaix Diagram for Water Al 2 O 3 + H 2 O AlO H + At equilibrium; K = (AlO 2- ) (H + ) 2 = For [AlO - 2 ] = 1 M, ph = 14.6 For [AlO - 2 ] = 1 M, ph = 8.6

55 Pourbaix Diagram for Al 55 The Pourbaix diagram shows the ranges of stability for all major phases of a metal in water The diagram shows graphical representations of electrochemical equilibra. Note that Al can corrode in low ph and high ph but not in neutral solutions. Two possible anodic reactions of Al (reactions 1 and 4) are below the water lines. Pourbaix diagram shows the stable phases: Metal Soluble ions Oxide films Ions con =1x10-6 M AL

56 Effect of Concentration 56 Ions con =1x10-6 M Different ions concentrations [AlO 2 - ]=10-2 M

57 Pourbaix diagram and Corrosion 57 Pourbaix diagram shows the stable phases: Metal Soluble ions Oxide films We call these regions as: Immune region (where the metal is stable and corrosion is not possible) Corrosion region (where soluble ions are stable and corrosion is possible) Passive region (where oxide films are stable and the metal is protected by a passive film)

58 Pourbaix Diagram for Iron 58 The shaded region shows the stable region for soluble ions where Fe can corrode. Two possible anodic reactions of Fe are below the water lines.

59 Pourbaix Diagram for Gold 59 Note that the reversible anodic potential for pure gold is above the oxygen line for all ph. Water can not oxidize gold. Thus, gold is immune from corrosion in water at all ph.

60 Pourbaix Diagram for Copper 60 Note that corrosion reactions of copper are above the deaerated water. Hence, copper does not corrode in deaerated water. However, copper does corrode in aerated water.

61 Pourbaix Diagram for Titanium 61 Ti is highly corrosion resistant due to its wide range of passivity.

62 Limitation of Pourbaix diagrams 62 A Pourbaix diagram is potential-ph diagram which shows all the thermodynamic equilibria for a given metal in pure water at 25 C. Engineers can control the potential or the ph of the solution to control corrosion. The vertical axis is varied by: Varying the oxidizer concentration in the solution Applying an external potential to the metal The ph can be varied by changing the acidity of the solution. Limitations: The diagrams are drawn for equilibrium condition, and practical corrosion is far from equilibrium The diagrams are for pure metals in dilute solutions but in practice there are other ions which may affect equilibria The diagram can tell us whether corrosion is possible or not (no indication for corrosion rates)

63 Cathodic Protection of Iron 63 Iron in neutral water (ph 7) develops a potential of -0.5 V, which is inside the corrosion region (Fe +2 ). To protect a piece of iron, we need to force the metal to be in the immune region (reduce the potential). The potential should be reduced to less than -0.6 V (vs. SHE) using an external current supply This is the basic principle of cathodic protection.

64 Pourbaix Diagram for Chromium 64 Cr is added to steel in large amounts (>11%) to make stainless steels. Cr has wide passive region which will make stainless steel more corrosion resistance than steel alloys.

65 Reference electrodes 65 The SHE is complicated setup. There are several other practical reference electrodes: The saturated calomel electrode (SCE) Hg 2 Cl 2 + 2e - 2 Hg + 2 Cl - e = log (Cl - ) e = V vs. SHE Mainly used in chloride containing solutions.

66 Reference electrodes 66 The Silver-silver chloride electrode AgCl + e - Ag + + Cl - e = ph The copper-saturated copper sulfate electrode Cu e- Cu e = log (Cu +2 ) Lead wire to voltmeter Brass washer Brass nut Threaded plastic cap Plastic tube Electrode (copper rod) Electrolyte (saturated copper sulfate solution) Undissolved copper sulfate crystals Porous plug Mud or moist soil

67 Conversion of potentials b/w different reference electrodes 67 Example: If the corrosion potential of steel in seawater is -0.5 V (SCE), what is the potential in the SHE: = V (SHE) in the Ag/AgCl: -0.5-( )= V (Ag/AgCl) Example: If the corrosion potential of Zn in seawater is -0.8 V (SHE), what is the potential in the SCE: = V (SCE) in the Ag/AgCl: = V (Ag/AgCl)

68 Electrolysis 68 We can use external electric current to drive an electrochemical reaction in the reverse of its spontaneous direction. The applied voltage must be greater than the spontaneous E cell. In electrolytic cell, the reduction reaction occurs at the ve terminal of the power supply and the anodic reaction at the +ve terminal. Galvanic Cell Electrolytic Cell