Electrochemical Properties of Materials for Electrical Energy Storage Applications

Transcription

1 Electrochemical Properties of Materials for Electrical Energy Storage Applications Lecture Note 2 March 4, 2015 Kwang Kim Yonsei Univ., KOREA kbkim@yonsei.ac.kr 39 Y O N Se I 126.9

2 Electrochemical Cell Electrode/electrolytic solution interface Electrode : A phase containing electrons Electrodics Ionics Electrolytic Solution: A phase containing ions Electron and Ion Charge Transfer Reaction at an electrode solution interface Electrodics: Concerns the region between an electronic and ionic conductor and the transfer of electric charge across it in terms of electrode potential, polarization, charge transfer kinetics Ionics: Concerns ions in solution in terms of material transport, ion distribution ion/solvent,and ion/ion interactions

3 Batteries What happens when a battery operates? - How is a potential difference between +ve and ve terminals developed? -Which reaction gives its unique potential value to a terminal (or electrode)? - What makes it possible to pump electrons from ve electrode to +ve electrode? - Which reaction produces electrons? - Which reaction consumes electrons? - What is a electrical work for such an electron transport from ve electrode to +ve electrode?

4 Reaction at each electrode (half cell reaction) Differing reactivities of metals of Mg and Cu: z + Me nh2o + ze ( Me) Me + nh2o Ions are surrounded by polar water molecular When metals react, they give away electrons and form positive ions. This particular topic sets about comparing the ease with which a metal does this to form hydrated ions in solution - for example, Mg 2+ (aq) or Cu2+ (aq). Half cell reaction written as a reduction reaction

5 Reaction at each electrode (half cell reaction) Differing reactivities of metals of Mg and Cu: When metals react, they give away electrons and form positive ions. This particular topic sets about comparing the ease with which a metal does this to form hydrated ions in solution - for example, Mg 2+ (aq) or Cu2+ (aq).

6 Oxidation and Reduction equilibrium reaction at a single electrode E (volts) The more positive the E value, the further the position of equilibrium lies to the right. That means that the more positive the E value, the more likely the substances on the left-hand side of the equations are to pick up electrons. A substance which picks up electrons from something else is an oxidising agent. The more positive the E value, the stronger the substances on the lefthand side of the equation are as oxidising agents. Chlorine gas is the strongest oxidising agent (E = v). A solution containing dichromate(vi) ions in acid is almost as strong an oxidising agent (E = v). None of these three are as strong an oxidising agent as Au 3+ ions (E = v).

7 Reaction at each electrode (half cell reaction) Differing reactivities of metals: When metals react, they give away electrons and form positive ions. This particular topic sets about comparing the ease with which a metal does this to form hydrated ions in solution - for example, Mg 2+ (aq) or Cu2+ (aq).

8 Oxidation and Reduction equilibrium reaction at a single electrode E (volts) The more positive the E value, the further the position of equilibrium lies to the right. That means that the more positive the E value, the more likely the substances on the left-hand side of the equations are to pick up electrons. A substance which picks up electrons from something else is an oxidising agent. The more positive the E value, the stronger the substances on the lefthand side of the equation are as oxidising agents. Chlorine gas is the strongest oxidising agent (E = v). A solution containing dichromate(vi) ions in acid is almost as strong an oxidising agent (E = v). None of these three are as strong an oxidising agent as Au 3+ ions (E = v).

9 Reaction at each electrode (half cell reaction) In the magnesium case, there is a lot of difference between the negativeness of the metal and the positiveness of the solution around it. In the copper case, the difference is much less. This potential difference could be recorded as a voltage - the bigger the difference between the positiveness and the negativeness, the bigger the voltage. Unfortunately, that voltage is impossible to measure! It would be easy to connect a voltmeter to the piece of metal, but how would you make a connection to the solution? By putting a probe into the solution near the metal? No - it wouldn't work!

10 Reaction at each electrode (half cell reaction) Voltmeter Can you measure the potential difference between the metal and the solution directly? How? V The potential difference between met al and the probe (usually another metal)

11 Reaction at each electrode (half cell reaction) Voltmeter Any probe you put in is going to have a similar sort of equilibrium happening around it. The best you could measure would be some sort of combination of the effects at the probe and the piece of metal you are testing M M z+ M

12 Reaction at each electrode (half cell reaction) Suppose you had an optical device for measuring heights some distance away, and wanted to use it to find out how tall a particular person was. Unfortunately, you can't see their feet because they are standing in long grass. Although you can't measure their absolute height, what you can do is to measure their height relative to the convenient post. Suppose that in this case, the person turned out to be 15 cm taller than the post. You could repeat this for a range of people... Although you don't know any of their absolute heights, you can usefully rank them in order, and do some very simple sums to work out exactly how much taller one is than another. person height relative to post (cm) C +20 A +15 B -15

13 Reaction at each electrode (half cell reaction) This turns out to be exactly what we need to do with the equilibria we started talking about. We don't actually need to know the absolute position of any of these equilibria. Going back to the magnesium and copper equilibria: All we need to know is that the magnesium equilibrium lies further to the left than the copper one. (equilibrium constant) We need to know that magnesium sheds electrons and forms ions more readily than copper does. That means that we don't need to be able to measure the absolute voltage between the metal and the solution. It is enough to compare the voltage measured with a standardised system called a reference electrode. The system used is called a standard hydrogen electrode.

14 Oxidation and Reduction equilibrium reaction at a single electrode E (volts) The more positive the E value, the further the position of equilibrium lies to the right. That means that the more positive the E value, the more likely the substances on the left-hand side of the equations are to pick up electrons. A substance which picks up electrons from something else is an oxidising agent. The more positive the E value, the stronger the substances on the lefthand side of the equation are as oxidising agents. Chlorine gas is the strongest oxidising agent (E = v). A solution containing dichromate(vi) ions in acid is almost as strong an oxidising agent (E = v). None of these three are as strong an oxidising agent as Au 3+ ions (E = v).

15 Reaction at each electrode (half cell reaction) Standard hydrogen electrode As the hydrogen gas flows over the porous platinum, an equilibrium is set up between hydrogen molecules and hydrogen ions in solution. The reaction is catalysed by the platinum. This is the equilibrium that we are going to compare all the others with. Standard conditions: -Hydrogen pressure is 1 bar (100 kpa). (You may find 1 atmosphere quoted in older sources.) -Temperature is 298 K (25 C). - All ions concentrations are taken as being 1 mol dm -3.

16 The copper is the more positive (less negative) electrode. The voltage measured would be 0.34 volts The voltmeter will show the hydrogen electrode as the negative one and the copper electrode as positive. Reaction at each electrode (half cell reaction) There is a major difference between the charge on the two electrodes - a potential difference which can be measured with a voltmeter. The voltage measured would be 2.37 volts and the voltmeter would show the magnesium as the negative electrode and the hydrogen electrode as being positive.

17 Reaction at each electrode (half cell reaction) voltmeter You may have noticed that the voltmeter was described as having a "high input resistance". Ideally, it wants to have an infinitely high input resistance. This is to avoid any flow of current through the circuit. If there was a low resistance in the circuit, electrons would flow from where there are a lot of them (around the magnesium, for example) to where there are less (on the hydrogen electrode). If any current flows, the voltage measured drops. In order to make proper comparisons, it is important to measure the maximum possible voltage in any situation. This is called the electromotive force or emf. The emf of a cell measured under standard conditions is given the symbol E cell. You read E as "E nought" or "E standard".

18 Reference electrodes Standard Hydrogen Electrode (SHE), or Normal Hydrogen Electrode Electrode (NHE) E 0 = 0 V ASSUMED Saturated Calomel Electrode (SCE) E 0 = V Silver-silver Chloride Electrode E 0 = V

19 Ideal polarizable electrode vs. ideal nonpolarizable electrode i i E E An ideal polarized electrode shows a very large change in potential upon the passage of a small current. Ideal polarizability is characterized by a horizontal region of an i-e curve Ideal nonpolarizable electrode is an electrode whose potential does not change upon passage of current. Nonpolarizability is characterized by a vertical region on an i-e curve.

20 Oxidation and Reduction equilibrium reaction at a single electrode E (volts) The more positive the E value, the further the position of equilibrium lies to the right. That means that the more positive the E value, the more likely the substances on the left-hand side of the equations are to pick up electrons. A substance which picks up electrons from something else is an oxidising agent. The more positive the E value, the stronger the substances on the lefthand side of the equation are as oxidising agents. Chlorine gas is the strongest oxidising agent (E = v). A solution containing dichromate(vi) ions in acid is almost as strong an oxidising agent (E = v). None of these three are as strong an oxidising agent as Au 3+ ions (E = v).

21 REDOX POTENTIALS Combining a zinc with a copper half cell So far, we have looked at combinations of a hydrogen electrode with the half cell we have been interested in. What happens if you combine a zinc half cell with a copper half cell? In the presence of a high resistance voltmeter The two equilibria which are set up in the half cells are: The negative sign of the zinc E value shows that it releases electrons more readily than hydrogen does. The positive sign of the copper E shows that it releases electrons less readily than hydrogen. That means that you can compare any two equilibria directly. For example, in this case you can see that the zinc releases electrons more readily than the copper does - the position of the zinc equilibrium lies further to the left than the copper equilibrium.

22 REDOX POTENTIALS Stripping everything else out of the diagram, and looking only at the build up of electrons on the two pieces of metal: Removing the voltmeter The high resistance of the voltmeter is deliberately designed to stop any current flow in the circuit. What happens if you remove the voltmeter and replace it with a bit of wire? Electrons will flow from where there are a lot of them (on the zinc) to where there are fewer (on the copper). The movement of the electrons is an electrical current. Zn Cu

23 REDOX POTENTIALS Electrons are flowing away from the zinc equilibrium. According to Le Chatelier's Principle, the position of equilibrium will move to replace the lost electrons. Electrons are being dumped onto the piece of copper in the copper equilibrium. According to Le Chatelier's Principle, the position of equilibrium will move to remove these extra electrons. Zn Cu If electrons continue to flow, the positions of equilibrium keep on shifting. The two equilibria essentially turn into two oneway reactions. The zinc continues to ionise, and the copper(ii) ions keep on picking up electrons.

24 REDOX POTENTIALS What occurs when you drop a piece of zinc into some copper(ii) sulphate solution? The blue color of the copper sulphate solution fades as the copper(ii) ions are converted into brown copper metal. The final solution contains zinc sulphate. (The sulphate ions are spectator ions.) You can add the two electron-half-equations above to give the overall ionic equation for the reaction The only difference in this case is that the zinc gives the electrons directly to the copper(ii) ions rather than the electrons having to travel along an external circuit.

25 Reaction at each electrode (half cell reaction) Cell conventions

26 Reaction at each electrode (half cell reaction) Remember that the standard electrode potential of a metal / metal ion combination is the emf measured when that metal / metal ion electrode is coupled to a hydrogen electrode under standard conditions.

27 Electrochemical Cell Anion : -ve ion Cation : +ve ion Anode : electrode for oxidation reaction Cathode : electrode for reduction reaction

28 Galvanic or Voltaic cells Chemical reaction : material transformation Electrochemical reaction : material transformation and electron flow Galvanic cell : electron as a product Use a chemical reaction (material transformation) to generate electrical energy Ex) discharging of a secondary battery Electrolytic cell: electron as a reactant Use electrical energy to drive a chemical reaction (material transformation) Ex) Charging of a secondary battery

29 Electrochemical Cell : Galvanic cell Electric load Electron flow Current flow Electron flow -ve +ve Zn Zn e Electron generation Anodic oxidation Cu e Cu Electron consumption Cathodic reduction

30 Electrochemical Cell : Electrolytic cell DC power supply Electron flow Current flow Electron flow -ve +ve Zn e Zn Electron consumption Cathodic reduction Cu Cu e Electron generation Anodic oxidation

31 Galvanic cell vs. Electrolytic cell Galvanic cell Electrolytic cell Electric load DC power supply Electron flow Current flow Electron flow Electron flow Current flow Electron flow -ve +ve -ve +ve Zn Zn e Cu e Cu Zn e Zn Cu Cu e Electron generation Electron consumption Electron consumption Electron generation Anodic oxidation Cathodic reduction Cathodic reduction Anodic oxidation Electrochemical cell determines the polarity of electrodes. DC power supply determines the polarity of electrodes.

32 Electrolytic cell I net = i c - i a Electrolytic cell Charge Electron flow DC power supply Current flow Electron flow E o Cu2+/Cu potential Cell voltage +ve Anodic polarization -ve Cathodic polarization -ve +ve E o Zn2+/Zn Current Change in electrode potential with current during galvanostatic discharge Zn e Zn Electron consumption Cathodic reduction Cu Cu e Electron generation Anodic oxidation DC power supply determines the polarity of electrodes.

33 Galvanic cell Galvanic cell Discharge I net = i c - i a Electric load Electron flow Current flow Electron flow -ve Cathodic polarization -ve +ve E o Cu2+/Cu potential Cell voltage Zn Zn e Electron generation Anodic oxidation Cu e Cu Electron consumption Cathodic reduction Electrochemical cell determines the polarity of electrodes. E o Zn2+/Zn Current Change in electrode potential with current during galvanostatic discharge +ve Anodic polarization

34 Electrochemical series equilibrium E (volts) Reduction reaction Equilibrium potential measured vs. SHE (Standard Hydrogen Electrode) The more negative the E value, the more the position of equilibrium lies to the left - the more readily the metal loses electrons. The more negative the value, the stronger reducing agent the metal is. The more positive the E value, the more the position of equilibrium lies to the right - the less readily the metal loses electrons, and the more readily its ions pick them up again. The more positive the value, the stronger oxidising agent the metal ion is.

35 Nernst Equation O + ne = R E = E o - (RT/nF) (a R /a O ), E o =? E = E o - (RT/nF) (C R /C O ), E o = formal potential If the system follows the Nernst equation, the electrode reaction is often said to be thermodynamically or electrochemically reversible (or nernstian). Reversibility of a process ; one s ability to detect the signs of disequilibrium Rate of change of force driving the observed process vs. speed with which the system can reestablish equilibrium If the perturbation applied to the system is small enough, or if the system can attain equilibrium rapidly enough compared to the measuring time, thermodynamic relation will apply.

36 Electrochemical Cell Anion : -ve ion Cation : +ve ion Anode : electrode for oxidation reaction Cathode : electrode for reduction reaction

37 Galvanic or Voltaic cells Chemical reaction : material transformation Electrochemical reaction : material transformation and electron flow Galvanic cell : electron as a product Use a chemical reaction (material transformation) to generate electrical energy Ex) discharging of a secondary battery Electrolytic cell: electron as a reactant Use electrical energy to drive a chemical reaction (material transformation) Ex) Charging of a secondary battery

38 Nernst Equation Nernst equation for an electrode reaction of O + ne = R as: G = G 0 + RT ln ((a (Reductant) /a (Oxidant) ) - G = zf E, - G 0 = zf E 0 E = E 0 - (RT/nF) ln (a (Reductant) /a (Oxidant) ) Other thermodynamic quantities can be derived from electrochemical measurement. G = H- T S, dg = VdP S dt S = -(δ G/δT) p S = nf (δe cell /δt) p H = G + T S = nf [T(δE cell /δt) p -E cell ] RT ln K rxn = - G 0 = zf E 0

39 The first law of thermodynamics The total energy of a system : Internal energy, U The absolute value of the total energy of a system cannot be known. For an isolated system, the total energy of the system remains constant. For an infinitesimal change, du = 0 When the total energy of a system changes, the energy change goes into either work or heat. The change in the internal energy for a process is equal to the work plus the heat involved during the process. Heat flow into the system: +ve, heat flow out of the system : -ve

40 Work

41 Free Energy, du = dq + dw dw = - P ext dv + dw net

42 Gibbs Free Energy H = U + PV dh = du + PdV + VdP

43 Gibbs Free Energy

44 Gibbs Free Energy dw = - P ext dv + dw net

45 Work When a force, F, acts on a particle, work is done on the particle in moving from point a to point b W a b b = F dl If the force is a conservative, then the work done can be expressed in terms of a change in potential energy a W a b = ( U U ) = U b a Also if the force is conservative, the total energy of the particle remains constant KE + PE = KE + a a b PE b

46 Electric Work Work Done by Uniform Electric Field Force on charge is F = q 0 E Work is done on the charge by field W = Fd = q a b 0 Ed The work done is independent of path taken from point a to point b because the Electric Force is a conservative force.

47 Electric Work and Electric Potential Energy The work done by the force is the same as the change in the particle s potential energy W a b = ( U U ) = U b a U b U a = b a F ds = qe uniform The work done only depends upon the change in position ( y y ) b a

48 Electric Potential Energy General Points 1) Potential Energy increases if the particle moves in the direction opposite to the force on it Work will have to be done by an external agent for this to occur 2) Potential Energy decreases if the particle moves in the same direction as the force on it

49 Electric Force and Electric work Suppose we have two charges q and q 0 separated by a distance r The force between the two charges is given by Coulomb s Law F = 1 4π ε qq r 0 We now displace charge q 0 along a radial line from point a to point b 0 2 The force is not constant during this displacement W a b r r = Frdr = r b a r b a 1 qq0 qq dr = 0 4π ε 2 0 r 4π ε0 1 r a 1 r b

50 Electric Potential Energy The work done is not dependent upon the path taken in getting from point a to point b The work done is related to the component of the force along the displacement F dr

51 Electric Potential Energy Looking at the work done we notice that there is the same functional at points a and b and that we are taking the difference W a b = qq0 π ε r a 1 r b We define this functional to be the potential energy 1 4π ε qq r U 0 = 0 The signs of the charges are included in the calculation The potential energy is taken to be zero when the two charges are infinitely separated

52 Electric Potential Energy Given several charges, q 1 q n, in place Now a test charge, q 0, is brought into position Work must be done against the electric fields of the original charges This work goes into the potential energy of q 0 We calculate the potential energy of q 0 with respect to each of the other charges and then Just sum the individual potential energies Remember - Potential Energy is a Scalar 1 q q PEq = 0 0 4π ε r i 0 i i

53 Electric Potential The potential energy of the test charge, q 0, was given by 1 q q PE = 0 q0 4π ε r i Notice that there is a part of this equation that would remain the same regardless of the test charge, q 0, placed at point a 0 i i The value of the test charge can be pulled out from the summation PE q 1 = q 0 0 4π ε i 0 q r i i

54 Electric Potential We define the term to the right of the summation as the electric potential at point a Potential Like energy, potential is a scalar. a = i 1 4πε We define the potential of a given point charge as being 0 q r i i Potential = V = 1 4πε 0 q r This equation has the convention that the potential is zero at infinite distance

55 Electric Potential Energy The potential at a given point Represents the potential energy that a positive unit charge would have, if it were placed at that point It has units of Volts Energy = charge = joules coulomb The Potential increases if you move in the direction opposite to the electric field The Potential decreases if you move in the same direction as the electric field

56 Electric Potential Energy The work done by the electric force in moving a test charge from point a to point b is given by = = b a b a b a l d E q dl F W 0 Dividing through by the test charge q 0 we have = b a b a l d E V V = b a a b l d E V V Rearranging so the order of the subscripts is the same on both sides

57 Electric Potential From this last result We get dv = E dl or V b dv dx V a = E b = E dl We see that the electric field points in the direction of decreasing potential a

58 Electric Potential Energy There is an additional unit that is used for energy in addition to that of joules A particle having the charge of e (1.6 x C) that is moved through a potential difference of 1 Volt has an increase in energy that is given by W = qv = joules = 1eV

59 Gibbs Free Energy dw = - P ext dv + dw net At equilibrium, dg = dw net = - E dq

60 Nernst Equation At equilibrium, dg = dw net = - E dq In materials electrochemistry, work done by the system on the surrounding in transferring a n moles of electrons across a potential difference E is expressed as a positive work G = -[-E (-nf)] = -nfe equilibrium E (volts)

61 Nernst Equation

62 Nernst Equation G = -nfe Take the expression for the Gibbs dependence on activity and turn this around for an expression in terms of the cell potential. G = G + RT ln Q = -nfe The relation between cell potential E and free energy gives n F E = n F E + RT ln Q Rearrange and obtain the Nernst Equation. E =E RT nf ln Q

63 Nernst Equation The Nernst equation relates the potential of a cell in its standard state to that of a cell not in its standard state. Zn e Zn(s) 2Zn e 2Zn(s) E = E o RT 2F ln Α Α Zn(s) Zn 2+ E E : intrinsic property like T and P = E o RT 4F ln Α Α 2 Zn(s) 2 Zn 2+ We know from Le Chatelier s principle that increasing the concentration of Zn 2+ should drive the reaction to the right. How does the potential of the half cell changes? The Nernst equation allows us to calculate this increase for the above half reaction as

64 Chemical Equilibrium in Gas Phase Mixture Le Chatelier s Principle LeChatlier s Principle states that a system at equilibrium when stressed (a stress could be an increase or decrease in temperature, pressure, volume, number of moles of reactants or products, etc.) will establish a new equilibrium in a way that will reduce the applied stress.

65 Nernst Equation Nernst equation for an electrode reaction of O + ne = R as: G = G 0 + RT ln (a (Reductant) /a (Oxidant) ) - G = zfe, - G 0 = zfe 0 E = E 0 - (RT/nF) ln (a (Reductant) /a (Oxidant) ) where R is the universal gas constant, T is the absolute temperature, n is the charge number of the electrode reaction (which is the number of moles of electrons involved in the reaction as written), and F is the Faraday constant. The charge on a mole of electrons can be calculated from Avogadro's number and the charge on a single electron. E = E 0 - ( /n) ln (a (Reductant) /a (Oxidant) ) As a (Reductant) increases, E decreases. (or increases to ve direction) As a (Oxidant) increases, E increases. (or increases to +ve direction)

66 Nernst Equation For the Daniel cell at 25C, Zn + Cu 2+ = Zn 2+ + Cu What happen to the cell potential as the reaction proceeds? When the reaction quotient is very small, the cell potential is positive and relatively large, because the reaction is far from equilibrium and the driving force behind the reaction should be relatively large. When the reaction quotient is very large, the cell potential is negative. This means that the reaction would have to shift back toward the reactants to reach equilibrium.

67

68 Φ (Galvani potential) Ψ (Volta potential) χ (surface potential)

69 Volta potential (ϕ) or outer potential is the part of total potential difference across metal-electrolyte interface. Defined as the work required to bring a unit point charge from infinity to a point just outside the surface of the phase by a distance of about 10 5 ~ 10 3 cm from surface. Outer potential is a measurable quantity.

70 Galvani potential (φ) or inner potential is the electrostatic potential which is actually experienced by a charged particle inside the phase. Defined as the work required to bring a unit point charge from infinity to a point inside the phase. Inner potential is not measurable.

71 The inner and outer potential differ by the surface potential (χ) Caused by an inhomogeneous charge distribution at the surface Can not be measured

72 Galvani potential and Electrochemical potential Zero energy charged particle in vacuum at infinite separation from a charged phase 1. Electrochemical potential (μ) work done when a charged particle is transferred from infinite separation in vacuum to the interior of a charged phase 2. Chemical potential (μ) work done when an uncharged particle is transferred from infinite separation in vacuum to the interior of an uncharged phase stripped from the charged surface layer 3. Inner potential (Galvani potential, φ) work done when a charged particle is transferred from infinite separation in vacuum across a surface shell which contains an excess charge and oriented dipoles

73 Electrochemical potential µ α + z e φ α = µ α i i 0 i Electrochemical potential z i is the charge on species i, φ, the inner potential, is the potential of phase α. 1) If z = 0 (species uncharged) 2) for a pure phase at unit activity 3) for species i in equilibrium between α and β. µ α µ α i = i µ α i = µ 0, α i µ α µ β i = i

74 Inner, outer and surface potential (1) Potential in vacuum: the potential of certain point is the work done by transfer unit positive charge from infinite to this point. (Only coulomb force is concerned). x φ = Fdx = εdx x F = eε ε - strength of electric field

75 (2) Potential of solid phase Electrochemical reaction can be simplified as the transfer of electr on from species in solution to inner part of an electrode charged sphere Vacuum, infinite This process can be divided into two separated steps ~ 10-7 m W 2 + µ W 1

76 (3) Inner, outer and surface potential W 2 + µ 10-6 ~ 10-7 m W 1 The work (W 1 ) done by moving a test charge from infinite to 10-6 ~ 10-7 m vicinity to the solid surface (only related to long-distance force) is outer potential. Outer potential also termed as Volta Potential (Ψ) is the potential measured just outside a phase. Moving unit charge from vicinity (10 6 ~10-7 m) into inner of the sphere overcomes surface potential (χ). Short-distance force takes effect. W 2 For hollow ball, µ can be excluded. χ arises due to the change in environment experienced by the charge (redistribution of charges and dipoles at the interface)

77 10-6 ~ 10-7 m W 2 W 1 The total electrical work done for moving unit charge to the inner charged sphere is W 1 + W 2 φ = (W 1 + W 2 ) / z e 0 = ψ + χ The electrostatic potential within a phase termed the Galvani potential or inner potential (φ). If short-distance interaction, i.e., chemical interaction, is taken into consideration, the total energy change during moving unite test charge from infinite to inside the sphere: chemical potential

78 infinite distance 10-6 ~10-7 m hollow Work function inner

79 work function the minimum energy (usually measured in electron volts) needed to remove an electron from a solid to a point immediately outside the solid surface or energy needed to move an electron from the Fermi energy level into vacuum.

80 The potential difference between the solution and metal: φ = φ S φ M φ: the inner or Galvani potentials of the two phases φ = χ + ψ ψ : the Volta potential; the long range coulombic forces near the electrode χ : surface potential; determined by short range effects of adsorbed ions and oriented water molecules. ψ can be measured directly. χ cannot. Galvani potential can only be measured relative to a refer ence electrode.

81 Cu(s) + 2 Ag + (aq) = Cu 2+ (aq) + 2 Ag(s), E o = 0.46 V Anodic oxidation Cathodic Reduction Cu/Cu 2+ //Ag + /Ag/Cu or Ag /Cu/Cu 2+ //Ag + /Ag Cathode half cell rxn : Ag + + e (Cu ) = Ag Anode half cell rxn : Cu e (Cu) = Cu Note that all half cell rxns are written in reduction rxn. Overall unit cell rxn = [Cathode half cell rxn] [Anode half cell rxn] Unit cell voltage = [Cathode half cell rxn potential] [Anode half cell rxn potential] 2Ag + + 2e (Cu ) + Cu = 2Ag + Cu e (Cu) Electrochemical equilibrium?

82 Cu(s) + 2 Ag + (aq) = Cu 2+ (aq) + 2 Ag(s), E o = 0.46 V Cu/Cu 2+ //Ag + /Ag/Cu Overall unit cell rxn = [Cathode half cell rxn] [Anode half cell rxn] Unit cell voltage = [Cathode half cell rxn potential] [Anode half cell rxn potential] 2Ag + + 2e (Cu ) + Cu = 2Ag + Cu e (Cu) Electrochemical equilibrium?