1. 2/18/2015 1 George Mason University General Chemistry 212 Chapter 21 Electrochemistry Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Change, 7 th edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material. Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

2. 2/18/2015 2 Electrochemistry Redox Reactions and Electrochemical Cells  Review of Oxidation Reduction Concepts  Half-Reaction Method for Balancing Redox Reactions  Electrochemical Cells Voltaic Cell: Using Spontaneous Reactions to Generate Electrical Energy  Construction and Operation  Cell Notation  Why Does the Cell Work

3. 2/18/2015 3 Electrochemistry Cell Potential: Output of a Voltaic Cell  Standard Cell Potentials  Strengths of Oxidizing and Reducing Agents Free Energy and Electrical Work  Standard Cell Potential  Effect of Concentration of E cell  Changes in E cell During Cell Operation  Concentration Cells Electrochemical Processes in Batteries  Primary (Nonrechargeable Batteries)  Fuel Cells

4. 2/18/2015 4 Electrochemistry Corrosion: A case of Environmental Electrochemistry  Corrosion of Iron  Protecting Against Corrosion Electrolytic Cells: Using electrical energy to drive Nonspontaneous Reactions  Construction and Operation  Predicting Electrolysis Products  Stoichiometry of Electrolytes

5. 2/18/2015 5 Electrochemistry  The study of the relationship between chemical change (reactions) and the flow of electrons (electrical work) Electrochemical Systems  Electrolytic – Work done by absorbing free energy from a source (passage of an electrical current through a solution) to drive a nonspontaneous reaction  Voltaic/Galvanic – Release of free energy from a spontaneous reaction to produce electricity (Batteries)

6. 2/18/2015 6 Electrochemistry Oxidation-Reduction Concepts Review  Oxidation – Loss of Electrons  Reduction – Gain of Electrons  Oxidizing Agent – Species that causes another species to be oxidized (lose electrons) Oxidizing agent is reduced (gains e - )  Reducing Agent – Species that cause another species to be reduced (gain electrons) Reducing agent is oxidized (loses e - )  Oxidation (e - loss) always accompanies Reduction (e - gain)  Total number of electrons gained by the atoms/ions of the oxidizing agent always equals the total number of electrons lost by the reducing agent

7. 2/18/2015 7 Electrochemistry

8. 2/18/2015 8 Electrochemistry Oxidation Number  A number equal to the magnitude of the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly  The oxidation number in a binary ionic compound equals the ionic charge  The oxidation number for each element in a covalent compound (or polyatomic ion) are assigned according to the relative attraction of an atom for electrons  See next slide for a summary of the rules for assigning oxidation numbers

9. 2/18/2015 9 Electrochemistry Rules for Assigning an oxidation Number (O.N.)

10. 2/18/2015 10 Electrochemistry Balancing Redox Reactions  Oxidation Number Method  Half-Reaction Method The balancing process must insure that: The number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent

11. 2/18/2015 11 Electrochemistry Oxidation Number Method  Assign oxidation numbers to all elements in the reaction  From changes in oxidation number of given elements, identify oxidized and reduced species  For each element that undergoes a change of oxidation number, compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number change (Draw tie-lines between these atoms)  Multiply one or both these number by appropriate factors to make the electrons lost equal to the electrons gained  Use factors as coefficients in reaction equation

12. 2/18/2015 12 Practice Problem Balance equation with Oxidation Number method: -2+5 0+1-2 +2+4+1 Loses 2e - Gains 1e - Accounts for the two electrons needed to balance the 2 electrons from the Copper oxidation

13. 2/18/2015 13 Electrochemistry Half-Reaction Method  Applicable to Acid or Base solutions  Does not usually require Oxidation Numbers (ON) Procedure  Divide the overall reaction into: Oxidation Half-Reaction Reduction Half-Reaction  Balance each half-reaction for atoms & charge  Multiply one or both reactions by some integer to make electrons gained equal to electrons lost  Recombine to given balanced redox equation

14. 2/18/2015 14 Electrochemistry Redox Half-Reaction Method – Example  Divide steps into Half-Reactions

15. 2/18/2015 15 Electrochemistry  Balance Atoms & Charges for Cr 2 O 7 2- / Cr 3+  Balance Atoms & Charges for I - / I 2 Add 7 Water molecules to balance Oxygen Add 14 H + ions on left to balance 14 H on right Add 6 electrons (e - ) on left to balance reaction charges No need to add H 2 O or H + Add 2 electrons (e - ) on right to balance reaction charges (6 electrons gained  this is the reduction reaction -6 + (+14) + (+12 -14) = +6

16. 2/18/2015 16 Electrochemistry Redox Half-Reaction Method – Example (con’t)  Multiply each half-reaction, if necessary, by an integer to balance electrons lost/gained 2 e - lost in oxidation reaction and 6 e - gained in reduction  Multiply oxidation half-reaction by 3 Add 2 half-reactions together

17. 2/18/2015 17 Electrochemistry Half-Reaction Method in a “Basic” solution Sodium Permanganate & Sodium Oxalate NaMnO 4 Na 2 C 2 O 4 Half-Reactions Multiply each reaction by appropriate integer

18. 2/18/2015 18 Electrochemistry Sodium Permanganate & Sodium Oxalate (con’t)  Add reactions  Add OH - to neutralize H +, balance H 2 O, and form “basic” solution

19. 2/18/2015 19 Electrochemistry Electrochemical Cells  Voltaic (Galvanic) Cells Use spontaneous reaction (  G < 0) to generate electrical energy Difference in Chemical Potential energy between higher energy reactants and lower energy products is converted to electrical energy to power electrical devices Thermodynamically - The system does work on the surroundings

20. 2/18/2015 20 Electrochemistry Electrochemical Cells  Electrolytic Cells Uses electrical energy to drive nonspontaneous reaction (  G > 0) Electrical energy from an external power supply converts lower energy reactants to higher energy products Thermodynamically – The surroundings do work on the system Examples – Electroplating and recovering metals from ores

21. 2/18/2015 21 Electrochemistry

22. 2/18/2015 22 Electrochemistry Electrochemical Cells  Cell notation is used to describe the structure of a voltaic (galvanic) cell  For the Zn/Cu cell, the cell notation is: Zn(s)  Zn 2+ (aq) Cu 2+ (aq)  Cu(s)  = phase boundary (solid Zn vs. Aqueous Zn 2+ ) = salt bridge  Anode reaction (oxidation) is left of the salt bridge  Cathode reaction (reduction) is right of the salt bridge  Half-cell components usually appear in the same order as in the half-reactions (Zn(s) + 2e -  Zn 2+ ).  Zinc solid loses 2 e - (oxidized) to produce zinc(II) at the negative ANODE  Copper(II) gains 2e - (reduced) to form copper metal at positive CATHODE

23. 2/18/2015 23 Electrochemistry Voltaic (Galvanic) Cells  Zinc metal (Zn) in solution of Cu ++ ions Construction of a Voltaic Cell  The oxidizing agent (Zn) and reducing agent (Cu 2+ ) in the same beaker will not generate electrical energy  Separate the half-reactions by a barrier and connect them via an external circuit (wire)  Set up salt bridge between chambers to maintain neutral charge in electrolyte solutions

24. 2/18/2015 24 Electrochemistry  Oxidation Half-Cell Anode Compartment – Oxidation of Zinc (An Ox) Zinc metal in solution of Zn 2+ electrolyte (ZnSO 4 ) Zn is reactant in oxidation half-reaction Conducts released electrons (e - ) out of its half-cell  Reduction Half-Cell Cathode Compartment – Reduction of Copper (Red Cat) Copper bar in solution of Cu 2+ electrolyte (CuSO 4 ) Copper metal is product in reduction half-cell reaction Conducts electrons into its half-cell

25. 2/18/2015 25 Electrochemistry Zinc-Copper Voltaic Cell

26. 2/18/2015 26 Electrochemistry Relative Charges on the Anode/Cathode electrodes  Electrode charges are determined by the source of the electrons and the direction of electron flow  Zinc atoms are oxidized (lose 2 e-) to form Zn 2+ at the anode Anode – negative charge (e - rich)  Released electrons flow to right toward cathode to be accepted by Cu 2+ to form Cu(s) Cathode – positive charge (e - deficient)

27. 2/18/2015 27 Electrochemistry Purpose of Salt Bridge  Electrons from oxidation of Zn leave neutral ZnSO 4 solution producing net positive charge  Incoming electrons to CuSO 4 solution would produce net negative charge in solution as copper ions are reduced to copper metal  Resulting charge imbalance would stop reaction  Salt bridge provides “liquid wire” allowing ions to flow through both compartments completing circuit  Salt bridge constructed of an inverted “U-tube” containing a solution of non-reacting Na + & SO 4 2- ions in a gel

28. 2/18/2015 28 Electrochemistry Active vs Inactive Electrodes  Active Electrodes Electrodes in Zn/Cu 2+ cell are active Zinc & Copper bars are components of the cell reactions Mass of Zn bar decreases as Zn 2+ ions in cell solution increase Mass of Copper bar increases as Cu 2+ ions accept electron to form more copper metal

29. 2/18/2015 29 Electrochemistry Active vs Inactive Electrodes  Inactive Electrodes In many Redox reactions, one or the other reactant/product is not capable of serving as an electrode Inactive electrodes - Graphite or Platinum  Can conduct electrons into and out of half- cells  Cannot take part in the half-reactions

30. 2/18/2015 30 Electrochemistry Voltaic Cell with Inactive Graphite Electrodes

31. 2/18/2015 31 Practice Problem A mercury battery, used for hearing aids and electric watches, delivers a constant voltage (1.35 V) for long periods. The half reactions are given below. Which half reaction occurs at the Anode and which occurs at the Cathode? What is the overall cell reaction? HgO(s) + H 2 O(l) + 2e -  Hg(l) + 2 OH - (aq) Zn(s) + 2 OH - (aq)  Zn(OH) 2 (s) + 2e - Ans:Reduction occurs at Cathode (Red Cat) Hg 2+ gains 2 e - (reduced) to form Hg Oxidation occurs at the Anode (An Ox) Zn loses 2 e - (oxidized) to form Zn 2+

32. 2/18/2015 32 Practice Problem Write the cell notation for a voltaic cell with the following cell reaction Ans: Anode (oxidation) is represented on left side of Cell notation Cathode (reduction) is represented on right side of cell notation

33. 2/18/2015 33 Practice Problem Write the cell reaction for the following voltaic cell Pt|H 2 (g) | H + (aq) ║ Br 2 (l) | Br - (aq)|Pt Note: Platinum (Pt) serves as a reaction site at the anode, but does not participate in the reaction Ans:

34. 2/18/2015 34 Electrochemistry Cell Potential  The movement of electrons is analogous to the pumping of water from one point to another Water moves from a point of high pressure to a point of lower pressure. Thus, a pressure difference is required The work expended in moving the water through a pipe depends on the volume of water and the pressure difference

35. 2/18/2015 35 Electrochemistry Cell Potential  Movement of Electrons An electric charge moves from a point of high electrical potential (high electrical pressure) to one of lower electrical potential The work expended in moving the electrical charge through a conductor depends on the potential difference and the amount of charge

36. 2/18/2015 36 Electrochemistry Cell Potential  Purpose of a voltaic cell is to convert the free energy of a spontaneous reaction into the kinetic energy of electrons moving through an external circuit (electrical energy)  Electrical energy is proportional to the difference in the electrical potential between the two cell electrodes Cell Potential

37. 2/18/2015 37 Electrochemistry Cell Potential  Positive Cell Potential – Electrons flow spontaneously from the negative electrode (Anode) to the positive electrode (Cathode)  Negative cell potential is associated with a “nonspontaneous” cell reaction  Cell potential for a cell reaction at equilibrium would be “0”  As with Entropy, there is a clear relationship between E cell, K, and  G

38. 2/18/2015 38 Electrochemistry Units of Cell Potential  The SI (metric) unit of electrical charge is the: Coulomb (C)  The SI (metric) unit of current is the: Ampere (A)  The SI (metric) unit of electrical potential is the: “Volt (V)”  By definition, the energy released by a potential difference of one volt moving between the anode and cathode of a voltaic cell releases 1 joule of work per coulomb of charge

39. 2/18/2015 39 Electrochemistry  The charge (F) that flows through a cell equals the number of moles of electrons (n) transferred times the charge of 1 mol of electrons

40. 2/18/2015 40 Electrochemistry  The total amount of a substance undergoing a ReDox reaction in a cell depends on the total charge flowing through the cell, which is a function of the current (amperes) and the time (seconds)  The ratio of the total charge and the charge per mole of electrons determines the number moles of substance that will react  Thus,

41. 2/18/2015 41 Electrochemistry How long in seconds would it take to deposit 38.93 g of Pb from an aqueous solution of Pb 2+ with a current of 1.679 A?

42. 2/18/2015 42 Electrochemistry Standard Cell Potential  E o cell – The potential measured at a specific temperature (298 K) with no current flowing and all concentrations in their “Standard States” 1 atm for gases 1 M for solutions Pure solids for electrodes

43. 2/18/2015 43 Electrochemistry Standard Electrode Half-Cell Potentials  E o half-cell – Potential associated with a given half-cell reaction (electrode compartment) when all components are in “Standard States”  Standard Electrode Potential for a half-cell reaction, whether anode (oxidation) or cathode (reduction) is written and presented in Appendix D as a “reduction”  Ex. would be written in the table as:

44. 2/18/2015 44 Electrochemistry Standard Electrode Half-Cell Potentials  Electrons flow spontaneously from Anode (negative) to Cathode (positive)  Cathode must have a more “Positive” E o half-cell than the Anode  For a “positive” E o cell ; i.e., a spontaneous reaction  The standard cell potential is the difference between the standard electrode potential of the “Cathode” (reduction) half-cell and the standard electrode potential of the “Anode” (oxidation) half-cell  Standard half-cell potentials are “intensive” properties, thus their values do NOT have to be adjusted for stoichiometry (# of moles)

45. 2/18/2015 45 Practice Problem Write out the overall equation for the cell reaction and determine the standard cell potential for the following galvanic cell [E o (Ag + /Ag) = 0.80 V; E o (Ni 2+ /Ni) = - 0.26 V]

46. 2/18/2015 46 Electrochemistry The Standard Hydrogen Electrode  Half-cell potentials are not absolute quantities  The values found in tables are determined relative to a “Standard”  The Standard Electrode potential is defined as zero (E o reference ) = 0.00  The “standard reference half-cell” is a standard “Hydrogen” electrode  Specially prepared Platinum electrode immersed in a 1 M aqueous solution of a strong acid through which H 2 gas at 1 atm is bubbled

47. 2/18/2015 47 Electrochemistry Reference Half-Cell and Unknown Half-Cell  The “Standard” electrode can act as either the “Anode” or the “Cathode”  Oxidation of H 2 (lose e - ) at anode half-cell and reduction of unknown at cathode half-cell  Reduction of H + (gain e - ) at cathode half-cell and oxidation of unknown at anode half-cell

48. 2/18/2015 48 Practice Problem Determine the standard electrode potential, E o zinc, using a voltaic cell consisting of the Zn/Zn 2+ half-reaction and the H + /H 2 half-reaction. E o cell = + 0.76 V Ans: Zinc is being oxidized (loses 2e - ) producing electrons at the negative anode; H + gains e - at positive cathode

49. 2/18/2015 49 Electrochemistry Relative Strength of Oxidizing and Reducing Agents The more positive the E o value, the more readily the reaction occurs  Strength of Oxidizing Agents – Cu 2+ > H + > Zn 2+  Strength of Reducing Agents – Zn > H 2 > Cu Oxidizing agents decrease in strength as the value of E o decreases, while the strength of the reducing agents increases as the value of E o decreases Cu 2+ is the stronger Oxidizing agent Zn metal (not the ion) is the stronger Reducing agent

50. 2/18/2015 50 Electrochemistry Table of Standard Electrode Potentials (The emf Series)  All Values are relative to the “standard hydrogen (reference) electrode  All reactions are written as “reductions” Selected Standard Electrode Potentials (298 o C)

51. 2/18/2015 51 Electrochemistry EMF Series  All Values are relative to the “standard hydrogen (reference) electrode  All reactions are written as “reductions”  F 2 is the strongest oxidizing agent (high, positive E o ) Fluorine is very electronegative with a high ionization potential (easily reduced by gaining electrons) By gaining e - it forms the weakest reducing agent, F -, which is very reluctant to lose electrons)  Li metal is strongest reducing agent (low, more negative E o ) Lithium has a Low ionization energy and is easily oxidized by losing electrons By losing e-, Lithium forms the weakest oxidizing agent, Li +

52. 2/18/2015 52 Electrochemistry Similarities – Acid/Base vs Redox  Acid Strength vs Base Strength using K a & K b values  Redox (Oxidizing agent vs Reducing agent) using Standard Electrode Potential (E o ) values  Appendix D (Table of Standard Electrode Potentials) The stronger oxidizing agent (species on left side of table) has a half-reaction with a larger more positive (less negative) E o than a species lower in the list The stronger reducing agent (species on the right side of table) has a half-reaction with a smaller (less positive) E o value than a species higher in the list A spontaneous reaction between a metal, acting as a reducing agent by losing electrons (oxidized) to be gained by the ion of another element (reduced) would occur if the E o potential of the metal is less than that of the metal ion.

53. 2/18/2015 53 Electrochemistry Writing Spontaneous Redox Reactions  A spontaneous reaction (E o cell > 0) will occur between an oxidizing agent and any reducing agent that lies below it in the table  Zn(s) (reducing agent (E o = -0.76 V) will react spontaneously with Cu 2+ (oxidizing agent) (E o = +0.34 V), which lies above Zn in the table  The Reduction of Cu 2+ to Cu will occur at the Cathode  The Oxidation of Zn to Zn 2+ will occur at the Anode  A spontaneous reaction will occur when the E o of the reaction at the Cathode (reduction) minus the E o of the reaction at the Anode (oxidation) is > 0 (Positive)

54. 2/18/2015 54 Electrochemistry Writing Spontaneous Redox Reactions  Combing half-cell reactions to form a “Net” reaction In the Standard Electrode Potential table (Appendix D), both reactions are written as “reductions” (e - gain) One of the reactions will have to be reversed so that the applicable reactants are on the left side of the net equation and the products on the right side The sign of E o for the reversed reaction need not be reversed Which reaction to reverse?  Scenario #1 - Net reaction is supplied  Balanced equation clearly indicates which species is oxidized and which species is reduced  Consult table to see which half-cell reaction was reversed (Con’t)

55. 2/18/2015 55 Electrochemistry Writing Spontaneous Redox Reactions (con’t)  Combing half-cell reactions to form a “Net” reaction Net reaction example

56. 2/18/2015 56 Electrochemistry Writing Spontaneous Redox Reactions (con’t)  Combing half-cell reactions to form a “Net” reaction Scenario #2 - Net reaction to be determined from half-cell reactions  Since the net reaction is not known, it is not clear which reaction occurs at a particular electrode  For a spontaneous reaction, the cathode [reduction] potential minus the anode [oxidation] potential must be “positive” (E o cell > 0)  To ensure this, the Anode Oxidation term in the E o cell equation must have an E o value less than the Cathode Reduction term

57. 2/18/2015 57 Electrochemistry Writing Spontaneous Redox Reactions (con’t)  Combing half-cell reactions to form a “Net” reaction Scenario #2 - Net reaction to be determined from half-cell reactions (con’t)  Since -0.14 < +0.80, the Sn 2+ + 2 e - reaction must be reversed (converted to the Anode Oxidation term)  Balance equations to account for reaction coefficients (e - lost = e - gained)

58. 2/18/2015 58 Practice Problem Consult the table of standard electrode potentials in your textbook in order to decide which one of the following reagents is capable of reducing I 2 (s) to I - (aq, 1 M) a. Br - (aq) b. Ag(s) c. Sn(s) d. Zn 2+ (aq, 1 M) e. Sn 4+ (aq,1 M) Br 2 (l) + 2e-  2Br - (aq) +1.07V Ag + (aq) + e -  Ag(s) +0.80V Sn 2+ (aq) + 2e -  Sn(s) -0.14V Zn 2+ (aq) + 2e -  Zn(s) -0.76V Sn 4+ (aq) + 2e -  Sn 2+ (aq) +0.13V I 2 (s) + 2e -  2I - (aq) +0.53V Reduction Form (con’t)

59. 2/18/2015 59 Practice Problem (con’t) The I 2 is being reduced – gaining electrons – at Cathode (Red Cat) The other solution must be capable of being oxidized at Anode (AnOx) Only answers a, b, c reflect oxidizable elements, i.e., can lose electrons a. The Br - solution undergoes oxidation (lose e - ) at the Anode Thus b. The Silver (Ag(s) undergoes oxidation (loses e - ) to form Ag + Thus: (con’t)

60. 2/18/2015 60 Practice Problem (con’t) c.The Sn(s) will undergo oxidation to form Sn 2+ d. The Zn 2+ solution is already oxidized (will not lose any more e - ) Thus: e. The Sn 4+ solution is already oxidized Thus:

61. 2/18/2015 61 Practice Problem Combine the following 3 half-cell reactions into 3 balanced spontaneous reactions; Calculate E o cell for each; rank the relative strengths of the oxidizing & reducing agents (A) (B) (C) 1.Combine & Balance half-reactions A & B Reverse B E o in (A) is more positive than E o in (B) (A – cathode reduction; B – anode oxidation) (A) (B) Con’t

62. 2/18/2015 62 Practice Problem (con’t) 2.Combine & Balance half-reactions A & C Reverse half-reaction (A) ( E o 1.23 > E o 0.96 ) (C – cathode reduction; A – anode oxidation) (C) (A) 3.Combine and Balance half-reactions B & C Reverse half-reaction B (1.23 > -0.23) (C – cathode reduction; B – anode oxidation) (C) (B) Con’t

63. 2/18/2015 63 Practice Problem (con’t) Overall Ranking of Oxidizing & Reducing Agents Rank Oxidizing & Reducing Agents Within Each Equation  Equation 1 (A+B): Oxidizing Agent – NO 3 - > N 2  Reducing Agent – N 2 H 5 + > NO  Equation 2 (C+A): Oxidizing Agent – MnO 2 > NO 3 -  Reducing Agent – NO > Mn +2  Equation 3 (C+B): Oxidizing Agent – MnO 2 > N 2  Reducing Agent – N 2 H 5 + > Mn +2

64. 2/18/2015 64 Electrochemistry Relative Reactivities of Metals  Metals that displace H 2 from acid If the E o cell for the reaction of H + is more positive for metal A than it is for metal B, metal A is a stronger reducing agent than metal B and a more active metal Metals Li through Pb (includes Fe) in the standard electrode potential list (appendix D) lie below H + and give positive E o cell when reducing H + to H 2, i.e., Hydrogen gas is released

65. 2/18/2015 65 Electrochemistry Relative Reactivities of Metals  Metals that cannot displace H 2 from acid Metals that lie above the standard hydrogen reference half-reaction cannot reduce H + from acids The E o cell for the reversed metal half-reaction is negative and the reaction does not occur The higher the metal in the list, the more negative is its E o cell for the reduction of H + to H 2, thus its reducing strength (and reactivity) is less Thus, Gold (Au 3+, E o = +1.5V) is less active than Silver (Ag +, E o = +0.8V) and does not release H 2 gas

66. 2/18/2015 66 Electrochemistry Relative Reactivities of Metals  Metals that displace H 2 from water Metals that lie below the half-cell reaction potential for water can displace H 2 from water In the reaction below the E value for water is not the standard state value listed in the table because in pure water, [OH - ] is 1.0 x 10 -7 M, not the standard state value of 1 M (-0.83 V) E cell > 0  Sodium displaces Hydrogen from water

67. 2/18/2015 67 Electrochemistry Relative Reactivities of Metals  Metals that can displace other metals from solution Any metal that is lower in the standard electrode half-cell list can reduce the ion of a metal that is higher in the list, thus displacing that metal from solution (See next slide and slide #48) E cell > 0 Zinc is the stronger reducing agent reducing Fe 2+ to Fe and displacing it from solution

68. 2/18/2015 68 Electrochemistry

69. 2/18/2015 69 Electrochemistry Free Energy and Electrical Work  Electrical Work Potential (E cell, in volts) times the charge E cell measured with no current flowing No energy lost to heating E cell voltage is maximum possible for cell Work is maximum possible Only reversible process can do maximum work Reversible process with no current flow:  Forward reaction if opposing potential is smaller  Reverse reaction if opposing potential is larger

70. 2/18/2015 70 Electrochemistry  Spontaneous Reaction –  G < 0  Spontaneous Reaction – E cell > 0  The voltaic cell loses energy as it does work on the surroundings; thus the work term (w max ) is negative (Recall slide # 39)

71. 2/18/2015 71 Electrochemistry Electrical Work (con’t)  Relate standard cell potential to equilibrium constant (K) of the redox reaction

72. 2/18/2015 72 Electrochemistry Summary Relationship between  G o E o cell K

73. 2/18/2015 73 Electrochemistry Effect of Concentration on Cell Potential  Most cells do not start with concentrations in their “standard” states  Recall:

74. 2/18/2015 74 Electrochemistry Changes in Potential During Cell Operation  The potential of a cell changes as the concentration of the cell components change

75. 2/18/2015 75 Electrochemistry

76. 2/18/2015 76 Electrochemistry Concentration Cells  In a cell composed of the same substance, but differing concentrations in the two half-cells, the two concentrations move to equilibrate producing electrical energy  The cell reaction is the “sum” of identical half-cell reactions written in opposite directions  The Standard Electrode Potentials (E o cell ) are both based on a 1 M solution (standard conditions), so they “cancel” each other, i. e., E o cell = 0  The non-standard cell potential, E cell, depends on the ratio of the two concentrations [A] dil / [A] conc = Q

77. 2/18/2015 77 Electrochemistry  How the Concentration Cell Works  The dilute solution is in the Anode compartment (oxidation) and the concentrated solution is in the Cathode compartment (Reduction) In the Anode (dilute) half-cell, Cu atoms give up 2 electrons and the resulting Cu 2+ ions enter the solution and make it more concentrated In the Cathode (conc) half-cell, Cu 2+ ions gain 2 electrons and the resulting Cu atoms plate out on the electrode, making the solution less concentrated In this type of Voltaic cell, the dilution continues until equilibrium is attained, i.e., E cell decreases until E cell = 0 (Q = K)

78. 2/18/2015 78 Electrochemistry Alkaline Battery Nickel-Metal Hydride (Ni-MH) Battery Lithium Ion Battery

79. 2/18/2015 79 Practice Problem Calculate E cell for a voltaic cell containing the following half- cells: [Zn 2+ ] = 0.010 M [H + ] = 2.5 M P H 2 = 0.30 atm Construct E o cell Calculate Q

80. 2/18/2015 80 Practice Problem What is the equilibrium constant for the following reaction [E o (Ce 4+ /Ce 3+ ) = 1.72 V E o (Cl 2 /Cl - ) = 1.36 V 2 Cl - (aq) + 2 Ce 4+ (aq)  Cl 2 (g) + 2 Ce 3+ (aq)

81. 2/18/2015 81 Practice Problem Write out the overall equation for the cell reaction and determine the standard cell potential for the following galvanic cell. [E o (Ag + /Ag) = 0.80 E o (Ni 2+ /Ni) = -0.26] Ni(s)|Ni 2+ (aq)║Ag + (aq)|Ag(s)

82. 2/18/2015 82 Practice Problem What is the maximum work you can obtain from 15.0 g of Ni in the galvanic cell shown in the previous problem when the E cell is 0.97 V? [E o (Ag + /Ag) = 0.80 E o (Ni 2+ /Ni) = -0.26] Ni(s)|Ni 2+ (aq)║Ag + (aq)|Ag(s)

83. 2/18/2015 83 Practice Problem What is the cell voltage (E cell ) for the following galvanic cell? Cd(s)|Cd 2+ (0.026 M)║Ni 2+ (0.00420 M)|Ni(s)

84. 2/18/2015 84 Practice Problem Construct a Voltaic Cell to determine the pH of an Unknown Solution Compartment #1 – Cathode consisting of Standard Hydrogen Electrode based on H 2 /H + half-cell reaction at standard conditions (H + – 1 M; H 2 – 1 atm) Compartment #2 – Anode consisting of same apparatus but dipping into a solution of unknown H + (pH) Although E o cell = 0, the individual half-cells differ in [H + ] and E cell is not = 0 Con’t

85. 2/18/2015 85 Practice Problem (Con’t)

86. 2/18/2015 86 Practice Problem What is the pH of the test solution when E cell = 0.612 V at 25 o C? Pt|H 2 (g)(1 atm)|H + (test sol’n)║AgCl(s),Ag(s)|Cl - (2.80 M)

87. 2/18/2015 87 Summary Equations

88. 2/18/2015 88 Summary Equations