Voltaic Cells. An Energizing Experience. Using the energy of a spontaneous redox reaction to do work. aka Galvanic Cells.

Transcription

1 Voltaic Cells aka Galvanic Cells Using the energy of a spontaneous redox reaction to do work. Chapter 20 An Energizing Experience 1

2 A piece of copper is dropped into an aqueous solution of zinc nitrate, and in an other beaker a piece of zinc is dropped into copper(ii) nitrate. 1. A reaction occurs in both situations 2. A reaction occurs in only one of the situations 2 2

3 Complete the possible reactions A piece of copper dropped into aqueous zinc nitrate Cu (s) + Zn(NO 3 ) 2(aq) A piece of zinc dropped into aqueous copper(ii) nitrate Zn (s) + Cu(NO 3 ) 2(aq) 3 3

4 Activity Series You probably remember using an activity series like this in first year chemistry. TABLE 4.5 Activity Series of Metals in Aqueous Solution Metal Lithium 1 2 Potassium Barium1 2 Calcium1 2 Sodium Magnesium Aluminum Manganese Zinc Chromium Iron Cobalt Nickel Tin Lead Hydrogen Copper Silver Mercury Platinum Gold Increasing Activity 4 4

5 A piece of copper is dropped into an aqueous solution of zinc nitrate, and in an other beaker a piece of zinc is dropped into copper(ii) nitrate. 1. A reaction occurs in both situations 2. A reaction occurs in only one of the situations Since one metal is more active than the other 5 5

6 Spontaneous or Not? A piece of copper dropped into aqueous zinc nitrate Cu (s) + Zn(NO 3 ) 2(aq) No Reaction This reaction did NOT occur Cu is LESS active than Zn. This reaction is NOT spontaneous aka thermodynamically unfavorable. A piece of zinc dropped into aqueous copper(ii) nitrate Zn (s) + Cu(NO 3 ) 2(aq) Zn(NO 3 ) 2(aq) + Cu (s) This reaction DID occur Zn is more active than Cu. Zn has a greater potential to lose its electrons than Cu This reaction IS spontaneous aka thermodynamically favorable. 6 6

7 Let s write balanced net ionic equations for this reaction A piece of zinc dropped into aqueous copper(ii) nitrate Zn (s) + Cu(NO 3 ) 2(aq) Zn(NO 3 ) 2(aq) + Cu (s) 7 7

8 Let s write balanced net ionic equations for this reaction A piece of zinc dropped into aqueous copper(ii) nitrate Zn (s) + Cu(NO 3 ) 2(aq) Zn(NO 3 ) 2(aq) + Cu (s) Zn + Cu 2+ Zn 2+ + Cu 8 8

9 This single replacement reaction is really made up of two half reactions, one oxidation, the other reduction. Net Ionic: Zn + Cu 2+ Zn 2+ + Cu The reaction above, separated into two half reactions: Oxidation: Zn Zn e Reduction: Cu e Cu 9 9

10 We can separate these two reactions and force the electrons to transfer through an external wire. The two half reactions each occur in their own compartment. Net Ionic: Zn + Cu +2 Zn +2 + Cu Zn Zn e Cu e Cu We can use this natural potential of the movement of electrons to do work. By moving the electrons through an external channel instead of allowing the electrons to transfer directly. This is called a voltaic (or galvanic) cell, aka battery

11 A schematic of the electrochemical cell The two half reactions occur each in their own compartment. No voltage on the meter. 0.0 Write the half reaction that occurs in each compartment

12 When the electrons start to flow, in what direction will they move through the wire? 1. Left to right 2. Right to left 0.0 Cu e Cu Zn Zn e 12 12

13 When the electrons start to flow, in what direction will they move through the wire? 1. Left to right 2. Right to left Electrons need to move from the zinc (since zinc atoms are oxidizing) to the copper electrode so that the copper ions can be reduced. Cu e Cu 0.0 Zn Zn e 13 13

14 Remember what ions are in the left compartment: Cu 2+ and 2 NO3 When the electrons start to flow, what will happen to the balance of charge in the left side solution? (the amount of + compared to ) 1. The solution will become more positive. 2. The solution will become less positive (more negative) 3. The charge balance will stay the same. Cu e Cu 0.0 Zn Zn e 14 14

15 Remeber what ions are in the left compartment: Cu 2+ and 2 NO3 When the electrons start to flow, what will happen to the balance of charge in the solution? (the amount of + compared to ) 1. The solution will become more positive. 2. The solution will become less positive (more negative) This is a fundamental problem, and will stop the reaction in its tracks, because the imbalance of charge will not allow any more electrons to move. Thus the voltage reads zero 3. The charge balance will stay the same. Cu e Cu 0.0 Zn Zn e 15 15

16 A schematic of the electrochemical cell We need to provide a way for the charge balance to be maintained. We provide a salt bridge. Usually a non-reacting salt like KNO3 or NaNO3. As the left solution loses Cu 2+ ions, thus less positive, K + ions flow into the left solution. Cu e Cu As the right solution gets more Zn 2+ ions in the solution, thus more positive, NO3 ions will flow into the right solution. 0.0 salt bridge Zn Zn e 16 16

17 Let s measure The two half reactions each occur in their own compartment. Net Ionic: Zn + Cu +2 Zn +2 + Cu Zn Zn e Cu e Cu We can run the electrons through a meter to measure the amount of voltage This voltage is a measure of the electrical potential difference between the two reactions involved, or the electric potential energy. The electromotive force, emf

18 Inspection of a Typical Voltaic Cell Zn Zn +2 Cu +2 Cu 18 18

19 Important Features of the Voltaic Cell One half-cell contains the chemicals that involve the oxidation half reaction. The other half-cell contains chemicals that involve the reduction half reaction. A solid electrical connector, called an electrode is placed in each half-cell compartment. This provides a means for the electrons to travel through the external pathway

20 Electrodes The electrode may or may not actually participate in the reaction. If the electrode is not participating, it is said to be an inert or passive electrode. In this reaction the Cu is a passive (inert) electrode If the electrode is participating (i.e. metal is a reactant), it is said to be an active electrode. Reduction occurs at the cathode. (Red Cat) Oxidation occurs at the anode. (An Ox) 20 20

21 As the reaction proceeds 1.10 salt bridge 1. The mass of both electrodes change. Cu e Cu Zn Zn e 2. The mass of only one electrod changes. 3. The mass of both electrodes stay the same

22 During the running of the cell, where does the mass of the zinc electrode go? 1.10 salt bridge 1. through the salt bridge and onto the Cu electrode Cu e Cu Zn Zn e 2. into the solution as Zn 2+ ions 3. across the external wire as electrons to hook up with Cu 2+ ions 22 22

23 Where does the mass of the zinc electrode go? 1.10 salt bridge 1. through the salt bridge and onto the Cu electrode Cu e Cu Zn Zn e 2. into the solution as Zn 2+ ions 3. across the external wire as electrons to hook up with Cu 2+ ions 23 23

24 As the reaction proceeds 1.10 salt bridge 1. The mass of both electrodes change. Cu e Cu Zn Zn e 2. The mass of only one electrode changes. 3. The mass of both electrodes stay the same

25 The mass of the copper electrode increases 1.10 salt bridge 1. the same that the Zn electrode decreases. 2. Cu e Cu Zn Zn e more than the Zn electrode decreases. 3. less than the Zn electrode decreases. 4. hmmm.it is impossible to know whether the mass change would be the same or different. 5. Actually the mass of the electrodes do NOT change at all due to the law of conservation of matter

26 The mass of the copper electrode increases 1.10 salt bridge 1. the same that the Zn electrode decreases. Cu e Cu Zn Zn e 2. more than the Zn electrode decreases 3. less than the Zn electrode decreases 4. hmmm.it is impossible to know whether the mass change would be the same or different. 5. Actually the mass of the electrodes do NOT change at all due to the law of conservation of matter

27 During the reaction, the cathode increased by 0.64 g, 0.01 mol of Cu atoms are deposited on the cathode salt bridge x electrons are transferred and the mass of Cu the anode would decrease e Cu by 0.64 g x electrons are transferred and the mass of the anode would decrease by 0.64 g Zn Zn e x electrons are transferred and the mass of the anode would decrease by 0.65 g x electrons are transferred and the mass of the anode would decrease by 0.65 g 27 27

28 During the reaction, the cathode increased by 0.64 g, mol of Cu atoms are deposited on the cathode salt bridge x electrons are transferred and the mass of Cu the anode would decrease e Cu by 0.64 g x electrons are transferred and the mass of the anode would decrease by 0.64 g Zn Zn e x electrons are transferred and the mass of the anode would decrease by 0.65 g x electrons are transferred and the mass of the anode would decrease by 0.65 g 28 28

29 The mass changes for the Cu/ Zn cell are not different by very much because the molar masses of copper and zinc are very similar, and every time one ion of Cu is reduced, one atom of Zn is oxidized V salt bridge Cu e Cu Al Al e If the Zn electrode were replaced by Al the molar masses are quite different. a different number atoms are oxidized / reduced

30 During the reaction, the cathode increased by 0.64 g, and the mass of the anode would decrease by V salt bridge g g g g Cu e Cu Al Al e g g g 30 30

31 During the reaction, the cathode increased by 0.64 g, and the mass of the anode would decrease by V salt bridge g g g g Cu e Cu Al Al e g g g g g 0.64gCu 1mol 63.55g 2Al 3Cu 24.3g 1mol = 0.163g 31 31

32 What about the e- and ions? The electrons flow from the anode to the cathode through the external pathway. This of course is the wire. If those electrons are run through an appliance, they can be used to do work

33 Movement of Electrons and Ions Electrons hop off the substance that is being oxidized and move through the wire. Electrons flow from the anode to the cathode through the wire that connects them. Electrons are presented at the cathode and join with the substance that is being reduced. Positive ions in solution move away from the anode and towards the cathode. Negative ions in solution move away from the cathode and towards the anode

34 Balance of Charge When the reaction starts, the flow of electrons would stop as the charge in the solution becomes imbalanced. There must be some method by which each cell can maintain a net charge of zero. A salt bridge or a semi-permeable membrane can provide the vehicle by which ions can flow into and out of the cells without allowing a quick mixing of the chemicals

35 Electromotive Force The driving force that makes these reactions go, is the potential of these electrons to move we call this the electromotive force aka: emf symbolized as E Eº (naught) means at standard conditions (1 atm, 1 M) E is measured in volts 1 V = 1 J/coulomb One volt is the potential difference required to give one Joule of energy to one coulomb of charge (a chunk of charge). Let s look at the reduction table which has recorded potentials measured at 25ºC

36 Let s calculate the theoretical potential for the Zinc/Copper cell. Use the reduction potential table to look up the Reduction Potential for each half reaction. Net Ionic: Zn + Cu +2 Zn +2 + Cu Zn e Zn Eºred = 0.76 V Cu e Cu Eºred = V One half reaction is oxidized. Flip the reaction = change the sign Zn Zn e Eºox = V Add Eºred + Eºox = Eºcell 36 36

37 Electromotive Force for a Voltaic Cell The Eº of the oxidation is best called Eº ox To determine Eº oxidation you should reverse the sign of the Eº reduction for the substance that is oxidized. To determine the Ecell, simply add the Eº reduction and Eº oxidation of the two half reactions. Your text book will show this equation: Eº cell = Eº red - Eº red (of the reduction that occurs at the cathode) (of the oxidation that occurs at the anode) Standard potentials are intensive because the Eº measures the potential energy per electrical charge. They are NOT affected by the stoichiometry used to balance the redox equation

38 Using emf to Predict Spontaneity The more positive the standard reduction potential, the greater the tendency for that reduction half-reaction to occur. A voltaic cell will be spontaneous if Eºcell is positive

39 LAD B3 Voltaic Cells 39

40 Remember... When the red is connected to the copper (reduction) the voltage is + When the black is connected to the copper, the voltage is The voltmeter is set up to detect a positive voltage when the red wire is connected to the reduction reaction. 1.1 V salt bridge Cu e Cu Zn Zn e 1.1 V salt bridge Cu e Cu Zn Zn e 40 40

41 Using the multimeters to measure volts Some have on off, some do not. Be sure the switch is on DC (not AC) Turn the dial (sometimes the on/off switch) to the 2 V setting. The red wire is the end at which reduction will occur. The black wire is where oxidation will occur. 2V 2V 41 41

42 Post LAD B4 SHE Standard Hydrogen Electrode 2H + Find the only voltage on the reduction potential table with a value of zero. 2e H 2( ) 2+ g

43 How to Measure Ered for Half-cells? We have seen how we can measure the potential of a cell (Ecell), but if reduction can t take place without oxidation, how are the halfcell potentials (Ered) measured? In our B4 LAD you sae that one half-reaction must assigned an arbitrary value (in our lab we chose Cu 2+ /Cu, and then all other halfreactions can be determined relative to that reference. Universally, the standard hydrogen electrode (SHE) has been arbitrarily assigned a half-cell potential (Ered) of zero

44 Write the half reaction that occurs at a hydrogen electrode when it serves as the cathode of a voltaic cell. What is standard about the Standard Hydrogen Electrode? What is the role of platinum foil in a standard hydrogen electrode? 44 44

45 Relative Reduction Potentials? 1atm H2 1 M H + H2 2H + + 2e 2H + + 2e H2 Since reduction only occurs while oxidation occurs at the same time, it is impossible to measure one without the other. To determine half reaction values, some half reaction had to be set as 0 The Standard Hydrogen Electrode (SHE) has been chosen as

46 Relative Reduction Potentials? When connected to a zinc/zinc nitrate half cell, the potential for the complete cell is 0.76 V We know that the Zn is the anode. By defining the reduction of H + as 0.0, we conclude that the oxidation of Zn is 0.76 V Thus the reduction of Zn 2+ will be V 46 46

47 Relative Reduction Potentials? When connected to a copper/copper nitrate half cell, the potential for the complete cell is 0.34 V This time we know that the copper is the cathode By defining the reduction of H + as 0.0, we conclude that the reduction of Cu 2+ is 0.34 V By connecting many different half cells together, all the potentials can be determined relative to each other

48 Mg(s) + Al 3+ (aq) Mg 2+ (aq) + Al(s) Write out the aluminum half-reaction and determine the voltage, Eº for the half reaction you write

49 Mg(s) + Al 3+ (aq) Mg 2+ (aq) + Al(s) Write out the aluminum half-reaction and determine the voltage, Eº for the half reaction you write. Al e Al Eºred = 1.66 V (J/coulomb) 49 49

50 Mg(s) + Al 3+ (aq) Mg 2+ (aq) + Al(s) Write out the magnesium half-reaction and determine the voltage, Eº for the half reaction you write

51 Mg(s) + Al 3+ (aq) Mg 2+ (aq) + Al(s) Write out the magnesium half-reaction and determine the voltage, Eº for the half reaction you write. Eºred = V (J/coulomb) Mg Mg e Eºox = V (J/coulomb) 51 51

52 What is the Eºcell Mg(s) + Al +3 (aq) Mg +2 (aq) + Al(s) Al e Al Eºred = 1.66 V (J/coulomb) Mg Mg e Eºox = V (J/coulomb) 52 52

53 What is the Eºcell Mg(s) + Al +3 (aq) Mg +2 (aq) + Al(s) Al e Al Eºred = 1.66 V (J/coulomb) Mg Mg e Eºox = V (J/coulomb) Eºred + Eºox = Eºcell = Eºcell = 0.71 V (J/coulomb) 53 53

54 Balance the equation Mg(s) + Al +3 (aq) Mg +2 (aq) + Al(s) 54 54

55 Balance the equation Mg(s) + Al 3+ (aq) Mg 2+ (aq) + Al(s) 3Mg(s) + 2Al 3+ (aq) 3Mg 2+ (aq) + 2Al(s) 55 55

56 What is the Eºcell 3Mg(s) + 2Al +3 (aq) 3Mg +2 (aq) + 2Al(s) Al e Al Eºred = 1.66 V (J/coulomb) Mg Mg e Eºox = V (J/coulomb) Eºred + Eºox = Eºcell = Eºcell = 0.71 V (J/coulomb) 56 56

57 Draw a sketch of a Galvanic cell at standard conditions to represent this reaction. 3Mg(s) + 2Al 3+ (aq) 3Mg 2+ (aq) + 2Al(s) 1. Which direction do e- flow? 2. Which is cathode? anode? 3. What else must be done to make the cell functional? 4. Which electrode gains mass? which loses? 57 57

58 Draw a Galvanic cell at standard conditions to represent this reaction. 3Mg(s) + 2Al 3+ (aq) 3Mg 2+ (aq) + 2Al(s) 1. Which direction do e- flow? 2. Which is cathode? anode? 3. What else must be done to make the cell functional? need a salt bridge bridge 4. Which electrode gains mass? which loses? the Mg anode loses mass, the Al cathode gains mass Al Al(NO3)3(aq) cathode salt bridge e- Mg Mg(NO3)2(aq) anode 58 58

59 You have copper, iron, and lead metals and their respective M 2+ nitrate solutions. What combination would make the cell with the highest voltage? 1. Identify the possible half reactions. 2. Determine the potential of the cell. 3. Write the balanced net ionic equation for the cell. 4. Which metal is the cathode? Anode? 5. Complete the sketch with solutions and labels on the electrodes. 6. Show the direction that electrons flow through the wire? 59 59

60 You have copper, iron, and lead metal. What combination would make the cell with the highest voltage? Cu e- Cu Eºred = 0.34 V Pb e- Pb Eºred = V Fe e- Fe Eºred = -.44 V 1. Identify the two half reactions. 2. Determine the potential of the cell. 3. Write the balanced net ionic equation for the cell. 4. Which metal is the cathode? Anode? 5. Complete the sketch with solutions and labels on the electrodes. 6. Show the direction that electrons flow through the wire? 60 60

61 You have copper, iron, and lead metal. What combination would make the cell with the highest voltage? e- Cu e- Cu Eºred = 0.34 V Pb e- Pb Eºred = V Fe e- Fe Eºred = -.44 V 1. Identify the two half reactions. 2. Determine the potential of the cell. 3. Write the balanced net ionic equation for the cell. 4. Which metal is the cathode? Anode? Cu(NO3)2(aq) Fe(NO3)2(aq) 5. Complete the sketch with solutions and labels on the electrodes. 6. Show the direction that electrons flow through the wire? Cu Fe 61 61

62 You have copper, iron, and lead metals and their respective M 2+ nitrate solutions. What combination would make the cell with the highest voltage? Cu e- Cu Eºred = 0.34 V Pb e- Pb Eºred = V Fe e- Fe Eºred = -.44 V 1. Identify the two half reactions. the oxidation of iron and reduction of copper 2. Determine the potential of the cell. Eºcell = 0.44 V V = 0.78 V 3. Write the balanced net ionic equation for the cell. Fe + Cu +2 Cu + Fe Which metal is the cathode? Anode? RED CAT: copper is the cathode, iron is the anode. 5. Complete the sketch with solutions and labels on the electrodes. 6. Show the direction that electrons flow through the wire? electrons flow from anode to cathode Cu Cu(NO3)2(aq) e- Fe Fe(NO3)2(aq) 62 62

63 You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal (Co 2+, Ni 2+ ). Write a balanced equation to represent the redox reaction that will produce the best cell potential, then draw a schematic diagram of the galvanic cell. 1. Which direction do e- flow? 2. Which is cathode? anode? 3. Label the charges of cathode and anode. 4. What else must be done to make the cell functional? 5. Which electrode gains mass? which loses? 63 63

64 You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal. Write the reaction that will produce the best cell potential, then draw a schematic diagram of the galvanic cell. Ag + + e Ag Eºred = V Ni e- Ni Eºred = 0.25 V Co e- Co Eºred = 0.28 V Al e- Al Eºred = 1.66 V 1. Which direction do e- flow? 2. Which is cathode? anode? 3. What else must be done to make the cell functional? 4. Which electrode gains mass? which loses? 64 64

65 You have electrodes of nickel, silver, aluminum and cobalt and a nitrate solution of each metal. Write the reaction that will produce the best cell potential, then draw a schematic diagram of the galvanic cell. 1. Which direction do e- flow? 2. Which is cathode? anode? 3. Label the charges of cathode and anode. 4. What else must be done to make the cell functional? need a salt bridge 5. Which electrode gains mass? which loses? the Al anode loses mass, the Ag cathode gains mass Al Al(NO3)3(aq) anode salt bridge e > Ag AgNO3(aq) cathode 65 65

66 Hydrogen Peroxide 66

67 Recall the H2O2 elephant toothpaste 1. Write the reduction half reaction for the decomposition of hydrogen peroxide

68 Recall the H2O2 elephant toothpaste 1. Look at the back of the orange table and write the reduction half reaction for the decomposition of hydrogen peroxide. H2O2 + 2H + + 2e 2H2O Eºred = V 68 68

69 Write the half-reaction for the oxidation of all 4 halide ions without using your reduction potential table. Record the oxidation potential as well

70 Write the half-reaction for the oxidation of all 4 halide ions without using your reduction potential table. Record the oxidation potential as well. 2F F2 + 2e 2Cl Cl2 + 2e 2Br Br2 + 2e 2I I2 + 2e Eºox = 2.87 V Eºox = 1.36 V Eºox = 1.07 V Eºox = 0.53 V 70 70

71 Which halide is easiest to oxidize? 1. 2F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

72 Which halide is easiest to oxidize? 1. 2F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

73 H2O2 + 2H + + 2e 2H2O Eºred = Which halide ion(s) should be able to be oxidized by hydrogen peroxide? Select all that apply. 1. 2F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

74 H2O2 + 2H + + 2e 2H2O Eºred = Which halide ion(s) should be able to be oxidized by hydrogen peroxide? Let s try it. 1. 2F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

75 The Cl and Br were unable to start the reduction, hmmmm. Use the back of your orange sheet to write a half reaction for the reduction of H2O2 and look up the Eºred 75 75

76 The Cl and Br were unable to start the reduction, hmmmm. Use the back of your orange sheet to write a half reaction for the reduction of H2O2 and look up the Eºred H2O2 O2 + 2H + + 2e Eºox = 0.68 V H2O2 + 2H + + 2e 2H2O Eºred = V 76 76

77 Reconsider the decomposition of hydrogen peroxide in light of the peroxide s ability to be both reduced and oxidized. Which halide ions can actually be oxidized during this decompostion? H2O2 O2 + 2H + + 2e - Eºox = H2O2 + 2H + + 2e - 2H2O Eºred = F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

78 Reconsider the oxidation of halide ions by hydrogen peroxide. Which halide ions can actually be oxidized by H2O2? H2O2 O2 + 2H + + 2e Eºox = H2O2 + 2H + + 2e 2H2O Eºred = F F2 + 2e Eºox = Cl Cl2 + 2e Eºox = Br Br2 + 2e Eºox = I I2 + 2e Eºox =

79 Decomposition of hydrogen peroxide. Using the back of your Reduction Potential Sheet, and write down a reaction for the reduction of hydrogen peroxide. Record the reduction potential 1. Then write the overall reaction. 2. Write the reduction half reaction. 3. Write the oxidation half reaction. 4. Caclculate Eºcell 79 79

80 The oxygen in hydrogen peroxide is both oxidized and reduced when hydrogen peroxide decomposes. 1. Write the overall reaction. 2H2O2 2H2O + O2 Eº = V 2. Write the reduction half reaction. H2O2 + 2H + + 2e 2H2O Eºred = 1.78 V 3. Write the oxidation half reaction. H2O2 O2 + 2H + + 2e Eºred = 0.68 V 80 80

81 Write the half-reaction for the oxidation of all 4 halide ions. Look up the Eºox for each halfreaction. 2F F2 + 2e 2Cl Cl2 + 2e 2Br Br2 + 2e 2I I2 + 2e 81 81

82 Which halide is easiest to oxidize? 1. 2F F2 + 2e 2. 2Cl Cl2 + 2e 3. 2Br Br2 + 2e 4. 2I I2 + 2e 82 82

83 Use the reduction potential table in the textbook to write the halfreaction for the oxidation of oxygen in hydrogen peroxide and look up the Eºoxid 83 83

84 The Effect of Concentration on Voltage 84

85 Calculate the voltage of the following cell V V Cu salt bridge Cu V V 4. impossible to determine 1 M Cu(NO3)2 1 M Cu(NO3)

86 Calculate the voltage of the following cell V V Cu salt bridge Cu V the voltage of the two half reactions cancel out 4. impossible to determine 1 M Cu(NO3)2 1 M Cu(NO3)2 Cu Cu e Cu e Cu Cu Cu e Cu e Cu no reaction 86 86

87 Calculate the voltage of the following cell V V V 4. greater than V 5. more than V but less than 0.0 V 6. less than 0.34 V but greater than 0.0 V 7. more negative than -0.34V (that is to say, smaller than -0.34V) V salt bridge Cu M 5 M Cu(NO3)2 Cu(NO3)2 Cu 87 87

88 Calculate the voltage of the following cell V V V 4. greater than V 5. more than V but less than 0.0 V 6. less than 0.34 V but greater than 0.0 V 7. more negative than -0.34V (that is to say, smaller than -0.34V) Cu 5 M Cu(NO3) M Cu(NO3)2 Cu This is called a Concentration Cell The voltage will be greater in magnitude than 0.34 or -0.34, the sign depends on how the voltmeter is hooked up

89 Calculate the voltage of the following cell V V V 4. greater than V 5. less than 1.10 V but greater than 0.0 V 6. more than V but less than 0.0 V 7. more negative than -1.10V (that is to say, smaller than V) 2 M Cu(NO3)2 Cu e Cu M Zn(NO3)2 Zn Zn e 89 89

90 Calculate the voltage of the following cell V V V 4. greater than V 5. less than 1.10 V but greater than 0.0 V 6. more than V but less than 0.0 V 7. more negative than -1.10V (that is to say, smaller than V) 2 M Cu(NO3)2 Cu e Cu M Zn(NO3)2 Zn Zn e 90 90

91 The voltage for this cell would be greater than 1.10 V Zn + Cu 2+ Zn 2+ + Cu Since the concentration of the reactant is greater than standard condition, and since the concentration of the product is less than standard conditions we can predict that the reaction has an even greater tendency to occur in the forward direction and is therefore greater than V 2 M Cu(NO3)2 Cu e Cu V 0.01 M Zn(NO3)2 Zn Zn e 91 91

92 The Nernst Equation The mathematics that supports the calculation of nonstandard Zn + Cu 2+ Zn 2+ + Cu The mathematics that supports the calculation of the voltage for nonstandard conditions. 2 M Cu(NO3) M Zn(NO3)2 Cu e Cu Zn Zn e 92 92

93 that s it for now