1. UNIT V Electrochemical Cells

2. ELECTROCHEMICAL CELLS http://www.bclearningnetwork.com/moodle/mod/resource/view.php?id=618

3. TERMINOLOGY Electrochemical cell – A device which converts chemical energy into electrical energy. Electrode – A conductor (usually a metal) at which a half-cell reaction (oxidation or reduction) occurs. Anode - The electrode at which oxidation occurs (A & O are both vowels). LEOA Cathode – The electrode at which reduction occurs. (R & C are both consonants) GERC

4. HALF-CELL REACTIONS Anode – Oxidation Half-Reaction Zn (s)  Zn 2+ (aq) + 2e - Metal atoms are changed to + ions. Metal dissolves and anode loses mass as the cell operates.

5. HALF-CELL REACTIONS Cathode – Reduction Half-Reaction Cu 2+ + 2e -  Cu Positive ions are changed to metal atoms. New metal is formed so the cathode gains mass.

6. FLOW OF ELECTRONS Flow of Electrons Since the anode loses e - ‘s, (LEOA) and the cathode gains e - ‘s (GERC) Electrons flow from the anode toward the cathode in the wire (conducting solid). Electrons flow from A  C in the wire.

7. FLOW OF ELECTRONS

8. SALT BRIDGE Flow of Ions in the Salt Bridge Salt bridge contains any electrolyte (conducting solution) Examples: Porous barrier which ions can pass through Most common – U-tube containing a salt such as KNO 3 or NaNO 3 Common things which contain electrolytes include potatoes, apples, oranges, lemons, frogs, people

9. SALT BRIDGE If there was no salt bridge, positive ions (cations) would build up at the anode. If there was no salt bridge, negative ions (anions) would build up at the cathode.

10. IDENTIFYING THE ANODE AND CATHODE Look at the reduction table All half-rxns are reversible (can go forward or backward) All are written as reductions (GERC) Their reverse would be oxidations (LEOA) The half-rxn with the greater potential to be reduced is higher on the table (higher reduction potential E o )

11. IDENTIFYING THE ANODE AND CATHODE So the Higher half-rxn Is the Cathode ( HIC ) (Notice Cu 2+ + 2e -  Cu is higher than Zn 2+ + 2e -  Zn so Cu gets to be the cathode ) Also notice that the Anode reaction Is Reversed ( AIR ) (Anode rx: Zn  Zn 2+ + 2e - )

12. CHECK YOUR UNDERSTANDING Metal/Ion Cathode (HIC) Cathode Half-Rxn Anode (AIR) Anode Half-Rxn Ag/Ag + Fe/Fe 2+ Ag (higher) Ag + + e -  Ag Fe (lower) Fe  Fe 2+ + 2e - Zn/Zn 2+ Pb/Pb 2+ Ni/Ni 2+ Al/Al 3+ Au/Au 3+ Ag/Ag + Mg/Mg 2+ H 2 /H + Co/Co 2+ Sn/Sn 2+

13. SUMMARY 1. Electrochemical cells convert __________ energy into __________ energy. 2. __________ is the electrode where oxidation occurs. 3. Electrons are __________ at the anode. 4. __________ is the electrode where reduction occurs. 5. In the half-rxn at the cathode, e - ‘s are on the __________ side of the equation. 6. Electrons flow from the __________ toward the __________ in the __________. 7. Cations ((+) ions) flow from the __________ beaker toward the __________ beaker through the __________. 8. Anions ((-) ions flow from the __________ beaker through the __________.

14. SUMMARY 9. The higher half-rxn on the table is the one for the __________ and is not reversed. 10. The lower half-rxn on the table is the one for the __________ and is reversed. 11. Electrons do not travel through the ____________________, only through the __________. 12. Ions (cations & anions) do not travel through the wire but only through the ____________________. 13. The salt bridge can contain any __________. 14. The anode will __________ (gains/loses) mass as it is __________ (oxidized/reduced). 15. The cathode will __________ mass as it is __________ (oxidized/reduced). Hebden Textbook p.217 Questions #34-35

15. STANDARD REDUCTION POTENTIALS Voltage: The tendency for e - ‘s to flow in an electrochemical cell. Note that a cell may have a high voltage even if no e - ‘s are flowing. It is the tendency (or potential) for e - ‘s to flow. Can also be defined as the potential energy per coulomb (where 1C = the charge carried by 6.25x10 18 e - ). 1 Volt = 1 Joule/Coulomb Reduction potential of half-cells The tendency of a half-cell to be reduced (take e - ‘s). Voltage only depends on the difference in potentials not the absolute potentials.

16. STANDARD REDUCTION POTENTIALS The voltage of a cell depends only on the difference in reduction potentials of the two half-cells. Both people spent $20 on the calculator. $ before buying calculator $ after buying calculator Difference Mrs. A $2000$1980$20 Mrs. B $50$30$20

17. STANDARD REDUCTION POTENTIALS Relative potentials of half-cells can only be determined by connecting with other half-cells and reading the voltage. Ex. How good a basketball team is can only be determined by playing with other teams and looking at points (scores).

18. STANDARD HALF-CELL A “ standard ” half-cell was arbitrarily chosen to compare other half-cells with. It was assigned a “ reduction potential ” of 0.000 V. 2H + (aq) + 2e - H 2(g) E o = 0.000 V Since potential depends on gas pressure, temperature, and [H + ], this half cell is assigned a value of 0.000 v at “ standard state ” which is 25 o C, 101.3 kPa and [H + ] =1.0 M.

19. STANDARD HALF-CELL E o = 0.000 V

20. STANDARD HALF-CELL The standard half-cell acts as an anode (LEOA) or cathode (GERC) depending on what it is connected to. For example, when the standard half-cell is connected to the Ag/Ag + half-cell: For this cell, the voltage is 0.80 volts with the electrons flowing toward Ag. Cathode half-rxn: Ag + + e -  Ag E o = ? Anode half-rxn: H 2  2H + + 2e - E o = 0.00 V Voltage of cell = 0.80 V Therefore the reduction potential of the of Ag/Ag + half-cell is +0.80 V.

21. STANDARD HALF-CELL Example The standard (H 2 /H + ) half-cell is connected to the Ni/Ni 2+ half-cell. Electrons are found to flow away from the nickel toward the H 2 /H + half-cell and the voltage (at 25 o C, 101.3 Kpa & 1.0 M solutions) is found to be 0.26 volts. Give the:Cathode Half-rx: ___________________ Anode Half-rx: _____________________ Determine the E o for the Ni/Ni 2+ half cell: _________________

22. HALF-CELL REACTIONS Now look at Standard Reduction Table. Notice that all half-rxns are written as reductions. The E o is the standard reduction potential for the species on the left side. Ex. Of the following combinations, find the one which gives the highest voltage? ______ (answer only…explain later) a) Ag + /Ag with Cu 2+ /Cu b) Pb 2+ /Pb with Ni + /Ni c) Ag + /Ag with Pb + /Pb d) Au 3+ /Au with Ni 2+ /Ni Which combo gives the lowest voltage? _______

23. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS 1. Find the two metals on the reduction table. Higher one Is the Cathode. ( HIC ) 2. Write the cathode half-rxn as is on the table (cathode  reduction). Include the reduction potential (E o ) beside it. 3. Reverse the anode reaction ( AIR ) (anode  oxidation). Reverse the sign on the E o (If (+)  (-) | If (-)  (+)).

24. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS 4. Multiply half-rxns by factors that will make e - ‘s cancel. DON’T multiply the E o ’s by these factors. 5. Add up half-rxns to get overall redox reaction. 6. Add up E o ’s (as you have written them) to get the initial voltage of the cell.

25. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS Example: A cell is constructed using nickel metal and 1M nickel (II) nitrate along with Fe metal and 1M iron (II) nitrate. a) Write the equation for the half-rxn at the cathode (with the E o ) ____________________________________ b) Write the equation for the half-rxn at the anode (with the E o ) _____________________________________

26. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS c) Write the balanced equation for the overall reaction (with the E o ) ____________________________________ d) What is the initial cell voltage ? __________V

27. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS Example: A cell is constructed using aluminum metal, 1M Al(NO 3 ) 3 and lead metal with 1M Pb(NO 3 ) 2. Write the overall redox reaction and find the initial cell voltage.

28. USING THE REDUCTION TABLE TO FIND THE INITIAL VOLTAGE OF ELECTROCHEMICAL CELLS Overall redox reaction _____________________________________ Initial cell voltage: ___________V

29. CHECK YOUR UNDERSTANDING Example: A student has 3 metals: Ag, Zn and Cu; three solutions: AgNO 3, Zn(NO 3 ) 2, and Cu(NO 3 ) 2, all 1M. She also has a salt bridge containing KNO 3 (aq) wires and a voltmeter. a. Which combination of 2 metals and 2 solutions should she choose to get the highest possible voltage? Metal: _________Solution: ____________

30. CHECK YOUR UNDERSTANDING b. Draw a diagram of her cell labeling metals, solutions, salt bridge, wires, and voltmeter.

31. CHECK YOUR UNDERSTANDING c. Write an equation for the half-rxn at the cathode. (with E o ) _________________________________________ d. Write an equation for the half-rxn at the anode (with E o ) _________________________________________ e. Write a balanced equation for the overall redox reaction in the cell (with E o ) __________________________________________

32. CHECK YOUR UNDERSTANDING f. The initial voltage of this cell is _________volts. g. In this cell, e - ‘s are flowing toward which metal? _________ in the _________. h. Positive ions are moving toward the _________ solution in the __________________ i. Nitrate ions migrate toward the _________ solution in the __________________. j. As the cell operates _________ metal is gaining mass _________ metal is losing mass

33. CHECK YOUR UNDERSTANDING The student now wants to find the combination of metals and solutions that will give the lowest voltage. k. Which metals and solution should she use? Metal ___________ Solution _____________

34. CHECK YOUR UNDERSTANDING l. Find the overall redox equation for this cell. ______________________________________ m. Find the initial cell voltage of this cell __________ volts.

35. CHECK YOUR UNDERSTANDING Example: Cr Metal X 1M Cr(NO 3 ) 3 1M X(NO 3 ) 2

36. CHECK YOUR UNDERSTANDING The voltage on the voltmeter is 0.45 volts. a. Write the equation for the half-reaction taking place at the anode. Include the E o. ____________________________________________E o : _________V b. Write the equation for the half-reaction taking place at the cathode. ____________________________________________E o : _________V

37. CHECK YOUR UNDERSTANDING c. Write the balanced equation for the redox reaction taking place as this cell operates. Include the E o. ____________________________________________ E o : __________ d. Determine the reduction potential of the ion X 2+. E o : __________V

38. CHECK YOUR UNDERSTANDING e. Toward which beaker (X(NO 3 ) 2 ) or (Cr(NO 3 ) 3 ) do NO 3 - ions migrate? ______________________ f. Name the actual metal “X” ________________________________

39. CHECK YOUR UNDERSTANDING Example: (optional) Ag Cd 1M AgNO 3 1M Cd(NO 3 ) 2

40. CHECK YOUR UNDERSTANDING The voltage on the voltmeter is 1.20 volts. a. Write the equation for the half-reaction taking place at the cathode. Include the E o. ____________________________________________E o : _________V b. Write the equation for the half-reaction taking place at the anode. ____________________________________________E o : _________V

41. CHECK YOUR UNDERSTANDING c. Write the balanced equation for the overall redox reaction taking place. Include the E o. ____________________________________________ E o : __________V d. Find the oxidation potential for Cd. E o : __________V e. Find the reduction potential for Cd 2+ : E o = __________ V

42. CHECK YOUR UNDERSTANDING f. Which electrode gains mass as the cell operates? _______ g. Toward which beaker (AgNO 3 or Cd(NO 3 ) 2 ) do K + ions move? _______ h. The silver electrode and AgNO 3 solution is replaced by Zn metal and Zn(NO 3 ) 2 solution. What is the cell voltage now? __________ Which metal now is the cathode? _________

43. CHECK YOUR UNDERSTANDING Example: (optional) Ni Cu 1M Ni(NO 3 ) 2 1M Cu(NO 3 ) 2

44. CHECK YOUR UNDERSTANDING a. Write the equation for the half-reaction taking place at the nickel electrode. Include the E o _________________________ E o = _______v b. Write the equation for the half-reaction taking place at the Cu electrode. Include the E o. __________________________ E o =_______v c. Write the balanced equation for the redox reaction taking place. __________________________ E o =_______V

45. CHECK YOUR UNDERSTANDING d. What is the initial cell voltage? __________________ e. Show the direction of electron flow on the diagram. f. Show the direction of flow of cations in the salt bridge.

46. THE OVERPOTENTIAL EFFECT Notice that the two water half-reactions have a dotted arrow leading away from them flags an exception to their behavior:  These two half-reactions are correctly positioned according to their E° values, but the arrows point to the correct position according to their reactive behaviors.  You must mentally reposition these two half- reactions at the head of the arrow before judging their reaction participation.

47. VOLTAGES AT NON-STANDARD CONDITIONS Note: When cells are first constructed, they are not at equilibrium. All the voltages calculated by the reduction table are initial voltages. As the cells operate, the concentrations of the ions change: For the cell: Cu(NO 3 ) 2 /Cu//Zn/Zn(NO 3 ) 2  cathode ½ reaction: Cu 2+ + 2e -  Cu E o = + 0.34 V  anode ½ reaction: Zn  Zn 2+ + 2e - E o = + 0.76 V  overall reaction: Cu 2+ + Zn  Cu + Zn 2+ E o = + 1.10 V

48. VOLTAGES AT NON-STANDARD CONDITIONS All electrochemical cells are exothermic. Initially:Cu 2+ + Zn ↔ Cu + Zn 2+ + energy Voltage = 1.10 V As the cell operates [Cu 2+ ] decreases (reactants used up), reaction is pushed left (reduction potential of Cu 2+ /Cu decreases) At the same time, [Zn 2+ ] increases (products formed), reaction is pushed left (reduction potential of Zn 2+ /Zn increases)

49. VOLTAGES AT NON-STANDARD CONDITIONS Cu 2+ + Zn ↔ Cu + Zn 2+ + energy Voltage < 1.10 V Eventually, the gap between the two half reactions is zero. That is, the reduction potential of Cu 2+ /Cu decreases and the reduction potential of Zn 2+ /Zn increases until both reactions have the same reduction potential. When the difference in reduction potentials is zero (E cell = 0), the reaction is at equilibrium.

50. VOLTAGES AT NON-STANDARD CONDITIONS Question: A cell is constructed using Cr/Cr(NO 3 ) 3 and Fe/Fe(NO 3 ) 2 with both solutions at 1.0 M and the temperature at 25 o C. a. Determine the initial cell voltage. b. What is the equilibrium cell voltage ?

51. VOLTAGES AT NON-STANDARD CONDITIONS c. Write the balanced equation for the overall reaction taking place. _________________________________________ d. Using the equation in (c), predict what will happen to the cell voltage when the following changes are made: i) More Cr(NO 3 ) 3 is added to the beaker to increase the [Cr 3+ ]. ii) The [Fe 2+ ] ions is increased.

52. VOLTAGES AT NON-STANDARD CONDITIONS iii) A solution is added to precipitate the Fe 2+ ions. iv) Cr 3+ ions are removed by precipitation. v) The surface area of the Fe electrode is increased. vi) The salt bridge is removed.

53. PREDICTING SPONTANEITY FROM E O OF A REDOX REACTION If E o for any redox (overall) reaction is > 0 (positive) the reaction is spontaneous. If E o is < 0 (negative) the reaction is non- spontaneous When a reaction is reversed the sign of E o changes.

54. PREDICTING SPONTANEITY FROM E O OF A REDOX REACTION Example: a. Find the standard potential (E o ) for the following reaction: 2MnO 4 - + 4H 2 O + 3Sn 2+  2MnO 2 + 8OH - + 3Sn 4+ b. Is this reaction as written (forward rxn) spontaneous? _____ c. Is the reverse reaction spontaneous? _____ E o = _____

55. PREDICTING SPONTANEITY FROM E O OF A REDOX REACTION Example a. Calculate E o for the reaction: 3N 2 O 4 + 2Cr 3+ + 6H 2 O  6NO 3 - + 2Cr + 12H + b. Is the forward rxn spontaneous? _________ The reverse rxn? ___________ Hebden Textbook p. 217 #35-45

56. PRACTICAL APPLICATIONS OF ELECTROCHEMICAL CELLS The Lead Acid Storage Battery cathode anode

57. THE LEAD-ACID STORAGE BATTERY Anode Half-Reaction: (d ischarging or operating) Pb (s) + HSO 4 - (aq)  Pb SO 4 (s) + H + (aq) + 2e - Or simplified: Pb  Pb 2+ + 2e - Cathode Half-Reaction: ( discharging or operating) Pb O 2 (s) + HSO 4 - (aq) + 3H + (aq) + 2e -  Pb SO 4 (s) + 2H 2 O (l) Or simplified: Pb 4+ + 2e -  Pb 2+

58. THE LEAD-ACID STORAGE BATTERY Overall Redox Reaction : (discharging or operating) Pb (s) + PbO 2(s) + 2H + (aq) + 2HSO 4 - (aq)  2PbSO 4 (s) + 2H 2 O (l) + electrical energy Or simplified: Pb + Pb 4+  Pb 2+ + Pb 2+

59. THE LEAD-ACID STORAGE BATTERY As the cell discharges the anode (Pb) and cathode (PbO 2 ) disintegrate and the white solid (PbSO 4 ) forms on both plates. Originally, [H + ] and [HSO 4 - ] are high. H 2 SO 4 is denser than H 2 O, therefore the density (specific gravity) of the electrolyte is high to start with. As the cell discharges, H 2 SO 4 (H + & HSO 4 - ) is used up and H 2 O is formed. Therefore, electrolyte gets less dense as the battery discharges. Condition of the battery can be determined using a “hydrometer” or battery tester. (Higher the float, the denser the electrolyte.)

60. THE LEAD-ACID STORAGE BATTERY Adding electrical energy to this reaction will reverse it ( recharging ). Charging Reaction: Electrical energy + 2PbSO 4(s) + H 2 O (l)  Pb (s) + PbO 2(s) +2H + (aq) + 2HSO 4 - (aq) The E o for the discharging rxn, is +2.04 volts. A typical car battery has _____ of these in (series/parallel) ____________ to give a total voltage = __________V.

61. THE ZINC-CARBON BATTERY Also called LeClanche-Cell, Common Dry Cell, or regular carbon battery

62. THE ZINC-CARBON BATTERY Cathode Half-Reaction : 2MnO 2 (s) + 2NH 4 + (aq) +2e -  2MnO(OH) (s) + 2NH 3 (aq) Or simplified: Mn 4+ + e -  Mn 3+ Anode Half-Reaction : Zn (s) + 4NH 3 (aq)  Zn(NH 3 ) 4 2+ + 2e - Or simplified: Zn (s)  Zn 2+ + 2e - Not rechargeable Doesn’t last too long …especially with large currents Fairly cheap

63. THE ALKALINE DRY CELL

64. Cathode Half-Reaction: 2 Mn O 2 + H 2 O +2e -  Mn 2 O 3 + 2OH - Or simplified: Mn 4+ + e -  Mn 3+ Anode Half-Reaction: Zn (s) +2OH -  Zn O + H 2 O +H 2 O +2e - Or simplified: Zn (s)  Zn 2+ + 2e - Operates under basic conditions Delivers much greater current Voltage remains constant Lasts longer More expensive

65. FUEL CELLS

66. Fuel cells are continuously fed fuel and they convert the chemical energy in fuel to electrical energy More efficient (70-80%) than burning gas or diesel to run generators (30-40%) No pollution - only produces water Can use H 2 and O 2 or hydrogen rich fuels (e.g. methane CH 4 ) and O 2 Used in space capsules – H 2 & O 2 in tanks. H 2 O produced used for drinking

67. FUEL CELLS Anode Half-Reaction: 2H 2(g) + 4OH - (aq)  4H 2 O (l) + 4e - Cathode Half-Reaction: O 2(g) + 2H 2 O (l) + 4e -  4OH - (aq) Overall reaction: 2H 2(g) + O 2(g)  2H 2 O (l)

68. THE BREATHALYZER TEST

69. After drinking, breath contains ethanol C 2 H 5 OH Acidified dichromate (at E o = 1.23 on the reduction table) will oxidize alcohol. The unbalanced formula equation is: C 2 H 5 OH + K 2 Cr 2 O 7 + H 2 SO 4  C H 3 C OOH + Cr 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O  K 2 Cr 2 O 7 is yellow (orange at higher concentrations)  Cr 2 (SO 4 ) 3 is green

70. THE BREATHALYZER TEST Exhaled air is mixed with standardized acidified dichromate Put in a spectrophotometer set at the wavelength of green light More alcohol produces more green Cr 2 (SO 4 ) 3 (green) Machine is calibrated with known concentration samples of alcohol to ensure accuracy

71. THE BREATHALYZER TEST Question: In the reaction above name The oxidizing agent _________ The reducing agent _________ The product of oxidation ________ The product of reduction ________