1. Chapter 18 Electrochemistry 2007, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Tro, Chemistry: A Molecular Approach2 Redox Reaction one or more elements change oxidation number all single displacement, and combustion, some synthesis and decomposition always have both oxidation and reduction split reaction into oxidation half-reaction and a reduction half-reaction aka electron transfer reactions half-reactions include electrons oxidizing agent is reactant molecule that causes oxidation contains element reduced reducing agent is reactant molecule that causes reduction contains the element oxidized

3. Tro, Chemistry: A Molecular Approach3 Oxidation & Reduction oxidation is the process that occurs when oxidation number of an element increases element loses electrons compound adds oxygen compound loses hydrogen half-reaction has electrons as products reduction is the process that occurs when oxidation number of an element decreases element gains electrons compound loses oxygen compound gains hydrogen half-reactions have electrons as reactants

4. Tro, Chemistry: A Molecular Approach4 Rules for Assigning Oxidation States rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

5. Tro, Chemistry: A Molecular Approach5 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3 –, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2

6. Tro, Chemistry: A Molecular Approach6 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority NonmetalOxidation StateExample FCF 4 H+1CH 4 O-2CO 2 Group 7ACCl 4 Group 6A-2CS 2 Group 5A-3NH 3

7. Tro, Chemistry: A Molecular Approach7 Oxidation and Reduction oxidation occurs when an atom’s oxidation state increases during a reaction reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction

8. Tro, Chemistry: A Molecular Approach8 Oxidation–Reduction oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent

9. Tro, Chemistry: A Molecular Approach9 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

10. Tro, Chemistry: A Molecular Approach10 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 ox agred ag +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction oxidation reduction red agox ag

11. 11 Common Oxidizing Agents

12. 12 Common Reducing Agents

13. Tro, Chemistry: A Molecular Approach13 Balancing Redox Reactions 1) assign oxidation numbers a)determine element oxidized and element reduced 2) write ox. & red. half-reactions, including electrons a)ox. electrons on right, red. electrons on left of arrow 3) balance half-reactions by mass a)first balance elements other than H and O b)add H 2 O where need O c)add H +1 where need H d)neutralize H + with OH - in base 4) balance half-reactions by charge a)balance charge by adjusting electrons 5) balance electrons between half-reactions 6) add half-reactions 7) check

14. Tro, Chemistry: A Molecular Approach14 Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Assign Oxidation States I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) Separate into half- reactions ox: red: Assign Oxidation States Separate into half- reactions ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s)

15. Tro, Chemistry: A Molecular Approach15 Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance half- reactions by mass ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass then O by adding H 2 O ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) Balance half- reactions by mass then H by adding H + ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) Balance half- reactions by mass in base, neutralize the H + with OH - ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) 4 H + (aq) + 4 OH  (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) 4 H 2 O (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) MnO 4  (aq) + 2 H 2 O (l)  MnO 2(s) + 4 OH  (aq)

16. Tro, Chemistry: A Molecular Approach16 Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance Half- reactions by charge ox: 2 I  (aq)  I 2(aq) + 2 e  red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) Balance electrons between half- reactions ox: 2 I  (aq)  I 2(aq) + 2 e  } x 3 red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) } x 2 ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq)

17. Tro, Chemistry: A Molecular Approach17 Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Add the Half- reactions ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq) tot: 6 I  (aq) + 2 MnO 4  (aq) + 4 H 2 O (l)  3 I 2(aq) + 2 MnO 2(s) + 8 OH  (aq) Check Reactant CountElement Product Count 6I6 2Mn2 12O 8H8 22 charge 22

18. Tro, Chemistry: A Molecular Approach18 Practice - Balance the Equation H 2 O 2 + KI + H 2 SO 4  K 2 SO 4 + I 2 + H 2 O

19. Tro, Chemistry: A Molecular Approach19 Practice - Balance the Equation H 2 O 2 + KI + H 2 SO 4  K 2 SO 4 + I 2 + H 2 O +1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2 oxidation reduction ox:2 I -1  I 2 + 2e -1 red:H 2 O 2 + 2e -1 + 2 H +  2 H 2 O tot2 I -1 + H 2 O 2 + 2 H +  I 2 + 2 H 2 O 1 H 2 O 2 + 2 KI + H 2 SO 4  K 2 SO 4 + 1 I 2 + 2 H 2 O

20. Tro, Chemistry: A Molecular Approach20 Practice - Balance the Equation ClO 3 -1 + Cl -1  Cl 2 (in acid)

21. Tro, Chemistry: A Molecular Approach21 Practice - Balance the Equation ClO 3 -1 + Cl -1  Cl 2 (in acid) +5 -2 -1 0 oxidation reduction ox:2 Cl -1  Cl 2 + 2 e -1 } x 5 red:2 ClO 3 -1 + 10 e -1 + 12 H +  Cl 2 + 6 H 2 O} x 1 tot10 Cl -1 + 2 ClO 3 -1 + 12 H +  6 Cl 2 + 6 H 2 O 1 ClO 3 -1 + 5 Cl -1 + 6 H +1  3 Cl 2 + 3 H 2 O

22. Tro, Chemistry: A Molecular Approach22 Electrical Current when we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time when we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time whether as electrons flowing through a wire or ions flowing through a solution

23. Tro, Chemistry: A Molecular Approach23 Redox Reactions & Current redox reactions involve the transfer of electrons from one substance to another therefore, redox reactions have the potential to generate an electric current in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring

24. Tro, Chemistry: A Molecular Approach24 Electric Current Flowing Directly Between Atoms

25. Tro, Chemistry: A Molecular Approach25 Electric Current Flowing Indirectly Between Atoms

26. Tro, Chemistry: A Molecular Approach26 Electrochemical Cells electrochemistry is the study of redox reactions that produce or require an electric current the conversion between chemical energy and electrical energy is carried out in an electrochemical cell spontaneous redox reactions take place in a voltaic cell aka galvanic cells nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

27. Tro, Chemistry: A Molecular Approach27 Electrochemical Cells oxidation and reduction reactions kept separate half-cells electron flow through a wire along with ion flow through a solution constitutes an electric circuit requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons through external circuit ion exchange between the two halves of the system electrolyte

28. Tro, Chemistry: A Molecular Approach28 Electrodes Anode electrode where oxidation occurs anions attracted to it connected to positive end of battery in electrolytic cell loses weight in electrolytic cell Cathode electrode where reduction occurs cations attracted to it connected to negative end of battery in electrolytic cell gains weight in electrolytic cell  electrode where plating takes place in electroplating

29. Tro, Chemistry: A Molecular Approach29 Voltaic Cell the salt bridge is required to complete the circuit and maintain charge balance

30. Tro, Chemistry: A Molecular Approach30 Current and Voltage the number of electrons that flow through the system per second is the current unit = Ampere 1 A of current = 1 Coulomb of charge flowing by each second 1 A = 6.242 x 10 18 electrons/second Electrode surface area dictates the number of electrons that can flow the difference in potential energy between the reactants and products is the potential difference unit = Volt 1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external circuit amount of force pushing the electrons through the wire is called the electromotive force, emf

31. Tro, Chemistry: A Molecular Approach31 Cell Potential the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode the cell potential under standard conditions is called the standard emf, E° cell 25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

32. Tro, Chemistry: A Molecular Approach32 Cell Notation shorthand description of Voltaic cell electrode | electrolyte || electrolyte | electrode oxidation half-cell on left, reduction half-cell on the right single | = phase barrier if multiple electrolytes in same phase, a comma is used rather than | often use an inert electrode double line || = salt bridge

33. Tro, Chemistry: A Molecular Approach33 Fe(s) | Fe 2+ (aq) || MnO 4  (aq), Mn 2+ (aq), H + (aq) | Pt(s)

34. Tro, Chemistry: A Molecular Approach34 Standard Reduction Potential a half-reaction with a strong tendency to occur has a large + half-cell potential when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction we select as a standard half-reaction the reduction of H + to H 2 under standard conditions, which we assign a potential difference = 0 v standard hydrogen electrode, SHE

35. Tro, Chemistry: A Molecular Approach35

36. Tro, Chemistry: A Molecular Approach36 Half-Cell Potentials SHE reduction potential is defined to be exactly 0 v half-reactions with a stronger tendency toward reduction than the SHE have a + value for E° red half-reactions with a stronger tendency toward oxidation than the SHE have a  value for E° red E° cell = E° oxidation + E° reduction E° oxidation =  E° reduction when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half- reactions to balance the equation

38. Tro, Chemistry: A Molecular Approach38

39. Tro, Chemistry: A Molecular Approach39 Ex 18.4 – Calculate E  cell for the reaction at 25  C Al (s) + NO 3 − (aq) + 4 H + (aq)  Al 3+ (aq) + NO (g) + 2 H 2 O (l) Separate the reaction into the oxidation and reduction half-reactions ox:Al (s)  Al 3+ (aq) + 3 e − red:NO 3 − (aq) + 4 H + (aq) + 3 e −  NO (g) + 2 H 2 O (l) find the E  for each half- reaction and sum to get E  cell E  ox = −E  red = +1.66 v E  red = +0.96 v E  cell = (+1.66 v) + (+0.96 v) = +2.62 v

40. Tro, Chemistry: A Molecular Approach40 Ex 18.4a – Predict if the following reaction is spontaneous under standard conditions Fe (s) + Mg 2+ (aq)  Fe 2+ (aq) + Mg (s) Separate the reaction into the oxidation and reduction half-reactions ox:Fe (s)  Fe 2+ (aq) + 2 e − red: Mg 2+ (aq) + 2 e −  Mg (s) look up the relative positions of the reduction half- reactions red: Mg 2+ (aq) + 2 e −  Mg (s) red: Fe 2+ (aq) + 2 e −  Fe (s) since Mg 2+ reduction is below Fe 2+ reduction, the reaction is NOT spontaneous as written

41. Tro, Chemistry: A Molecular Approach41 the reaction is spontaneous in the reverse direction Mg (s) + Fe 2+ (aq)  Mg 2+ (aq) + Fe (s) ox:Mg (s)  Mg 2+ (aq) + 2 e − red:Fe 2+ (aq) + 2 e −  Fe (s) sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

42. Tro, Chemistry: A Molecular Approach42 Practice - Sketch and Label the Voltaic Cell Fe(s)  Fe 2+ (aq)  Pb 2+ (aq)  Pb(s), Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.

43. Tro, Chemistry: A Molecular Approach43 ox: Fe(s)  Fe 2+ (aq) + 2 e − E  = +0.45 V red: Pb 2+ (aq) + 2 e −  Pb(s) E  = −0.13 V tot: Pb 2+ (aq) + Fe(s)  Fe 2+ (aq) + Pb(s) E  = +0.32 V

44. Tro, Chemistry: A Molecular Approach44 Predicting Whether a Metal Will Dissolve in an Acid acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H + (aq) metals whose ion reduction reaction lies below H + reduction on the table will dissolve in acid

45. Tro, Chemistry: A Molecular Approach45 E° cell,  G° and K for a spontaneous reaction one the proceeds in the forward direction with the chemicals in their standard states  G° < 1 (negative) E° > 1 (positive) K > 1  G° = −RTlnK = −nFE° cell n is the number of electrons F = Faraday’s Constant = 96,485 C/mol e −

46. Tro, Chemistry: A Molecular Approach46 Example 18.6- Calculate  G° for the reaction I 2(s) + 2 Br − (aq) → Br 2(l) + 2 I − (aq) since  G° is +, the reaction is not spontaneous in the forward direction under standard conditions Answer: Solve: Concept Plan: Relationships: I 2(s) + 2 Br − (aq) → Br 2(l) + 2 I − (aq)  G , (J) Given: Find: E° ox, E° red E° cell  G° ox: 2 Br − (aq) → Br 2(l) + 2 e − E° = −1.09 v red: I 2(l) + 2 e − → 2 I − (aq) E° = +0.54 v tot: I 2(l) + 2Br − (aq) → 2I − (aq) + Br 2(l) E° = −0.55 v

47. Tro, Chemistry: A Molecular Approach47 Example 18.7- Calculate  at 25°C for the reaction Cu (s) + 2 H + (aq) → H 2(g) + Cu 2+ (aq) since  < 1, the position of equilibrium lies far to the left under standard conditions Answer: Solve: Concept Plan: Relationships: Cu (s) + 2 H + (aq) → H 2(g) + Cu 2+ (aq)  Given: Find: E° ox, E° red E° cell  ox: Cu (s) → Cu 2+ (aq) + 2 e − E° = −0.34 v red: 2 H + (aq) + 2 e − → H 2(aq) E° = +0.00 v tot: Cu (s) + 2H + (aq) → Cu 2+ (aq) + H 2(g) E° = −0.34 v

48. Tro, Chemistry: A Molecular Approach48 Nonstandard Conditions - the Nernst Equation  G =  G° + RT ln Q E = E° - (0.0592/n) log Q at 25°C when Q = K, E = 0 use to calculate E when concentrations not 1 M

49. Tro, Chemistry: A Molecular Approach49 E  at Nonstandard Conditions

50. Tro, Chemistry: A Molecular Approach50 Example 18.8- Calculate E cell  at 25°C for the reaction 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + Cu 2+ (aq) + 4 H 2 O (l) units are correct, E cell > E° cell as expected because [MnO 4 − ] > 1 M and [Cu 2+ ] < 1 M Check: Solve: Concept Plan: Relationships: 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + Cu 2+ (aq) + 4 H 2 O (l) [Cu 2+ ] = 0.010 M, [MnO 4 − ] = 2.0 M, [H + ] = 1.0 M E cell Given: Find: E° ox, E° red E° cell E cell ox: Cu (s) → Cu 2+ (aq) + 2 e − } x 3E° = −0.34 v red: MnO 4 − (aq) + 4 H + (aq) + 3 e − → MnO 2(s) + 2 H 2 O (l) } x 2 E° = +1.68 v tot: 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + Cu 2+ (aq) + 4 H 2 O (l)) E° = +1.34 v

51. Tro, Chemistry: A Molecular Approach51 Concentration Cells it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different the difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than the less concentrated electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

52. Tro, Chemistry: A Molecular Approach52 when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow Concentration Cell when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) Cu(s)  Cu 2+ (aq) (0.010 M)  Cu 2+ (aq) (2.0 M)  Cu(s)

53. Tro, Chemistry: A Molecular Approach53 LeClanche’ Acidic Dry CellDry Cell electrolyte in paste form ZnCl 2 + NH 4 Cl  or MgBr 2 anode = Zn (or Mg) Zn(s)  Zn 2+ (aq) + 2 e - cathode = graphite rod MnO 2 is reduced 2 MnO 2 (s) + 2 NH 4 + (aq) + 2 H 2 O(l) + 2 e -  2 NH 4 OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.5 v expensive, nonrechargeable, heavy, easily corroded

54. Tro, Chemistry: A Molecular Approach54 Alkaline Dry Cell same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste anode = Zn (or Mg) Zn(s)  Zn 2+ (aq) + 2 e - cathode = brass rod MnO 2 is reduced 2 MnO 2 (s) + 2 NH 4 + (aq) + 2 H 2 O(l) + 2 e -  2 NH 4 OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.54 v longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

55. Tro, Chemistry: A Molecular Approach55 Lead Storage Battery 6 cells in series electrolyte = 30% H 2 SO 4 anode = Pb Pb(s) + SO 4 2- (aq)  PbSO 4 (s) + 2 e - cathode = Pb coated with PbO 2 PbO 2 is reduced PbO 2 (s) + 4 H + (aq) + SO 4 2- (aq) + 2 e -  PbSO 4 (s) + 2 H 2 O(l) cell voltage = 2.09 v rechargeable, heavy

56. Tro, Chemistry: A Molecular Approach56 NiCad Battery electrolyte is concentrated KOH solution anode = Cd Cd(s) + 2 OH -1 (aq)  Cd(OH) 2 (s) + 2 e -1 E 0 = 0.81 v cathode = Ni coated with NiO 2 NiO 2 is reduced NiO 2 (s) + 2 H 2 O(l) + 2 e -1  Ni(OH) 2 (s) + 2OH -1 E 0 = 0.49 v cell voltage = 1.30 v rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

57. Tro, Chemistry: A Molecular Approach57 Ni-MH Battery electrolyte is concentrated KOH solution anode = metal alloy with dissolved hydrogen oxidation of H from H 0 to H +1 M∙H(s) + OH -1 (aq)  M(s) + H 2 O(l) + e -1 E° = 0.89 v cathode = Ni coated with NiO 2 NiO 2 is reduced NiO 2 (s) + 2 H 2 O(l) + 2 e -1  Ni(OH) 2 (s) + 2OH -1 E 0 = 0.49 v cell voltage = 1.30 v rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad

58. Tro, Chemistry: A Molecular Approach58 Lithium Ion Battery electrolyte is concentrated KOH solution anode = graphite impregnated with Li ions cathode = Li - transition metal oxide reduction of transition metal work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode rechargeable, long life, very light, more environmentally friendly, greater energy density

59. Tro, Chemistry: A Molecular Approach59

60. Tro, Chemistry: A Molecular Approach60 Fuel Cells like batteries in which reactants are constantly being added so it never runs down! Anode and Cathode both Pt coated metal Electrolyte is OH – solution Anode Reaction: 2 H 2 + 4 OH – → 4 H 2 O(l) + 4 e - Cathode Reaction: O 2 + 4 H 2 O + 4 e - → 4 OH –

61. Tro, Chemistry: A Molecular Approach61 Electrolytic Cell uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction must be DC source the + terminal of the battery = anode the - terminal of the battery = cathode cations attracted to the cathode, anions to the anode cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized some electrolysis reactions require more voltage than E tot, called the overvoltage

62. Tro, Chemistry: A Molecular Approach62

63. Tro, Chemistry: A Molecular Approach63 electroplating In electroplating, the work piece is the cathode. Cations are reduced at cathode and plate to the surface of the work piece. The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution

64. Tro, Chemistry: A Molecular Approach64 Electrochemical Cells in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode in voltaic cells, anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge in electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery

65. Tro, Chemistry: A Molecular Approach65 Electrolysis electrolysis is the process of using electricity to break a compound apart electrolysis electrolysis is done in an electrolytic cell electrolytic cells can be used to separate elements from their compounds generate H 2 from water for fuel cells recover metals from their ores

66. Tro, Chemistry: A Molecular Approach66 Electrolysis of Water

67. Tro, Chemistry: A Molecular Approach67 Electrolysis of Pure Compounds must be in molten (liquid) state electrodes normally graphite cations are reduced at the cathode to metal element anions oxidized at anode to nonmetal element

68. Tro, Chemistry: A Molecular Approach68 Electrolysis of NaCl (l)

69. Tro, Chemistry: A Molecular Approach69 Mixtures of Ions when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E° red when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E° ox

70. Tro, Chemistry: A Molecular Approach70 Electrolysis of Aqueous Solutions Complicated by more than one possible oxidation and reduction possible cathode reactions reduction of cation to metal reduction of water to H 2  2 H 2 O + 2 e -1  H 2 + 2 OH -1 E° = -0.83 v @ stand. cond. E° = -0.41 v @ pH 7 possible anode reactions oxidation of anion to element oxidation of H 2 O to O 2  2 H 2 O  O 2 + 4e -1 + 4H +1 E° = -1.23 v @ stand. cond. E° = -0.82 v @ pH 7 oxidation of electrode  particularly Cu  graphite doesn’t oxidize half-reactions that lead to least negative E tot will occur unless overvoltage changes the conditions

71. Tro, Chemistry: A Molecular Approach71 Electrolysis of NaI (aq) with Inert Electrodes possible oxidations 2 I -1  I 2 + 2 e -1 E° = −0.54 v 2 H 2 O  O 2 + 4e -1 + 4H +1 E° = −0.82 v possible reductions Na +1 + 1e -1  Na 0 E° = −2.71 v 2 H 2 O + 2 e -1  H 2 + 2 OH -1 E° = −0.41 v possible oxidations 2 I -1  I 2 + 2 e -1 E° = −0.54 v 2 H 2 O  O 2 + 4e -1 + 4H +1 E° = −0.82 v possible reductions Na +1 + 1e -1  Na 0 E° = −2.71 v 2 H 2 O + 2 e -1  H 2 + 2 OH -1 E° = −0.41 v overall reaction 2 I − (aq) + 2 H 2 O (l)  I 2(aq) + H 2(g) + 2 OH -1 (aq)

72. Tro, Chemistry: A Molecular Approach72 Faraday’s Law the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs charge that flows through the cell = current x time

73. Tro, Chemistry: A Molecular Approach73 Example 18.10- Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au 3+ (aq) + 3 e − → Au(s) units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e − Check: Solve: Concept Plan: Relationships: 3 mol e − : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Given: Find: t(s), ampcharge (C)mol e − mol Aug Au

74. Tro, Chemistry: A Molecular Approach74 Corrosion corrosion is the spontaneous oxidation of a metal by chemicals in the environment since many materials we use are active metals, corrosion can be a very big problem

75. Tro, Chemistry: A Molecular Approach75 Rusting rust is hydrated iron(III) oxide moisture must be present water is a reactant required for flow between cathode and anode electrolytes promote rusting enhances current flow acids promote rusting lower pH = lower E° red

76. Tro, Chemistry: A Molecular Approach76

77. Tro, Chemistry: A Molecular Approach77 Preventing Corrosion one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode  galvanized nails

78. Tro, Chemistry: A Molecular Approach78 Sacrificial Anode