1. Electrochemistry

2. Oxidation – Reduction Reactions  Consider the reaction of Copper wire and AgNO 3 (aq) AgNO 3 (aq) Ag(s) Cu(s)

3. Oxidation – Reduction Reactions  If you leave the reaction a long time the solution goes blue!  The blue is due to Cu 2+ (aq)  If you leave the reaction a long time the solution goes blue!  The blue is due to Cu 2+ (aq)

4. Oxidation-Reduction Reactions  So when we mix Ag + (aq) with Cu (s) we get Ag(s) and Cu 2+ (aq)  Ag + (aq) + 1e -  Ag (s)  Cu (s)  Cu 2+ (aq) + 2e -  The electrons gained by Ag + must come from the Cu 2+  Can’t have reduction without oxidation (redox)  Each Cu can reduce 2 Ag + 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e - 2Ag + (aq) + 2e - + Cu (s)  2Ag (s) + Cu 2+ (aq) + 2e -  So when we mix Ag + (aq) with Cu (s) we get Ag(s) and Cu 2+ (aq)  Ag + (aq) + 1e -  Ag (s)  Cu (s)  Cu 2+ (aq) + 2e -  The electrons gained by Ag + must come from the Cu 2+  Can’t have reduction without oxidation (redox)  Each Cu can reduce 2 Ag + 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e - 2Ag + (aq) + 2e - + Cu (s)  2Ag (s) + Cu 2+ (aq) + 2e - lose electrons = oxidation gain electrons = reduction

5. Redox Cu/Ag Cu Ag + E electron flow

6. Redox Cu/Ag Cu 2+ Ag E ΔE = e.E cell e = charge on an electron E cell = Voltage in a electrochemical cell If we could separate the two reactions we could use the energy gained by the e to do work

7. Redox Cu/Ag Cu 2+ Ag E The maximum amount of energy available to do work is a definition of the free energy Δ G Free energy for redox rxn cell voltage # electrons in redox rxn F = Faraday’s constant = 96485 C/mol

8. ΔE cell : the cell potential  In a redox reaction electrons are transferred to a more stable state  Most of the free energy of the reaction is due to these electrons  Can we access this energy?  Yes, by conducting the two half reactions in separate cells 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e -  This is called the electrochemical cell  The cell potential can be measured in such a cell with a voltmeter  In a redox reaction electrons are transferred to a more stable state  Most of the free energy of the reaction is due to these electrons  Can we access this energy?  Yes, by conducting the two half reactions in separate cells 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e -  This is called the electrochemical cell  The cell potential can be measured in such a cell with a voltmeter

9. Electrochemical/Voltaic Cell Cu/Ag 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e -

10. 10 Electric Current Flowing Directly Between Atoms

11. 11 Electric Current Flowing Indirectly Between Atoms

12. 12 Electrochemistry and 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 (galvanic cell)  nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy  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 (galvanic cell)  nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

13. 13 Electrodes  Anode (donates electrons to the cathode)  electrode where oxidation occurs  In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal  anions attracted to it  connected to positive end of battery in electrolytic cell  loses weight in electrolytic cell  Cathode (attracts electrons from the anode)  electrode where reduction occurs  In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal  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  Anode (donates electrons to the cathode)  electrode where oxidation occurs  In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal  anions attracted to it  connected to positive end of battery in electrolytic cell  loses weight in electrolytic cell  Cathode (attracts electrons from the anode)  electrode where reduction occurs  In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal  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

14. 14 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 (the potential for an electric field to cause an electrical current)  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  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 (the potential for an electric field to cause an electrical current)  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

15. 15 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  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

16. 16 Cell Notation  shorthand description of Voltaic cell  electrode | electrolyte || electrolyte | electrode  oxidation half-cell on left (anode), reduction half-cell on the right (cathode)  single | = phase barrier  if multiple electrolytes in same phase, a comma is used rather than |  often use an inert electrode  double line || = salt bridge  shorthand description of Voltaic cell  electrode | electrolyte || electrolyte | electrode  oxidation half-cell on left (anode), reduction half-cell on the right (cathode)  single | = phase barrier  if multiple electrolytes in same phase, a comma is used rather than |  often use an inert electrode  double line || = salt bridge

17. 17 Fe(s) | Fe 2+ (aq) || MnO 4 - (aq), Mn 2+ (aq), H + (aq) | Pt(s)

18. Keeping track of electrons  Redox reactions involve the transfer of electrons from the donor to the acceptor  The donor loses electrons and is oxidized  The acceptor acquires electrons and is reduced  To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has  We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom  Redox reactions involve the transfer of electrons from the donor to the acceptor  The donor loses electrons and is oxidized  The acceptor acquires electrons and is reduced  To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has  We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom 18

19. 19 Redox Reaction  one or more elements change oxidation number  all single displacement (A+BX--> AX+B), 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  one or more elements change oxidation number  all single displacement (A+BX--> AX+B), 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

20. 20 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  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

21. 21 Assigning Oxidation Numbers 1.The sum of the oxidation numbers Q of all the atoms in a compound/ion at up to the charge on the compound/ion. This means …. a.free elements have an oxidation state = 0  Q Na = 0 and Q Cl = 0 in 2 Na(s) + Cl 2 (g) b.monatomic ions have an oxidation state equal to their charge  Q Na = +1 and Q Cl = -1 in NaCl c.The sum of the oxidation numbers of all the atoms in a neutral compound is 0 d.the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion 2.(a) Group I metals have an oxidation state of +1 in all their compounds  Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds  Mg = +2 in MgCl 2 3.F has an oxidation number Q F = -1 4.H has an oxidation number Q H = +1 5.O has an oxidation number Q O = -2 1.The sum of the oxidation numbers Q of all the atoms in a compound/ion at up to the charge on the compound/ion. This means …. a.free elements have an oxidation state = 0  Q Na = 0 and Q Cl = 0 in 2 Na(s) + Cl 2 (g) b.monatomic ions have an oxidation state equal to their charge  Q Na = +1 and Q Cl = -1 in NaCl c.The sum of the oxidation numbers of all the atoms in a neutral compound is 0 d.the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion 2.(a) Group I metals have an oxidation state of +1 in all their compounds  Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds  Mg = +2 in MgCl 2 3.F has an oxidation number Q F = -1 4.H has an oxidation number Q H = +1 5.O has an oxidation number Q O = -2

22. 22 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  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 rule 1rule 4 rule 5

23. 23 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  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

24. 24 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 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

25. 25 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 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

26. 26 Common Oxidizing Agents

27. 27 Common Reducing Agents

28. 28 Balancing Redox Reactions 1.assign oxidation numbers determine element oxidized and element reduced 2.write ox. & red. half-reactions, including electrons 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 balance charge by adjusting electrons 5.balance electrons between half-reactions 6.add half-reactions 7.check 1.assign oxidation numbers determine element oxidized and element reduced 2.write ox. & red. half-reactions, including electrons 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 balance charge by adjusting electrons 5.balance electrons between half-reactions 6.add half-reactions 7.check

29. 29 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)

30. 30 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)

31. 31 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)

32. 32 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 8-8-charge8-8-

33. 33 Practice - Balance the Equation H 2 O 2 + KI + H 2 SO 4  K 2 SO 4 + I 2 + H 2 O

34. 34 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

35. 35 Practice - Balance the Equation ClO 3 -1 + Cl -1   Cl 2 (in acid)

36. 36 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

37. 37 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  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

38. 38

39. 39 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  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

41. 41

42. 42 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 o for each half-reaction and sum to get E o cell E o ox = −E o red = +1.66 v E o red = +0.96 v E o cell = (+1.66 v) + (+0.96 v) = +2.62 v

43. 43 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) -2.37V red: Fe 2+ (aq) + 2 e −  Fe (s) -0.44V since Mg 2+ reduction is below Fe 2+ reduction, the reaction is NOT spontaneous as written

44. 44 the reaction is spontaneous in the reverse direction tot: 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

45. 45 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.

46. 46 ox: Fe(s)  Fe 2+ (aq) + 2 e − E ox = +0.45 V red: Pb 2+ (aq) + 2 e −  Pb(s) E red = −0.13 V tot: Pb 2+ (aq) + Fe(s)  Fe 2+ (aq) + Pb(s) E cell = +0.32 V

47. 47 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  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

48. 48 Electrochemical Cell Summary A device for either harnessing the electrical power of a redox reaction (battery), or a device for using electricity to induce non-spontaneous redox reactions (electrolytic cell) The cell consists of two half cells, one where oxidation occurs, and one where reduction occurs Each half cell consists of an electrode and an electrolyteThe electrode which loses electrons we call the anode, and the electrode that gains electrons we call the cathode The anode and cathode are connect by a wire through which electrons can flow, and the electrolytes in each half cell are connected by a salt bridge through which cations and anions can flow salt bridge e-e- anode Zn (s)--> Zn 2+ (aq)+ 2e - cathode Cu 2+ (aq)+ 2e - --> Cu(s )

49. 49 Electrochemical Cell Summary salt bridge e-e- anode Zn (s)--> Zn 2+ (aq)+ 2e - cathode Cu 2+ (aq)+ 2e - --> Cu(s ) The differing stability of reactants, (Zn(s), Cu 2+ (aq)), and products (Zn 2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low). This manifests as a potential difference E cell, across the electrodes. Where -qE cell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode The cell potential is related to the free energy of the reaction according to the relation  G cell = -nFE cell The cell potential can be calculated knowing the standard reduction potentials. These can be used to find E o red for the reaction at the cathode, and E o ox (= - E o red ). Then E o cell = E o ox + E o red Zn 2+ (aq) + 2e - --> Zn(s)-0.76V Cu 2+ (aq) + 2e - --> Cu(s) E red =0.34V Zn(s) --> Zn 2+ (aq) + 2e - E ox = 0.76V E cell =1.1 V E cell = 0.76V+0.34V = 1.1V

50. 50 E° cell, ΔG° and K  for a spontaneous reaction  one proceeds in the forward direction with the chemicals in their standard states  ΔG° < 1 (negative)  E° > 1 (positive)  K > 1  ΔG° = −RTlnK = −nFE° cell  n is the number of electrons  F = Faraday’s Constant = 96,485 C/mol e −  for a spontaneous reaction  one proceeds in the forward direction with the chemicals in their standard states  ΔG° < 1 (negative)  E° > 1 (positive)  K > 1  ΔG° = −RTlnK = −nFE° cell  n is the number of electrons  F = Faraday’s Constant = 96,485 C/mol e −

51. 51 The Relationship between ΔG o and E o cell  Remember Δ G o is the maximum work the system can do  E o cell is the cell potential energy (standard emf) per unit of charge  Since the potential energy is the maximum amount of work that can be done on the surrounding we can write The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write  Remember Δ G o is the maximum work the system can do  E o cell is the cell potential energy (standard emf) per unit of charge  Since the potential energy is the maximum amount of work that can be done on the surrounding we can write The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write

52. 52 Example 18.6- Calculate ΔG° for the reaction I 2(s) + 2 Br − (aq) → Br 2(l) + 2 I − (aq) since ΔG° 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) ΔG , (J) Given: Find: E° ox, E° red E° cell Δ 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

53. 53 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

54. 54 Nonstandard Conditions - the Nernst Equation  ΔG = ΔG° + 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  ΔG = ΔG° + 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

55. 55 E° at Nonstandard Conditions

56. 56 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) + 3Cu 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) + 3Cu 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

57. 57 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 till the concentrations are equal  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 till the concentrations are equal

58. 58 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)

59. 59 LeClanche’ Acidic Dry CellDry Cell 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  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

60. 60 Alkaline Dry Cell  same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste  anode = Zn (or Mg) Zn(s) + 2OH -  Zn(OH) 2 (s) + 2 e -  cathode = graphite  MnO 2 is reduced 2 MnO 2 (s) + 2 H 2 O(l) + 2 e -  2 Mn(O)OH(s) + 2 OH - (aq)  Overall reaction Zn(s)+2MnO 2 (s)+2H 2 O(l)  Zn(OH) 2 (s) + 2MnO(OH)(s)  cell voltage = 1.54 V  longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc  same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste  anode = Zn (or Mg) Zn(s) + 2OH -  Zn(OH) 2 (s) + 2 e -  cathode = graphite  MnO 2 is reduced 2 MnO 2 (s) + 2 H 2 O(l) + 2 e -  2 Mn(O)OH(s) + 2 OH - (aq)  Overall reaction Zn(s)+2MnO 2 (s)+2H 2 O(l)  Zn(OH) 2 (s) + 2MnO(OH)(s)  cell voltage = 1.54 V  longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

61. 61 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  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

62. Tro, Chemistry: A Molecular Approach 62 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  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

63. Tro, Chemistry: A Molecular Approach 63 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  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

64. 64 Lithium Ion Battery  electrolyte is concentrated KOH solution  anode = graphite impregnated with Li ions  cathode = Li - transition metal oxide  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  electrolyte is concentrated KOH solution  anode = graphite impregnated with Li ions  cathode = Li - transition metal oxide  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

65. 65

66. 66 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 –  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 –

67. 67 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  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

68. 68

69. 69 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 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

70. 70 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  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

71. 71 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  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

72. 72 Electrolysis of Water

73. 73 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  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

74. 74 Electrolysis of NaCl (l)

75. 75 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  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

76. 76 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  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

77. 77 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 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 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) overall reaction 2 I − (aq) + 2 H 2 O (l)  I 2(aq) + H 2(g) + 2 OH -1 (aq)

78. 78 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  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

79. 79 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

80. 80 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  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

81. 81 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  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

82. 82

83. 83 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  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

84. 84 Sacrificial Anode Zn  Zn 2+ + 2e - 0.76V Fe  Fe 2+ + 2e - 0.45V

85. The Final 85 Long Questions on the Following: Kinetics Method of Initial Rates to find rate law Arrhenius plot to find Activation energy E a and the collision factor A Acid-Base Titration of a Weak Acid Calculate the initial pH of the acid given K a Calculate the pH in the buffer region using Henderson- Hasselbalch Calculate the pH at the end point using K b Galvanic and Electrolytic Cell An extra credit question on kinetics mechanisms (worth 15% of the final) Multiple Choice Redo for Electrochemistry A short (optional multiple choice on electrochemistry for those who want a second chance)