ELECTROCHEMISTRY

1. ELECTROCHEMISTRY Electrochemistry involves the relationship between electrical energy and chemical energy. OXIDATION-REDUCTION REACTIONS SPONTANEOUS REACTIONS Can extract electrical energy from these. Examples: voltaic cells, batteries. NON-SPONTANEOUS REACTIONS Must put in electrical energy to make them go. Examples: electrolysis, electrolytic cells. QUANTITATE REACTIONS

2. OXIDATION-REDUCTION Oxidation = loss of electrons. Occurs at ANODE An oxidizing agent is a substance that causes oxidation (and is itself reduced). Reduction = gain of electrons. At CATHODE A reducing agent is a substance that causes reduction (and is itself oxidized). LEO goes GER LEO: Loss of Electrons = Oxidation GER: Gain of Electrons = Reduction

3. Rules for determining Oxidation States • For pure elements, oxidation state = 0 • (examples: S8, P4, O3, F2, K, Be) • In compounds, some elements have “common” oxidation numbers: • Assign others by difference (using charge on ion) Alkali metals (Na+, K+..) +1 Alkaline earths (Mg2+, Ba2+…) +2 F = -1, O = -2 (but -1 in peroxides) Cl, Br, I = -1 (except if bonded to O or halogen) H = +1 or -1 (depends what it is bonded to)

4. BALANCING REDOX REACTIONS • 1. Write incomplete half-reactions. • 2. Balance each half-reaction separately. • a. Balance atoms undergoing redox. • b. Balance remaining atoms • i. Add H2O to balance oxygens. • ii. Add H+ to balance hydrogens. • 3. Balance charges by adding electrons. • Multiply half-reactions to cancel electrons. • 5. Add the half-reactions. • 6. In basic solutions, add OH to neutralize H+

5. Half Reactions The half-reactions forSn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)Sn2+(aq)  Sn4+(aq) +2e-2Fe3+(aq) + 2e- 2Fe2+(aq)Oxidation:electrons are products.Reduction:electrons are reagents.

6. Balancing Redox Reactions In Acid solution.:C + HNO3 NO2 + CO2 + H2OBasic solution.:PbO2 + Cl + OH  Pb(OH)3 + ClO

7. VOLTAIC CELL The energy released in a spontaneous redox reaction is used to perform electrical work.Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit.Voltaic cells are spontaneous.If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.Zn0(s) + Cu2+(aq)  Zn2+(aq) + Cu0(s)

8. E0cell = 1.10 V

9. Voltaic cells consist ofAnode: Zn(s)  Zn2+(aq) + 2e-Cathode: Cu2+(aq) + 2e- Cu(s) The two solid metals are the electrodes (cathode and anode).Salt bridge: (used to complete the electrical circuit): Anions and cations move to compensate excess charge.

10. Zn is oxidized to Zn2+ and 2e. The electrons flow to the anode where they are used in the reduction reaction.We expect the Zn electrode to lose mass and the Cu electrode to gain mass.“Rules” of voltaic cells:1. At the anode electrons are products.Oxidation occurs at the anode.2. At the cathode electrons are reagents.Reduction occurs at the cathode3. Electrons cannot swim!

11. VOLTAIC CELLS Cell voltage (EMF or Ecell) is the measure of rxn spontaneity Ecell:Intensive property, energy per electronCell voltage depends on: the individual species undergoing oxidation and reduction and their concentrationsThe more spontaneous a reaction, the higher the voltage (more positive) the higher the Keq the more negative the G0

12. STANDARD POTENTIAL FOR AN ELECTROCHEMICAL CELL The standard potential for an electrochemical cell is the potential (voltage) generated when reactants and products of a redox reaction are in their standard states. Standard States: T = 25°C. Gases, P = 1 atm. [Solutes] = 1M Solids, liquids = pure

13. The potential associated with the half-reaction. • Rules for half-cell potentials: • The sum of two half-cells potentials in a cell equals the overall cell potential: • E°cell = E°1/2(oxid) + E°1/2(reduc) • 2. For any half-reaction: • E°1/2(oxid) =  E°1/2(reduc) • 3. Standard half-cell is a hydrogen electrode: • H2(g,1atm) = 2H+ (aq, 1M) + 2e • E°1/2(oxid) = E°1/2(reduc) = 0 V HALF-CELL POTENTIAL

14. E0cell = 0.76 V

15. E0cell = 1.10 V

16. E01/2 (Zn  Zn2+) = ? E01/2 (Cu2+ Cu) = ?

18. Oxidizing and Reducing Agents The more positive Ered the stronger the oxidizing agent on the left.The more negative Ered the stronger the reducing agent on the right.A species that is higher and to the left will spontaneously oxidize one that is lower and to the right in the table.Example: F2 will oxidize H2 or Li;Ni2+ will oxidize Al(s).

20. Are the following reactions spontaneous? If so, evaluate E0cell Cu(s) + Cl2(g)  Cu2+(aq) + 2Cl(aq) 2Cu2+(aq) + 2H2O  Cu(s) + O2(g) + 4H+(aq)

21. RELATIONSHIP BETWEEN G AND E G = nFE At Standard State:G° = nFE° n = number of electrons transferred in a balanced redox reaction F= Faraday = 96,500 coulomb/mole e- 1 coulomb = 1 Amp-sec 1 J = 1 Amp-sec-V = 1 coulomb-V 1 F = 96,500 J/V-mole e

22. Example: Is this reaction spontaneous? Cd(s) + 2H+ Cd2+ + H2 (g) Cd2+ + 2e Cd(s) Eored= 0.403V 2H+ + 2e H2 Eored = 0 Reaction goes as written (reduction) for more positive redox couple. H+ cathode: 2H+ + 2e H2 Cd anode: Cd  Cd2+ + 2e Eocell = Eored (cathode) + Eooxid (anode) Eocell = 0 + 0.403 = + 0.403 Go = nFEo=  2 x 96,500 J/V-mol x 0.403V =  8.3 x 104 = 83 kJ/mol Yes: the reaction is spontaneous!

23. For electrochemical cell at equilibrium: G = 0 G° = 2.303 RT log KeqG° = nFE°E° = 2.303 RT log Keq nFR = 8.314 J/K-moleF = 96,500 J/V-mole eAt 25°C = 298K:E°= (0.0592) log Keq n

24. Effect of Concentration Standard states: 1M solution, 1 atm gas pressure What if concentrations are different?

25. Effect of Concentration So for a half reaction: aA + bB + n e cC + dD

26. Examples What is Ecell for a fuel cell running in air (PO2 = 0.2 atm), at pH = 2, with PH2 = 1 atm? What is the half cell potential of the Ag/Ag+ redox couple (E0 = +0.799 V) in a 1 M NaCl solution that contains solid AgCl (Ksp = 1.1 x 10-10)? O2(g) + 4H+(aq) + 4e 2H2O(l) Eo = +1.229V 2H+ + 2e H2 Eo = 0

27. Batteries Lead/Acid Battery DURING DISCHARGE Anode: Pb(s) + SO42(aq)  PbSO4(s) + 2e Cathode: PbO2(s) + SO42(aq) + 4H+ + 2e PbSO4(s) + 2H2O Overall: Pb(s) + PbO2(s) + 2H2SO4 2PbSO4(s) + 2H2O

28. Lead-Acid Battery

29. Batteries DRY CELL Anode: Zn(s) Zn(s)  Zn2+(aq) + 2e Cathode: NH4Cl + MnO2 + graphite paste 2NH4+(aq) + 2MnO2(s) + 2e  Mn2O3(s) + 2NH3(aq) + H2O In ALKALINE CELL, NH4Cl is replaced by KOH. Provides up to 50% more energy. Zn is used as a powder mixed with the electrolyte.

30. Dry Cell Alkaline Battery

31. Batteries Rechargeable Nickel-Cadmium Battery Anode: Cd metal Cd(s) + 2OH(aq)  Cd(OH)2(s) + 2e Cathode: NiO2(s) NiO2(s) + 2H2O + 2e  Ni(OH)2(s) + 2OH(aq) Overall: Cd(s) + NiO2(s) + 2H2O  Cd(OH)2(s) + Ni(OH)2(s)

32. Batteries FUEL CELLS H2-O2 Fuel Cell - expensive but light. Used in spacecraft. Anode: 2H2(g) + 4OH(aq)  4H2O(l) + 4e Cathode: O2(g) + 2H2O(l) + 4e  4OH(aq) Overall: 2H2(g) + O2(g)  2H2O(l) E0cell = 1.23 V

33. Corrosion Differential aeration mechanism

34. Corrosion CATHODIC PROTECTION OF IRON

35. Corrosion CATHODIC PROTECTION OF IRON

36. ELECTROLYSIS Electrolysis: Driving non-spontaneous reactions by applying electrical energy. An electrolysis cell consists of two electrodes in either aqueous solution (of ions) or in a molten salt e.g. molten NaCl. The anode is where oxidation occurs. Anions migrate to the anode and lose electrons. The cathode is where reduction occurs. Cations migrate to the cathode and gain electrons.

37. ELECTROLYSIS OF MOLTEN NaCl Cathode: 2Na+ + 2e 2Na(l) Anode:2 Cl Cl2(g) + 2e ___________________________ 2 Na+ + 2 Cl 2Na(l) + Cl2(g)

38. ELECTROLYSIS OF NaCl, cont. • Electrolysis of aqueous NaCl: • Cathode: H2O + 2e H2(g) + 2OH • Anode: 2 Cl Cl2(g) + 2e • ___________________________________ • 2 H2O + 2 Cl H2(g) + Cl2(g) + 2OH • so • 2 Na+ + 2 OH = 2 NaOH is left behind • It is easier to reduce H2O than Na+ • The easiest (least non-spontaneous) reaction happens

39. ELECTROLYSIS - examples What products will form when an aqueous solution of ZnBr2 is electrolyzed? What products will form when aqueous AgNO3 is electrolyzed?

40. ELECTROLYSIS OF AQUEOUS Na2SO4 Cathode: 4H2O + 4e 2H2(g) + 4OH Anode: 2H2O  O2(g) + 4H+ + 4e _____________________________ 6H2O  2H2(g) + O2(g) + 4H+ + 4OH or 2H2O  2H2(g) + O2(g)

41. COMMERCIAL APPLICATIONS OF ELECTROLYSIS Production of metals – Na, Al. Purification of Metals – Cu. Electroplating.

42. Purification of Copper Cathode: thin sheet of pure copper Cu2+ + 2e Cu(s) Anode: impure copper Cu(s)  Cu2+ + 2e As the reaction proceeds, Cu moves from anode to cathode.

43. 1 mole of e- = 1 Faraday = 96,500 Coulombs = charge on 1 mole of e- 1 Ampere = 1 coulomb/second 1 coulomb = 1 Amp-sec Electromotive Force (EMF) force that cases electrons to flow (voltage) 1 Watt = 1 Amp-Volt 1 Joule = 1 coul-Volt = 1 Amp-sec-Volt = 1 Watt-sec 1 kW-hour = (1000 Watt)(3600 sec) = 3.6 x 106 Watt-sec = 3.6 x 106 Joules ELECTROLYSIS CALCULATIONS

44. ELECTROLYSIS CALCULATION Electrolysis gives 1.00g of Cu from CuSO4 Reaction is: Cu2+ + 2e Cu How many Faradays (F) of charge are required? How many Coulombs is this?

45. ELECTROLYSIS CALCULATION If 1.00g of Cu is obtained in 1 hour, how many amps are required? If 2 amps were used, how long would it take to produce 1g?