1. Electrochemistry Applications of Redox Reactions

2. Review  Oxidation reduction reactions involve a transfer of electrons.  OIL- RIG  Oxidation Involves Loss  Reduction Involves Gain

3. Oxidation Numbers  Free elements = 0  Monatomic ions = charge of ion  Oxygen in most compounds = 2-  Hydrogen in most compounds = 1+  The sum of all oxidation numbers in a compound must = 0  The sum of all oxidation numbers in a polyatomic ion = charge of ion

4. Oxidizing and Reducing Agents  The oxidizing agent is the reactant that contains the reduced species in the reaction  The reducing agent is the reactant that contains the oxidized species in the reaction

5. Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. 2PbS + 3O 2 ⇁ 2 PbO + 2SO 2 2+ 2- 0 2+ 2- 4+ 2- Oxidized= sulfur Reduced = oxygen OA = oxygen gas RA = lead(II) sulfide

6. Try another one: Iron metal reacts with nitric acid to produce iron (III) nitrate and hydrogen gas. Fe + HNO 3 → Fe(NO 3 ) 3 + H 2 0 1+ 5+ 2- 3+ 5+ 2- 0 Oxidized = iron Reduced = hydrogen OA = nitric acid RA = iron metal

7. Try another one: 8H + +MnO 4 - + 5Fe +2  Mn +2 + 5Fe +3 +4H 2 O 1+ 7+ 2- 2+ 2+ 3+ 1+ 2- Oxidized = iron Reduced = manganese OA = permanganate ion RA = iron (II) ion

8. Electrochemistry  A study of the relationship between chemical reactions and electron movement  Spontaneous reactions can be used to produce electrical energy – Galvanic or Voltaic Cell = battery.  Nonspontaneous reactions can be forced to occur by applying electrical energy from an outside source – Electrolytic Cell = battery recharger or electroplating.

9. Applications 8H + +MnO 4 - + 5Fe +2  Mn +2 + 5Fe +3 +4H 2 O Break the reaction into half reactions:  Ox: 5Fe +2 → 5Fe +3 + 5e -  Red: 8H + + MnO 4 - + 5e - → Mn +2 + 4H 2 O Electrons are produced in the oxidation and consumed in the reduction.

10. Continued…  When the two half reactions are together in the same container no work can be done outside of the reaction  The half reactions must be physically separated so the electrons are forced to move through an outside conductor

11. H + MnO 4 - Fe +2  Connected this way the reaction starts but stops immediately because charge builds up. e-e- e-e- e-e-

12. H + MnO 4 - Fe +2 Galvanic Cell A Salt Bridge allows current to flow by balancing charge in each beaker

13. H + MnO 4 - Fe +2 e-e- e-e- Cl - A salt bridge can be a glass tube packed with cotton or a gel that is soaked in any electrolyte solution. The solution should contain ions that will not interfere with the reaction, and usually contains spectator ions that are those in the reaction. KCl (aq)

14. H + MnO 4 - Fe +2 Porous Disk l Instead of a salt bridge

15. Example Sketch the galvanic cell for the reaction: Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) 1)#s: 0 2+ 2+ 0 2)Half reactions: Ox: Zn → Zn 2+ + 2 e - Red: Cu 2+ + 2 e - → Cu 3) cell shorthand: Zn│ Zn 2+ ║ Cu 2+ │Cu

16. oxidationreduction e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode Semipermeable membrane Zn → Zn 2+ + 2 e - Cu 2+ + 2 e - → Cu Zn Cu SO 4 2- Zn 2+ Cu 2+ SO 4 2- V

17. Cell Potential  Oxidizing agent pulls the electron.  Reducing agent pushes the electron.  The push or pull (“driving force”) is called the cell potential (E cell ) or emf (electromotive force)  Unit is the volt(V) = joules of energy/coulombs of charge

18. oxidationreduction e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode Zn → Zn 2+ + 2 e - Cu 2+ + 2 e - → Cu Zn Cu SO 4 2- Zn 2+ Cu 2+ SO 4 2- 1.101 V Standard conditions: 25 C, 1 atm, 1 M

19. Zn +2 SO 4 -2 1 M HCl Anode 0.76 V 1 M ZnSO 4 H + Cl - H 2 in Cathode Zn + 2 HCl → ZnCl 2 + H 2 Pt Zn│Zn2+ ║ H + │H 2 │Pt

20. 1 M HCl H + Cl - H 2 in Standard Hydrogen Electrode  This is the reference all other oxidations are compared to  E º = 0 V º indicates standard states of : 25ºC 1 atm 1 M solutions.

21. Cell Potential  Consider the Zn-Cu Cell shown earlier Zn│ Zn 2+ ║ Cu 2+ │Cu  The total cell potential is the sum of the potential at each electrode.  We can look up reduction potentials in a table for each half reaction, but the voltage value for the oxidation half reaction will need to be reversed, so the sign must be changed.

22. Standard Reduction Potentials Zn 2+ + 2 e - → Zn E = -0.762 V Cu 2+ + 2 e - → Cu E = +0.339 V Flip the Zn cell to show oxidation: E = +0.762 V Sum the half cell potentials: E cell = 1.101 V

23. Cell Potential  A positive cell potential indicates a spontaneous chemical reaction.  ∆G = - n F E (F = 96485 C/mol e-)  Recall that spontaneous reactions have negative free energies.

24. Reduction potentials for half reactions: Higher positive reduction potential means:  more easily electron is added  More easily reduced  Better oxidizing agent (Ex: F 2 = +2.87 V)  Larger negative reduction potential means:  more likely to lose electrons than gain  More easily oxidized  Better reducing agent (Ex: Li + = - 3.05 V)

25. Cell Potential  Determine the cell potential for a galvanic cell based on the redox reaction and sketch the cell (show e- flow and ion flow through the salt bridge)  Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) Cu(s)  Cu +2 (aq) + 2 e- E = -0.34 V 2Fe +3 (aq) + 2e-  2Fe +2 (aq) E = +0.77 V E cell = +0.43 V

26. Cu 2+ Fe +3 Cu(s)  Cu +2 (aq)  Fe +3 (aq)│Fe +2 (aq)  Pt(s) Fe +2 Cu Fe Pt NO 3 - anode cathode NO 3 -

27. In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________. a. cathode, oxidation b. anode, reduction c. anode, oxidation d. cathode, reduction

28. Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd|Cd 2+ || Cu 2+ |Cu a. Cu 2+ + Cd → Cu + Cd 2+ b. Cu + Cd → Cu 2+ + Cd 2+ c. Cu 2+ + Cd 2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu 2+

29. Galvanic Cell  The reaction always runs spontaneously in the direction that produced a positive cell potential.  complete description: 1)Cell shorthand 2)Cell potential 3)Direction of flow of electrons and ions 4)Designation of anode and cathode 5)Nature of all the components- electrodes and ions in solution for each compartment (half cell)

30. Practice  Completely describe the galvanic cell based on the following half-reactions under standard conditions.  MnO 4 - + 8 H + +5e -  Mn +2 + 4H 2 O E º=1.51 V  Fe +3 +3e -  Fe(s) E º=0.036V

31. Potential, Work and  G – UIL!  emf = potential (V) = work (J) / charge(C)  Ɛ cell = work done by system / charge = -w/q  -w = q Ɛ cell  Faraday = 96,485 C/mol e -  q = nF = moles of e - x charge/mole e -  w = -q Ɛ cell = -nF Ɛ cell =  G

32. Cell Potential at Nonstandard Conditions When a cell isn’t at standard conditions, LeChâtelier’s Principle can be used to qualitatively determine if the expected voltage will be lower or higher than the standard cell potential. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) What is the standard cell potential? Predict if E cell will be greater or less than Eº cell if: [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M +0.48 V less more

33. The Nernst Equation

34. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) When [Al +3 ] = 5.0 M and [Mn +2 ] = 5.0 M at 25 C: Standard cell voltage = +0.48 V E = 0.48 – (-0.0069) = 0.49 V

35. Corrosion  Rusting is just spontaneous oxidation of the metal solid.

36. Preventing Corrosion  Painting/Sealing to keep out air and water.  Galvanizing - Putting on a zinc coating Zinc has a lower reduction potential, so it is more easily oxidized than many other metals.  Alloying with metals that form oxide coats.  Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

37. Electrolysis Cells  Basically this involves running a galvanic cell backwards…  It can be used to force a nonspontaneous reaction to occur (normally negative cell potential).  Used for electroplating, recharging a battery, electrolysis of water into hydrogen and oxygen, reclaiming metals from ores, and separating mixtures of ions.

38. 1.0 M Zn +2 e-e- Anode Cathode 1.10 V Zn Cu 1.0 M Cu +2 Galvanic cell Zn│ Zn 2+ ║ Cu 2+ │Cu

39. 1.0 M Zn +2 e-e- e-e- Anode Cathode battery >1.10V Zn Cu 1.0 M Cu +2 Electrolytic cell Cu│ Cu 2+ ║ Zn 2+ │Zn

40. Electrolysis of water From standard reduction chart: O 2 + 4 H + + 4 e - → 2 H 2 O E = 1.23 From standard reduction chart: 2 H 2 O + 2 e - → H 2 + 2 OH - E = -0.83 This is the cathodic reaction Reverse the reaction so that water Is a reactant (oxidation): 2 H 2 O → O 2 + 4 H + + 4 e - E = -1.23 This will be the anodic reaction Net rxn: 2 H 2 O → 2 H 2 + O 2 What occurs when water undergoes electrolysis? Identify the potential oxidation and reduction reactions that can occur with water as a reactant:

41. Electrolysis of molten NaCl From table: Na + + e - → Na (s) E = -2.71 cathode 2 Cl - → Cl 2 + 2 e - E = -1.36 anode From standard reduction table: Cl 2 + 2 e - → 2 Cl - E = 1.36 Reverse to show the chloride ions as reactants: Net rxn: 2 NaCl (aq) → 2 Na + Cl 2 What occurs when liquid phase salts undergo electrolysis? Identify the potential oxidation and reduction reactions that can occur with the free molten ions as a reactants: Na metal produced Cl 2 gas produced

42. Electrolysis of aqueous NaCl 2 Cl - → Cl 2 + 2 e - E = -1.36 Na + + e - → Na (s) E = -2.71 2 H 2 O → O 2 + 4 H + + 4 e - E = -1.23 2 H 2 O + 2 e - → H 2 + 2 OH - E = -0.83 What occurs when dissolved salt undergoes electrolysis? Identify the potential oxidation and reduction reactions that can occur with sodium ions, chloride ions, and water as reactants: Potential oxidations: Potential reductions: The reactions that require the least voltage will occur…water will be reduced and since both oxidations are close in voltage, the chloride will be oxidized.

43. Electroplating  Involves forcing metallic ions in solution to be reduced to form the metallic element.  Amperage in the electrolysis cell is a measure of the applied electrical current. Measured in amps (A) which are coulombs per second.

44. Calculations How long must 5.00 amp current be applied to produce 15.5 g of Ag from AgNO 3 (aq)?

45. Calculations Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

46. What current would be needed to plate 5.00 g Fe from an aqueous solution of Fe(NO 3 ) 3 if the cell runs for 10.0 hr? Calculations

47. AP Practice A student places a copper electrode in a 1 M solution of CuSO 4 and in another beaker places a silver electrode in a 1 M solution of AgNO 3. A salt bridge composed of Na 2 SO 4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt. (a) Draw a diagram of this cell. (b) Describe what is happening at the cathode (Include any equations that may be useful.) (c) Describe what is happening at the anode (d) Write the balanced net ionic reaction (e) Write the cell notation (f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH 3 ) + (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu 2+ (aq) concentration in the cell after the ammonia has been added?