2. Electrochemistry

3. Electron Transfer Reactions Electron transfer reactions are oxidation- reduction or redox reactions. Results in the generation of an electric current (electricity) or be caused by imposing an electric current. Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

4. 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 19.1 Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 002+2-

5. Terminology for Redox Reactions OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel) REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized. OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel) REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized.

6. You can’t have one… without the other! Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons. You can’t have 2 oxidations only or 2 reductions only in the same equation. Reduction has to occur at the cost of oxidation LEO the lion says GER! o s e l e c t r o n s x i d a t i o n a i n l e c t r o n s e d u c t i o n GER!

7. Another way to remember OIL RIG x i d a t i o n s o s e e d u c t i o n s a i n

8. Review of Oxidation numbers The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1. 4.4

9. 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = -2H = +1 3x(-2) + 1 + ? = -1 C = +4 Oxidation numbers of all the atoms in HCO 3 - ? 4.4

10. Balancing Redox Equations 19.1 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr 3+ +6+3 Reduction: Fe 2+ Fe 3+ +2+3 3.Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2- 2Cr 3+

11. Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe 3+ + 1e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6.If necessary, equalize the number of electrons in the two half- reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe 3+ + 6e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 19.1

12. Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6Fe 2+ 6Fe 3+ + 6e - Oxidation: Reduction: 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 19.1 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation. You should combine H + and OH - to make H 2 O.

13. Reactivity Series for Metals

14. To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. CHEMICAL CHANGE ---> ELECTRIC CURRENT This is accomplished in a GALVANIC, VOLTAIC, or an ELECTROCHEMICAL cell. A group of such cells is called a battery. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

15. Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction - +

17. The potential difference between the two electrodes of a voltaic cell provides is callled, cell voltage electromotive (causing electron motion) force (emf) cell potential It’s denoted as E cell and is measured in Volts (V). This difference is the driving force that pushes electrons through the external circuit.

18. For any cell rxn happening spontaneously, the cell potential will be positive.

19. Galvanic Cells 19.2 Line notation: Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anodecathode

20. Standard Electrode Potentials Standard reduction potential (E 0 red ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M, 25°C and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) is a reference cell to determine E 0 red of the half cells. 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

21. The superscript ° indicates standard state conditions.

22. Whenever we assign a potential to a half-reaction, we write the reaction as a reduction.

23. Standard Electrode Potentials 19.3 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm)

24. 19.3 E 0 is for the reaction as written The more positive E 0 red, the greater the driving force for reduction. The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

25. 19.3 E 0 = 0 + 0.76 V = 0.76 V cell Standard emf (E 0 ) cell Zn 2+ (1 M) + 2e - Zn E 0 = -0.76 V E 0 = E H /H + E Zn /Zn cell 00 + 2 Standard Electrode Potentials E 0 = E cathode + E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) If the reaction is backwards, be sure to flip the sign! +2 So E o Zn/Zn = + 0.76 V +2

26. Standard Electrode Potentials 19.3 Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = E cathode + E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu + E H /H+ 2+ 2 00 0 0.34 = E Cu /Cu + - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

27. In a voltaic cell,the cathode reaction is always the one that has the more positive value for E 0 red since the greater driving force of the cathode half rxn will force the anode rxn to occur “in reverse,” as an oxidation.

28. What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 = E cathode + E anode cell 00 E 0 = -0.40 + (+0.74) cell E 0 = 0.34 V cell 19.3

29. Factors affecting the cell potential: Specific rxns that occur at the cathode and anode Concentrations of reactants and products temperature

30. Effect of concentration on cell EMF As a voltaic cell discharges, concentrations of reactants and products change. The emf drops until E=0, at which the cell is “dead”, the reactants and products are at equilibrium. Any effect that will shift the cell reaction to the right will increase E cell.

31. When is a good time to measure the cell potential of a cell that is constructed in the laboratory?

32. How will you calculate the emf of a voltaic cell generated at non- standard conditions? Use the Nernst equation (Walther Nernst, 1864-1941, a German chemist)

33. The Effect of Concentration on Cell Emf Nernst equation At 298 K 19.5 - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E =

34. n: the number of electrons transferred in the rxn F: Faraday’s constant (96,500 J/V-mol) R = constant (8.31 J/mole K) Q: Rxn quotient, determined in the same w/ Kc except that the concentrations used in Q are those that exist in the rxn mixture at a given moment. E: non-standard voltaic cell potential

35. Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) 2e - + Fe 2+ 2Fe Cd Cd 2+ + 2e - Oxidation: Reduction: n = 2 E 0 = -0.44 + -(-0.40) E 0 = -0.04 V E 0 = E Fe /Fe + E Cd /Cd 00 2+ - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.04 VE = 0.010 0.60 E = 0.013 E > 0Spontaneous 19.5

36. CONCENTRATION CELLS

37. The cell operates until the concentrations of reacting ions in two compartments become equal, at which point the cell has reached equilibrium and is “dead.”

38. Anode: Cu  Cu 2+ (.01M) + 2e - Cathode: Cu 2+ (1M) + 2e -  Cu Overall: Cu 2+ (1M)  Cu 2+ (.01M) At 298K; E= 0-.0592/2 log.01/1 =.0592V

39. How can you increase the cell potential of a concentration cell? By making the difference between concentrations of the ions higher.

40. 2) Temperature Temp ↑, voltaic cell potential ↓

41. Electrolysis Voltaic cells are based on spontaneous rxns. It’s possible to cause a nonspontaneous rxn to occur by using a source of electrical energy.This process is called electrolysis. Electrolysis occur in an electrolytic cell.

42. 19.8 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. electrolytic cell An electrolytic cell consists of two electrodes in a molten salt or solution. Source of direct electrical current: pushes e - into cathode & pulls e - from anode (Where reduction occurs) (Where oxidation occurs) -+ - + The Downs Cell for the Electrolysis of Molten Sodium Chloride

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. (a) A Standard Galvanic Cell (b) A Standard Electrolytic Cell based on a spontaneous rxn: Zn + Cu 2+  Zn 2+ + Cu Zn + Cu 2+  Zn 2+ + Cu Chemical energy turns into electrical energy electrical energy turns into chemical energy

44. By means of the movement of both anions and cations, the electrical current is carried in the electrolyte.

45. Electrolysis of molten salts Overall electrolytic cell rxn: AlCl 3(l)  Al(s) + 3/2Cl 2(g) C(graphite) electrode Al deposits C(graphite) electrode Al +3 Molten AlCl 3 Cathode:Al +3 (l)+3e -  Al(s)

46. In the electrolysis of molten salts: at the anode:at the cathode: The elemental form of the anion is obtained The elemental form of the cation is obtained Because of high melting points of metals, electrolysis of molten salts requires very high temperatures.

47. Exercise Which substances will be obtained at anode and cathode? C(graphite) electrode C(graphite) electrode Molten Al 2 O 3

48. This method has made it possible to obtain reactive metals such as Al from their ores.

49. Exercise Which substances will first be obtained at anode and cathode? C(graphite) electrode C(graphite) electrode Molten MgCl 2 & KF

50. answer Tendency to lose e: K>Mg At cathode: Mg solid will be obtained. Tendency to gain e: F>Cl At the anode: Cl 2 gas will be obtained.

51. Electrolysis of Water 19.8