1. Electrochemistry
Chapter 19

2. 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.

3. 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 2+ 2-
2Mg (s) + O2 (g) MgO (s)
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)
19.1

4. Terminology for Redox Reactions
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.
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.

5. 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!
ose
lectrons
xidation
ain
lectrons
eduction
GER!

6. Another way to remember
OIL RIG s s
xidation
ose
eduction
ain

7. Review of Oxidation numbers
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.
4.4

8. HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +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.
Group IA metals are +1, IIA metals are +2 and fluorine is always –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.
HCO3-
Oxidation numbers of all the atoms in HCO3- ?
O = -2
H = +1
3x(-2) ? = -1
C = +4
4.4

9. Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Write the unbalanced equation for the reaction in ionic form.
Fe2+ + Cr2O Fe3+ + Cr3+
Separate the equation into two half-reactions.
Fe Fe3+ +2 +3
Oxidation:
Cr2O Cr3+ +6 +3
Reduction:
Balance the atoms other than O and H in each half-reaction.
Cr2O Cr3+
19.1

10. Balancing Redox Equations
For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O Cr3+ + 7H2O
14H+ + Cr2O Cr3+ + 7H2O
Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe Fe3+ + 1e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O
If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe Fe3+ + 6e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O
19.1

11. Balancing Redox Equations
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.
Oxidation:
6Fe Fe3+ + 6e-
Reduction:
6e- + 14H+ + Cr2O Cr3+ + 7H2O
14H+ + Cr2O Fe Fe3+ + 2Cr3+ + 7H2O
Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
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 H2O.
19.1

12. CHEMICAL CHANGE ---> ELECTRIC CURRENT
To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.
This is accomplished in a GALVANIC, VOLTAIC, or an ELECTROCHEMICAL cell.
A group of such cells is called a battery.

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

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

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

17. Galvanic Cells Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Line notation:
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
cathode
19.2

18. A common dry cell battery (for watches, calculators etc.)

19. NiO2 , H2O, OH- are reacting materials.
Anode: Cd, Cathode:Ni

20. Mercury battery, used in calculators

21. Standard Electrode Potentials
Standard reduction potential (E0red) 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.
Reduction Reaction
2e- + 2H+ (1 M) H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE) is a reference cell to determine E0red of the half cells.

22. The superscript ° indicates standard state conditions.

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

24. Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e-
Cathode (reduction):
2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)
19.3

25. E0 is for the reaction as written
The more positive E0red, the greater the driving force for reduction.
The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0
19.3

26. Standard Electrode Potentials
Standard emf (E0 )
cell
E0 = Ecathode + Eanode
cell
If the reaction is backwards, be sure to flip the sign!
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
E0 = EH /H + EZn /Zn
cell + +2 2
Zn2+ (1 M) + 2e Zn E0 = V
So Eo Zn/Zn = V +2
E0 = V = 0.76 V
cell
19.3

27. Standard Electrode Potentials
E0 = 0.34 V
cell
E0 = Ecathode + Eanode
cell
Ecell = ECu /Cu + E H /H+ 2+ 2
0.34 = ECu /Cu + - 0 2+
ECu /Cu = 0.34 V 2+
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm) H+ (1 M) + 2e-
Cathode (reduction):
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
19.3

28. In a voltaic cell,the cathode reaction is always the one that has the more positive value for E0red since the greater driving force of the cathode half rxn will force the anode rxn to occur “in reverse,” as an oxidation.

29. Cd is the stronger oxidizer
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e Cd (s) E0 = V
Cd is the stronger oxidizer
Cd will oxidize Cr
Cr3+ (aq) + 3e Cr (s) E0 = V
Anode (oxidation):
Cr (s) Cr3+ (1 M) + 3e-
x 2
Cathode (reduction):
2e- + Cd2+ (1 M) Cd (s)
x 3
2Cr (s) + 3Cd2+ (1 M) Cd (s) + 2Cr3+ (1 M)
E0 = Ecathode + Eanode
cell
E0 = (+0.74)
cell
E0 = 0.34 V
cell
19.3

30. Spontaneity of Redox Reactions
n = number of moles of electrons in reaction
F = 96,500 J
V • mol
DG0 = -nFEcell
= 96,500 C/mol
19.4

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

32. 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.

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

34. The Effect of Concentration on Cell Emf
DG0 = -nFE
Nernst equation
At 298 - V n
ln Q E
E = - V n
log Q E
E =
19.5

35. n: the number of electrons transferred in the rxn
F: Faraday’s constant (96,500 J/V-mol)
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

36. Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Oxidation:
Cd Cd2+ + 2e-
n = 2
Reduction:
2e- + Fe Fe
E0 = EFe /Fe + ECd /Cd 2+ 2+
E0 = (-0.40) - V n
ln Q E
E =
E0 = V - V 2 ln
-0.04 V
E =
0.010
0.60
E = 0.013
E > 0
Spontaneous
19.5

37. Calculate the emf at 298 K generated by the cell containing K2Cr2O7 and H2SO4 solutions in one compartment and a solution of KI in another compartment(a metallic conductor that will not react w/ either solution is suspended in each solution) when [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0M, and [Cr3+] = 1.0x10-5M.
Cr2O72− + 14H+ + 6I−  2Cr3+ + 7H2O + 3I2

38. Cr2O72− + 14H+ + 6e−  2Cr3+ + 7H2O E° = 1. 33V I2 + 2e−  2I− E° = 0
Cr2O72− + 14H+ + 6e−  2Cr3+ + 7H2O E° = 1.33V I2 + 2e−  2I− E° = 0.54V Answer: .89V

39. CONCENTRATION CELLS

40. 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.”

41. Anode: Cu  Cu2+ (.01M) + 2e-
Cathode: Cu2+ (1M) + 2e-  Cu
Overall: Cu2+ (1M)  Cu2+ (.01M)
At 298K;
E= /2 log .01/1 = .0592V

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

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

44. 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.

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

46. (a) A Standard Galvanic Cell (b) A Standard Electrolytic Cell
based on a spontaneous rxn:
Zn + Cu2+  Zn2+ + Cu
Zn + Cu2+ Zn2+ + Cu
electrical energy turns into chemical energy
Chemical energy turns into electrical energy
Copyright©2000 by Houghton Mifflin Company. All rights reserved.

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

48. Electrolysis of molten salts
Overall electrolytic cell rxn: AlCl3(l) Al(s) + 3/2Cl2(g)
C(graphite)
electrode
C(graphite)
electrode
Al deposits
Cathode:Al+3(l)+3e-Al(s)
Al+3
Molten AlCl3

49. 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.

50. Exercise Which substances will be obtained at anode and cathode?
C(graphite)
electrode
C(graphite)
electrode
Molten MgCl2 & KF

51. answer Tendency to lose e: K>Mg
At cathode: Mg solid will be obtained.
Tendency to gain e: F>Cl
At the anode: Cl2 gas will be obtained.

52. Electrolysis of Water
19.8

53. Electrolysis of aqueous solutions
We have to also consider whether the water is oxidized (to form O2) or reduced (to form H2) rather than the ions of the salt.
Possible reactants: Na+, Cl-, H2O

54. Tendency to be oxidized at anode:
2Cl---->Cl2 + 2e Eo=+1.36V
2H2O---> O2 + 4H+ + 4e- Eo =1.23V
- Cl- will be oxidized at the anode since the activation energy of Cl- is lower.So, it is kinetically favored.
Tendency to be oxidized at anode:
Anions of halogens (except F-)> H2O > SO4-2, CO3-2, NO3-
At cathode:
2  H2O + 2 e−  H2(g) + 2 OH− Eo= -0.83V
Na+ +  e−  Na(s) Eo= -2.71V
-Hydrogen gas will be produced at the cathode since the reduction of water is more favorable.
Tendency to be reduced at cathode:
Cations of semi/Noble metals > H2O > cations of IA &IIA metals

55. The electrolysis of aqueous NaCl(brine) is an important for the production of chlorine and NaOH.

56. Exercise
If we electrolyze the aqueous solutions of CuCl2, K2SO4 , KBr, and NaOH is separate containers,which substances will be obtained at anode and cathode for each solution?

57. answer
Anode Cathode CuCl2 - Cl2(g) Cu(s) K2SO4 - O2(g) H2(g) KBr - Br2(l) H2(g) NaOH  O2(g) H2(g)

58. Quantitative aspects of electrolysis
Faraday’s Law: the amount of a substance reduced/oxidized at each electrode during electrolysis is directly proportional to the quantity of electricity (the number of electrons) passed into the cell.(Mass α Q)
(Faradays)
Q (charge in Coulombs) = I (current in Amperes) x time (sec)
Q = I x t
1 mole e- = 96,500 C = 1 Faraday (F)
1 amp = 1 Coulomb / sec
19.8

59. How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours?
Anode:
2Cl- (l) Cl2 (g) + 2e-
Cathode:
2 mole e- = 1 mole Ca
Ca2+ (l) + 2e Ca (s)
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
mol Ca = 0.452 C s
x 1.5 hr x 3600 s hr
96,500 C
1 mol e- x
2 mol e-
1 mol Ca x
= mol Ca
= 0.50 g Ca
19.8

60. Quantitative aspects of electrolysis
The steps relating the quantity of electrical charge used in electrolysis to the amounts of substances oxidized/reduced:
(Faradays)
19.8

61. exercise
How many seconds does it take to produce 11.2 L of hydrogen gas measured at 760 mm-Hg and 27 °C by electrolysis of water using a current of 1.5 A?

62. answer
59,000s

63. exercise
Two electrolytic cells including molten MgCl2 and salt of X+n ions separately are connected in series.When some electrical current is passed through these cells for a while, 7.2 g solid Mg is collected in the first electrolytic cell and 64.8 g of solid X is collected in the second one.What is the formula of the compound that is formed between X+n ions and sulfate ions? (Mg: 24g/mol, X: 108g/mol)

64. answer
n=1

65. In the electrolytic cells connected in series:
The amount of electrical current that passes through the cells is the same. Therefore, the number of moles of electrons exchanged in the electrolytic cells is the same.

66. w/o the moisture, corrosion can’t happen since it acts as a kind of salt bridge.
Fe2O3.nH2O
Corrosion rxn are spontaneous redox rxns in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
19.7

67. Cathodic Protection of an Iron Storage Tank
19.7

68. Electroplating
In electroplating, electrodes participate in electrolysis.
Electroplating uses electrolysis to deposit a thin layer of one metal on another metal in order to improve beauty or resistance to corrosion. e-
cathode

69. Charging a Battery
When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal.
In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.

70. Batteries Dry cell Leclanché cell Zn (s) Zn2+ (aq) + 2e- Anode:
Cathode:
2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
19.6

71. Batteries Mercury Battery Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Cathode:
HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
19.6

72. Batteries Lead storage battery Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Cathode:
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4
19.6

73. Solid State Lithium Battery
Batteries
Solid State Lithium Battery
19.6

74. Batteries
A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
2H2 (g) + 4OH- (aq) H2O (l) + 4e-
Cathode:
O2 (g) + 2H2O (l) + 4e OH- (aq)
2H2 (g) + O2 (g) H2O (l)
19.6

75. Chemistry In Action: Dental Filling Discomfort
Hg2 /Ag2Hg V 2+
Sn /Ag3Sn V 2+
Sn /Ag3Sn V 2+