1. Electrochemistry
Chapter 20

2. Redox Reactions
Reduction – Oxidation, or redox, involves the transfer of electrons
Reduction – gain of electrons
Oxidation – loss of electrons

3. Redox Reactions LEO the lion goes GER Lose Electrons Oxidation
Gain Electrons Reduction

4. Redox Reactions OIL RIG Oxidation Is Losing (OIL)
Reduction Is Gaining (RIG)

5. Oxidation Numbers (States)
Positive, negative or neutral values assigned to an atom to keep track of the number of electrons lost or gained.
Charge

6. Oxidation Number Rules
Elements alone = 0
Monatomic ion = charge
Compound Σ atoms = 0
Polyatomic ion Σ atoms = charge

7. Common Oxidation Numbers
Group 1 +1
Group 2  +2
Group 13  +3
Group 15  -3
Group 16  -2
Group 17  -1
Some exceptions to each above

8. Oxidation Number Exceptions
H in metal hydrides
H-1 when paired with a metal
NaH, LiH
O in peroxide
O2-2 (O-1)
H2O2
2 nonmetals
More electronegative element will be negative
OF2 F-1 O+2

9. Equations Net Ionic Half Reactions
Shows only the ions involved in the redox reaction, not spectator ions
Half Reactions
Only shows one element and how many electrons are gained or lost (Regents)

10. Half Reactions Zn + CuSO4  Cu + ZnSO4 Zn + Cu+2  Cu + Zn+2
Net Ionic
Zn  Zn e-
Oxidation
Cu e-  Cu
Reduction

11. Electrochemical Cells
Voltaic (Chemical)
Spontaneous reaction based on potential difference between metals
Electrolytic
Non-spontaneous reaction using an outside power source

12. Electrodes Cathode – electrode where reduction takes place
Red Cat
Anode – electrode where oxidation takes place
An Ox

13. Homework
Read Chapter 20 Sections 1 and 2
( )

15. Balancing Reactions
The number of electrons lost must equal the number of electrons gained
Example:
2Na + ZnCl2  Zn + 2NaCl
Zn e-  Zn
2(Na Na + + e- )

16. Balancing Redox Reactions in Acidic solution
Balance all elements except H, O
Balance O by adding H2O
Balance H by adding H+
Balance charge by adding e-
Multiply half reactions to balance electrons lost/gained
Combine & cancel like species

17. Example PbO2(s) + I-(aq) → Pb+2(aq) + I2(g) 2 I- → I2 + 2e- 2e- +
Oxidation
2e- +
4H+ +
PbO2 → Pb+2
+ 2H2O
Reduction
2e- + 4H+ + 2I- + PbO2 → Pb+2 + I2 + 2H2O + 2e-
4H+ + 2I- + PbO2 → Pb+2 + I2 + 2H2O

18. Balancing Redox Reactions in Basic solution
Follow balancing for acidic solution
Add OH- to both sides equally to balance out H+
Combine H+ and OH- to make H2O
Combine and cancel like species

19. Example MnO4-(aq) + NO2-(aq) → MnO2(s) + NO3-(aq) 4OH- + 2( 3e- +
Reduction
4OH- + 2(
3e- +
3e- + 4H2O +
4H+ +
MnO4- → MnO2
+ 2H2O
+ 4OH- ) 3(
2OH- +
H2O +
NO2- → NO3-
+ 2H2O + 2e-
+ 2H+
+ 2e-
+ 2OH- )
Oxidation
6e- + 6OH- + 11H2O + 2MnO4- + 3NO2- → 2MnO2 + 3NO H2O + 8OH- + 6e-
H2O + 2MnO4- + 3NO2- → 2MnO2 + 3NO3- + 2OH-

20. Homework (Re)Read Chapter 20.2 Answer questions 20, 22

22. Electrochemical Cells
Voltaic (Galvanic)
Spontaneous reaction based on potential difference between metals
Electrolytic
Non-spontaneous reaction using an outside power source

23. Electrodes Cathode – electrode where reduction takes place
Red Cat
Anode – electrode where oxidation takes place
An Ox

24. Spontaneous Reaction
Some metals are more likely to ionize than another metals
Lose electrons (oxidize)
Activity Series from Regents
When 2 metals are paired together, one will lose and one will gain electrons

25. Voltaic Cell

28. Potential Difference (V)
Amount of work done per unit charge as a charged particle is moved between points
From Physics
Units V = J/C
𝑉= 𝑊 𝑞

29. Electromotive force (emf)
The difference in the potential between the 2 metals is what drives the electrons to move
From high potential to low potential

30. Half Reaction Potential
Measured with a standard hydrogen electrode to determine the potential of each half reaction
Reported as Reduction Potentials
Table 20.1
Appendix E

32. Cell Potential, E Difference between potential of cathode and anode
Cell emf
𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝐹𝑖𝑛𝑎𝑙 − 𝐸 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝐶𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸 𝐴𝑛𝑜𝑑𝑒

33. Standard Cell Potential, E°
Ecell when at standard conditions
Concentration = 1M
Standard emf
𝐸° 𝑐𝑒𝑙𝑙 = 𝐸° 𝐶𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸° 𝐴𝑛𝑜𝑑𝑒

34. Example Ecell Zn + Cu+2  Zn+2 + Cu Cu+2 + 2e-  Cu E = +0.34V
Zn+2 + 2e-  Zn E = -0.76V
Ecell = (-0.76) = 1.10V

35. Example Ecell 2Al + 3I2  2Al+3 + 6I- I2 + 2e-  2I- E = +0.54V
Al+3 + 3e-  Al E = -1.66V
Ecell = (-1.66) = 2.20V

36. Homework (Re)Read Chapter 20.3-20.4 Answer questions 6, 34, 36, 38

38. Spontaneous ΔG° = – RT∙ln(K) A reaction is spontaneous if: ΔG < 0

39. Spontaneous ΔG = – nFE ΔG° = – nFE° Relating E to ΔG
n = number of electrons transferred in reaction

40. Faraday Constant Electron = 1.6x10-19C 1 mole of electrons = 96,485 C
F = 96,485 C/mol e-

41. Example ΔG° = – nFE° ΔG° = – (2)(96485)(1.10) ΔG° = – 212 kJ
Zn + Cu+2  Zn+2 + Cu
E°cell = 1.10V
ΔG° = – nFE°
ΔG° = – (2)(96485)(1.10)
ΔG° = – 212 kJ

42. Example ΔG° = – nFE° ΔG° = – (6)(96485)(2.20) ΔG° = – 1274 kJ
2Al + 3I2  2Al+3 + 6I-
E°cell = 2.20V
ΔG° = – nFE°
ΔG° = – (6)(96485)(2.20)
ΔG° = – 1274 kJ

43. Electricity Review What is current?
Amount of charges flowing per unit time
Unit : A = C/s
𝐼= 𝑞 𝑡

44. Example
Suppose an electrochemical cell runs for 15 minutes with a current of 0.10A. For the following reaction, how much would the mass of the cathode increase by?
Zn + Cu+2  Zn+2 + Cu
9.3𝑥 10 −4 𝑚𝑜𝑙 𝑒 − 2 𝑒 − =4.7x 10 −4 𝑚𝑜𝑙
𝑞=𝐼𝑡= 0.1𝐴 900𝑠 =90𝐶
(90𝐶) (96485𝐶/𝑚𝑜𝑙) =9.33𝑥 10 −4 𝑚𝑜𝑙 𝑒 −
4.7𝑥 10 − =0.030 𝑔

45. Homework (Re)Read Chapter 20.5 Answer questions #50, 52, 54, 56
Quiz Wed

47. Non-Standard Conditions
∆𝐺= ∆𝐺°+𝑅𝑇𝑙𝑛𝑄
−𝑛𝐹𝐸=−𝑛𝐹𝐸°+𝑅𝑇𝑙𝑛𝑄
𝐸=𝐸°− 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄

48. Non-Standard Conditions
𝐸=𝐸°− 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄
𝐸=𝐸°− 2.303𝑅𝑇 𝑛𝐹 𝑙𝑜𝑔𝑄
𝐸=𝐸°− 𝑛 𝑙𝑜𝑔𝑄

49. Nernst Equation
𝐸=𝐸°− 𝑛 𝑙𝑜𝑔𝑄

50. Non-Standard Conditions
𝐸=𝐸°− 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]

51. Concentration Cells 𝐸=𝐸°− 0.0592 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]
Potential difference can be provided by a difference in concentration
𝐸=𝐸°− 𝑛 𝑙𝑜𝑔 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]

52. Concentration Cells
E°cell = 0
[reactants]<[products]

54. Electrolysis
Process of using electrical current to drive a non-spontaneous reaction
Electrolytic Cell
Oxidation at anode
Reduction at cathode
E (–)
Electroplating
Electropolishing

55. Electrolysis Can be used to separate compounds into their elements
2H2O  2H2 + O2