1. Electric Power Chemical Reactions
Electric power conversion in electrochemistry
Electrolysis / Power consumption
Electric Power
Chemical Reactions
Electrochemical battery / Power generation

2. Sign or polarity of electrodes
Cell
Construction - +
battery
power
source e- e-
conductive
medium
(-)
(+)
vessel
inert
electrodes
Sign or polarity of electrodes

3. Let’s examine the electrolytic cell for molten NaCl.
What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)?
Na+
Cl-
Let’s examine the electrolytic cell for molten NaCl.

4. Observe the reactions at the electrodes
Molten NaCl
Observe the reactions at the electrodes - +
battery
Cl2 (g) escapes
Na (l)
NaCl (l)
Na+
Cl-
Na+
Cl-
(-)
(+)
electrode half-cell
electrode half-cell
Cl-
Na+
Na+ + e-  Na
2Cl-  Cl e-

5. At the microscopic level
Molten NaCl
At the microscopic level - +
battery e-
NaCl (l)
cations
migrate
toward
(-)
electrode
anions
migrate
toward
(+)
electrode
Na+
Cl-
Na+ e-
Cl-
(-)
(+)
anode
cathode
Cl-
Na+
2Cl-  Cl e-
Na+ + e-  Na

6. Molten NaCl Electrolytic Cell
cathode half-cell (-)
REDUCTION Na+ + e-  Na
anode half-cell (+)
OXIDATION 2Cl-  Cl e-
overall cell reaction
2Na Cl-  2Na Cl2
X 2
Non-spontaneous reaction!

7. Definitions: CATHODE REDUCTION occurs at this electrode ANODE
OXIDATION occurs at this electrode

8. Will the half-cell reactions be the same or different?
What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)?
Na+
Cl-
H2O
Will the half-cell reactions be the same or different?

9. Water Complications in Electrolysis
In an electrolysis, the most easily oxidized and most easily reduced reaction occurs.
When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown.
Electrode Ions ... Anode Rxn Cathode Rxn E°
Pt (inert) H2O H2O(l)+ 2e- g H2(g)+ 2OH-(aq) V
H2O H2O(l) g 4e H+(g) + O2(g) V
Net Rxn Occurring:
2 H2O g 2 H2(g)+ O2 (g) E° = V

10. What could be reduced at the cathode?- +
Aqueous NaCl
battery
power
source e- e-
NaCl (aq)
What could be reduced at the cathode?
Na+
Cl-
(-)
(+)
H2O
cathode
different half-cell
anode
2Cl-  Cl e-

11. Aqueous NaCl Electrolytic Cell
possible cathode half-cells (-)
REDUCTION Na+ + e-  Na
2H2O + 2e-  H OH-
possible anode half-cells (+)
OXIDATION 2Cl-  Cl e-
2H2O  O H e-
overall cell reaction
2Cl H2O  H2 + Cl OH-

12. For every electron, an atom of silver is plated on the electrode.
Ag e-  Ag e-
Electrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO3 (aq) under standard conditions, will deposit silver at the rate of
g Ag/sec
Ag+ Ag
1 amp = g Ag/sec

13. Faraday’s Law Q = It 1 coulomb = 1 amp-sec = 0.001118 g Ag
The mass deposited or eroded from an electrode depends on the quantity of electricity.
Quantity of electricity – coulomb (Q)
Q is the product of current in amps times time in seconds
Q = It
time in seconds
coulomb
current in amperes (amp)
1 coulomb = 1 amp-sec = g Ag

14. molemetal depends on the half-cell reaction
Ag e-  Ag
1.00 mole e- = 1.00 mole Ag = g Ag
g Ag/mole e-
g Ag/coul
= 96,485 coul/mole e-
1 Faraday (F )
mole e- = Q/F
mass = molemetal x MM
molemetal depends on the half-cell reaction

15. Examples using Faraday’s Law
How many grams of Cu will be deposited in 3.00 hours by a current of 4.00 amps?
Cu e-  Cu
The charge on a single electron is x coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e-.

18. 21-8 Industrial Electrolysis Processes
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General Chemistry: Chapter 21

19. A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode. e- - +
battery - + - + - + e- e- e-
1.0 M Au+3
1.0 M Zn+2
1.0 M Ag+
Au e-  Au
Zn e-  Zn
Ag+ + e-  Ag

20. Paper moisturized with NaCl solution
Volta’s battery (1800)
Alessandro Volta
Paper moisturized with NaCl solution Cu Zn

21. Galvanic cells and electrodes
4/6/2017
Galvanic cells and electrodes
To sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.
Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution.
Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential).
Lesson 9
NEEP 423
Test

22. with Galvanic Cells anode oxidation cathode reduction spontaneous
redox reaction
19.2

23. Electrodes are passive (not involved in the reaction)
Olmsted Williams
Electrodes are passive (not involved in the reaction)

24. Observe the electrodes to see what is occurring.
Cell
Construction
Salt bridge –
KCl in agar
Provides conduction
between half-cells
Observe the electrodes to see what is occurring. Cu Zn
1.0 M CuSO4
1.0 M ZnSO4

25. - + What about half-cell reactions?
What about the sign of the electrodes? - +
Why?
cathode half-cell
Cu e-  Cu
anode half-cell
Zn  Zn e- Cu
plates out or deposits on electrode
Zn electrode erodes
or dissolves
What happened at each electrode? Cu Zn
1.0 M CuSO4
1.0 M ZnSO4

26. Now replace the light bulb with a volt meter.- +
1.1 volts
cathode half-cell
Cu e-  Cu
anode half-cell
Zn  Zn e- Cu Zn
1.0 M CuSO4
1.0 M ZnSO4

27. We need a standard electrode to make measurements against!
The Standard Hydrogen Electrode (SHE)
H2 input
1.00 atm
25oC
1.00 M H+
1.00 atm H2 Pt
Half-cell
2H e-  H2
inert
metal
EoSHE = 0.0 volts
1.00 M H+

28. How do we calculate Standard Redox Potentials?
We must compare the half reactions to a standard
What is that standard?
2 H3O+(aq) + 2e-  H2(g) H2O(l) E°= 0.00 V
This is called the standard hydrogen electrode or SHE
Now that we have a standard, we can calculate standard
redox potential by using the table of standard redox potentials

29. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
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

30. Oxidizing and Reducing Agents
Cell EMF
Oxidizing and Reducing Agents
Copyright 1999, PRENTICE HALL
Chapter 20

31. Now let’s combine the copper half-cell with the SHE
Eo = v +
0.34 v
cathode half-cell
Cu e-  Cu
anode half-cell
H2  2H e-
H atm
KCl in agar Cu Pt
1.0 M CuSO4
1.0 M H+

32. Now let’s combine the zinc half-cell with the SHE
Eo = v -
0.76 v
anode half-cell
Zn  Zn e-
cathode half-cell
2H e-  H2
H atm
KCl in agar Pt Zn
1.0 M ZnSO4
1.0 M H+

33. Assigning the Eo
Write a reduction half-cell, assign the voltage measured, and the sign of the electrode to the voltage.
Al+3 + 3e-  Al Eo = v Zn+2 + 2e-  Zn Eo = v 2H+ + 2e-  H2 Eo = 0.00 v Cu+2 + 2e-  Cu Eo = Ag+ + e-  Ag Eo = v
Increasing activity

34. Measuring Standard Reduction Potential
anode
cathode
cathode
anode
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General Chemistry: Chapter 21

35. 2Cr (s) + 3Cd2+ (1 M)  3Cd (s) + 2Cr3+ (1 M)
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

36. Calculating the cell potential, Eocell, at standard conditions
H2O with O2
Consider a drop of oxygenated water on an iron object Fe
Fe e-  Fe Eo = v
reverse 2x
Fe  Fe e- -Eo = v
O2 (g) H2O e-  4 OH- Eo = v
2Fe + O2 (g) H2O  2Fe(OH)2 (s) Eocell= v
This is corrosion or the oxidation of a metal.

37. E=? -Concentration -Related half reaction kinetic e- +2Fe3+  Fe E0 = e- +2H+  H2 E0= Fe +H+ Fe !!!!!!!!!!!!!!!!!!!!!!!! =================================================================================================== e- +Fe 3+  Fe 2+ E0= e +Fe2+  Fe E0=-0.44 Fe Fe Fe ================================================= Cu Cu Cu

38. Equilibrium Constants in Redox Reactions
Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.
DG° = –nFE°cell
and DG° = –RT ln Keq
therefore –RT ln Keq = –nFEocell
RT ln Keq RT
E°cell = ––––––––– = –––– ln Keq
nF nF
R and F are constant, therefore at 298 K: V
E°cell = –––––––– ln Keq n

39. 1) 2e +Fe2+  Fe E0=-0. 440 2) e +Fe3+  Fe2+ E0=-0
1) 2e +Fe2+  Fe E0= ) e +Fe3+  Fe2+ E0= e +Fe3+  Fe E0=??? ===============

40. 1)G0=-2(-0. 440)F=+0. 880F 2)G0=-1(+0. 771)F=-0
1)G0=-2(-0.440)F=+0.880F 2)G0=-1(+0.771)F=-0.771F F G0 =-nfE0=+0.109F=3FE0 E)=-0.036

41. Effect of Concentration on Cell EMF
A voltaic cell is functional until E = 0 at which point equilibrium has been reached.
The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction.
The Nernst Equation
The Nernst equation relates emf to concentration using
and noting that

42. and the previous relationship: DGo = -nFEocell
from thermodynamics:
DGo = RT log K
and the previous relationship:
DGo = -nFEocell
-nFEocell = RT log K
at 25oC: Eocell = log K n
where n is the number of electrons for the balanced reaction

43. What happens to the electrode potential if conditions are not at standard conditions?
The Nernst equation adjusts for non-standard conditions
For a reduction potential: ox + ne  red
in general: E = Eo – RT ln (red)
nF (ox)
at 25oC: E = Eo log (red)
n (ox)
Calculate the E for the hydrogen electrode where 0.50 M H+ and 0.95 atm H2.

44. Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)
1) An example:
Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)
According to the reduction potentials:
2 e- + Ni Ni(s) V
2 e- + Sn Sn(s) V
One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.

45. Ni(s) + Sn2+34 Ni2+ + Sn(s) 0.090 V = Eo
Ni(s) e- + Ni V
2 e- + Sn Sn(s) V
Ni(s) + Sn2+34 Ni2+ + Sn(s) V = Eo

46. E) Calculation of the equilibrium constant
1) at equilibrium E = 0 ; Q = ____
From the Nernst Equation:
For the cell:
Ni(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)

48. Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s)
2) An example:
Sn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s)
According to the reduction potentials:
2 e- + Pb Pb(s) E0= V
2 e- + Sn Sn(s) E0=-0.136V
One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!

49. E=E0-0.059/2log[Sn2+]/[Pb2+] E=-0.079

50. Free Energy and the Cell Potential
Cu  Cu e- -Eo = 2x
Ag+ + e-  Ag Eo = v
Cu Ag+  Cu Ag
Eocell= v
DGo = -nFEocell
where n is the number of electrons for the balanced reaction
What is the free energy for the cell?
1F = 96,500 J/v

51. Electrolysis of Copper
A net reaction of zero, yet a process does take place.
A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions.

52. Cu+Cu2+(1.0 M)Cu2+(0.1M)+Cu E=E0-0.059/2Log(0.1/1.0) =+0.0296

53. Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)
Chemistry 140 Fall 2002
Concentration Cells
Two half cells with identical electrodes but different ion concentrations.
Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)
2 H+(1 M) + 2 e- → H2(g, 1 atm)
H2(g, 1 atm) → 2 H+(x M) + 2 e-
2 H+(1 M) → 2 H+(x M)
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General Chemistry: Chapter 21

54. Concentration Cells Ecell = Ecell° - log Q n 0.0592 V
2 H+(1 M) → 2 H+(x M)
Ecell = Ecell° log n V x2 12
Ecell = log 2 V x2 1
Ecell = V log x
Ecell = ( V) pH
Prentice-Hall © 2002
General Chemistry: Chapter 21

55. F)The pH meter is a special case of the Nernst Equation
1/2 H2  H+ + 1 e-
Q = [H+]

56. pH = - log [H+]

57. E = Eo pH
The scale on the pH meter is marked off so that a change of 1 pH unit equals volts or 59.2 millivolts.

58. The pH Meter
In practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!
The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH.
A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode.

59. Comparison of Electrochemical Cells
galvanic
electrolytic
need
power
source
produces electrical current
two
electrodes
anode (-)
cathode (+)
anode (+)
cathode (-)
conductive medium
salt bridge
vessel
DG > 0
DG < 0

60. Corrosion Corrosion of Iron
Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.
Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l).
Anode: Fe(s)  Fe2+(aq) + 2e-.
Dissolved oxygen in water usually causes the oxidation of iron.
Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).
Oxidation occurs at the site with the greatest concentration of O2.
Copyright 1999, PRENTICE HALL
Chapter 20

61. 21-6 Corrosion: Unwanted Voltaic Cells
In neutral solution:
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
EO2/OH- = V
2 Fe(s) → 2 Fe2+(aq) + 4 e-
EFe/Fe2+ = V
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)
Ecell = V
In acidic solution:
O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = V
Prentice-Hall © 2002
General Chemistry: Chapter 21

62. Corrosion
Corrosion of Iron
Copyright 1999, PRENTICE HALL
Chapter 20

63. Corrosion of Iron

64. Corrosion of Iron 2Fe3+(aq) + 4H2O(l)  Fe2O3·H2O(s) + 6H+(aq)
Rust formation:
4Fe2+(aq) + O2(g) + 4H+(aq)  4Fe3+(aq) + 2H2O(l)
2Fe3+(aq) + 4H2O(l)  Fe2O3·H2O(s) + 6H+(aq)

65. Prevention of Corrosion
Cover the Fe surface with a protective coating
Paint
Tin Zn
Galvanized iron

66. Corrosion Preventing the Corrosion of Iron
Corrosion can be prevented by coating the iron with paint or another metal.
Galvanized iron is coated with a thin layer of zinc.
Zinc protects the iron since Zn is the anode and Fe the cathode:
Zn2+(aq) +2e-  Zn(s), Ered = V
Fe2+(aq) + 2e-  Fe(s), Ered = V
With the above standard reduction potentials, Zn is easier to oxidize than Fe.
Copyright 1999, PRENTICE HALL
Chapter 20

67. Corrosion Protection Prentice-Hall © 2002
General Chemistry: Chapter 21

68. Cathodic Protection
galvanized steel (Fe)

69. Cathodic Protection
(anode)
(cathode)
(electrolyte)

70. Cathodic Protection
In cathodic protection, an iron object to be protected is connected to a chunk of an active metal.
The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.
Water heaters often employ a magnesium anode for cathodic protection.