1. Chapter 17
Electrochemistry

2. Contents Galvanic cells Standard reduction potentials
Cell potential, electrical work, and free energy
Dependence of cell potential on concentration
Batteries
Corrosion
Electrolysis
Commercial electrolytic processes

3. Oxidation reduction reactions
17.1 Galvanic Cells
Oxidation reduction reactions
Oxidation reduction reactions
involve a transfer of electrons.
Oxidation Involves Loss of electrons
Increase in the oxidation number
Reduction Involves Gain electrons
Decrease in the oxidation number

4. 8H++MnO4-+ 5Fe+2 ® Mn+2 + 5Fe+3 +4H2O
Example
8H++MnO4-+ 5Fe ® Mn+2 + 5Fe+3 +4H2O
If we break the reactions into half reactions.
8H++MnO4-+5e- ® Mn+2 +4H2O (Red)
5(Fe+2 ® Fe+3 + e- ) (Ox)
Electrons are transferred directly.
This process takes place without doing useful work

5. When the compartments of the two beakers are connected as shown the reaction starts
Current flows for an instant then stops
No flow of electrons in the wire, Why?
Current stops immediately because charge builds up. H+
MnO4-
Fe+2

6. Galvanic Cell H+ Solutions must be connected so ions can flow to
keep the net charge in each compartment zero
Salt Bridge allows ions to flow without extensive mixing in order to keep net charge zero. Electrons flow through the wire from reductant to oxidant
Oxidant
reductant H+
MnO4-
Fe+2

7. Porous Disk H+
MnO4-
Fe+2

8. e- e- e- e- Anode Cathode e- e- Fe2+ Reducing Agent
Oxidizing Agent MnO4-

9. Electrochemical Cells Spontaneous redox reaction
_______
__________
_______
__________
Electrochemical Cells
Spontaneous redox reaction
19.2

10. Thus a Galvanic cell is a device in which a chemical energy is changed to electrical energy
The electrochemical reactions occur at the interface between electrode and solution where the electron transfer occurs
Anode: the electrode compartment at which oxidation occurs
Cathode: the electrode compartment at which reduction occurs

11. Cell Potential Oxidizing agent pulls the electrons
Reducing agent pushes the electrons
The total push or pull (“driving force”) is called the cell potential, Ecell
Also called the electromotive force (emf)
Unit is the volt(V)
= 1 joule of work/coulomb of charge
Measured with a voltmeter

12. Measuring the cell potential
Can we measure the total cell potential??
A galvanic cell is made where one of the two electrodes is a reference electrode whose potential is known.
Standard hydrogen electrode (H+ = 1M
and the H2 (g) is at 1 atm) is used as a
reference electrode and its potential was assigned to be zero at 25 0C.

13. Standard Hydrogen Electrode
This is the reference all other oxidations are compared to
Eº = 0
(º) indicates standard states of 25ºC, 1 atm, 1 M solutions.
1 atm H2
H+ Cl-
1 M HCl

14. 0.76 H2
Cathode
Anode
H+ Cl-
Zn+2 SO4-2
1 M ZnSO4
1 M HCl

15. 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)

16. 17.2 Standard Reduction Potentials, E
The E values corresponding to reduction half-
reactions with all solutes at 1M and all gases at 1
atm.
E can be measured by making a galvanic cell in
which one of the two electrodes is the Standard
Hydrogen electrode, SHE, whose E = 0 V
The total potential of this cell can be measured
experimentally
However, the individual electrode potential can
not be measured experimentally.
Why?

17. Eº cell = EºZn® Zn2+ + Eº H+ ® H2
If the cathode compartment of the cell is SHE, then the half reaction would be
2H+ + 2e  H2 (g); Eo = 0V
And the anode compartment is Zn metal in Zn2+, (1 M) then the half reaction would be
Zn  Zn2+ + 2e
The total cell potential measured experimentally was found to be V
Thus, V was obtained as a result of this calculation:
Eº cell = EºZn® Zn2+ + Eº H+ ® H2
0.76 V V V

18. Standard Reduction Potentials
The E values corresponding to reduction half-
reactions with all solutes at 1M and all gases at
1 atm. can be determined by making them half
cells where the other half is the SHE.
E0 values for all species were determined as
reduction half potentials and tabulated. For
example:
Cu2+ + 2e  Cu E = 0.34 V
SO42 + 4H+ + 2e  H2SO3 + H2O E = 0.20 V
Li+ + e-  Li E = V

19. Some Standard Reduction Potentials
Li+ + e- ---> Li v
Zn e- ---> Zn v
Fe e- ---> Fe v
2 H+(aq) e- ---> H2(g) v
Cu e- ---> Cu v
O2(g) H+(aq) e- ---> 2 H2O(l) v
F e- ---> 2 F v

20. Standard Reduction Potentials at 25°C

21. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
The more negative E0 the greater the tendency for the substance to be oxidized
Under standard-state conditions, any species on the left of a given half-reaction will react spontaneously with a species that appears on the right of any half-reaction located below it in the table

22. 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

23. Can Sn reduce Zn2+ under standard-state conditions?
Look up the Eº values in in the table of reduction potentials
Zn e- ---> Zn(s) v
Sn e- ---> Sn v
How do we find the answer?
Look up the Eº values in in the table of reduction potentials
\Which reactions in the table will reduce Zn2+(aq)?

24. Standard cell potential
Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)
The total standard cell potential is the sum of the potential at each electrode.
Eº cell = EºZn® Zn2+ + Eº Cu+2 ® Cu
We can look up reduction potentials in a table.
One of the reactions must be reversed, in order to change its sign.

25. Standard Cell Potential
Determine the cell potential for a galvanic cell based on the redox reaction.
Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V 
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
Cu(s) ® Cu+2(aq)+2e Eº = V
2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V 
Eo cell = EoFe3+ ®Eo Fe2+ + EoCu®Eo Cu2+
Eo cell = (-0.34) = o.43 V

26. The total reaction: Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V
2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V 
Cu(s) + 2Fe+3(aq) Cu Fe2+
Eºcell = V

27. Line Notation Solid½Aqueous½½Aqueous½solid
Anode on the left½½Cathode on the right
Single line different phases.
Double line porous disk or salt bridge.
Zn(s)½Zn2+(aq)½½Cu2+½Cu
If all the substances on one side are aqueous, a platinum electrode is indicated.
For the last reaction
Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

28. Complete description of a Galvanic Cell
The reaction always runs spontaneously in the direction that produces a positive cell potential.
Four parameters are needed for a complete description:
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the components- electrodes and ions

29. Exercise
Describe completely the galvanic cell based on the following half-reactions under standard conditions.
MnO H+ +5e- ® Mn+2 + 4H2O Eº=1.51
Fe+3 +3e- ® Fe(s) Eº=0.036V
Write the total cell reaction
Calculate Eo cell
Define the cathode and anode
Draw the line notation for this cell

30. 17.3 Cell potential, electrical work and free energy
The work accomplished when electrons are transferred through a wire depends on the “push” (thermodynamic driving force) behind the electrons
The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit
emf = potential difference (V)
= work (J) / Charge(C) =

31. The work done by the system has a
–ve sign
Potential produced as a result of doing a work should have a +ve sign
The cell potential, E, and the work, w, have opposite signs.
Relationship between E and w can be expressed as follows:
E = work done by system / charge
( )

32. Charge is measured in coulombs. Thus,
-w = qE
Faraday = 96,485 C/mol e-
q = nF = moles of e- x charge/mole e-
w = -qE = -nFE = DG
Thus, DG = -nFE and
DGo = -nFEo

33. Potential, Work, DG and spontaneity
DGº = -nFE º
if E º > 0, then DGº < 0 spontaneous
if E º < 0, then DGº > 0 nonspontaneous
In fact, the reverse process is spontaneous.

34. Spontaneity of Redox Reactions
DG = -nFEcell
n = number of moles of electrons in reaction
F = 96,500 J
V • mol
DG0 = -nFEcell
= 96,500 C/mol
DG0 = -RT ln K
= -nFEcell
Ecell = RT nF
ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)
ln K = = V n
ln K
Ecell = V n
log K
Ecell

35. Spontaneity of Redox Reactions
If you know one, you can calculate the other…
If you know K, you can calculate Eº and Gº
If you know Eº, you can calculate Gº

36. Spontaneity of Redox Reactions
Relationships among G º, K, and Eºcell

37. Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) Al(s) + 3Mg+2(aq)
Oxidation:
3 (Mg Mg2+ + 2e-)
n = ?
Reduction:
2(3e- + Al Al)
E0 = Ered + Eox
cell
DG0 = -nFEcell
DG0 = -nFEcell
= ___ X (96,500 J/V mol) X ___ V
DG0 = _______ kJ/mol

38. 17.4 Dependence of Cell Potential on Concentration
Qualitatively: we can predict direction of change in E from LeChâtelier pinciple
2Al(s) + 3Mn+2(aq) Al+3(aq) + 3Mn(s); Eo cell = 0.48 V
Predict if Ecell will be greater or less than Eºcell for the following cases:
if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
An increase in conc. of reactants would favor forward reaction thus increasing the driving force for electrons; i.e. Ecell becomes > Eo cell

39. Concentration Cell: both compartments contain same
components but at different concentrations
Half cell potential are not identical
Because the Ag+ Conc. On both sides are not
same
Eright > Eleft
To make them equal, [Ag+]
On both sides should same
Electrons move from left to
right

40. Effect of Concentration on Cell Emf
The Nernst Equation
Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE
-nFE = -nFE0 + RT ln Q
E = E0 -
ln Q RT nF
Nernst equation
At 298K - V n
ln Q E
E = - V n
log Q E
E =

41. When equilibrium is reached Q = K ; Ecell = 0
The Nernst Equation
As reactions proceed concentrations of products increase and reactants decrease.
When equilibrium is reached
Q = K ; Ecell = 0
and G = 0 (the cell no longer has
the ability to do work)

42. Predicting spontaneity using Nernst equation
Qualitatively: we can predict the direction of change in E from Lechatelier principle
Find Q
Calculate E
E > 0; the reaction is spontaneous to the
right
E < 0; the reaction is spontaneous to the
left

43. 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 = ____________
E ___ 0
________________

44. Exercise- p. 843
Determine the cell potential at 25oC for the following cell, given that
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
[Mn2+] = 0.50 M; [Al3+]=1.50 M; E0 cell = 0.4
Always we have to figure out n from the balanced equation
2(Al(s)+ ® Al+3(aq) + 3e-)
3(Mn+2(aq) + 2e- ® Mn(s))
n = 6 - V n
log Q E
E =

45. Calculation of Equilibrium Constants
for redox reactions
At equilibrium, Ecell = and Q = K.
Then,
at 25 oC

46. What is the equilibrium constant for the following reaction at 250C
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = V n
ln K
Ecell
Oxidation:
2Ag Ag+ + 2e-
n = ___
Reduction:
2e- + Fe Fe
E0 = EFe /Fe+ EAg /Ag 2+ +
E0 = –0.80= V V
x n E0
cell
exp
K = V
x 2
-1.24 V
= exp
E0 = V
K = ________________

47. Batteries are Galvanic Cells
Lead-Storage Battery
A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V.
Cathode: PbO2 on a metal grid in sulfuric acid:
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- 
PbSO4(s) + 2H2O(l)
Anode: Pb:
Pb(s) + SO42-(aq)  PbSO4(s) + 2e-

48. Lead storage battery Anode: Pb (s) + SO2- (aq) PbSO4 (s) + 2e-
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

49. Lead-Storage Battery
The overall electrochemical reaction is
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq) 
2PbSO4(s) + 2H2O(l)
for which
Ecell = Ered(cathode) - Ered(anode)
= ( V) - ( V)
= V.
H2SO4 is consumed while the battery is discharging
H2SO4 is 1.28g/ml and must be kept
Water is depleted thus the battery should be topped off always

50. Dry cell Batteries Anode: Zn (s) Zn2+ (aq) + 2e- 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)

51. Dry Cell Battery
Anode: Zn cap:
Zn(s)  Zn2+(aq) + 2e-
Cathode: MnO2, NH4Cl and C paste:
2NH4+(aq) + 2MnO2(s) + 2e-  Mn2O3(s) + 2NH3(aq) + 2H2O(l)
Total reaction:
Zn + NH4+ +MnO2 ® Zn2+ + NH3 + H2O
This cell produces a potential of about 1.5 V.
The graphite rod in the center is an inert cathode.

52. Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)
Alkaline Cell Battery
For an alkaline battery, NH4Cl is replaced with KOH.
Anode: oxidation of Zn
Zn(s) + 2OH-  ZnO + H2O + 2e-
Cathode: reduction of MnO2.
2MnO2 + H2O + 2e-  Mn2O3 + 2OH-
Total reaction
Zn(s) MnO2(s) ---> ZnO(s) + Mn2O3(s)
It lasts longer because Zn anode corrodes
less rapidly than under acidic conditions.

53. Alkaline Battery

54. Nickel-Cadmium (Ni-Cad) Battery
Anode: Cd(s) + 2OH Cd(OH)2 + 2e-
Cathode: NiO2 + 2H2O + 2e ` Ni(OH)2 + 2OH-
NiO2 + Cd + 2H2O Cd(OH)2 +Ni(OH)2
NiCad v/cell
The products adhere to the
electrodes thus the battery can
be recharged indefinite number
of times.

55. 2H2 (g) + O2 (g) 2H2O (l) Fuel Cells
A fuel cell is a galvanic 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)

56. 17.6 Corrosion Rusting - spontaneous oxidation of metals.
Most metals used for structural purposes have reduction potentials that are less positive than O2 . (They are readily oxidized by O2)
Fe+2 +2e- ® Fe Eº= V
O2 + 2H2O + 4e- ® 4OH Eº= 0.40 V
When a cell is formed from these two half reactions a cell with +ve potential will be obtained
Au, Pt, Cu, Ag are difficult to be oxidized (noble metals)
Most metals are readily oxidized by O2 however, this process develops a thin oxide coating that protect the internal atoms from being further oxidized.
Al that has Eo = -1,7V is easily oxidized. Thus, it is used for making the body of the airplane.

57. e- Electrochemical corrosion of iron Water Rust
Salt speeds up process by increasing conductivity
Water
Cathodic area
Rust
Anodic area
Iron dissolves forming a pit e-
Fe Fe+2 + 2e-
Anodic reaction
O2 + 2H2O + 4e OH-
cathodic reaction
Fe2+ (aq) + O2(g) + (4-2n) H2O(l) ®
2F2O3(s).nH2O (s)+ 8H+(aq)

58. Fe on the steel surface is oxidized (anodic regions)
Fe ® Fe+2 +2e Eº= V
e-’s released flow through the steel to the areas that have O2 and moisture (cathodic regions). Oxygen is reduced
O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V
Thus, in the cathodic region Fe+2 will react with O2
The total reaction is:
Fe2+ (aq) + O2(g) + (4-2n) H2O(l) ®
2F2O3(s).nH2O (s)+ 8H+(aq)
Thus, iron is dissolved to form pits in steel
Moisture must be present to act as the salt bridge
Steel does not rust in the dry air
Salts accelerates the process due to the increase in conductivity on the surface

59. Preventing of Corrosion
Coating to keep out air and water.
Galvanizing - Putting on a zinc coat
Fe Fe e- Eoox = 0.44V
Zn Zn2+ + 2e- Eoox = 0.76 V
Zn has a more positive oxidation potential than Fe, so it is more easily oxidized.
Any oxidation dissolves Zn rather than Fe
Alloying is also used to prevent corrosion. stainless steel contains Cr and Ni that make make steel as a noble metal
Cathodic Protection - Attaching large pieces of an active metal like magnesium by wire to the pipeline that get oxidized instead. By time Mg must be replaced since it dissolves by time

60. Cathodic Protection of an Underground Pipe

61. Cathodic Protection of an Iron Storage Tank

62. 17.7 Electrolysis Running a galvanic cell backwards.
Put a voltage bigger than the galvanic potential and reverse the direction of the redox reaction.
Electrolysis: Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
That is causing a nonspontaneous reaction to occur
It is used for electroplating.

63. 1.10 e- e- Zn Cu 1.0 M Cu+2 1.0 M Zn+2 Anode Cathode
Galvanic cell based on spontaneous reaction:
Zn + Cu Zn Cu Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Anode
Cathode

64. e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode Electrolytic cell
A battery >1.10V e-
Zn Cu Zn + Cu2+ Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Cathode
Anode

65. Electrolytic Cell
Galvanic Cell

66. Calculating plating
How much chemical change occurs with the flow of a given current for a specified time?
Determine quantity of electrical charge in coulombs
Measure current, I (in amperes) per a period of time
1 amp = 1 coulomb of charge per second
coulomb of charge = amps X seconds =
q = I x t
q/nF = moles of metal
Mass of plated metal can then be calculated

67. Exercise
Calculate mass of Cu that is plated out when a current of 10.0 amps is passed for 30.0 min through a solution of Cu2+

70. Excercise
How long must 5.00 amp current be applied to produce 10.5 g of Ag from Ag+?

72. Electroplating
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?

73. How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
#g Cr =

74. (45 min)(60 sec)
#g Cr =
(1 min)

75. How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45) (60 sec) (25 amp)
#g Cr =
(1)

76. (45)(60 sec)(25 amp)(1 C)
#g Cr =
(1) (1 amp sec)

77. Faraday’s constant
(45)(25)(60)(1 C)(1 mol e-)
#g Cr =
(1)(1)(96,500 C)

78. (45)(60)(25)(1)(1 mol e-)(1 mol Cr)
#g Cr =
(1)(1)(96,500) (6 mol e-)

79. (45)(60)(25)(1)(1 mol e-)(52 g Cr)
(1)(1)(96,500) (1 mol Cr)

80. (45)(60)(25)(1)(1 mol e-)(52 g Cr)
Electroplating
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45)(60)(25)(1)(1 mol e-)(52 g Cr)
#g Cr =
(1)(1)(96,500)(6 mol e-)
= 58 g Cr

81. Michael Faraday Lecturing at the Royal Institution Before Prince Albert and Others (1855)

82. Electrolysis of Water

83. The Electrolysis of Water Produces Hydrogen Gas at the Cathode (on the Right) and Oxygen Gas at the Anode (on the Left)

84. Electrolysis of water

91. Other uses of electrolysis
Separating mixtures of ions.
More positive reduction potential means the reduction reaction proceeds forward.
We want the reverse.
Most negative reduction potential is easiest to plate out of solution.

92. 17.8 Commercial electrolytic processes

93. A Schematic Diagram of an Electrolytic Cell for Producing Aluminum by the Hall-Heroult Process.
To reduce mp of Al
From 2000 to 1000

94. The Downs Cell for the Electrolysis of Molten Sodium Chloride

95. The Mercury Cell for Production of Chlorine and Sodium Hydroxide

96. Metal Plating