This document discusses electrolysis and Faraday's law of electrolysis. It begins by introducing electrolytic cells and predicting the products of electrolysis. It then explains that according to Faraday's law, the amount of substance produced at each electrode is directly proportional to the quantity of electricity passing through the cell. The document provides an example calculation applying Faraday's law to determine the current needed in an electrolysis process. It also discusses some key aspects of Faraday's law, such as defining the faraday as one mole of electrons.

1. 21-1
ELECTROCHEMISTRY II
• ELECTROLYTIC CELLS
• PREDICTION OF PRODUCTS
• FARADAY’S LAW
• Chapter 21.4-21.7 Silberberg

2. 21-2
Chapter 21
Electrochemistry:
Chemical Change and Electrical Work

3. 21-3
Electrochemistry:
Chemical Change and Electrical Work
21.4 Free Energy and Electrical Work
21.5 Electrochemical Processes in Batteries
21.6 Corrosion: An Environmental Voltaic Cell
21.7 Electrolytic Cells: Using Electrical Energy to Drive
Nonspontaneous Reactions

4. 21-4
Goals & Objectives
• See the following Learning Objectives on
pages 969-970.
• Understand these Concepts:
• 21.13-16; 19-23.
• Master these Skills:
• 21.5-13.

5. 21-5
Products of Electrolysis
Electrolysis is the splitting (lysing) of a substance by the
input of electrical energy.
During electrolysis of a pure, molten salt, the cation will
be reduced and the anion will be oxidized.
During electrolysis of a mixture of molten salts
- the more easily oxidized species (stronger reducing agent) reacts at
the anode, and
- the more easily reduced species (stronger oxidizing agent) reacts at
the cathode.

6. 21-6
Sample Problem 21.8 Predicting the Electrolysis Products of a
Molten Salt Mixture
PROBLEM: A chemical engineer melts a naturally occurring mixture of
NaBr and MgCl2 and decomposes it in an electrolytic cell.
Predict the substance formed at each electrode, and write
balanced half-reactions and the overall cell reaction.
PLAN: We need to determine which metal and nonmetal will form more
easily at the electrodes. We list the ions as oxidizing or reducing
agents.
If a metal holds its electrons more tightly than another, it has a
higher ionization energy (IE). Its cation will gain electrons more
easily, and it will be the stronger oxidizing agent.
If a nonmetal holds its electrons less tightly than another, it has a
lower electronegativity (EN). Its anion will lose electrons more
easily, and it will be the reducing agent.

7. 21-7
Sample Problem 21.8
SOLUTION:
Possible oxidizing agents: Na+, Mg2+
Possible reducing agents: Br-, Cl-
Mg is to the right of Na in Period 3. IE increases from left to right across
the period, so Mg has the higher IE and gives up its electrons less
easily. The Mg2+ ion has a greater attraction for e- than the Na+ ion.
Br is below Cl in Group 7A. EN decreases down the group, so Br
accepts e- less readily than Cl. The Br- ion will lose its e- more easily, so
it is more easily oxidized.
Mg2+(l) + 2e- → Mg(l) [cathode; reduction]
2Br-(l) → Br2(g) + 2e- [anode; oxidation]
The overall cell reaction is: Mg2+(l) + 2Br-(l) → Mg(l) + Br2(g)

8. 21-8
Electrolytic Cells
• The Electrolysis of Molten Potassium
Chloride
– produces liquid potassium at one electrode
– produces gaseous chlorine at the other
• Diagram this cell using inert electrodes
– write the electrode reactions
– label the electrodes
– indicate the direction of electron flow

9. 21-9

10. 21-10
Electrolysis of Aqueous Salt Solutions
When an aqueous salt solution is electrolyzed
- The strongest oxidizing agent (most positive electrode potential) is
reduced, and
- The strongest reducing agent (most negative electrode potential) is
oxidized.
Overvoltage is the additional voltage needed (above
that predicted by E° values) to produce gases at metal
electrodes.
Overvoltage needs to be taken into account when
predicting the products of electrolysis for aqueous
solutions.
Overvoltage is 0.4 – 0.6 V for H2(g) or O2(g).

11. 21-11
• Cations of less active metals (Au, Ag, Cu, Cr,
Pt, Cd) are reduced to the metal.
• Cations of more active metals are not reduced.
H2O is reduced instead.
• Anions that are oxidized, because of
overvoltage from O2 formation, include the
halides, except for F-.
• Anions that are not oxidized include F- and
common oxoanions. H2O is oxidized instead.
Summary of the Electrolysis of Aqueous Salt Solutions

12. 21-12
Sample Problem 21.9 Predicting the Electrolysis Products of
Aqueous Salt Solutions
PROBLEM: What products form at which electrode during electrolysis of
aqueous solution of the following salts?
(a) KBr (b) AgNO3 (c) MgSO4
PLAN: We identify the reacting ions and compare their electrode
potentials with those of water, taking the 0.4 – 0.6 V overvoltage
into account. The reduction half-reaction with the less negative
E° occurs at the cathode, while the oxidation half-reaction with
the less positive E° occurs at the anode.
Despite the overvoltage, which makes E for the reduction of water
between -0.8 and -1.0 V, H2O is still easier to reduce than K+, so
H2(g) forms at the cathode.
SOLUTION:
(a) KBr K+(aq) + e- → K(s) E° = -2.93
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V

13. 21-13
(b) AgNO3 Ag+(aq) + e- → Ag(s) E° = 0.80 V
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V
Sample Problem 21.9
2Br-(aq) → Br2(l) + 2e- E° = 1.07 V
2H2O(l) → O2(g) + 4H+(aq) + 4e- E° = 0.82 V
The overvoltage makes E for the oxidation of water between 1.2
and 1.4 V. Br- is therefore easier to oxidize than water, so Br2(g)
forms at the anode.
As the cation of an inactive metal, Ag+ is a better oxidizing agent
than H2O, so Ag(s) forms at the cathode.
NO3
- cannot be oxidized, because N is already in its highest (+5)
oxidation state. Thus O2(g) forms at the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e-

14. 21-14
(c) MgSO4 Mg2+(aq) + 2e- → Mg(s) E° = -2.37 V
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V
Sample Problem 21.9
Mg2+ is a much weaker oxidizing agent than H2O, so H2(g) forms at
the cathode.
SO4
2- cannot be oxidized, because S is already in its highest (+6)
oxidation state. Thus O2(g) forms at the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e-

15. 21-15
Electrolytic Cells
• The Electrolysis of Aqueous Potassium
Chloride Solution
– produces gaseous hydrogen at one electrode
and the solution becomes basic at this electrode
– produces gaseous chlorine at the other
• Diagram this cell using inert electrodes
– write the electrode reactions
– label the electrodes
– indicate the direction of electron flow

16. 21-16

17. 21-17
Electrolytic Cells
• The Electrolysis of Aqueous Potassium
Sulfate Solution
– produces hydrogen at one electrode and the
solution becomes basic at this electrode
– produces gaseous oxygen at the other and the
solution becomes acidic at this electrode
• Diagram this cell using inert electrodes
– write the electrode reactions
– label the electrodes
– indicate the direction of electron flow

18. 21-18

19. 21-19
Electrode Products
• Generalization:
– In all electrolytic cells the most easily
reduced species is reduced and the most
easily oxidized species is oxidized.

20. 21-20
Electrode Products
• Prediction of Electrode Products
• Anode
– Cl-, Br-, I- -> Cl2, Br2, I2 respectively
– all other anions produce O2

21. 21-21
Electrode Products
• Prediction of Electrode Products
• Cathode
– uses the Activity Series or Electromotive Series
– neutral solutions:
• above Zn -> H2
• Zn and below -> metal
– acid solutions:
• H and above -> H2
• below H -> metal

22. 21-22
Activity Series
• A Partial Activity Series of the Elements
– Li
– Na See page 132 in McMurry
– Al
– Zn
– Cr See page 171 in Silberberg
– Ni 5th edition
– H
– Ag See page 165 in Silberberg
– Pt 6th edition
– Au

23. 21-23
Electrode Products
• Predict the products expected for each of
the following electrolytic cells:
• Anode Cathode
• NaBr(molten)
– Na lies above Zn
• MgI2(aq)
– Mg lies above Zn

24. 21-24
Electrode Products
• Predict the products expected for each of
the following electrolytic cells:
• Anode Cathode
• ZnSO4(aq)
• Ni(NO3)(aq,H+)
– Ni lies above H
• CuSO4(aq)
– Cu lies below Zn

25. 21-25
Figure 21.28 The electrolysis of water.
Oxidation half-reaction
2H2O(l) → 4H+(aq) + O2(g) + 4e-
Reduction half-reaction
2H2O(l) + 4e- → 2H2(g) + 2OH-(aq)
Overall (cell) reaction
2H2O(l) → H2(g) + O2(g)

26. 21-26
Table 21.3 Some Ions Measured with Ion-Specific Electrodes
Species Detected Typical Sample
NH3/NH4
+ Industrial wastewater, seawater
CO2/HCO3
- Blood, groundwater
F- Drinking water, urine, soil, industrial
stack gases
Br- Grain, plant tissue
I- Milk, pharmaceuticals
NO3
- Soil, fertilizer, drinking water
K+ Blood serum, soil, wine
H+ Laboratory solutions, soil, natural
waters

27. 21-27
Figure 21.14 Minimocroanalysis.
A microelectrode records electrical impulses of a single neuron in a
monkey’s visual cortex. The electrical potential of a nerve cell is due to
the difference in concentration of [Na+] and [K+] ions inside and outside
the cell.

28. 21-28
Faraday’s Law of Electrolysis
• The amount of substance undergoing
chemical reaction at each electrode during
electrolysis is directly proportional to the
amount of electricity passing through the
electrolytic cell.
• The unit faraday refers to the amount of
electricity that reduces one equivalent of a
species at the cathode and oxidizes one
equivalent at the anode.

29. 21-29
Michael Faraday
• Michael Faraday,
(September 21, – August
25, 1867) was an English
chemist and physicist who
contributed to the fields of
electromagnetism and
electrochemistry..

30. 21-30
Andre- Marie Ampere
• André-Marie Ampère
(January 20, 1775 – June
10, 1836), was a French
physicist who is generally
credited as one of the main
discoverers of
electromagnetism. The SI
unit of measurement of
electric current, the ampere,
is named after him

31. 21-31
Stoichiometry of Electrolysis
Faraday’s law of electrolysis states that the amount of
substance produced at each electrode is directly
proportional to the quantity of charge flowing through
the cell.
The current flowing through the cell is the amount of
charge per unit time. Current is measured in amperes.
Current x time = charge

32. 21-32
Figure 21.29 A summary diagram for the stoichiometry of
electrolysis.
MASS (g)
of substance
oxidized or
reduced
AMOUNT (mol)
of substance
oxidized or
reduced
CHARGE
(C)
CURRENT
(A)
AMOUNT (mol)
of electrons
transferred
M (g/mol)
balanced
half-reaction
Faraday
constant
(C/mol e-)
time (s)

33. 21-33
Sample Problem 21.10 Applying the Relationship Among Current,
Time, and Amount of Substance
PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by
electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for
the plating, what current is needed?
PLAN: To find the current, we divide charge by time, so we need to find
the charge. We write the half-reaction for Cr3+ reduction to get
the amount (mol) of e- transferred per mole of Cr. We convert
mass of Cr needed to amount (mol) of Cr. We can then use the
Faraday constant to find charge and current.
divide by M
mass (g) of Cr needed
mol of Cr
3 mol e- = 1 mol Cr
divide by time in s
mol e- transferred Charge (C)
1 mol e- = 9.65x104 C
current (A)

34. 21-34
Sample Problem 21.10
SOLUTION:
Cr3+(aq) + 3e- → Cr(s)
0.86 g Cr x 1 mol Cr
52.00 g Cr
x 3 mol e-
1 mol Cr
= 0.050 mol e-
Charge (C) = 0.050 mol e- x 9.65x104 C
1 mol e-
= 4.8x103 C
Current (A) =
charge (C)
time (s)
=
4.8x103 C
12.5 min
x 1 min
60 s
= 6.4 C/s = 6.4 A

35. 21-35
Faraday’s Law of Electrolysis
• A faraday corresponds to the gain or loss
of one mole of electrons.
• Thus
– one faraday = 6.022x1023 electrons
– one faraday = one mole of electrons

36. 21-36
Faraday’s Law of Electrolysis
• A coulomb is defined as the amount of
charge that passes a given point when a
current of one ampere(amp) flows for
one second.
– one coulomb = amp x sec
• Or
– one amp = one Coulomb per second(C/s)

37. 21-37
Faraday’s Law of Electrolysis
• A faraday is equivalent to 96,487
coulombs or more frequently 9.65x104C
– one faraday = one mole of electron
– one faraday = 9.65x104C
– 9.65x104C = one mole of electrons

38. 21-38
Faraday’s Law of Electrolysis
• Determine the mass of palladium
produced by the reduction of
palladium(II) ions to palladium metal
during the passage of 3.20 amperes of
current through a solution of
palladium(II) sulfate for 30.0 minutes.
• Determine the volume of oxygen
(measured at STP) produced in this cell.

39. 21-39

40. 21-40

41. 21-41
Faraday’s Law of Electrolysis
• Determine the products, and amounts of
each expected, when an acidic solution
of NiSO4 is electrolyzed by a current of
5.46amps for 2.00 hours. Ni lies above H
on the activity series.

42. 21-42

43. 21-43

44. 21-44

45. 21-45
Faraday’s Law of Electrolysis
• Determination of the charge on an ion by
electrolysis
– An aqueous solution of a palladium salt was
electrolyzed for 1.00 hours with a current of
1.50amperes. This produced 2.977g of Pd
metal at the cathode. Determine the charge
on the palladium ion in this salt.

46. 21-46

47. 21-47

48. 21-48
Corrosion is the process whereby metals are oxidized to
their oxides and sulfides.
Corrosion: an Environmental Voltaic Cell
The rusting of iron is a common form of corrosion.
- Rusting requires moisture, and occurs more quickly at low pH, in
ionic solutions, and when the iron is in contact with a less active
metal.
- Rust is not a direct product of the reaction between Fe and
O2, but arises through a complex electrochemical process.

49. 21-49
The Rusting of Iron
Fe(s) → Fe2+(aq) + 2e- [anodic region; oxidation]
O2(g) + 4H+(aq) + 4e- → 2H2O(l) [cathodic region; reduction]
The loss of iron:
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) [overall]
The rusting process:
2Fe2+(aq) + ½O2(g) + (2 + n)H2O(l) → Fe2O3·nH2O(s) + 4H+(aq)
Overall reaction:
H+ ions are consumed in the first step, so lowering the pH increases the
overall rate of the process. H+ ions act as a catalyst, since they are
regenerated in the second part of the process.

50. 21-50
Figure 21.22 The corrosion of iron.

51. 21-51
Figure 21.23 Enhanced corrosion at sea.
The high ion concentration of seawater enhances the corrosion of
iron in hulls and anchors.

52. 21-52
Figure 21.24 The effect of metal-metal contact on the corrosion
of iron.
Fe in contact with Cu corrodes
faster.
Fe in contact with Zn does not
corrode. The process is known
as cathodic protection.

53. 21-53
Figure 21.25 The use of sacrificial anodes to prevent iron corrosion.
In cathodic protection, an active metal, such as zinc, magnesium, or
aluminum, acts as the anode and is sacrificed instead of the iron.

54. 21-54
Electrolytic Cells
An electrolytic cell uses electrical energy from an
external source to drive a nonspontaneous redox
reaction.
Cu(s) → Cu2+(aq) + 2e- [anode; oxidation]
Sn2+(aq) + 2e- → Sn(s) [cathode; reduction]
Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s) E°cell = -0.48 V and ΔG° = 93 kJ
As with a voltaic cell, oxidation occurs at the anode and
reduction takes place at the cathode.
An external source supplies the cathode with electrons,
which is negative, and removes then from the anode,
which is positive. Electrons flow from cathode to anode.

55. 21-55
Figure 21.26 The tin-copper reaction as the basis of a voltaic and
an electrolytic cell.
voltaic cell
Sn(s) → Sn2+(aq) + 2e-
Cu2+(aq) + 2e- → Cu(s)
Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq)
electrolytic cell
Cu(s) → Cu2+(aq) + 2e-
Sn2+(aq) + 2e- → Sn(s)
Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

56. 21-56
Figure 21.27 The processes occurring during the discharge and
recharge of a lead-acid battery.
VOLTAIC (discharge)
ELECTROLYTIC (recharge)
Switch

57. 21-57
Table 21.4 Comparison of Voltaic and Electrolytic Cells
Cell Type DG Ecell
Electrode
Name Process Sign
Voltaic
Voltaic
< 0
< 0
> 0
> 0
Anode
Cathode
Oxidation
Reduction
-
+
Electrolytic
Electrolytic
> 0
> 0
< 0
< 0
Anode
Cathode
Oxidation
Reduction -
+

58. 21-58
Zinc Air Hearing Aid Batteries
• Zinc–air batteries are electro-chemical
batteries powered by oxidizing zinc with
oxygen from the air. These batteries have high
energy densities and are relatively inexpensive
to produce. Sizes include very small button
cells for hearing aids

59. 21-59
Zinc Air Hearing Aid Batteries
• Here are the chemical equations for the
zinc–air cell:
• Anode: Zn + 4OH– → Zn(OH)4
2– + 2e–
• Cathode: 1/2 O2 + H2O + 2e– → 2OH–
• Overall: 2Zn + O2 → 2ZnO (E0 = 1.59 V)