1. Chapter 18 Electrochemistry
Presented by: JYOTI KIRAN PGT (Chemistry)
K.V. NAGROTA ,JAMMU REGION

2. Redox Reaction one or more elements change oxidation number
all single displacement, and combustion,
some synthesis and decomposition
always have both oxidation and reduction
split reaction into oxidation half-reaction and a reduction half-reaction
aka electron transfer reactions
half-reactions include electrons
oxidizing agent is reactant molecule that causes oxidation
contains element reduced
reducing agent is reactant molecule that causes reduction
contains the element oxidized

3. Oxidation & Reduction oxidation is the process that occurs when
oxidation number of an element increases
element loses electrons
compound adds oxygen
compound loses hydrogen
half-reaction has electrons as products
reduction is the process that occurs when
oxidation number of an element decreases
element gains electrons
compound loses oxygen
compound gains hydrogen
half-reactions have electrons as reactants

4. Rules for Assigning Oxidation States
rules are in order of priority
free elements have an oxidation state = 0
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
monatomic ions have an oxidation state equal to their charge
Na = +1 and Cl = -1 in NaCl
(a) the sum of the oxidation states of all the atoms in a compound is 0
Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

5. Rules for Assigning Oxidation States
(b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion
N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
(a) Group I metals have an oxidation state of +1 in all their compounds
Na = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their compounds
Mg = +2 in MgCl2

6. Rules for Assigning Oxidation States
in their compounds, nonmetals have oxidation states according to the table below
nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example F -1
CF4 H +1
CH4 O -2
CO2
Group 7A
CCl4
Group 6A
CS2
Group 5A -3
NH3

7. Oxidation and Reduction
oxidation occurs when an atom’s oxidation state increases during a reaction
reduction occurs when an atom’s oxidation state decreases during a reaction
CH O2 → CO2 + 2 H2O –
oxidation
reduction

8. Oxidation–Reduction oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
the reactant that reduces an element in another reactant is called the reducing agent
the reducing agent contains the element that is oxidized
the reactant that oxidizes an element in another reactant is called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent

9. Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O

10. Identify the Oxidizing and Reducing Agents in Each of the Following
red ag
ox ag
3 H2S + 2 NO3– H+ ® 3 S + 2 NO + 4 H2O
MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O
reduction
oxidation
ox ag
red ag
reduction
oxidation

11. Common Oxidizing Agents

12. Common Reducing Agents

13. Balancing Redox Reactions
assign oxidation numbers
determine element oxidized and element reduced
write ox. & red. half-reactions, including electrons
ox. electrons on right, red. electrons on left of arrow
balance half-reactions by mass
first balance elements other than H and O
add H2O where need O
add H+1 where need H
neutralize H+ with OH- in base
balance half-reactions by charge
balance charge by adjusting electrons
balance electrons between half-reactions
add half-reactions
check

14. Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Assign Oxidation States
Separate into half-reactions
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Assign Oxidation States
I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
Separate into half-reactions
ox:
red:

15. Balance half-reactions by mass
Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance half-reactions by mass
then O by adding H2O
ox: 2 I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s) + 2 H2O(l)
Balance half-reactions by mass
ox: 2 I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Balance half-reactions by mass
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
Balance half-reactions by mass
in base, neutralize the H+ with OH-
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
MnO4(aq) + 2 H2O(l)  MnO2(s) + 4 OH(aq)
Balance half-reactions by mass
then H by adding H+
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)

16. Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Balance Half-reactions by charge
ox: 2 I(aq)  I2(aq) + 2 e
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq)
Balance electrons between half-reactions
ox: 2 I(aq)  I2(aq) + 2 e } x3
red: MnO4(aq) + 2 H2O(l) + 3 e  MnO2(s) + 4 OH(aq) }x2
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)

17. Add the Half-reactions ox: 6 I(aq)  3 I2(aq) + 6 e
Ex 18.3 – Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
Add the Half-reactions
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)
Check
Reactant
Count
Element
Product 6 I 2 Mn 12 O 8 H 2
charge

18. Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O

19. Practice - Balance the Equation H2O2 + KI + H2SO4 ® K2SO4 + I2 + H2O
oxidation
reduction
ox: 2 I-1 ® I2 + 2e-1
red: H2O2 + 2e H+ ® 2 H2O
tot 2 I-1 + H2O2 + 2 H+ ® I2 + 2 H2O
1 H2O2 + 2 KI + H2SO4 ® K2SO4 + 1 I2 + 2 H2O

20. Practice - Balance the Equation ClO3-1 + Cl-1 ® Cl2 (in acid)

21. Practice - Balance the Equation ClO3-1 + Cl-1 ® Cl2 (in acid)
oxidation
reduction
ox: 2 Cl-1 ® Cl2 + 2 e-1 } x5
red: 2 ClO e H+ ® Cl2 + 6 H2O} x1
tot 10 Cl ClO H+ ® 6 Cl2 + 6 H2O
1 ClO Cl H+1 ® 3 Cl2 + 3 H2O

22. Electrical Current
when we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time
when we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time
whether as electrons flowing through a wire or ions flowing through a solution

23. Redox Reactions & Current
redox reactions involve the transfer of electrons from one substance to another
therefore, redox reactions have the potential to generate an electric current
in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring

24. Electric Current Flowing Directly Between Atoms

25. Electric Current Flowing Indirectly Between Atoms

26. Electrochemical Cells
electrochemistry is the study of redox reactions that produce or require an electric current
the conversion between chemical energy and electrical energy is carried out in an electrochemical cell
spontaneous redox reactions take place in a voltaic cell
aka galvanic cells
nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

27. Electrochemical Cells
oxidation and reduction reactions kept separate
half-cells
electron flow through a wire along with ion flow through a solution constitutes an electric circuit
requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons
through external circuit
ion exchange between the two halves of the system
electrolyte

28. Electrodes Anode electrode where oxidation occurs
anions attracted to it
connected to positive end of battery in electrolytic cell
loses weight in electrolytic cell
Cathode
electrode where reduction occurs
cations attracted to it
connected to negative end of battery in electrolytic cell
gains weight in electrolytic cell
electrode where plating takes place in electroplating

29. Voltaic Cell
the salt bridge is required to complete the circuit and maintain charge balance

30. Current and Voltage
the number of electrons that flow through the system per second is the current
unit = Ampere
1 A of current = 1 Coulomb of charge flowing by each second
1 A = x 1018 electrons/second
Electrode surface area dictates the number of electrons that can flow
the difference in potential energy between the reactants and products is the potential difference
unit = Volt
1 V of force = 1 J of energy/Coulomb of charge
the voltage needed to drive electrons through the external circuit
amount of force pushing the electrons through the wire is called the electromotive force, emf

31. Cell Potential
the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential
the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode
the cell potential under standard conditions is called the standard emf, E°cell
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions

32. Cell Notation shorthand description of Voltaic cell
electrode | electrolyte || electrolyte | electrode
oxidation half-cell on left, reduction half-cell on the right
single | = phase barrier
if multiple electrolytes in same phase, a comma is used rather than |
often use an inert electrode
double line || = salt bridge

33. Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)

34. Standard Reduction Potential
a half-reaction with a strong tendency to occur has a large + half-cell potential
when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur
we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction
we select as a standard half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 v
standard hydrogen electrode, SHE

36. Standard Cell

37. Half-Cell Potentials
SHE reduction potential is defined to be exactly 0 v
half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red
half-reactions with a stronger tendency toward oxidation than the SHE have a  value for E°red
E°cell = E°oxidation + E°reduction
E°oxidation = E°reduction
when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation

40. Separate the reaction into the oxidation and reduction half-reactions
Ex 18.4 – Calculate Ecell for the reaction at 25C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the reaction into the oxidation and reduction half-reactions
ox: Al(s)  Al3+(aq) + 3 e−
red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the E for each half-reaction and sum to get Ecell
Eox = −Ered = v
Ered = v
Ecell = (+1.66 v) + (+0.96 v) = v

41. Separate the reaction into the oxidation and reduction half-reactions
Ex 18.4a – Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
Separate the reaction into the oxidation and reduction half-reactions
ox: Fe(s)  Fe2+(aq) + 2 e−
red: Mg2+(aq) + 2 e−  Mg(s)
look up the relative positions of the reduction half-reactions
red: Fe2+(aq) + 2 e−  Fe(s)
since Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written

42. the reaction is spontaneous in the reverse direction
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
ox: Mg(s)  Mg2+(aq) + 2 e−
red: Fe2+(aq) + 2 e−  Fe(s)
sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

43. Practice - Sketch and Label the Voltaic Cell Fe(s) ½ Fe2+(aq) ½½ Pb2+(aq) ½ Pb(s) , Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.

44. ox: Fe(s)  Fe2+(aq) + 2 e− E = +0.45 V
red: Pb2+(aq) + 2 e−  Pb(s) E = −0.13 V
tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s) E = V

45. Predicting Whether a Metal Will Dissolve in an Acid
acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+(aq)
metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid

46. E°cell, DG° and K for a spontaneous reaction DG° = −RTlnK = −nFE°cell
one the proceeds in the forward direction with the chemicals in their standard states
DG° < 1 (negative)
E° > 1 (positive)
K > 1
DG° = −RTlnK = −nFE°cell
n is the number of electrons
F = Faraday’s Constant = 96,485 C/mol e−

47. Example 18.6- Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given:
Find:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
DG°, (J)
Concept Plan:
Relationships:
E°ox, E°red
E°cell
DG°
Solve:
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v
red: I2(l) + 2 e− → 2 I−(aq) E° = v
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 v
Answer:
since DG° is +, the reaction is not spontaneous in the forward direction under standard conditions

48. Example 18.7- Calculate K at 25°C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given:
Find:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K
Concept Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v
red: 2 H+(aq) + 2 e− → H2(aq) E° = v
tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 v
Answer:
since K < 1, the position of equilibrium lies far to the left under standard conditions

49. Nonstandard Conditions - the Nernst Equation
DG = DG° + RT ln Q
E = E° - (0.0592/n) log Q at 25°C
when Q = K, E = 0
use to calculate E when concentrations not 1 M

50. E at Nonstandard Conditions

51. Example Calculate Ecell at 25°C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
Given:
Find:
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
[Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Concept Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 v
red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = v
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)) E° = v
Check:
units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M

52. Concentration Cells
it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different
the difference in energy is due to the entropic difference in the solutions
the more concentrated solution has lower entropy than the less concentrated
electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution
oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode
reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

53. Concentration Cell
when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s)

54. LeClanche’ Acidic Dry Cell
electrolyte in paste form
ZnCl2 + NH4Cl
or MgBr2
anode = Zn (or Mg)
Zn(s) ® Zn2+(aq) + 2 e-
cathode = graphite rod
MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-
® 2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.5 v
expensive, nonrechargeable, heavy, easily corroded

55. Alkaline Dry Cell
same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste
anode = Zn (or Mg)
Zn(s) ® Zn2+(aq) + 2 e-
cathode = brass rod
MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-
® 2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.54 v
longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

56. PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-
Lead Storage Battery
6 cells in series
electrolyte = 30% H2SO4
anode = Pb
Pb(s) + SO42-(aq) ® PbSO4(s) + 2 e-
cathode = Pb coated with PbO2
PbO2 is reduced
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-
® PbSO4(s) + 2 H2O(l)
cell voltage = 2.09 v
rechargeable, heavy

57. NiCad Battery electrolyte is concentrated KOH solution anode = Cd
Cd(s) + 2 OH-1(aq) ® Cd(OH)2(s) + 2 e-1 E0 = 0.81 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 v
cell voltage = 1.30 v
rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

58. Ni-MH Battery electrolyte is concentrated KOH solution
anode = metal alloy with dissolved hydrogen
oxidation of H from H0 to H+1
M∙H(s) + OH-1(aq) ® M(s) + H2O(l) + e-1 E° = 0.89 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 v
cell voltage = 1.30 v
rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad

59. Lithium Ion Battery electrolyte is concentrated KOH solution
anode = graphite impregnated with Li ions
cathode = Li - transition metal oxide
reduction of transition metal
work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode
rechargeable, long life, very light, more environmentally friendly, greater energy density

61. Fuel Cells
like batteries in which reactants are constantly being added
so it never runs down!
Anode and Cathode both Pt coated metal
Electrolyte is OH– solution
Anode Reaction:
2 H2 + 4 OH–
→ 4 H2O(l) + 4 e-
Cathode Reaction:
O2 + 4 H2O + 4 e-
→ 4 OH–

62. Electrolytic Cell
uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction
must be DC source
the + terminal of the battery = anode
the - terminal of the battery = cathode
cations attracted to the cathode, anions to the anode
cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized
some electrolysis reactions require more voltage than Etot, called the overvoltage

64. electroplating In electroplating, the work piece is the cathode.
Cations are reduced at cathode and plate to the surface of the work piece.
The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution

65. Electrochemical Cells
in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode
in voltaic cells,
anode is the source of electrons and has a (−) charge
cathode draws electrons and has a (+) charge
in electrolytic cells
electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery
electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery

66. Electrolysis
electrolysis is the process of using electricity to break a compound apart
electrolysis is done in an electrolytic cell
electrolytic cells can be used to separate elements from their compounds
generate H2 from water for fuel cells
recover metals from their ores

67. Electrolysis of Water

68. Electrolysis of Pure Compounds
must be in molten (liquid) state
electrodes normally graphite
cations are reduced at the cathode to metal element
anions oxidized at anode to nonmetal element

69. Electrolysis of NaCl(l)

70. Mixtures of Ions
when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode
least negative or most positive E°red
when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode
least negative or most positive E°ox
Tro, Chemistry: A Molecular Approach

71. Electrolysis of Aqueous Solutions
Complicated by more than one possible oxidation and reduction
possible cathode reactions
reduction of cation to metal
reduction of water to H2
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = stand. cond.
E° = pH 7
possible anode reactions
oxidation of anion to element
oxidation of H2O to O2
2 H2O ® O2 + 4e-1 + 4H+1 E° = stand. cond.
E° = pH 7
oxidation of electrode
particularly Cu
graphite doesn’t oxidize
half-reactions that lead to least negative Etot will occur
unless overvoltage changes the conditions

72. Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations
2 I-1 ® I2 + 2 e-1 E° = −0.54 v
2 H2O ® O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1 ® I2 + 2 e-1 E° = −0.54 v
2 H2O ® O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions
Na+1 + 1e-1 ® Na0 E° = −2.71 v
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = −0.41 v
possible reductions
Na+1 + 1e-1 ® Na0 E° = −2.71 v
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l) ® I2(aq) + H2(g) + 2 OH-1(aq)

73. Faraday’s Law
the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs
charge that flows through the cell = current x time

74. Example Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e− → Au(s)
Given:
Find:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Concept Plan:
Relationships:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Solve:
Check:
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

75. Corrosion
corrosion is the spontaneous oxidation of a metal by chemicals in the environment
since many materials we use are active metals, corrosion can be a very big problem

76. Rusting rust is hydrated iron(III) oxide moisture must be present
water is a reactant
required for flow between cathode and anode
electrolytes promote rusting
enhances current flow
acids promote rusting
lower pH = lower E°red

78. Preventing Corrosion
one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment
paint
some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding
another method to protect one metal is to attach it to a more reactive metal that is cheap
sacrificial electrode
galvanized nails

79. Sacrificial Anode

80. Sample Calculations

81. COMMERCIAL PRODUCTION OF CHEMICALS
Power is measured in watts and has basic units of joules/second.
Power is calculated from the equation
P = V I, (J/C) (C/S) = J/S
A kilowatt-hour is an energy unit and is equal to 3.60x106J:
3600 S/hr times W/kW

82. COUNTING ELECTRONS
Current, I, is measured in amperes and has basic units of coulombs/second.
Faraday's Law tells us that the current times the time can be used to calculate the mass of material oxidized or reduced.
F, Faraday's constant, is 9.65x104C/mol e-.
These unit factor problems change: current x time ==> coulombs ==> moles e- ==> moles oxidized or reduced species ==> mass

83. Let’s do an example first.
COUNTING ELECTRONS
The balanced half-reaction provides the conversion from moles electrons to moles of species sought.
Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours.
Let’s do an example first.

84. Quantitative Aspects of Electrochemistry
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- ---> Ag(s)
1 mol e- ---> 1 mol Ag
If we could measure the moles of e-, we could know the quantity of Ag formed.
But how to measure moles of e-?

85. Quantitative Aspects of Electrochemistry
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- ---> Ag(s)
1 mol e- ---> 1 mol Ag
If we could measure the moles of e-, we could know the quantity of Ag formed.
But how to measure moles of e-?
Current =
charge passing
time I
(amps) =
coulombs
seconds

86. Quantitative Aspects of Electrochemistry
Current =
charge passing
time I
(amps)
coulombs
seconds
But how is charge related to moles of electrons?
Charge on 1 mol of e- =
(1.60 x C/e-)(6.02 x 1023 e-/mol)
= 96,500 C/mol e- = 1 Faraday

87. Quantitative Aspects of Electrochemistry
(amps) =
coulombs
seconds
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited?
Solution
(a) Calculate the charge
Coulombs = amps x time
= (1.5 amps)(15.0 min)(60 s/min) = C

88. Quantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited?
Solution
(a) Charge = 1350 C
(b) Calculate moles of e- used
1 mol e-
1350 C • =
mol e-
96,
500 C
(c) Calculate quantity of Ag
1 mol Ag
.0140 mol e- • =
mol Ag or 1.51 g Ag
1 mol e-

89. Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?
Solution
a) 454 g Pb = mol Pb
b) Calculate moles of e-
2 mol e- 2 . 19
mol Pb • =
4.38 mol e-
1 mol Pb
c) Calculate charge
4.38 mol e- • 96,500 C/mol e- = 423,000 C

90. Quantitative Aspects of Electrochemistry
Solution
a) 454 g Pb = mol Pb
b) Mol of e- = 4.38 mol
c) Charge = 423,000 C
d) Calculate time
Time ( s ) =
Charge (C) I
(amps)
Time ( s ) =
423,
000 C
1.50 amp
282,
000 s
About 78 hours

91. COUNTING ELECTRONS
Remember:
The balanced half-reaction provides the conversion from moles electrons to moles of species sought.
Now solve the problem:
Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours.

92. COUNTING ELECTRONS 23,760 C* 1/96500*1/3*27/1 2.21 g Al
Remember:
The balanced half-reaction provides the conversion from moles electrons to moles of species sought.
Now solve the problem:
Calculate the grams of aluminum formed at the cathode if a current of 4.40 A flows for 1.50 hours.
23,760 C* 1/96500*1/3*27/1
2.21 g Al