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Chapter 18 Electrochemistry

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Title: Chapter 18 Electrochemistry


1
Chapter 18Electrochemistry
Chemistry II
2
Redox Reaction
  • one or more elements change oxidation number
  • all single displacement, and combustion,
  • some synthesis and decomposition

3
Redox Reaction
  • always have both oxidation and reduction
  • split reaction into oxidation half-reaction and a
    reduction half-reaction
  • aka e- transfer reactions
  • half-reactions include e-
  • oxidizing agent is reactant molecule that causes
    oxidation
  • contains element reduced
  • reducing agent is reactant molecule that causes
    reduction
  • contains the element oxidized

4
Oxidation Reduction
  • oxidation
  • ox number of an element increases
  • element loses e-
  • compound adds O
  • compound loses H
  • half-reaction has e- as products
  • reduction
  • ox number of an element decreases
  • element gains e-
  • compound loses O
  • compound gains H
  • half-reactions have e- as reactants

5
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

6
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

7
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 -1 CCl4
Group 6A -2 CS2
Group 5A -3 NH3
8
Oxidation and Reduction
  • oxidation occurs when an atoms oxidation state
    increases during a reaction
  • reduction occurs when an atoms oxidation state
    decreases during a reaction

CH4 2 O2 ? CO2 2 H2O -4 1
0 4 2 1 -2
9
OxidationReduction
  • oxidation and reduction must occur simultaneously
  • if an atom loses electrons another atom must take
    them
  • reactant that reduces an element in another
    reactant reducing agent
  • the reducing agent contains the element that is
    oxidized
  • reactant that oxidizes element in another
    reactant oxidizing agent
  • the oxidizing agent contains the element that is
    reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na is oxidized, Cl
is reduced Na is the reducing agent, Cl2 is the
oxidizing agent
10
Identify the Oxidizing and Reducing Agents in
Each of the Following
  • 3 H2S 2 NO3 2 H ? 3 S 2 NO 4 H2O
  • MnO2 4 HBr ? MnBr2 Br2 2 H2O

11
Identify the Oxidizing and Reducing Agents in
Each of the Following
ox ag
red ag
  • 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
  • MnO2 4 HBr MnBr2 Br2 2 H2O

1 -2 5 -2 1 0
2 -2 1 -2
red ag
ox ag
4 -2 1 -1 2 -1 0
1 -2
12
Common Oxidizing Agents
13
Common Reducing Agents
14
Balancing Redox Reactions
  • 1. assign oxidation numbers
  • determine element oxidized and element reduced
  • 2. write ox. red. half-reactions, including e-
  • ox. electrons on right, red. electrons on left of
    arrow
  • 3. balance half-reactions by mass
  • first balance elements other than H and O
  • add H2O where need O
  • add H where need H
  • neutralize H with OH- in base
  • 4. balance half-reactions by charge
  • balance charge by adjusting e-
  • 5. balance e- between half-reactions
  • 6. add half-reactions
  • 7. check

15
Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Assign Oxidation States I?(aq) MnO4?(aq) ? I2(aq) MnO2(s)
Separate into half-reactions ox red
Assign Oxidation States
Separate into half-reactions ox I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
16
Ex 18.3 Balance the equationI?(aq)
MnO4?(aq) ? I2(aq) MnO2(s) in basic solution
Balance half-reactions by mass ox I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
Balance half-reactions by mass ox 2 I?(aq) ? I2(aq) red MnO4?(aq) ? MnO2(s)
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 then H by adding H ox 2 I?(aq) ? I2(aq) red 4 H(aq) MnO4?(aq) ? MnO2(s) 2 H2O(l)
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)
17
Ex 18.3 Balance the equationI?(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)
18
Ex 18.3 Balance the equationI?(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 Count
6 I 6
2 Mn 2
12 O 12
8 H 8
8? charge 8?
19
Practice - Balance the Equation H2O2 KI
H2SO4 K2SO4 I2 H2O
20
Practice - Balance the Equation H2O2 KI
H2SO4 K2SO4 I2 H2O
1 -1 1 -1 1 6 -2 1 6 -2
0 1 -2
oxidation
reduction
ox 2 I- I2 2e- red H2O2 2e- 2 H 2
H2O tot 2 I- H2O2 2 H I2 2 H2O
H2O2 2 KI H2SO4 K2SO4 I2 2 H2O
21
Practice - Balance the EquationClO3- Cl- Cl2
(in acid)
22
Practice - Balance the Equation ClO3-
Cl- ? Cl2 (in acid)
5 -2 -1 0
oxidation
reduction
ox 2 Cl- ? Cl2 2 e- x5 red 2 ClO3- 10 e-
12 H ? Cl2 6 H2O x1 tot 10 Cl- 2 ClO3-
12 H ? 6 Cl2 6 H2O
ClO3- 5 Cl- 6 H ? 3 Cl2 3 H2O
23
Electrical Current
  • current of a liquid in a stream, amount of
    water that passes by in a given period of time
  • electric current amount of electric charge that
    passes a point in a given period of time
  • whether as e- flowing through a wire or ions
    flowing through a solution

24
Redox Reactions Current
  • redox reactions involve the transfer of e- 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

25
Electric Current Flowing Directly Between Atoms
26
Electric Current Flowing Indirectly Between Atoms
27
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

28
Electrochemical Cells
  • redox reactions kept separate
  • half-cells
  • e- flow in a wire along and ion flow in solution
    constitutes an electric circuit
  • requires a conductive metal or graphite electrode
    to allow the transfer of e-
  • through external circuit
  • ion exchange between the two halves of the system
  • electrolyte

29
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

30
Voltaic Cell
the salt bridge is required to complete the
circuit and maintain charge balance
31
Current and Voltage
  • e- that flow through the system per second is
    the current
  • unit Ampere
  • 1 A of current 1 Coulomb of charge per second
  • 1 A 6.242 x 1018 e-/sec.
  • Electrode surface area dictates the number of e-
    that can flow

32
Current and Voltage
  • 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

33
Cell Potential
  • the difference in potential energy between the
    anode the cathode in a voltaic cell is called the
    cell potential
  • 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, Ecell
  • 25C, 1 atm for gases, 1 M concentration of
    solution
  • Ecell EOX ERED

34
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, double line salt
    bridge
  • if multiple electrolytes in same phase, a comma
    is used rather than
  • often use an inert electrode

35
Fe(s) Fe2(aq) MnO4?(aq), Mn2(aq), H(aq)
Pt(s)
36
Standard Reduction Potential
  • a half-reaction with a strong tendency to occur
    has a large half-cell potential
  • two half-cells are connected, e- 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
  • 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

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38
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 Ered
  • half-reactions with a stronger tendency toward
    oxidation than the SHE have a ? value for Ered
  • Ecell Eoxidation Ereduction
  • Eoxidation ?Ereduction
  • 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
  • When Ecell gt 0 reaction may be spontaneous

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41
Ex 18.4 Calculate E?cell for the reaction at
25?CAl(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 1.66 v E?red 0.96 v E?cell E?ox E?red E?cell (1.66 v) (0.96 v) 2.62 v
42
Ex 18.4a Predict if the following reaction is
spontaneous under standard conditionsFe(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 E? half-reactions E?cell E?ox E?red 0.45 -2.37 -1.92 since E?cell -ve the reaction is NOT spontaneous as written Mg2 reduction is below Fe2 reduction, the reaction is NOT spontaneous as written
43
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
44
Practice - Sketch and Label the Voltaic
CellFe(s) Fe2(aq) Pb2(aq) Pb(s) , Write
the Half-Reactions and Overall Reaction, and
Determine the Cell Potential under Standard
Conditions.
45
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?
0.32 V Spontaneous
46
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

All have ve Eox
Ecell Eox 0 ve
47
Ecell, ?G and K
  • for a spontaneous reaction
  • one that proceeds in the forward direction with
    the chemicals in their standard states
  • ?G lt 1 (negative)
  • E gt 1 (positive)
  • K gt 1
  • ?G -RTlnK -nFEcell
  • n is the number of electrons
  • F Faradays Constant 96,485 C/mol e-

48
Example 18.6- Calculate ?G for the
reactionI2(s) 2 Br-(aq) ? Br2(l) 2 I-(aq)
I2(s) 2 Br-(aq) ? Br2(l) 2 I-(aq) DG, (J)
Given Find
Concept Plan Relationships
ox 2 Br-(aq) ? Br2(l) 2 e- E -1.09 v
Solve
red I2(l) 2 e- ? 2 I-(aq) E 0.54 v
tot I2(l) 2Br-(aq) ? 2I-(aq) Br2(l) E
-0.55 v
since DG is , the reaction is not spontaneous
in the forward direction under standard conditions
Answer
49
Ecell, ?G and K
  • ?G -RTlnK -nFEcell
  • Ecell RT x ln K
  • nF
  • R 8.314 J/mol.K
  • lnK 2.303log K
  • F 96,485 C/mol e-
  • Ecell 0.0592 logK
  • n

50
Example 18.7- Calculate K at 25C for the
reactionCu(s) 2 H(aq) ? H2(g) Cu2(aq)
Cu(s) 2 H(aq) ? H2(g) Cu2(aq) K
Given Find
Concept Plan Relationships
ox Cu(s) ? Cu2(aq) 2 e- E -0.34 v
Solve
red 2 H(aq) 2 e- ? H2(aq) E 0.00 v
tot Cu(s) 2H(aq) ? Cu2(aq) H2(g) E
-0.34 v
since K lt 1, the position of equilibrium lies far
to the left under standard conditions
Answer
51
Nonstandard Conditions - the Nernst Equation
  • Relationship between Ecell (nonstandard) and
    Ecell (standard)
  • ?G ?G RT ln Q
  • Subs. ?G -nFEcell into above eqn.
  • -nFEcell -nFEcell -RTlnQ (divide by -nF)
  • Ecell Ecell - (RT/nF) log Q
  • R 8.314 J/mol.K, (RT/nF)lnQ (0.0592/n)logQ
  • Ecell E - (0.0592/n) logQ
  • Called the Nernst equation

52
Nonstandard Conditions - the Nernst Equation
  • Ecell Ecell - (0.0592/n) log Q at 25C
  • 1. when Q 1 (std. conditions) Ecell Ecell
  • 2. At equilibrium,Q K,
  • Ecell Ecell - (0.0592/n) log K and
    (0.0592/n) log K Ecell
  • Ecell 0
  • Potential reaches zero as concentrations approach
    equilibrium
  • Used to calculate E when concentrations not 1 M

53
E? at Nonstandard Conditions
Reactant conc. gt standard conditions Product
conc. lt standard conditions reaction shifts
right
54
Example 18.8- Calculate Ecell at 25C for the
reaction3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2
MnO2(s) 3 Cu2(aq) 4 H2O(l)
3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2 MnO2(s)
Cu2(aq) 4 H2O(l) Cu2 0.010 M, MnO4-
2.0 M, H 1.0 M Ecell
Given Find
Concept Plan Relationships
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 1.68 v
tot 3 Cu(s) 2 MnO4-(aq) 8 H(aq) ? 2 MnO2(s)
Cu2(aq) 4 H2O(l)) E 1.34 v
units are correct, Ecell gt Ecell as expected
because MnO4- gt 1 M and Cu2 lt 1 M
Check
55
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

56
Concentration Cell
Cu2(aq) 2e- ? Cu(s) 0.34V Cu(s) ? Cu2(aq)
2e- -0.34V Cu2(aq) Cu(s) ? Cu(s)
Cu2(aq) Ecell Ered Eox 0V
Cu(s)? Cu2(aq) (1 M) ?? Cu2(aq) (1 M)? Cu(s)
57
Concentration Cell
Cell potential Ecell calculated using Nernst
eqn. Ecell E - (0.0592/n) logQ Ecell E -
(0.0592/n) log(OX/RED) 0.068V
Cu(s)? Cu2(aq) (0.010 M) ?? Cu2(aq) (2.0 M)?
Cu(s)
58
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

59
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

60
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

61
NiCad Battery
  • electrolyte is concentrated KOH solution
  • anode Cd
  • Cd(s) 2 OH-(aq) ? Cd(OH)2(s) 2 e- E0 0.81 v
  • cathode Ni coated with NiO2
  • NiO2 is reduced
  • NiO2(s) 2 H2O(l) 2 e- ? Ni(OH)2(s) 2OH-
    E0 0.49 v
  • cell voltage 1.30 v
  • rechargeable, long life, light however
    recharging incorrectly can lead to battery
    breakdown

62
Ni-MH Battery
  • electrolyte is concentrated KOH solution
  • anode metal alloy with dissolved hydrogen
  • oxidation of H from H0 to H
  • MH(s) OH-(aq) ? M(s) H2O(l) e- E 0.89
    v
  • cathode Ni coated with NiO2
  • NiO2 is reduced
  • NiO2(s) 2 H2O(l) 2 e- ? Ni(OH)2(s) 2OH- E0
    0.49 v
  • cell voltage 1.30 v
  • rechargeable, long life, light, more
    environmentally friendly than NiCad, greater
    energy density than NiCad

63
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

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65
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

66
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

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68
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
69
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

70
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

71
Electrolysis of Water
72
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

73
Electrolysis of NaCl(l)
74
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 Ered
  • 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 Eox

75
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- ? H2 2 OH- E -0.83 v _at_ stand.
    cond.
  • E -0.41 v _at_ pH 7
  • possible anode reactions
  • oxidation of anion to element
  • oxidation of H2O to O2
  • 2 H2O ? O2 4e- 4H E -1.23 v _at_ stand.
    cond.
  • E -0.82 v _at_ pH 7
  • oxidation of electrode
  • particularly Cu
  • graphite doesnt oxidize
  • half-reactions that lead to least negative Etot
    will occur
  • unless overvoltage changes the conditions

76
Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations 2 I- ? I2 2 e- E -0.54
v 2 H2O ? O2 4e- 4H E -0.82 v
possible oxidations 2 I- ? I2 2 e- E -0.54
v 2 H2O ? O2 4e- 4H E -0.82 v
possible reductions Na e- ? Na0 E -2.71
v 2 H2O 2 e- ? H2 2 OH- E -0.41 v
possible reductions Na e- ? Na0 E -2.71
v 2 H2O 2 e- ? H2 2 OH- E -0.41 v
overall reaction 2 I-(aq) 2 H2O(l) ? I2(aq)
H2(g) 2 OH-(aq)
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Faradays 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

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Example 18.10- 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)
3 mol e- 1 mol Au, current 5.5 amps, time
25 min mass Au, g
Given Find
Concept Plan Relationships
Solve
units are correct, answer is reasonable since 10
A running for 1 hr 1/3 mol e-
Check
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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

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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 Ered

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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

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Sacrificial Anode
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