Chapter Nineteen
Electrochemistry
1
Electrochemistry
The study of chemical reactions through
electrical circuits.
Monitor redox reactions by controlling electron
transfer
REDOX: Shorthand for “REDuction-OXidation”
Use redox reactions to experimentally measure:
Reaction progress (kinetics)
Composition (equilibrium constants)
Energy changes (thermodynamics)
2
Oxidation-Reduction Reactions
Oxidation-reduction reactions (REDOX reaction) occur when
electrons are transferred from one reactant to another during a
chemical reaction. There is a change in oxidation number for
both substances
Oxidation Number: Theoretical charge on an ion
Oxidation is the process where the oxidation number increases.
Electrons are lost from the substance
Reduction is the process where the oxidation number decreases.
Electrons are gained by the substance
Oxidation and reduction always accompany each other;
Neither can occur alone
3
Redox Reaction: Magnesium Burning
Oxidation: Mg (s)  Mg2+ + 2eReduction: 1/2O2(g) + 2e- O2Reaction: Mg (s) + 1/2O2(g) MgO(s)
4
LEO the lion says GER
LEO
Lose
Electrons
Oxidation
GER
Gain
Electrons
Reduction
5
Oxidation Number Rules: See Chapter 4
The rule earlier in the list always takes
precedence.
1) ON = 0 for a compound or ionic charge for an ion
2)ON = +1 for IA elements and
ON = +2 2A elements
3) ON= -2 for oxygen
4)ON= -1 for 7A elements
If both elements in 7A, then the one higher in the list
is -1
5)ON = -2 for 6A elements other than oxygen
6) ON = -3 for 5A elements (very shaky!!!)
6
Common Oxidation Numbers
Must be able to determine charges on each
element
7
Balancing Redox Reactions:
Balance Elements
1. Write equation for the ions in acid solution:
Fe2+ + Cr2O72-  Fe3+ + Cr3+
2. Divide equation into 2 half reactions
Fe2+  Fe3+
Cr2O72-  2Cr3+
3. Balance all elements except H and O
Fe2+  Fe3+
Cr2O72-  2Cr3+
4. Balance O with H2O
Fe2+  Fe3+
Cr2O72-  2Cr3+ + 7H2O
5. Balance H with H+
Fe2+  Fe3+ Cr2O72- + 14H+ 2Cr3+ + 7H2O
8
Balancing Redox Reactions: Charge
6. Balance charge
Fe2+  Fe3+ + 1e-
6e- + Cr2O72- + 14H+  2Cr3+ + 7H2O
Oxidation: lose electron Reduction: gain electrons
7. Multiply by an integer to equalize # electrons
6Fe2+  6Fe3+ + 6e-
6e- + Cr2O72- + 14H+  2Cr3+ + 7H2O
8. Add reactions together and cancel like species
6Fe2+ + 6e- + Cr2O72- + 14H+  2Cr3+ + 7H2O+ 6Fe3+ + 6e-
9. Check balance of final # of atoms and charge
6Fe2+ + Cr2O72- + 14H+  2Cr3+ + 7H2O+ 6Fe3+
(12+) + (2-) + (14+) = +24
(6+) + (0) + (18+) = +24
9
Balancing Redox Reactions: Basic Reactions
Need OH- instead of H+ in final equation
1. Balance as if the reaction is in acidic solution.
6Fe2+ + Cr2O72- + 14H+  2Cr3+ + 7H2O+ 6Fe3+
2. Add OH- to both sides to match H+
14OH- + 6Fe2+ + Cr2O72- + 14H+  2Cr3+ + 7H2O+ 6Fe3++14OH-
3. Combine OH- and H+ to make water
14H2O + 6Fe2+ + Cr2O72-  2Cr3+ + 7H2O+ 6Fe3++14OH-
4. Cancel water from both sides
7H2O + 6Fe2+ + Cr2O72-  2Cr3+ + 6Fe3++14OH5. Check to see if balanced
10
Galvanic Cells
11
Parts of an Electrochemical Cell
Ionic Solutions
Provide ions to transfer charge
Solution + Electrode= Half-cell
Electrodes
Anode: oxidation occurs
Cathode: reduction occurs
Salt bridge.
Keeps 2 half-cells connected
Ions flow, but solution doesn’t
Metal wires
Connect the electrodes to the terminals of the voltmeter.
Provide means of transporting electrons between electrodes
Voltmeter
Measures the electron flow in the system
12
Galvanic Cell: Daniell Cell
13
Types of Cells
Voltaic or Galvanic Cell
Net oxidation/reduction reaction is spontaneous
Convert energy to useful work
Batteries
Electrolytic Cell
Net redox reaction is non-spontaneous
Work is done on the cell, energy must be
supplied: Recharge a battery, electroplating
14
Standard Reduction Potentials
15
Cell Diagram
Reaction: Zn(s) + Cu2+  Zn2+ + Cu
Cu2+(aq) | Cu(s)
Zn(s) | Zn2+(aq) ||
anode
solution
salt bridge
solution
cathode
Single bar (|)
Divides reduced from oxidized
Show phase differences
Double bars (||)
Represent salt bridge
Divides redox half reactions
Goes left to right
Anode written first
Remainder in order of actual cell
16
Cell Diagram
Zn(s)
anode
| Zn2+(aq)
solution
||
Cu2+(aq) | Cu(s)
salt bridge
solution
cathode
Anode (oxidation) reaction: Zn(s)Zn2+(aq) + 2e–
Solid zinc (Zn(s)) oxidized to Zn2+(aq)
Solid zinc is the physical electrode
Cathode (reduction) reaction: Cu2+(aq) + 2e–Cu(s)
Cu2+(aq) is reduced to solid copper (Cu(s))
Solid copper is the physical electrode
Net Reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
17
Standard Electrode Potentials
Standard Cell Potentials
Eocell = Eoreduction + Eooxidation
Standard Hydrogen Electrode (SHE)
2H+(aq) + 2e– H2(g)
Standard Electrode Potential, Eo.
[H+] = 1.00 M
PH2= 1.00atm
T = 298K
Eo= 0.000V
Calculate potential of oxidation half reactions(Eooxid)
Eooxid= Eocell - Eored = Eocell -0.000V = Eocell
18
Spontaneous: Galvanic Cell
Pt | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)
H2(g) + Cu2+(aq) 2H+(aq) + Cu(s)
Cell Potential:
Eocell = Eored + Eooxid
Eocell = +.340V
Anode Reaction:
H2(g)  2H+(aq) + 2e–
Eooxid = 0.000 V
Cathode Reaction:
Cu2+(aq) + 2e– Cu(s)
Eored = +.340V
+ Eocell; Reaction is spontaneous as written
(Galvanic)
19
Nonspontaneous: Electrolytic Cell
Pt | H2(g) | H+(aq) || Zn2+(aq) | Zn(s)
H2(g)+ Zn2+(aq)2H+(aq) + Zn(s)
Cell Potential:
Eocell = Eored + Eooxid
Anode Reaction:
H2(g)  2H+(aq) + 2e–
Cathode Reaction:
Zn2+(aq) + 2e– Zn(s)
Eocell = -0.763V
Eooxid = 0.00 V
Eored = -.763V
- Eocell: Reaction nonspontaneous
(electrolytic)
20
Calculating Cell Potentials
Daniell cell: Cu2+(aq) + Zn(s)Cu(s) + Zn2+(aq)
Cell Reactions
Reduction: Cu2+(aq) + 2e–Cu(s)
Oxidation: Zn(s)  Zn2+(aq) + 2e–
From Reduction Tables
Cu2+
Zn2+
Eored = + 0.34 V
Eored = – 0.76V, so Eooxid = + 0.76 V
Cell potential, Eocell
Eocell= Eored + Eooxid = +0.34V + (+0.76 V)
Eocell= +1.10V
21
25 oC
For a Spontaneous Reaction
Reduction reaction is the more positive, higher in table
Oxidation reaction is lower in table and needs to be reversed
22
Redox Rules
1. E0 is for the reaction as written
2. The more positive E0 the greater the tendency for
the substance to be reduced
3. Half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
4. Changing stoichiometric coefficients of a half-cell
reaction does not change E0
Intensive Property: Amount doesn’t matter
23
Will Br2(l) spontaneously oxidize Fe (aq)?
If so, what are the net cell reaction and the
Eocell?
Figure out what is being asked:
Oxidize: Cause to lose e-: Fe2+(aq)  Fe3+(aq)
Write and balance chemical equation:
2Fe2+(aq) + Br2(l)  2Fe3+(aq) + 2 Br– (aq)
Determine half-reactions and get Eo from table:
Br2(l) + 2e– 2Br–(aq)
Fe2+(aq) Fe3+(aq) + e–
Fe3+(aq) + e– Fe2+(aq
reduction
Eo = +1.07 V
oxidation, needs to be reversed
Eo = + 0.77 V
Eo = –0.77 V
Cell Potential
Eocell= Eoreduction+ Eooxidation= +1.07V + (-0.77V)= +0.30V
Eocell>0 Spontaneous reaction
24
Will I–(aq) reduce Cr3+(aq) to the free metal.
If so, what are the net reaction and the Eocell?
Figure out what is being asked:
Reduce: Cause something to gain e-: Cr3+(aq) + 3e- Cr(s)
Write and balance chemical equation:
2Cr3+(aq) + 6I–(aq)  2Cr(s) + 3I2(s)
Cell Reactions
Reverse:
Oxidation:
Reduction:
I2(s) + 2e– 2I–(aq)
2I–(aq)  I2(s) + 2e–
Cr3+(aq) + 3e– Cr(s)
Eo = +0.54 V
Eo = -0.54 V
Eo = -0.74 V
Cell Potential
Eocell= Eoreduction + Eooxidation = (-0.74) + (-.54V) = -1.28V
Eocell <0
Non-spontaneous Reaction
 NR
25
Thermodynamics
of Redox Reactions
26
Gibb’s Energy and Cell Potentials
Electrical work, w
w (joules)= Total cell charge (coulombs) x cell
potential (V)
Total cell charge = # mol electrons(n) x
(coulombs/mol e-s)
Faraday’s constant, F= 96485 Coulombs/1 mol e-s
Gibb's Energy:G
G =wmax = –nFEcell
At standard conditions: Go = –nFEocell = -RTlnK
(Voltaic cells are spontaneous, Ecell must be
positive)
Cell potential, Ecell
27
Summarizing the Important Relationships
∆G°
K
E°cell
Spontaneity
Direction
(-)
>1
(+)
Spontaneous
More products
0
1
0
At equilibrium
Products= reactants
(+)
<1
(-)
Nonspontaneous
More reactants
28
Find Keq at 25 oC for the reaction;
Sn(s) + 2Cu2+(aq) Sn2+(aq) + 2Cu+(aq)
Balance net reaction:
Sn(s) + 2Cu2+(aq)  Sn2+(aq) + 2Cu+(aq)
Cell Reactions
Oxidation (rev): Sn(s) Sn2+(aq) + 2e–
Reduction : 2Cu2+(aq) + 2e-  2Cu+(aq)
Eoox= + 0.14V
Eored= + 0.15 V
Link Eocell and K and fill in constants
Eocell = (RT/nF) lnKeq
R = 8.314 J/mol K
T = 25°C = 298K
Solve for Keq
ln Keq = 22.6
lnKeq= Eocell(nF/RT)
n=2
Eocell= +0.29 V
F = 96485 coul/mol
Keq = e22.6 = 7x109
29
Concentration
and Cell Potential
30
The Nernst Equation:
Cell Potentials at Nonstandard Conditions
From Thermodynamics:
G = Go + RT ln Q
Linking Thermodynamics & Electrochemistry
G = –nFE and Go = –nFEo
Substitute into first equation to remove G :
(–nF)E = (–nF)Eo + RT ln Q
Divide by –nF to get the Nernst Equation:
E = Eocell - (RT/nF) ln Q
Can now change temperature and concentrations
31
Concentration Cells:
Daniell Cell at Nonstandard Conditins
Electrochemical cell that does not have 1M solutions
Find the EMF at the following conditions:
1. [Cu2+] = 1.00 M, [Zn2+] = 1.0×10–9 M
2. [Cu2+] = 0.10 M, [Zn2+] = 0.90 M
Daniell Cell
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq) Eocell = +1.10 V
Nernst Equation
E = Eocell - (RT / nF) ln Q
Q = [Cu(s)] [Zn2+(aq)] / [Zn(s)] [Cu2+(aq)] = [Zn2+/Cu2+]
Plugging in from equation
E = Eocell - (RT / nF) ln [Zn2+/Cu2+]
32
Concentration Cells:
Daniell Cell at Nonstandard Conditions
Solve for E
2
8.314 J / molK * 298 K [ Zn ]
ln
E  1.10V 
2
2mol * 96485coul / mol Cu ]
Conditions of First Cell:
[Cu2+] = 1.00 M, [Zn2+] =1.0×10–9 M
E = 1.37 V
Conditions of Second Cell
[Cu2+] = 0.10 M [Zn2+] = 0.90 M
E = 1.07 V
33
Changing Zinc Concentration
After the initial drop, concentration has a
minimal effect on voltage
34
Galvanic Cells:
Batteries & Corrosion
35
The Dry Cell (Disposable Batteries)
Portable electronic devices
Zn(s) | ZnCl2(aq), NH4Cl(aq) || Mn2O3(aq) | MnO2(s)|C(s)
Oxidation:
Zn  Zn2+ (aq)+ 2eReduction: 2MnO2 + 2NH4+ + 2e-  Mn2O3 (s) + 2NH3
+ H2O
Cell Reaction: Zn+ 2MnO2+ 2NH4+  Zn2++ Mn2O3+
2NH3+ H2O
Cell Voltage
Calculated:+ 1.36V
Produced: + 1.5V
Irreversible Reaction
Cell is “dead” when
reactants are used up
36
Button Batteries: Mercury Battery
Pacemakers, hearing aids, watches
Zn | ZnO, OH- || HgO, OH- | Hg(s) | Fe(s)
Oxidation:
Zn + 2OH-  ZnO + H2O +2eReduction:
HgO + H2O + 2e-  Hg + 2OHCell Reaction:
Zn+ HgO + H2O  ZnO + Hg
Cell Voltage
Constant OH- composition
More constant voltage
Reported voltage: + 1.5V
Irreversible Reaction
Cell is “dead” when
reactants are used up
37
Lead Storage Battery
Car and boat batteries
Pb | H2SO4 (38%) || H2SO4 (38%) | PbO2
Oxidation:
Pb + SO42- PbSO4 +2eReduction:
PbO2 + 4H++ SO42- + 2e-  PbSO4 + 2H2O
Cell Reaction: Pb + PbO2 + 4H++ SO42-  2PbSO4 + 2H2O
Cell Voltage
Voltage: + 2.0V
Usually 6 cells=12V
Reversible Reaction
Reaction reverses when
engine is running
Battery recharges
Electrolytic cell reaction
38
Fuel Cells
Galvanic cells: Continually renew reactants
C-Ni (catalyst) | H2 | OH- (aq)|| OH- (aq)| O2 | C-Ni (catalyst)
Oxidation:
Reduction:
Cell Reaction:
2H2 + 4OH- 4H2O +4eO2 + 2H2O + 4e-  4OH2H2 + O2  2H2O
Cell Voltage
Voltage: + 1.23V
Nonreversible Reaction
Need fresh reactants
Removal of products
Need electrocatalysts
39
Corrosion:
Deterioration of metal through an
electrochemical process
Oxidation of metals: Rust, tarnish, “patina”
Rust: Fe2O3•xH2O
2Fe + O2 + 4H+  2Fe2+ + 2H2O
E=1.67V
4Fe2+ + O2+ (4+2x)H2O  2Fe2O3•xH2O + 8H+
E>0
Cell Reaction
Spontaneous reactions
Salts increase rate
Cathodic protection
More reactive metal protects (Zn)
40
Electrolysis
and
Electrometallurgy
41
Electrolytic Cells
Non-spontaneous processes driven by the
application of an external power supply
Side reactions from solvent and dissolved
ions
Determine reactions that are most
spontaneous
Least Negative Ecell
42
A 1 M solution of potassium iodide is electrolyzed
under acidic conditions.What are the products?
Dissociation Reaction:
Resulting Products:
KI(aq)  K+(aq) + I–(aq)
K+, I-, H+, H2O
Cation Reduction:
Acid Reduction to H2 :
Anion Oxidation:
H2O Oxidation to O2:
K+(aq) + e–K(s)
2H+(aq) + 2e– H2(g)
2I–(aq)  I2(s) + 2e–
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Half-Cell Potentials from table
K+(aq) + e–K(s)
2I–(aq)  I2(s) + 2e–
2H+(aq) + 2e– H2(g)
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Eo = –2.92 V
Eo = - 0.54 V
Eo = 0.00 V
Eo = -1.23 V
Reduction
Oxidation
Reduction
Oxidation
43
A 1 M solution of potassium iodide is
electrolyzed under acidic conditions.
What are the products?
Less Negative Reactions
Reduction:
Oxidation:
2H+(aq) + 2e– H2(g)
2I–(aq)  I2(s) + 2e–
Eo = 0.00 V
Eo = - 0.54 V
Net Reaction
2H+(aq) + 2I–(aq)  I2(s) + H2(g)
Standard Potential
Eo = 0.00 + (– 0.54) = – 0.54V
need at least .54V to start reaction
Products
Solid Iodine:I2(s) and Hydrogen gas: H2(g)
44
Producing Products by Electrolysis
Electroplating
Coating one metal onto another
Silver or gold over iron or steel
Cheaper
Product often more durable
Purification of copper
Impure copper anode
More reactive impurities oxidized (ions)
Less reactive impurities drop to bottom
Gold, silver, etc. now separated
Copper built up on cathode 99.5% pure
45
Quantitative Electrolysis
Use Electrolytic Cells to find
stoichiometry
Measure the charge passed through the cell
The current (amps, A) is the rate of charge flow
1amp = 1 coulomb per second.
nF = At
n = number of moles of electrons
F = Faraday's constant = 96485 C/mol
A = current in amps = 1 coulomb/sec
t = time in seconds
46
Quantitative Electrolysis of Water: 2H2OO2 + 2H2
Water is electrolyzed in a cell at 25 mA for 15 minutes. How
many mL of oxygen gas are produced at 1 atm & 25 °C?
A  25mAx
60 sec
1A
 9.0 x10 2 sec
 2.5 x10  2 A t  15 min x
1 min
1000mA
nF  At  (2.5 x10 2 C / s )(9.0 x10 2 sec)  22.5C
n
molO2 
V
C
22.5C

 2.3 x10  4 molelectrons
F 96485C / mol
1moleO 2
x 2.3 x10  4 molelectrons  5.8 x10 5 molO 2
4molelectrons
nRT
Latm
 (5.8 x10 5 molO 2 )( 0.0821
)( 298 K )  0.0014 L  1.4mL
P
molK
47