21 Electrochemistry
Electricity is such a mystery till
we understand chemistry.
Plain facts and sciences of
electrochemistry continue to 21 Electrochemistry
be important in technologies.
1
Electricity
Ancient people noticed electricity
1746 B. Franklin demonstrated lightening as electric effect and
performed the kite experiment in 1751. Two people tried to repeat his kite
flying experiment were killed by thunder.
1767 L. Galvani inserted two different metals in frog fluid and constructed
a electric cell
1800 A. Volta substituted frog fluid; made batteries, consisted of several
cells.
1802 G. Romagnosi noticed magnetism related to electricity
Michael Faraday 1791-1867 discovered many theories of electricity and
magnetism
21 Electrochemistry
2
Galvani
Luigi Galvani
(1737-1798):
erroneously
concluded that the
frog's nervous
system generated
an electrical
charge, his work
stimulated much
research into the
electrochemistry.
The depiction of his
laboratory 
21 Electrochemistry
3
Electrochemistry
A. Volta (1745-1827) experimented with different
materials, and made voltaic piles (batteries)
William Nicholson (1753-1815) observed bubbles
forming on the surfaces of metals submerged in
water when they are connected to a voltaic pile
Humphry Davy (1778-1829) observe electrolysis
of water and metal salts. Following that, …
Michael Faraday (1791-1860) studied
electrolysis, and discovered the relationship
between charges and chemical stoichiometry
21 Electrochemistry
4
Electrons
qe = –1.60217733e-19 C
F = 96485 C
me = 0.00054856 amu
= 9.1093897e-31 kg
spin = ½ (two state)
magnetic moment
= 9.284770e–24 J/tesla
Voltaic piles (batteries) made the following study possible
W. Crookes (1832-1919) observed cathode rays in low-pressure tubes.
1897: J.J. Thomson determined the charge to mass ratio (e– / me) of
cathode rays (electrons).
1916 R. Millikan (1868-1953) measured the amount of charge of e–.
21 Electrochemistry
5
Redox reactions and electrons
Energy drives chemical reactions.
Redox reactions involve the transfer of electrons.
Loss of electron (increase oxidation state) is oxidation, (Leo).
Gain of electron (decrease oxidation state) is reduction, (ger).
Zn  Zn2+ + 2 e–
leo
Cu2+ + 2 e –  Cu ger
net:
Zn + Cu2+  Cu + Zn2+ redox
Chemical energy in redox reactions may be convert to electric
energy by applying electrochemistry.
In the meantime, we should learn to balance the redox reaction
equations.
21 Electrochemistry
6
Galvanic Cell
A galvanic cell consists of two different metals inserted into a solution of
an electrolyte (salt, acid or base), simulated
Representation:
Zn | Zn2+ || Cu2+ | Cu
21 Electrochemistry
7
Assign oxidation states
0 for any element
–3 NH3
–2 N2H4
–1 NH2OH
0 N2
+1 N2O
+2 NO
+3 NO2
–
1 for H in compounds, but –1 for LiH, NaH, etc
– 2 for O in compounds, but –1 for H2O2, Na2O2
–1 Cl–
+1 for alkali metals, +2 for alkaline earth metals
0 Cl2
The oxidation states of other elements are then
assigned to make the algebraic sum of the
oxidation states equal to the net charge on the
molecule or ion.
+1 ClO–
+3 ClO2–
+4 ClO2
+5 ClO3–
+4 NO2
+7 ClO4–
+5 NO3–
21 Electrochemistry
8
Half reaction equations
Oxidation and reduction can be written as
half-reaction equations such as
Steps to balance half reaction
Zn  Zn2+ + 2 e–
leo
Assign oxidation number
ger
Figure out what is oxidized or
reduced.
Cu2+ + 2 e –  Cu
net: Zn + Cu2+  Cu + Zn2+
redox
Demonstrate how to balance these
Fe2+  Fe3+ + __ e–
Add electrons according to
oxidation number change
C2O42-  2 CO2 + __ e –
Balance charge with H+ (acid) or
OH – (base)
MnO4 – + __ e–  Mn2+
Balance atoms with H2O
Cr2O72– + __ e– + __ H+  2 Cr3+ + __ H2O
21 Electrochemistry
9
More half-reaction equations
2 I–  I2 + __ e–
ClO2 + __ OH–  ClO3- + __ e– + H2O (in basic solution)
2 S2O32–  S4O62- + __ e–
HS(=S)O3–  S + __ e- + HSO4–
__ H3O+ + __ e –  H2(g) + __ H2O
H2O2 + __ e –  2 H2O
ClO2 + __ e –  ClO2–
NO3– + __ e –  NH4+
21 Electrochemistry
10
Electrochemical Series
An electrochemical series is a list of metals in order of decreasing
strength as reductant, or increasing strength as oxidant.
An example of an activity series of metals based on the Standard
Potentials given would be:
K > Ba > Ca > Na > Mg > Al > Mn > Zn > Fe > Ni > Sn > Pb > Cu > Ag
In this series the most active metal is potassium (K) and the least active
metal is silver (Ag)
Reactions and cells are illustrated in 21-1 of Text (PHH).
21 Electrochemistry
11
Constructing half cells
A half cell consists of an oxidizing and its oxidized species
Zn | Zn2+
Cu | Cu2+
Pt | H2 | H+
(Pt as conductor)
Pt |Fe2+ , Fe3+
Cl– | Cl2 | Pt
Explain the cell convention
and reactions of cells.
21 Electrochemistry
Student cell set from School
Master Science $30
12
Galvanic cells
Using corns for a galvanic cells is illustrated by an Internet site:
(schoolnet.ca/general/electric-club/e/page9.html)
This picture illustrates a way
to make a pact of battery
using coins of different metals.
Apply the principles you
have learned regarding
electrochemical series and
electrolytes to make such a
pile will be an interesting
exercise.
21 Electrochemistry
13
Cell convention
Oxidation takes place always at the anode
Zn (s)  Zn2+ (aq) + __ e– Zn(s) | Zn2+(aq)
Fe2+ (aq)  Fe3+ (aq) + e–
Pt | Fe2+ , Fe3+
H2 (g)  2 H+ (aq) + __ e–
Pt | H2(g) | H+(aq)
Reduction takes place at the cathode
Cu2+ (aq) + __ e–  Cu (s) Cu2+ | Cu(s)
Cl2 (g) + __ e–  Cl– (aq)
Cl2(g) | Cl –(aq) | Pt
Fe3+ (aq) + __ e –  Fe2+ (aq) Fe3+(aq), Fe2+(aq)| Pt
2 H+ (aq) + __ e–  H2 (g)
H+(aq) | H2(g)| Pt
Concentration (1.1 M) and pressure (0.9 atm) are
also included in cell notations.
21 Electrochemistry
14
Electric energy and work
Electric energy or electric work = charge * potential difference
W=q*V
(1 J = 1 Coulomb Volt, C V)
compare W = m g h
The Faraday constant F is the charge for one mole of electrons,
F = 96485 C;
F / NA = 96485 C / 6.022e23 = 1.602177e-19 C, charge per e–
The maximum chemical energy of the cell that can be converted to
electric work is the Gibbs free energy change, G
G = – n F E (n F = q is the charge)
electromotive force (emf, or V)
number of electrons in the reaction equation
n F is the charge q
See slides in
15
21 Electrochemistry
16 Equilibria
Standard cell emf’s and electrode potentials
The standard cell emf is the emf of a voltaic cell operating under
standard conditions (1 M, 1 atm, 25oC etc). E.g.
Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s)
Eo = 1.10 V
Mg (s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s)
Eo = 2.90 V
The absolute potential of the electrode cannot be determined. Only
relative potential can be measured. The standard reduction potential is
measured against a SHE, for which Eo = 0.0000 V
Zn (s) | Zn2+ (aq, 1 M) || H+ (aq, 1 M) | H2 (1 atm)(g)
Zn = Zn2+ + 2 e–
(oxidation, displace H+ possible)
Zn2+ + 2 e–  Zn
21 Electrochemistry
(reduction)
Eo = 0.76 V
Eo = 0.76 V
Eo = – 0.76 V
16
Gibb’s Free Energy in a Cell
How much energy is available for the cell
Zn | Zn2+ || Ag+ | Ag
operating at standard condition when one mole of Zn is consumed ?
Solution
Zn = Zn2+ + 2 e
2 Ag+ + 2e = 2 Ag
Zn + 2 Ag+ = Zn2+ + 2 Ag
Eo
0.762 V
0.799 V (from table)
Eo = 1.561 V
Go = – n F E = – 2 * 96485 C * 1.561 V
= 301226 J (1J = 1 C V)
= 301.2 kJ
How much silver is consumed?
21 Electrochemistry
How much energy is available
if 6.5 g of Zn is consumed?
17
Table of standard reduction potential
Reaction E o (V)
Li+ + e– = Li (s)
Na+ + e– = Na (s)
Mg2+ + 2 e– = Mg (s)
Zn2+ + 2 e– = Zn (s)
(reference) 2 H+ + 2 e– = H2 (g)
(reference) H2 (g) = 2 H+ + 2 e–
Cu2+ + 2 e– = Cu (s)
Cu+ + e– = Cu (s)
I2 (s) + 2e– = 2 I– (aq)
Br2 (l) + 2 e– = 2 Br– (aq)
Cl2 (g) + 2 e– = 2 Cl– (aq)
F2 (g) + 2 e– = 2 F– (aq)
– 3.04
– 2.71
– 2.38
– 0.76
0.000
0.000
0.34
0.52
0.54
1.07
1.36
2.87
21 Electrochemistry
Standard cell potentials
E o
Cell
Li | Li+ || Cu2+ | Cu
____
Mg | Mg2+ || I2 | I– | Pt
____
Zn | Zn2+ || Br2 | Br– | Pt ____
Cu | Cu2+ || Zn2+ | Zn
____
Which is not spontaneous?
See 21-2
18
Table of standard reduction potential
Reaction E o (V)
F2 (g) + 2 e– = 2 F– (aq)
Cl2 (g) + 2 e– = 2 Cl– (aq)
Br2 (l) + 2 e– = 2 Br– (aq)
I2 (s) + 2e– = 2 I– (aq)
Cu+ + e– = Cu (s)
Cu2+ + 2 e– = Cu (s)
(reference) H2 (g) = 2 H+ + 2 e–
(reference) 2 H+ + 2 e– = H2 (g)
Zn2+ + 2 e– = Zn (s)
Mg2+ + 2 e– = Mg (s)
Na+ + e– = Na (s)
Li+ + e– = Li (s)
2.87
1.36
1.07
0.54
0.52
0.34
0.000
0.000
– 0.76
– 2.38
– 2.71
– 3.04
21 Electrochemistry
The listing order in the table
may be different in different
text books. However, the
principles and methods of
application remain the same.
This is the order given on the
Exam Data Sheet, that is
different from the text.
19
Strength of oxidation
The ability of a chemical to oxidize is its ability to take electrons from
other species,
Oxidizing agent + n e  Reduced species
Strength of oxidation of an oxidizing agent is measured by its reduction
potential.
Similarly, strength of reduction of a reducing agent is measured by its
oxidation potential.
Oxidized species  Reducing agent + n e
Be able to order the species according to oxidizing strength.
_____
21 Electrochemistry
20
Reaction direction and emf
What is the emf for the reaction,
Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)?
Solution:
Know what data to look for
Zn2+ + 2 e  Zn Eo = – 0.76 V
Fe3+ + e  Fe2+ Eo = + 0.77 V +
2 Fe2+  2 Fe3+ + 2 e Eo = – 0.77 V
Zn2+ (aq) + 2 Fe2+ (aq)  Zn (s) + 2Fe3+ (aq) Eo = – 1.53 V
non-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical
The reverse reaction is spontaneous,
Zn (s) + 2Fe3+ (aq)  Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 V
Zn | Zn2+ || Fe3+ | Fe2+ | Pt
21 Electrochemistry
21
Free energy and emf
What is the free-energy change for the cell,
Zn | Zn2+ (aq) 1 M || Ag+ (aq) 1 M | Ag?
Solution: Reduction potential required,
Zn  Zn2+ + 2e
Eo = 0.76
Ag+ + e  Ag
2 Ag+ + 2e  2Ag
Cell reaction
Eo = 0.80
Eo = 0.80
2 Ag+ + Zn  2Ag + Zn2+
Eo = 1.56 V
Go = – n F Eo = – 2 * 96485 * 1.56
= – 3.01e5 J or – 302 kJ
Condition for spontaneous
reaction is – G or + E.
Negative indicate energy is released.
The free energy for the cell is –301 kJ per mole of Zn, what is the emf?
21 Electrochemistry
22
General cell emf
G o is the standard energy change. G is for non-standard conditions.
Text uses Ecell instead of E
G = – n F E
Similarly, E o is the standard emf whereas E is general emf.
G = G o + R T ln Q
E = Eo – R T / n F ln Q
reaction quotient
When a system is at equilibrium (Q = K), G = 0. Therefore,
G = G o + R T ln K = 0 E = Eo – R T / n F ln K = 0 equilibrium constant
 G o = – R T ln K
E o =
RT
or G o = – ln(10) R T log K
E o =
2.303 R T
21 Electrochemistry
/ n F ln K
/ n F log K
At 298 K
0.0592
E o = ———— log K
n
23
The Nernst equation
For a general reaction,
aA+bB=cC+dD
RT
[C]c [D]d
E = E o – ——— ln ————
nF
[A]a [B]b
This Nernst equation correlates cell emf with [ ] or reactivities
of reactants and products as well as T
Units for [ ]:
mol L-1 for aqueous solution,
atm (for gas),
and constant for solid and liquid.
21 Electrochemistry
See 21-4
24
Evaluating E
RT
[C]c [D]d
E = E o – ——— ln ————
nF
[A]a [B]b
At 300 K, evaluate the cell emf for
Zn | Zn2+ (0.100 M) || H+ (0.200 M) | H2 (1.111 atm) | Pt
Solution:
Look up:
Zn  Zn2+ (aq) + 2e,
2 H+ (aq) + 2 e  H2 (g),
(R = 8.314 J mol-1 K-1, F = 96485 C mol-1)
E o = 0.76 V
E o = 0.00 V
E o = 0.76 V
The reaction is
Zn(s) + 2 H+ (aq)  Zn2+ (aq) + H2(g)
8.314 J mol-1 K-1 * 300 K
(0.100) (1.111)
E = 0.76 – —————————— ln ——————
2 * 96485 C mol-1
= 0.76 – 0.0129 * (1.02)
= 0.76 – 0.013 = 0.75 V
21 Electrochemistry
(0.200)2
See example 19.12
25
Concentration cell
Problem: At 298 K, evaluate the emf of the cell
Cu | Cu2+ (0.10 M) | | Cu2+ (1.0 M) | Cu
Cu(s)  Cu2+(0.1 M) + 2e;
Cu2+ (1.0 M) + 2e  Cu(s)
Solution:
The standard emf (Eo = 0.00)for
Cu | Cu2+ || Cu2+ | Cu
The reaction is actually Cu (s) + Cu2+ (1.0 M) = Cu2+ (0.1 M) + Cu (s)
R T [Cu2+]
R T 0.10
E = 0.00 – ——- ln ——— = – –––– ln –––––
2F
[Cu2+]
2F
1.00
8.3145 * 298 1.0
When 2 [ ]’s are
= + –––––––––– ln –––– = 0.0295 V
equal, E = 0
2*96485
0.1
The voltage is purely due to concentration difference. Solutions in the
two compartment try to become equal.
26
21 Electrochemistry
See p. 841
Equilibrium Constant K and Eocell
Calculate the solubility product of AgCl from data of standard cell
Solution: Look up desirable data
Ag+Cl– (s) + e  Ag0(s) + Cl– E° = 0.2223 V
Ag+ (aq) + e  Ag (s) E° = 0.799 V
Ag (s)  Ag+ (aq) + e E° = – 0.799 V
Get the desirable eq’n
AgCl (s)  Ag+ (aq) + Cl– (aq) E° = – 0.577 V
Ksp = [Ag+][Cl– ]
Show that for this cell
–
0.577
log Ksp =
/ 0.0592 = – 9.75
Ag | Ag+, 1 M || Cl–, 1 M | AgCl | Ag
Ksp = 10– 9.75 = 1.8e–10
Eo = – 0.577 V
but for this cell
Ag | AgCl |Cl–, 1 M || Ag+, 1 M | Ag
Eo = +0.577 V
At 298 K
0.0592
E o = ———— log K
n
21 Electrochemistry
See p. 837
27
Evaluate free-energy change
Evaluate G o for the reaction
Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s)
Solution:
Required to look up:
Eo V
Ag+ + e = Ag
Zn2+ + 2 e = Zn
2 Ag+ + 2 e  2Ag
Zn  Zn2+ + 2 e
Zn (s) + 2 Ag+ (aq)  Zn2+(aq) + 2 Ag (s)
See slides 17 and 22
Write the cell for this rxn
0.80
– 0.76
0.80 +
+ 0.76
1.56 (= E o)
G o = – n F
E o
= – 2 * 96485 C * 1.56 V
= – 301000 J
= – 301 kJ
21 Electrochemistry
28
Summary of thermodynamics
Chemical energy
Ho, So
Stoichiometry
Go
=
Ho –
T
So
n
Reaction quotient &
equilibrium constant
G = G o + R T ln Q
G o = – R T ln K
21 Electrochemistry
Electric energy
Go = – n F Eo
Reaction quotient &
equilibrium constant
E = E o – R T/ n F ln Q
E o = R T/ n F ln K
29
insight from cold denaturation and
a two-state water structure
By Tsai CJ, Maizel JV Jr, Nussinov R.
…The exposure of non-polar surface reduces the entropy and
enthalpy of the system, at low and at high temperatures. At low
temperatures the favorable reduction in enthalpy overcomes the
unfavorable reduction in entropy, leading to cold denaturation. At high
temperatures, folding/unfolding is a two-step process: in the first, the
entropy gain leads to hydrophobic collapse, in the second, the
reduction in enthalpy due to protein-protein interactions leads to the
native state. The different entropy and enthalpy contributions to the
Gibbs energy change at each step at high, and at low, temperatures
can be conveniently explained by a two-state model of the water
structure….
21 Electrochemistry
30
Biochem Mol Biol. 2002;37(2):55-69
pH and emf
Consider the cell,
Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt
From table data,
Zn2+ + 2e = Zn
Eo = – 0.76
2 H+ + 2 e = H2
Zn = Zn2+ + 2e
Zn + 2H+ = Zn2+ + H2
Eo = 0.00
Eo = 0.76
Eo = 0.76
RT
[Zn2+] PH2
E = Eo – —— ln ————
2F
[H+]2
= Eo + 0.0592 log [H+]
At 298 K (pH meters)
0.76 – E
pH = —————
0.0592
= 0.76 – 0.0592 pH
21 Electrochemistry
31
pH electrodes
pH Range: 0-14
Temp. Range:
0-100 C
Internal Ref: ROSS
Junction: Ceramic
Dimensions: 120 mm x 12 mm
Slope: 92 - 102%
Temp. Accuracy: 0.5 C
Catalog Number: 8202BN
(BNC Connector, 1 meter cable)
21 Electrochemistry
32
Ion selective electrode
More research has gone into pH
measurements. Nernst started it.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Concentration;
Temperature;
Electrode surface conditions;
Number of charges of ions (8);
Stirring (6);
Suspension (7);
Zwitterionic nature, net charge density;
Anything changing ionic adsorption;
Isoelectric nature of surface material;
The Nernst equation deals only with concentration and temperature
21 Electrochemistry
33
Battery technology
By use: automobile, flash light, radio, computer,
camera, watch, emergency equipment, artificial
heart machine, pace makers, hearing aids,
calculators, … (portable energy)
By type: alkaline, dry, wet, storage, rechargeable,
etc
Aluminum for battery
By chemistry: alkaline, carbon zinc, dry, cell, lithium,
manufacture
lithium ion, lithium polymer, NiCd, etc.
By material: anode material, cathode material,
electrode, etc.
21 Electrochemistry
See 21-5
34
Lead Storage Battery for Autos
Anode – Negative plate
Pb + SO42 PbSO4 + 2eSeparator
Cathode –
Positive plate
A 12-V battery
consists of 6
such cells
PbO2 + 4H+ + SO42- + 2e PbSO4 + 2 H2O
H2SO4
Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O
21 Electrochemistry
35
Figure 18.6: Schematic of one cell of the lead battery.
21 Electrochemistry
Copyright© by Houghton Mifflin Company. All rights reserved.
3610
A mercury battery.
21 Electrochemistry
37
Corrosion:
Unwanted Voltaic Cells
Fe(s)  Fe2+ (aq) + 2e–
O2 + H2O (l) + 4e–  4 OH– (aq)
2 Fe(s) + O2 + H2O
 2 Fe2+ (aq) + 4 OH– (aq)
What are effective corrosion
prevention methods?
Coating
Use sacrifice electrode
21 Electrochemistry
See 21-6
38
Cathodic protection of an
underground pipe.
21 Electrochemistry
39
Ion displacement reactions (corrosion)
Reaction E o (V)
What metal will react
with certain ions?
Zn + 2 Ag+  Zn2+ + 2 Ag
Zn + 2 Cu2+  Zn2+ + Cu
Zn2+ + 2 Ag  Zn + 2 Ag+
Zn2+ + Cu  Zn + 2 Cu2+
See 21-1
Li+ + e– = Li (s)
Na+ + e– = Na (s)
Mg2+ + 2 e– = Mg (s)
Zn2+ + 2 e– = Zn (s)
(reference) 2 H+ + 2 e– = H2 (g)
(reference) H2 (g) = 2 H+ + 2 e–
Cu2+ + 2 e– = Cu (s)
Cu+ + e– = Cu (s)
I2 (s) + 2e– = 2 I– (aq)
Ag+ (aq) + e- = Ag (s)
Br2 (l) + 2 e– = 2 Br– (aq)
Cl2 (g) + 2 e– = 2 Cl– (aq)
F2 (g) + 2 e– = 2 F– (aq)
21 Electrochemistry
– 3.04
– 2.71
– 2.38
– 0.76
0.000
0.000
0.34
0.52
0.54
0.80
1.07
1.36
2.87
40
Electrolysis of molten salts
Eo=-1.36 V; 2 Cl–  Cl2 + 2e–
e
oxidation
2 Cl–  Cl2 + 2e–
2 Na+ + 2 e–  2 Na
Net
2 NaCl  2 Na + Cl2
Eo=-2.71V;
2Na+ + 2e–  2Na
Battery
e
C
A
T
H
O
D
E
A
N
O
D
E
805o C
reduction
Molten salts consists of Na+ and Cl– ions
Charges required to produce 1 mole Cl2 and 2 moles Na = 2F
Energy required = 2 F E;
21 Electrochemistry
(E > 4.07 V)
See 21-7
41
Electrometallurgy of Sodium
21 Electrochemistry
42
Electrolysis of NaCl solution
2 Cl– = Cl2 + 2e–
Cl2 + H2O = HCl + ½ O2
oxidation
Battery
e
e
2Na+ + 2e– = 2Na
2Na + 2H+ = H2 + 2Na+
C
A
T
H
O
D
E
A
N
O
D
E
reduction
Salt solution consists of Na+ and Cl– ions
21 Electrochemistry
43
Refining Copper by Electrolsis
Copper can be purified by electrolysis. Raw copper is oxidized
Cu = Cu2+ + 2e
and purer copper deposited on to the cathode from a solution
containing CuSO4
Cu2+ + 2e = Cu
If a current of 2 amperes pass through the cell, how long will it take to
deposit 5.00 g of copper on the cathode? (1 ampere = 1 C s–1)
Solution
5.00
2*96485 C 1 s
------ mol ----------- ------ = work out your answer
65.5
1 mol 2 C
_______
21 Electrochemistry
New
44
Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al2O3,
remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some
Al metal by reducing Al2O3 with potassium vapore.
In 1886, two young men electrolyzed molten cryolite Na3AlF6 (melting point 1000° C), but did
not get aluminum.
Production of aluminum
Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and
obtained Al metal. This is the Hall process.
AlF63– + 3 e–  Al + 6 F– . . . Cathode
2 Al2OF62– + C(s) + 12 F–  + 4 AlF63– + CO2 + 4 e– . . . Anode
2 Al2O3 + 3 C  4 Al + 3 CO2 . . . Overall cell reaction
Charge required for each mole Al = 3 F
Energy required = 3 F E
21 Electrochemistry
45
Electrometallurgy of Aluminum
21 Electrochemistry
46
Electrolysis of acid solution
H2O = ½ O2 + 2e– + 2 H+ Battery
e
oxidation
A
N
O
D
E
e
2H+ + 2e– = H2
C
A
T
H
O
D
E
reduction
Solutions containing H+ and SO42– ions
Charges required to produce 1 mole H2 and ½ moles O2 = 2F
Energy required = 2 F E
21 Electrochemistry
47
Electrolysis of H2SO4 solution
Pure water is not a good electric conductor. In the presence of
electrolytes, water can be decomposed by electrolysis.
On the other hand, electrolysis of electrolyte solutions may reduce H+
and oxidize O2– in H2O.
In an H2SO4 solution,
cathode reductions are
2 H2O (l) + 2 e– = H2 (g) + 2 OH–
(same as 2H+ + 2e– = H2)
Anode oxidation:
2 H2O (l) = 4 e– + O2 (g) + 4 H+
2 SO42– = [SO3O–OSO3]2– + 2 e–
21 Electrochemistry
E o = – 1.23 V (observed)
E o = – 2.01 V (not observed)
48
Electrolysis of H2SO4 solution
E o = – 1.23 V
E o = – 2.01 V
2 H2O (l) = 4 e– + O2 (g) + 4 H+
2 SO42– = [SO3O–OSO3]2– + 2 e–
Battery
e
oxidation
A
N
O
D
E
e
2H+ + 2e = H2
reduction
C
A
T
H
O
D
E
Solution consists of H+ and SO42– ions
21 Electrochemistry
49
Electroplating of metals
Galvanizing
Zn2+ + 2 e–  Zn
onto metal surface
Copper purification
Cu2+ + 2 e–  Cu
onto pure Cu electrode
Silver plating
Ag+ + e–  Ag
onto metal surface
Over a half century of extensive and
innovative research has made us one of the
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Since 1971, Cal-Aurum has provided electroplating
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Miller specializes in plating metals such as
magnesium, aluminum, zinc, copper, powdered
metals, steel and various other substrates.
21 Electrochemistry
50
Summary
The 20th century belongs to electrons. They continue affecting our lives the 21st century.
Chemistry studies the drama played by electrons, and electrochemistry is the finale.
Energy directs and produces the show, but you set the magic stage for a great
performance.
Leo and Ger tell electrons to get in and out of your stage, and you must
skillfully provide paths to balance the flow.
Cells are the stages for the performance, you must construct,
represent, figure out the potentials, and control the show.
Chemical reaction, equilibrium, (acid, base, heterogeneous, and
complex formation) and electrochemistry guide us using simple rules.
Apply rules you have learned in Chem1235 to understand what is
happening around you and may your live be full of happiness.
21 Electrochemistry
51
Skills for Electrochemistry (review)
Make up a Daniel cell using Pb and Ag as the
electrodes. Draw a diagram for it.
Use short notation to represent the cell for the
spontaneous reaction
Write half reaction equations for both cathode and
anode and explain the reactions
Write balanced redox equations
Calculate emf for a nonstandard cell and its energy
Calculate equilibrium constant K from Eo
21 Electrochemistry
52