Transcript Balancing Redox Reactions

Chapter 18 Electrochemistry
The batteries we use to power
portable computers and other
electronic devices all rely on
redox reactions to generate
an electric current. Redox
reactions are central to the
development of small, light,
long-lasting power sources.
To a large extent, the future
development of technology
depends on the capabilities
of these power sources. In
this chapter, we see what is
involved in using chemical
reactions to generate
electricity.
Assignment for Chapter 18
18.2; 18.6; 18.17; 18.27
Electrochemistry in Biological Systems
Ion transport
Muscle contraction
Neuron excitation
Psychological activities
Electrochemistry
• Electrochemistry is a branch of chemistry
that deals with the use of spontaneous
chemical reactions to produce electricity
and the use of electricity to bring about
spontaneous chemical change.
Why electricity and chemical reaction may be connected?
Because both involve electrons! The motion or transfer of
electrons!
Proton transfer: acid-base reactions
Electron transfer: redox reactions
Half-Reactions
• Chemical equations showing the changes involved
only in oxidation or reduction.
Oxidation half-reaction:
2
Mg(s)  Mg (aq)  2e
Reduction half-reaction:
Fe (aq)+3e  Fe(s)
2+
-

Example
The oxidation of iron(II) to iron(III):
Fe (aq)  Fe (aq)+e
2+
3+
-
The reduction of copper(II) to copper metal:
Cu (aq)+2e  Cu(s)
2+
-
Exercise:
Aluminum metal is oxidized to Al3+ in aqueous solution:
Al(s)  Al (aq)+3e
3+
-
Balancing Redox Reactions
MnO4-(aq) + H2C2O4 (aq)  Mn2+ (aq) + CO2 (g)
!
• The chemical equation of a reduction halfreaction is added to that of an oxidation
half-reaction to form the chemical equation
for the overall redox reaction.
Use H2O and H+ and OH- as “extra” chemicals
Balancing Redox Reactions
in acidic solutions
MnO4-(aq) + H2C2O4 (aq)  Mn2+ (aq) + CO2 (g)
!
Reduction: MnO4- Mn2+ (aq)
Oxidation: H2C2O4 (aq)  CO2 (g)
Balance all elements except H and O:
Reduction: MnO4- Mn2+ (aq)
Oxidation: H2C2O4 (aq)  2CO2 (g)
Balance O using water and H using H+:
Reduction: MnO4- + 8H+ Mn2+ (aq)+4H2O
Oxidation: H2C2O4 (aq)  2CO2 (g) + 2H+
Balancing Redox Reactions
Balance Charge:
Reduction: MnO4- + 8H+ +5e- Mn2+ (aq)+4H2O
Oxidation: H2C2O4 (aq)  2CO2 (g) + 2H+ + 2eReduction x 2 + Oxidation x 5
2MnO4- + 16H+ +10e- 2Mn2+ (aq)+8H2O
5H2C2O4 (aq)  10CO2 (g) + 10H+ + 10e2MnO4- (aq)+ 5H2C2O4 (aq) +6H+ (aq) 
2Mn2+ (aq)+8H2O(l) + 10CO2 (g)
Figure 18.1 A schematic diagram of how to balance
a redox equation by balancing the half-reactions
separately and then combining them.
Example
• Write down the balanced net ionic reaction for the
reaction: Cu+HNO3Cu2++NO.
Cu+H++NO3-Cu2++NO
CuCu2+
2NO3-2NO + 4H2O
CuCu2+
8H++2NO3-2NO + 4H2O
CuCu2++2e-
8H++2NO3-+6e-2NO + 4H2O
x3
3Cu(s)+2NO3-(aq)+8H+(aq)3Cu2+(aq)+2NO(g)+4H2O(l)
Exercise
MnO4++H2SO3  HSO4- + Mn2+.
Write down the balanced net ionic reaction.
Balancing Redox Reactions
in basic solutions
Br- (aq)+ MnO4-(aq) MnO2(s)+BrO3-(aq)
Br- BrO3-
MnO4-(aq) MnO2(s)
Br- + 3H2OBrO3-
MnO4-(aq) MnO2(s) +2H2O
Br- + 3H2O+6OH-BrO3-+6H2O
MnO4-(aq)+4H2O MnO2(s) +2H2O+4OHBr- + 6OH-BrO3-+3H2O +3eMnO4-(aq)+2H2O MnO2(s) +4OH-+6e2MnO4-(aq)+ Br-(aq) +H2O(l) MnO2(s)+BrO3-(aq)+2OH-(aq)
Electrochemical Cell
• A device in which an electric current is
either produced by a spontaneous chemical
reaction or is used to bring about a
nonspontaneous reaction.
A galvanic cell is an electrochemical cell in which a spontaneous
chemical reaction is used to generate an electric current.
Figure 18.2 In an
electrochemical cell, a reaction
takes place in two separate
negative
regions. Oxidation occurs at
one electrode, and the
electrons released travel
through the external circuit to
the other electrode, where they
cause reduction. The site of
oxidation is called the anode,
and the site of reduction is
called the cathode.
positive
Any two objects that have different
(first) ionization energies may function
as a cell.
- 1.234 +
- 0.02 +
Figure 18.3(a) When a bar of zinc is placed in a beaker of copper(II)
sulfate solution, copper is deposited on the zinc and the blue copper
(II) ions are gradually replaced by colorless zinc ions. (b) The residue
in the beaker is copper metal. No more copper ions can be seen in
solution.
Zn(s)+Cu (aq)  Zn (aq)+Cu(s)
2+
2+
Figure 18.4 The reaction shown in Fig. 18.3 takes
place all over the surface of the zinc as electrons
are transferred to the Cu2 ions in solution.
Cu 2+ (aq)+2e-  Cu(s)
Zn(s)  Zn (aq)+2e
2+
-
Figure 18.5 The Daniell cell consists of copper and zinc electrodes
dipping into solutions of copper(II) sulfate and zinc sulfate, respectively.
The two solutions make contact through the porous pot, which allows
ions to pass through to complete the electrical circuit.
Zn(s)|Zn 2+ Cu 2+ |Cu(s)
Zn(s)  Zn 2+ (aq)+2e-
Cu (aq)+2e  Cu(s)
2+
-
Electrodes and Cell Diagram
Zn(s)|Zn 2+ Cu 2+ |Cu(s)
+
H (aq)|H 2 (g)|Pt(s)
3+
2+
Fe (aq),Fe (aq) |Pt(s)
2+
2+
Zn(s)|Zn (aq)|Cu (aq) |Cu(s)
Figure 18.6 This cell is typical of galvanic cells used in the
laboratory. The two electrodes are connected by an external circuit
and a salt bridge. The latter completes the electrical circuit within the
cell.
Pt(s)|Fe3+ (aq),Fe2+ (aq) ||Cu 2+ (aq) |Cu(s)
Figure 18.7 The cell potential is measured with an electronic voltmeter, a
device that draws negligible current so that the composition of the cell
does not change during the measurement. The display shows a positive
value when the  terminal of the meter is connected to the cathode of the
galvanic cell.
2+
2+
Zn(s)|Zn (aq)||Cu (aq) |Cu(s)
Cell Potential
E = 1.1 V
The cell potential
• An indication of the electron-pulling and –
pushing power of the cell reactions; cell
reactions at equilibrium generate zero
potential.
Figure 18.8 Electrons produced by oxidation leave a galvanic cell at the
anode (), travel through the external circuit, and reenter the cell at the
cathode (), where they cause reduction. The circuit is completed inside
the cell by migration of ions through the salt bridge. A salt bridge is
unnecessary when the two electrodes share a common electrolyte.
positive
negative
Figure 18.9 This schematic picture of a galvanic cell indicates the
identities of the anode and cathode, displays the oxidation and reduction
half-reactions, and shows the direction of electron flow.
Describing a galvanic cell and identifying the cell reaction
Hg2+ (aq)+2e-  2Hg(l) 2Hg(l)+2Cl- (aq)  Hg 2Cl2 (s)+2e-
 Hg (aq)+2Cl (aq)  Hg 2Cl2 (s)
2+
-
Hg(l)|Hg 2 Cl2 (s)|HCl(aq)||Hg (NO ) (aq) |Hg(l)
2
3 2
(KCl gel)
Exercise: Describing a galvanic cell and identifying the cell reaction
(Assume platinum electrode is used)
H2 (g)  2H+ (aq)+2e-
Co3+ (aq)+e-  Co2+ (aq)
H2 (g)+2Co (aq)  2H (aq)+ 2Co (aq)
3+
+
+
3+
2+
2+
Pt(s)|H 2 (g)||H (aq)||Co (aq),Co (aq) |Pt(s)
Figure 18.12 The cell potential can be thought of as being the difference
of the two reduction potentials produced by the two electrodes. The cell
potential is positive if the cathode has a higher potential than the anode.
Cell potential and electrode potential
standard cell potential:
E 0  E 0 (cathode)  E 0 (anode)
Fe(s)|Fe2+ (aq)||Ag + (aq) | Ag(s):
E 0  1.24 V
Hydrogen standard potential
2H +2e  H2 (g): E  0 V
+
-
0
Zn(s)|Zn (aq)||H (aq) | H 2 (g)|Pt(s): E  0.76 V
2+
+
0
E 0  E 0 (H + / H 2 )  E 0 (Zn 2+ / Zn)  0.76 V
 E 0 (Zn 2+ / Zn)  0.76 V
Figure 18.13 The variation of standard potentials in the main groups of
the periodic table. Note that the most negative values occur in the s block
and the most positive values occur close to fluorine.
Table 18-1 p856
Example: deducing the standard potential of an electrode
The standard potential of a Zn2+/Zn electrode is -0.76 V and
the standard potential of the cell
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) is 1.10 V.
What is the standard potential of Cu2+/Cu electrode?
E 0 (Cu 2+ /Cu)  E 0 (Zn 2+ /Zn)  1.10 V
E (Cu / Cu )  1.10 V  E (Zn / Zn)
 1.10 V - 0.76 V=+0.34 V
0
2+
0
2+
Exercise: deducing the standard potential of an electrode
The standard potential of a Fe2+/Fe electrode is -0.44 V and
the standard potential of the cell
Fe(s)|Fe2+(aq)||Pb2+(aq)|Pb(s) is 0.31 V.
What is the standard potential of Pb2+/Pb electrode?
E 0 (Pb2+ /Pb)  E 0 (Fe2+ /Fe)  0.31 V
E ( Pb / Pb)  0.31 V+E ( Fe / Fe)
 0.31 V  0.44 V =-0.13 V
0
2+
0
2+
Figure 18.14 The significance of standard potentials. Only couples with
negative standard potentials (and hence lying below hydrogen) can
reduce hydrogen ions to hydrogen gas. The reducing power increases as
the standard potential becomes more negative.
E 0 (Cu 2+ / Cu)=+0.34 V
E (Cu /Cu)  E (H /H)  0.34 V
0
2+
0
+
Cu (aq)+H 2 (g)  Cu(s)+H (aq)
2+
+
E 0 (Zn2+ / Zn)=-0.76 V
E (Zn /Zn)  E (H /H)  0.76 V
0
2+
0
+
Zn(s)+H (aq)  Zn (aq)+H2 (g)
+
2+
Figure 18.15 Although aluminum has a negative standard potential, signifying that
it can be oxidized by hydrogen ions (as in the hydrochloric acid, left), nitric acid
(right) stops reacting with it as soon as an impenetrable layer of aluminum oxide
has formed on its surface. This resistance to further reaction is termed
passivation of the metal.
E 0 (Al3+ /Al)=-1.66 V
E (Al /Al)  E (H /H)  1.66 V
0
3+
0
+
Al(s)+H (aq)  Al (aq)+H 2 (g)
+
3+
18.4 Cell Potential, Electrical Work, and Free Energy
Work
 Work is never the maximum possible if any current
is flowing.
 In any real, spontaneous process some energy is
always wasted – the actual work realized is always
less than the calculated maximum.
Basic Concepts
7.) Electric Potential (E)

Combining definition of electrical charge and potential
 G   work   E  q
Relation between free energy
difference and electric potential
difference:
q  nF
 G   nFE
Describes the voltage that can be generated by a chemical reaction
Figure 18.16 The relation between
the standard potential of a reaction
(reactants, purple; products, yellow)
and the equilibrium constant.
For a (half) reaction (Mn+ + ne-  M),
the higher the reduction potential (the
more negative ΔGo), the more easily it
proceeds.
Galvanic Cells
1.) Galvanic or Voltaic cell

Example: Calculate the voltage for the following chemical reaction
G = -150kJ/mol of Cd
n – number of moles of electrons
Solution:
G   nFE  E  
G
nF
 150  10 3 J
E
 0.777 J  0.777 V
C
C 

( 2 mol ) 9.649  10 4

mol 

18.4 Cell Potential, Electrical Work, and Free Energy
Maximum Cell Potential
 Directly related to the free energy difference
between the reactants and the products in the cell.
 ΔG° = –nFE°
 F = 96,485 C/mol e–
Example 18.5 Calculating G for a Cell Reaction
Using the data in Table 18.1, calculate G for the
reaction
Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)
Is this reaction spontaneous?
Example 18.5 Calculating G for a Cell Reaction
The half-reactions are
We can calculate G from the equation
G  G o  RTlnQ
G  -nFE
E  Eol - RT
nF lnQ
Example 18.5 Calculating G for a Cell Reaction
Since two electrons are transferred per atom in the
reaction, 2 moles of electrons are required per mole
of reactants and products. Thus n = 2 mol e-, F =
96,485 C/mol e-, and
= 0.78 V = 0.78 J/C.
Therefore,
C 
J

G  (2 mol e )  96, 485
0.78 
 
mol e 
C

 1.5 105 J

The process is spontaneous, as indicated by both
the negative sign of G and the positive sign of
Example 18.5 Calculating G for a Cell Reaction
This reaction is used industrially to deposit copper
metal from solutions resulting from the dissolving of
copper ores.
18.5 Dependence of Cell Potential on Concentration
A Concentration Cell
18.5 Dependence of Cell Potential on Concentration
Nernst Equation
E cell  Eocell - RT
nF lnQ
The relationship between cell potential and
concentrations of cell components
At 25°C:
0.0591
E = E 
log Q 
n
At equilibrium (E=0, Q=K):
0.0591
E=
log  K 
n
aA + ne- » bB
Eo
b
A
RT
E  Eo 
ln B
a
nF AA
b
0.05916
V
[
B
]
 E  Eo 
log
n
[ A]a
Eo and the Equilibrium Constant
1.)
A Galvanic Cell Produces Electricity because the Cell Reaction is
NOT at Equilibrium


Concentration in two cells change with current
Concentration will continue to change until Equilibrium is reached
E = 0V at equilibrium
Battery is “dead”
Consider the following ½ cell reactions:
aA + ne- » cC
dD + ne- » bB
E+o
E-o
Cell potential in terms of Nernst Equation is:
E cell  E   E  
E o
0.05916
[ C ] c  o 0.05916
[ B ]b

log
 E 
log
a
d

n
n
[ A]
[
D
]

Simplify:
E cell  ( E o

E o
0.05916
[C ]c [ D ]d
)
log
n
[ A]a [ B ]b




18.5 Dependence of Cell Potential on Concentration
CONCEPT CHECK!
Explain the difference between E and E°.
When is E equal to zero?
When the cell is in equilibrium ("dead" battery).
When is E° equal to zero?
E  is equal to zero for a concentration cell.
18.5 Dependence of Cell Potential on Concentration
EXERCISE!
A concentration cell is constructed using two nickel
electrodes with Ni2+ concentrations of 1.0 M and
1.00 × 10-4 M in the two half-cells.
Calculate the potential of this cell at 25°C.
0.118 V
1
E  0.0591
log
 0.118
2
110-4
o
18.5 Dependence of Cell Potential on Concentration
CONCEPT CHECK!
You make a galvanic cell at 25°C containing:
 A nickel electrode in 1.0 M Ni2+(aq)
 A silver electrode in 1.0 M Ag+(aq)
Sketch this cell, labeling the anode and cathode,
showing the direction of the electron flow, and
calculate the cell potential.
1.03 V
1.03 V
NO3-
K+
Ni
Ag
Ni2
+
E Ag/Ag  0.78
o
NO3Ag+
E Ni/Ni2  -0.25
o
Example 18.7 The Effects of Concentration on
For the cell reaction
predict whether
is larger or smaller than
for the following cases.
a. [Al3+] = 2.0 M, [Mn2+] = 1.0 M
b. [Al3+] = 1.0 M, [Mn2+] = 3.0 M
E cell  E cell o
0.0591
n
logQ  E cell o
0.0591
n
[Al3 ]2
log [Mn 2 ]3
Example 18.7 The Effects of Concentration on
Solution
a. A product concentration has been raised above 1.0
M. This will oppose the cell reaction and will cause
to be less than
(
< 0.48 V).
E cell  E cell o
0.0591
n
log
22
13
b. A reactant concentration has been increased above
1.0 M, and
will be greater than
(
>
0.48 V).
E cell  E cell o
0.0591
n
log
12
33
Investigating Matter 18.1 (a)
A glass electrode in a protective plastic sleeve
(left) is used to measure pH. It is sometimes
used in conjunction with a calomel electrode
(right) in pH meters such as this one.
Table 18-2 p869
Calomel reference electrodes
 The potential of the calomel electrode is known vs
the SHE. This is used as the reference electrode
in the measurement of pH
Hg 2Cl2 (s)  2e  2Hg (l )  2Cl

 The other electrode in a pH probe is a glass
electrode which has a Ag wire coated with AgCl
dipped in HCl(aq). A thin membrane separates the
HCl from the test solution
Eo and the Equilibrium Constant
1.)
A Galvanic Cell Produces Electricity because the Cell Reaction is
NOT at Equilibrium
Since Eo=E+o- E-o:
0.05916
[C ]c [ D ]d
E cell  E 
log
n
[ A]a [ B ]b
o
At equilibrium Ecell =0:
Definition of
equilibrium constant
Eo 
0.05916
log K
n
K  10
nE o
0.05916
at 25oC
at 25oC
Investigating Matter 18.1 (b)
This durable portable pH meter can be used for quick
measurements of pH in the field. Its accuracy is not as high as
that of a laboratory pH meter.
The pH meter is an electrochemical cell
 Overall cell potential is proportional to pH
Ecell  0.06V  pH   Eref
pH 
Ecell  Eref
0.06V
 In practice, a hydrogen electrode is
impractical
18.6 Batteries
One of the Six Cells in a 12–V Lead Storage Battery
Anode (negative plate) reaction:
Pb(s) + HSO−4(aq) → PbSO4(s) + H+(aq) + 2e−
Cathode (positive plate) reaction:
PbO2(s) + HSO−4(aq) + 3H+ (aq) + 2e−
→ PbSO4(s) + 2H2O(l)
The total reaction:
Pb(s) + PbO2(s) + 2H2SO4(aq) →
2PbSO4(s) + 2H2O(l)
Figure 18.18 One cell of a leadacid battery like those used in
automobiles. A lead-acid battery
is an example of a secondary cell.
It needs to be charged before it
can produce a current. The
electrolyte is dilute sulfuric acid.
18.6 Batteries
A Common Dry Cell Battery
Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e− [E° = -1.28 V]
2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq) [E° = +0.15 V]
The total reaction:
Zn(s) + 2MnO2(s) ⇌ ZnO(s) + Mn2O3(s)
[E°cell = +1.43 V]
Figure 18.17 A commercial dry cell
consists of a graphite cathode in a
zinc container; the latter acts as the
anode. The container is filled with a
moist paste of NH4Cl, MnO2, finely
divided carbon, and an inert filler
such as starch. In the cell reaction,
manganese(IV) is reduced to
manganese(III) and zinc metal is
oxidized to Zn2 ions.
The Rechargeable Lithium Ion Battery
The cathode (marked +) half-reaction is:
The anode (marked -) half reaction is:
The overall reaction has its limits. Overdischarge supersaturates
lithium cobalt oxide, leading to the production of lithium oxide,
possibly by the following irreversible reaction:
The Disposable Lithium Battery (1.5 – 3.7 V)
3
2Li(s) → 2Li+ + 2e− [E° = 3.0 V]
2MnO2(s) + H2O(l) + 2e− → Mn2O3(s) + 2OH−(aq) [E° = +0.15 V]
https://en.wikipedia.org/wiki/Lithium_battery
18.6 Batteries
 Mercury batteries use either pure mercury(II) oxide (HgO)—also called mercuric oxide—or a
mixture of HgO with manganese dioxide (MnO2) as the cathode. Mercuric oxide is a nonconductor, so some graphite is mixed with it; the graphite also helps prevent collection of
mercury into large droplets. The half-reaction at the cathode is:
 HgO + H2O + 2e− → Hg + 2OH−
 with a standard potential of +0.0977 V.
 The anode is made of zinc (Zn) and separated from the cathode with a layer of paper or other
porous material soaked with electrolyte; this is known as a salt bridge. Two half-reactions occur
at the anode. The first consists of an electrochemical reaction step:
 Zn + 4OH− → Zn(OH)4−2 + 2e−
 followed by the chemical reaction step:
 Zn(OH)4−2 → ZnO + 2OH− + H2O
 yielding an overall anode half-reaction of:
 Zn + 2OH− → ZnO + H2O + 2e−
 The overall reaction for the battery is:
 Zn + HgO → ZnO + Hg
 In other words, during discharge, zinc is oxidized (loses electrons) to become zinc oxide (ZnO)
while the mercuric oxide gets reduced (gains electrons) to form elemental mercury. A little extra
mercuric oxide is put into the cell to prevent evolution of hydrogen gas at the end of life.
Fuel cells – a battery with a
difference
 Reactants are not contained within a sealed
container but are supplied from outside
sources
anode : 2H 2 ( g )  4OH  (aq)  4H 2O(l )  4e
cathode : O2 ( g )  2H 2O(l )  4e  4OH  (aq)
overall : 2H 2 ( g )  O2 ( g )  2H 2O(l )
18.6 Batteries (New Generation)
Schematic of the Hydrogen-Oxygen Fuel Cell
Anode Reaction: 2H2 + 2O2− → 2H2O + 4e−
Cathode Reaction: O2 + 4e− → 2O2−
Overall Cell Reaction: 2H2 + O2 → 2H2O
18.7 Corrosion
 Process of returning metals to their natural state –
the ores from which they were originally obtained.
 Involves oxidation of the metal.
Figure 18.22 The mechanism of rust formation. (a) Oxidation of
the iron occurs at a point out of contact with the oxygen of the air,
and the surface of the metal acts as an anode in a tiny galvanic
cell. (b) Further oxidation of Fe2 to Fe3 results in the deposition
of rust on the surface.
18.7 Corrosion
The Electrochemical Corrosion of Iron
18.7 Corrosion
Corrosion Prevention
 Application of a coating (like paint or metal plating)
 Galvanizing
 Alloying
 Cathodic Protection
 Protects steel in buried fuel tanks and pipelines.
18.7 Corrosion
Cathodic Protection
Fig. 18.24 In the cathodic
protection of a buried
pipeline, or other large
metal construction, the
artifact is connected to a
number of buried blocks of
metal, such as magnesium
or zinc. The sacrificial
anodes (the magnesium
block in this illustration)
supply electrons to the
pipeline (the cathode of the
cell), thereby preserving it
from oxidation.
Figure 18.21 Iron nails stored
in oxygen-free water (left) do
not rust because the oxidizing
power of water itself is weak.
When oxygen is present (as a
result of air dissolving in the
water, right), oxidation is
thermodynamically
spontaneous and rust soon
forms.
18.8 Electrolysis
 Forcing a current through a cell to produce a chemical
change for which the cell potential is negative.
Galvanic cell
Electrolytic cell
Electrolysis
 Electrolysis of a molten salt using inert electrodes
 Signs of electrodes:
 In electrolysis, anode is positive because electrons are removed
from it by the battery
 In a galvanic cell, the anode is negative because is supplies
electrons to the external circuit
Anode : 2Cl  (l )  Cl2 ( g )  2e
Cathode : 2 Na  (l )  2e  2 Na(l )
Overall :
2 Na  (l )  2Cl  (l )  2 Na(l )  Cl2 ( g )
Figure 18-26 p886
Electrolysis in aqueous solutions – a
choice of process
 There are (potentially) competing processes in the
electrolysis of an aqueous solution
 Cathode
Cathode:2 Na (l)  2e  2 Na(l)...E   2.71V
Cathode : 2H 2O(l )  2e  H 2 ( g )  2OH  (aq)...E  0.83V
 Anode
Anode : 2Cl  (l )  Cl2 ( g )  2e...E  1.36V
Anode : 2H 2O(l )  O2 ( g )  4H   4e...E  1.23V
(See later, overvoltage)
Thermodynamics or kinetics?
 On the basis of thermodynamics we choose
the processes which are favoured energetically
（processes with lowest potential）
Anode : 2H 2O(l )  O2 ( g )  4H   4e...E  1.23V

Cathode : 2H 2O(l )  2e  H 2 ( g )  2OH (aq)...E  0.83V
 But…chlorine is evolved at the anode
Electrolysis of water
 In aqueous solutions of most salts or acids or bases
the products will be O2 and H2
Cathode : 2H 2O(l )  2e  H 2 ( g )  2OH  (aq)...E  0.83V

Anode : 2H 2O(l )  O2 ( g )  4H  4e...E  1.23V
Other applications of electrolysis
Figure 18.28 A schematic
picture showing the
electrolytic process for
refining copper. The anode
is impure copper. It
undergoes oxidation, and
the Cu2 ions so produced
migrate to the cathode,
where they are reduced to
pure copper metal. A similar
arrangement is used for
electroplating objects.
Plating Order
The higher the reduction potential of the half reaction, the easier
the plating occurs.
Cathode : M i
Mj
mj
mi 
(aq )  m i e -  M i (s) E i
(aq )  m je -  M j (s) E j
o
o
If E i  E j , then M i forms before M j as power volt age increases.
o
o
Electrolysis of Mixtures of Ions
Ag  (l )  e-  Ag ...E  0.8 V
Cu 2 (l )  2e-  Cu ...E  0.34 V
Zn 2 (l )  2e-  Zn...E  -0.76 V
The larger the reduction potential, the more negative the free energy
change for the half reaction, the more easily it proceeds.
Therefore, as the electrolysis voltage increases gradually from very
low, the metals plate out in this order (silver being the first to deposit):
Ag   Cu 2  Zn 2
Electrolysis of Mixtures of Ions
Ce4  4e -  Ce ...E   1.7 V

2
VO 2  2H   e -  VO 2  H 2O ...E   1.0 V
Fe3  e-  Fe2 ...E   0.77 V
Oxidizing order (plating-out order):
4

Ce  VO 2  Fe3
Be aware of overvoltage
In the electrolysis of aqueous solution of NaCl, the half reactions
at the anode are:
Anode : 2Cl  (l )  Cl2 ( g )  2e...E  1.36V
Anode : 2H 2O(l )  O2 ( g )  4H   4e...E  1.23V
It seems oxygen will be produced before chlorine.
In reality, however, this is no so – chlorine is produced first.
Why so? That is because of overvoltage – caused by the electrode
—solution interface that blocks the transfer of electrons from
solution to the atoms on electrode.
Figure 18.29 A schematic representation of the
stoichiometric relations that are used to calculate the
amount of product formed by electrolysis or the amount
of time current must flow to produce a given product.
Anode : aA  pR n   qS  ne Cathode : bC  me  rP  sQ
-

18.8 Electrolysis
Stoichiometry of Electrolysis
 How much chemical change occurs with the flow of
a given current for a specified time?
current and time quantity of charge
moles of electrons moles of analyte
grams of analyte
18.8 Electrolysis
Stoichiometry of Electrolysis
 current and time quantity of charge
Coulombs of charge = amps (C/s) × seconds (s)
 quantity of charge moles of electrons
mol e

= Coulombs of charge 
1 mol e

96, 485 C
18.8 Electrolysis
CONCEPT CHECK!
An unknown metal (M) is electrolyzed. It took
52.8 sec for a current of 2.00 amp to plate
0.0719 g of the metal from a solution containing
M(NO3)3.
What is the metal?
gold (Au)
Solution in detail
The charge: 2 A x 52.8 s = 105.6 C
which is
105.6 C / F = 105.6/96485 mol for a monovalent
metal.
From the formula M(NO3)3, M is a trivalent metal. Therefore, the
number of moles for M is
105.6/96485/3 mol
The atomic weight therefore is
0.0719/(105.6/96485/3) g/mol = 197.1 g/mol.
18.8 Electrolysis
CONCEPT CHECK!
Consider a solution containing 0.10 M of each of
the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+.
Predict the order in which the metals plate out as
the voltage is turned up from zero.
Cu2+, Pb2+, Sn2+, Ni2+, Zn2+
Do the metals form on the cathode or the anode?
Explain.
The plating order is the order of reduction potential of the electrode
(Table 18.1)
Example 18.11 Electroplating
How long must a current of 5.00 A be applied to a
solution of Ag+ to produce 10.5 g silver metal?
Example 18.11 Electroplating
Solution
In this case, we must use the steps given earlier in
reverse:
1 mol Ag
10.5 g Ag 
 9.73 102 mol Ag
107.868 g Ag
Example 18.11 Electroplating
Each Ag+ ion requires one electron to become a
silver atom:
Ag+ + e  Ag
Thus 9.73  102 mole of electrons is required, and
we can calculate the quantity of charge carried by
these electrons:
9.73 10
2
96, 485 C
3
mol e 

9.39

10
C

mol e

Example 18.11 Electroplating
The 5.00 A (5.00 C/s) of current must produce 9.39
 103 C of charge. Thus
C

3
5.00

(time,
in
s)

9.39

10
C


s

9.39 103 s
Time 
s  1.88 103 s  31.3 min
5.00
Example 18.11 describes only the half-cell of interest. There also must
be an anode at which oxidation is occurring.
18.9 Commercial Electrolytic Processes





Production of aluminum
Purification of metals
Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
Electrolysis for Producing Aluminum
Al 3  3e -  Al ...E   -1.66 V
2H 2O(l )  2e-  H 2 ( g )  2OH  ...E  0.83V
Aluminum metal would not be plated out of an aqueous solution of Al3+.
Alumina (Al2O3) has high melting point (2050 oC), too hot for electrolysis.
Hall and Heroult (1886) found that if alumina is
mixed with Na3AlF6, the melting point drops to
1000 oC, low enough for electrolysis.
3
Cathode ：AlF6  3e-  Al  6FAnode ：2Al 2OF6  12F-  C  4AlF6  CO2  4e2-
3-
Overall ：2Al 2O3  3C  4Al  3CO2
Charles Martin Hall (1863-1914)
18.9 Commercial Electrolytic Processes
Producing Aluminum by the Hall-Heroult Process
Miscellaneous applications
Figure 18-24 p884
18.9 Commercial Electrolytic Processes
The Downs Cell for the Electrolysis of Molten Sodium
Chloride
Figure 18.30 In the Downs
process, molten sodium
chloride is electrolyzed with
a graphite anode (at which
the Cl ions are oxidized to
chlorine) and a steel
cathode (at which the Na
ions are reduced to
sodium). The sodium and
chlorine are kept apart by
the hoods surrounding the
electrodes. Calcium chloride
is present to lower the
melting point of sodium
chloride to a more
economical value.
Figure 18.25 A schematic
picture of the electrolytic cell
used in the Dow process for
magnesium. The electrolyte is
molten magnesium chloride.
As the current generated by
the external source passes
through the cell, magnesium
metal is produced at the
cathode and chlorine gas is
produced at the anode.
Figure 18.26 In this schematic picture of an electrolysis experiment, the
electrons emerge from a galvanic cell at its anode () and enter the
electrolytic cell at its cathode (), where they bring about reduction.
Electrons are drawn out of the electrolytic cell through its anode () and
into the galvanic cell at its cathode (). If the cell reaction in the galvanic
cell is more strongly spontaneous than the reaction in the electrolytic
cell is nonspontaneous, then the overall process is spontaneous. This
experiment is an example of one reaction driving another to which it is
coupled.
Figure 18.27 Michael Faraday (1791–1867) giving a public
lecture on chemistry at the Royal Institution in London.
Figure 18.20 The electric eel (Electrophorus electricus)
lives in the Amazon. The average potential difference it
produces along its length (1 m) is about 700 V.
Figure 18.23 Metal girders are galvanized by immersion
in a bath of molten zinc.
Figure 18.31 Chromium plating lends decorative flair as well as
protection to the steel of this motorcycle. Large quantities of electricity
are needed to plate chromium because six electrons are required to
produce each atom of chromium.
Case Study 18 (a) One of the three alkali fuel cells used on the
space shuttle. Although only one cell is needed to provide lifesupport, electricity, and drinking water, shuttle flight rules require
that all three be functioning.
Case Study 18 (b) High-pressure hydrogen tanks run across the
top of this bus provided by Ballard Power Systems for testing
hydrogen-oxygen fuel cells in Chicago. The bus is pollution free
and can go 250 miles before needing to be refueled.
History of Electrochemistry
https://en.wikipedia.org/wiki/History_of_electrochemistry
History of Battery
https://en.wikipedia.org/wiki/History_of_the_battery
Introduction to Fuel Cells
https://www.youtube.com/watch?v=cQ8dFM4GFxk
Nobel Prize in Chemistry 2007
https://www.youtube.com/wath?v=j3FecuuZ_kg
http://www.nobelprize.org/nobel_prizes/chemistry/laureates/2007/p
opular-chemistryprize2007.pdf
Assignment for Chapter 18
18.2; 18.6; 18.17; 18.27