Electrochemistry
Chapter 17
Slide 1
Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial
production of
chemicals such as Cl2,
NaOH, F2 and Al
• Biological redox
reactions,photosynthesis
6CO2 + 6H2O --> C6H12O6 + 6O2
C6H12O6 + O2 --> 6CO2 + 6H2O +Energy
Slide 2
Redox Reactions
01
Slide 3
Redox Reactions
01
Redox reaction are those involving the oxidation
and reduction of species.
LEO – Loss of Electrons is Oxidation .
GER –Gain of Electrons Is Reduction .
Oxidation and reduction must occur together.
They cannot exist alone.
Slide 4
Redox Reactions
02
Oxidation Half-Reaction: Zn(s)  Zn2+(aq) + 2 e–.
• The Zn loses two electrons to form Zn2+.
•
Slide 5
Redox Reactions
03
Reduction Half-Reaction: Cu2+(aq) + 2 e–  Cu(s)
• The Cu2+ gains two electrons to form copper.
•
Slide 6
Electrochemical Cells
anode
oxidation
cathode
reduction
spontaneous
redox reaction
Slide 7
Redox Reactions
•
04
Overall: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Slide 8
Electromotive Force (emf)
Water only spontaneously
flows one way in a waterfall.
• Likewise, electrons only
spontaneously flow one way
in a redox reaction—from
higher to lower potential
energy.
•
Slide 9
Electromotive Force (emf)
The potential difference between the anode and
cathode in a cell is called the electromotive force
(emf).
• It is also called the cell potential, and is designated
Ecell.
•
Slide 10
Cell Potential
Cell potential is measured in volts (V).
Volt, a potential difference between two points which results to a current of one
ampere through a resistance of one ohm
C (Coulomb):
The amount of charge transferred when a current of 1 ampere (A)
Flows for one second.
Slide 11
Cell Potentials and Free-Energy Changes
for Cell Reactions
1J=1Cx1V
joule
SI unit of energy
volt
SI unit of electric potential
coulomb
Electric charge
1 coulomb is the amount of charge transferred
when a current of 1 ampere flows for 1 second.
J
1V=1
C
Slide 12
Slide 13
Electrochemical Cells
01
•
Electrodes: are usually metal strips/wires
connected by an electrically conducting wire.
•
Salt Bridge: is a U-shaped tube that contains a gel
permeated with a solution of an inert electrolyte.
•
Anode: is the electrode where oxidation takes
place, (-).
•
Cathode: is the electrode where reduction takes
place, (+) terminal
Slide 14
Slide 15
Cells Notation
02
Anode Half-Cell || Cathode Half-Cell
Electrode | Anode Soln || Cathode Soln | Electrode
Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)
Slide 16
Electrochemical Cells
02
Anode Half-Cell || Cathode Half-Cell
Electrode | Anode Soln || Cathode Soln | Electrode
Fe(s) | Fe2+(aq) || Fe3+(aq), | Pt(s)
Slide 17
Write the cell notation for:
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Slide 18
Electrochemical Cell
Potentials
•
The standard half-cell potentials are determined
from the difference between two electrodes.
•
The reference point is called the standard hydrogen
electrode (S.H.E.) and consists of a platinum
electrode in contact with H2 gas (1 atm) and
aqueous H+ ions (1 M).
•
The standard hydrogen electrode is assigned an
arbitrary value of exactly 0.00 V.
Slide 19
Slide 20
Cu2+ (aq) + 2eCu (s) E0 = 0.34 V
∆G˚ = –nFE˚
Slide 21
Electrochemical Cells
---
06
Slide 22
•
E0 is for the reaction as
written
•
The more positive E0 the
greater the tendency for the
substance to be reduced
∆G˚ = –nFE˚` F = 96,500 C/mole
n = number of moles of electrons in reaction
•
The half-cell reactions are
reversible
•
The sign of E0 changes
when the reaction is
reversed
•
Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0
Slide 23
Electrochemical Cells
•
03
The standard potential of any galvanic cell is the
sum of the standard half-cell potentials for the
oxidation and reduction half-cells.
E°cell = E°oxidation + E°reduction
•
Standard half-cell potentials are always tabulated
as a reduction process. The sign must be changed
for the oxidation process.
Slide 24
Electrochemical Cells
•
When selecting two half-cell reactions the more
negative value, or smaller E° will form the
oxidation half-cell.
•
Consider the reaction between zinc and silver:
Ag+(aq) + e–  Ag(s)
Zn2+(aq) + 2 e–  Zn(s)
•
07
E° = 0.80 V
E° = – 0.76 V
Therefore, zinc forms the oxidation half-cell:
2+(aq) + 2 e–
0 (–0.76
0 V) V = 1.56 V
0 E°
Zn(s)

Zn
=
–
E
=
0.76
V+
0.80
0
0
0
cell
Ecell = E°Ox + E°Red
Slide 25
Write the cell rection and calculate the standard emf of an
electrochemical cell made of a Cd electrode in a 1.0 M
Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3
solution?
Cd2+ (aq) + 2e-
Cd (s) E0 = -0.40 V Cd is the stronger oxidizer
Cr3+ (aq) + 3e-
Cr (s)
Anode (oxidation):
Cd will oxidize Cr
E0 = -0.74 V
Cr3+ (1 M) + 3e- x 2
Cr (s)
Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)
Cd (s)
x3
3Cd (s) + 2Cr3+ (1 M)
E°cell = E°oxidation + E°reduction
0 = 0.74 V + (-0.4 V)
Ecell
0 = 0.34 V
Ecell
Slide 26
Spontaneity of a Reaction
01
•
The value of E˚cell is related to the thermodynamic
quantities of ∆G˚ and K.
•
The value of E˚cell is related to ∆G˚ by:
∆G˚ = –nFE˚cell
F = 96,500 C/mole
•
The value of K is related to ∆G˚ by:
∆G˚ = –RT ln K
0
DG0 = -RT ln K = -nFEcell
Slide 27
Spontaneity of Redox Reactions
ΔGº= -nFEcell
DG0
0
-nFEcell
=
n = number of moles of electrons in reaction
J
F = 96,500
= 96,500 C/mol ( see slide 4)
V • mol
0
DG0 = -RT ln K = -nFEcell
0
Ecell
(8.314 J/K•mol)(298 K)
RT
ln K =
ln K
=
nF
n (96,500 J/V•mol)
0.0257 V
ln K
n
0
Ecell
xn
K = exp
0.0257 V
0
Ecell
=
0
Ecell
=
ln K = 2.3035 Log K
0.0592 V
log K
n
Slide 28
Spontaneity of Redox Reactions
∆G˚ = –RT ln K
0
Ecell
=
0.0257 V
ln K
n
---
Slide 29
What is the equilibrium constant for the following reaction
at 250C? Fe2+ (aq) + 2Ag (s)
Fe (s) + 2Ag+ (aq)
0
Ecell
=
0.0257 V
ln K
n
Oxidation:
2Ag+
2Ag
Reduction: 2e- + Fe2+
Fe
+
2en=2
E0 = -0.80 V
E0 = -0.45V
E°cell = E°oxidation + E°reduction
E0 = (-0.80) +( -0.45)
E0 = -1.25 V
0
Ecell
xn
-1.25 V x 2
= exp
K = exp
0.0257 V
0.0257 V
K = 5.67 x 10-43
Slide 30
Cell Emf Under Nonstandard Conditions
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE 0
-nFE = -nFE0 + RT ln Q
If we divide both sides by “-nF”
E = E0 -
RT
ln Q
nF
At 298 K
E = E0 -
0.0257 V
ln Q
n
ln Q = 2.3035 Log Q
E = E0 -
0.0592 V
log Q
n
Nernst equation
Slide 31
Will the following reaction occur spontaneously at 250C if
[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?
Fe2+ (aq) + Cd (s)
Fe (s) + Cd2+ (aq)
Oxidation:
Reduction:
Cd
2e- + Fe2+
Cd2+ + 2en=2
Fe
E0 = 0.40 V
E0 = -0.45 V
E°cell = E°oxidation + E°reduction
E0 = 0.40 V + (-0.45 V)
E0 = -0.05 V
ΔG < 0
0.0257 V
ln Q
n
0.010
0.0257 V
ln
E = -0.05 V 2
0.60
E = 0.0026
E = E0 -
DG = -nFEcell
E>0
Spontaneous
Slide 32
Write Cell Notation and Calculate the
Cell Potential for the Following Cell
0.0592 V
log Q
E= n
E = 0.088 V
E0
Slide 33
E=
E0
0.0592V
log Q
n
Concentration Cells
•
Notice that the Nernst equation implies that a cell could be
created that has the same substance at both electrodes.
 would be 0, but Q would not.
• For such a cell, Ecell
• Therefore, as long as the concentrations
are different, E will not be 0.
Slide 34
Slide 35
Slide 36
Slide 37
Slide 38
Slide 39
Slide 40
Slide 41
Slide 42
Nernst equation
E=
E0
0.0592 V
log Q
n
Calculate the concentration of H+ in the
following if the Cell voltage is 0.7 V .
Zn(s)|Zn2+(aq, 1 M)|| H+(aq, ?)|H2(g, 1 atm)|Pt
E °red (Zn/Zn2+)= 0.76 V.
H+ = 0.1 M
Slide 43
Nernst Equation Could be Applied to Half Cell Potential
•
A particularly important use of the Nernst equation
is in the electrochemical determination of pH.
Pt | H2 (1 atm) | H+ (? M) || Reference Cathode
Ecell = EH2  H+ + Eref
•
The Nernst equation can be applied to the
half-reaction:
H2(g)  2 H+(aq) + 2 e–
Slide 44
The Nernst Equation
•
For the half-reaction:
0.0592 V
0
log Q
E
=
E
+
–
n
H2(g) 2 H (aq) + 2 e

2
+
H
o
0.0592
V
E H 2  H+ = E H2  H + –
log
n
PH
2
•
E° = 0 V for this reaction
(standard hydrogen
electrode). According to the problem, P is 1 atm.

EH 2  H+
05
H2
2
0.0592
V
+




0
.
0592
V
pH
=0 V–
log
H
n
Ecell  0.0592VpH  Eref
Slide 45
The Nernst Equation
•
07
The overall potential is given by:
Ecell  0.0592 V pH  Eref
•
Which rearranges to give an equation for the
determination of pH:
Ecell  Eref
 pH
0.0592 V
Slide 46
Replacing Standard Hydrogen Electrode With
Glass Electrode (Ag/AgCl wire in dilute HCl)
Slide 47
pH meter
•
A higher cell potential indicates a higher pH,
therefore we can measure pH by measuring Ecell.
•
A glass electrode (Ag/AgCl wire in dilute HCl) with
a calomel reference is the most common
arrangement.
Glass: Ag(s) + Cl–(aq)  AgCl(s) + e–
E° = –0.22 V
Calomel: Hg2Cl2(s) + 2 e–  2 Hg(l) + 2 Cl–(aq)
E° = 0.28 V
Slide 48
pH Electrode
09
• The glass pH probe is constructed as follows:
Ag(s) | AgCl(s) | HCl(aq) | glass | H+(aq) || reference
•The difference in [H+] from one side of the glass
membrane to the other causes a potential to develop,
which adds to the measured Ecell.
Ecell  Eref
 pH
0.0592 V
Slide 49
The following cell has a potential of 0.28 V at 25°C:
Pt(s) | H2 (1 atm) | H+ (? M) || Pb2+ (1 M) | Pb(s)
What is the pH of the solution at the anode?
•
H2(g) + Pb2+(aq)  2 H+(aq) + Pb(s)
•
Eoref = - 0.13 V ( Please see page 698);
Ecell  Eref
pH 
0.0592 V
pH 
0.28V  (0.13V )
0.0592V
 6.9
Slide 50
Batteries
Lead storage
battery
Slide 51
Batteries
Lead Storage Battery
Anode:
Cathode:
Overall:
Pb(s) + HSO41-(aq)
PbO2(s) + 3H1+(aq) + HSO41-(aq) + 2ePb(s) + PbO2(s) + 2H1+(aq) + 2HSO41-(aq)
PbSO4(s) + H1+(aq) + 2ePbSO4(s) + 2H2O(l)
2PbSO4(s) + 2H2O(l)
Batteries
Dry cell
Leclanché cell
Anode:
Cathode:
Zn (s)
2NH+4 (aq) + 2MnO2 (s) + 2e-
Zn (s) + 2NH4 (aq) + 2MnO2 (s)
Zn2+ (aq) + 2eMn2O3 (s) + 2NH3 (aq) + H2O (l)
Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
1.5 V but deteriorates to 0.8 V with use
Slide 53
Batteries
A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning
Anode:
Cathode:
2H2 (g) + 4OH- (aq)
O2 (g) + 2H2O (l) + 4e2H2 (g) + O2 (g)
4H2O (l) + 4e4OH- (aq)
2H2O (l)
Slide 54
© 2003 John Wiley and Sons Publishers
Figure 7.13: The essentials of a typical fuel cell.
Slide 55
Replacing KOH with Proton ExchangeMembrane
Slide 56
Proton Exchange Membrane (PEM) Fuel Cell
Anode side:
2H2 => 4H+ + 4eCathode side:
O2 + 4H+ + 4e- => 2H2O
Net reaction:
2H2 + O2 => 2H2O
Slide 57
Corrosion
01
•
Corrosion is the oxidative deterioration of metal.
•
25% of steel produced in USA goes to replace steel
structures and products destroyed by corrosion.
•
Rusting of iron requires the presence of BOTH
oxygen and water.
•
Rusting results from tiny galvanic cells formed by
water droplets.
Slide 58
Corrosion
Corrosion: The oxidative deterioration of a metal.
Slide 59
Corrosion
4Fe+2(aq) + O2(g)+ 4H+(aq)
2Fe+3(aq) + 4H2O(l)
4Fe+3(aq) + 2H2O
Fe2O3.H2O + 6H+
Slide 60
…Corrosion Prevention: Galvanizing
Slide 61
Corrosion Prevention
03
•
Galvanizing: is the coating of iron with zinc. Zinc is
more easily oxidized than iron, which protects and
reverses oxidation of the iron.
•
Cathodic Protection: is the protection of a metal
from corrosion by connecting it to a metal (a
sacrificial anode) that is more easily oxidized.
Attaching a magnesium stake to iron will corrode
the magnesium instead of the iron
Slide 62
Electrolysis
01
Slide 63
Electrolysis of Water
Slide 64
Electrolysis
01
•
Electrolysis: is the process in which electrical
energy is used to drive a nonspontaneous chemical
reaction.
•
An electrolytic cell is an apparatus for carrying
out electrolysis.
•
Processes in an electrolytic cell are the reverse of
those in a galvanic cell.
Slide 65
Electrolysis
•
Electrolysis of Water: Requires an electrolyte
species, that is less easily oxidized and reduced
than water, to carry the current.
•
Anode: Water is oxidized to oxygen gas.
05
2 H2O(l)  O2(g) + 4 H+(aq) + 4 e–
•
Cathode: Water is reduced to hydrogen gas.
4 H2O(l) + 4 e–  2 H2(g) + 4 OH–(aq)
Slide 66
Electrolysis Applications
•
01
Manufacture of Sodium (Downs Cell):
Bp (NaCl) = 801o C
Bp (NaCl-CaCl2) = 580oC
Slide 67
Electrolysis
07
•
Quantitative Electrolysis: The amount of
substance produced at an electrode by
electrolysis depends on the quantity of charge
passed through the cell.
•
Reduction of 1 mol of sodium ions requires 1 mol
of electrons to pass through the system.
•
The charge on 1 mol of electrons is 96,500
coulombs.
Slide 68
Electrolysis
•
08
To determine the moles of electrons passed, we
measure the current and time that the current
flows:
Charge (C) = Current (A) x Time (s)
•
Because the charge on 1 mol of e– is 96,500 C, the
number of moles of e– passed through the cell is:

–
1
mole
e
Moles of e – = Charge (C) 
96,500 C
Slide 69
Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
Slide 70
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through the
cell for 1.5 hours?
Anode:
Cathode:
2Cl- (l)
Ca2+ (l) + 2eCa2+ (l) + 2Cl- (l)
Cl2 (g) + 2eCa (s)
Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
C
s 1 mol e- 1 mol Ca
0.452
x 1.5 hr x 3600 x
x
s
hr 96,500 C 2 mol e= 0.0126 mol Ca
= 0.50 g Ca
Slide 71
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb,
how long will the battery last?
Solution
a) 454 g Pb = 2.19 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
Slide 72
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the
battery last?
Solution
a)
454 g Pb = 2.19 mol Pb
b)
Mol of e- = 4.38 mol
c)
Charge = 423,000 C
d)
Calculate time
Time (s) =
Charge (C)
Time (s) =
I (amps)
423,000 C
= 282,000 s
About 78 hours
1.50 amp
Slide 73
Electrolysis Applications
•
01
Manufacture of Sodium (Downs Cell):
Bp (NaCl) = 801o C
Bp (NaCl-CaCl2) = 580oC
Slide 74
Electrolysis Applications
•
02
Manufacture of Cl2 and NaOH (Chlor–Alkali):
Chlorine bleach: Cl2 + 2 NaOH → NaCl + NaClO + H2O
Slide 75
•
Manufacture of Aluminum (Hall–Heroult):
Mixture mp. 1000 C
Al2O3 mp. 2045 C
Anode: (positive electrode)
C(s) + 2O2-(l) ---> CO2(g) + 4eCathode: (negative electrode)
Bauxite:
Al3+(l) + 3e- ---> Al(l)
Al2O3 + SiO2 + TiO2 + Fe2O3
Hot NaOH used to dissolve alumnum
Comounds and other materials separated
by filration
Overall Reaction:
2Al2O3(l) + 3C(s) ---> 4Al(l) + 3CO2(g)
Slide 76
•
Electrorefining and Electroplating:
Slide 77
Given the following reaction, which is
true?
Cu(s) + 2 Ag+(aq)
1.
2.
3.
4.
5.
Cu2+(aq) + 2 Ag(s)
E° = +0.46 V
Plating Ag onto Cu is a spontaneous
process.
Plating Cu onto Ag is a spontaneous
process.
Plating Ag onto Cu is a nonspontaneous
process.
Plating Cu onto Ag is a nonspontaneous
process.
Energy will have to be put in for the
reaction to proceed.
Given the following reaction, which is
true?
Cu(s) + 2 Ag+(aq)
1.
2.
3.
4.
5.
Cu2+(aq) + 2 Ag(s)
E° = +0.46 V
Plating Ag onto Cu is a spontaneous
process.
Plating Cu onto Ag is a spontaneous
process.
Plating Ag onto Cu is a nonspontaneous
process.
Plating Cu onto Ag is a nonspontaneous
process.
Energy will have to be put in for the
reaction to proceed.
Based on the standard reduction
potentials, which metal would not provide
cathodic protection to iron?
1.
2.
3.
4.
Magnesium
Nickel
Sodium
Aluminum
Electrochemical Cells
---
06
Slide 81
Correct Answer:
1.
2.
3.
4.
Magnesium
Nickel
Sodium
Aluminum
In order to provide cathodic
protection, the metal that is
oxidized while protecting the
cathode must have a more negative
standard reduction potential. Here,
only Ni has a more positive
reduction potential (0.26 V) than
Fe2+ (0.44 V) and cannot be used
for cathodic protection.
Ni2+ is electrolyzed to Ni by a current of
2.43 amperes. If current flows for 600 s,
how much Ni is plated (in grams)?
(AW Ni = 58.7 g/mol)
1.
2.
3.
4.
0.00148 g
0.00297 g
0.444 g
0.888 g
Correct Answer:
1.
2.
3.
4.
i  t  FW
mass 
nF
0.00148 g
0.00297 g mass  2.43 A  (600. s)  (58.7 g/mol)
(2  96,500 C/mol)
0.444 g
0.888 g
mass  0.444 g
Alkaline Dry-Cell
•
04
Alkaline Dry-Cell: Modified Leclanché cell which
replaces NH4Cl with NaOH or KOH.
Anode: Zinc metal can on outside of cell.
Zn(s) + 2 OH–(aq)  ZnO(s) + H2O(l) + 2 e–
Cathode: MnO2 and carbon black paste on graphite.
2 MnO2(s) + H2O(l) + 2 e–  Mn2O3(s) + 2 OH–(aq)
Electrolyte: NaOH or KOH, and Zn(OH)2 paste.
Cell Potential: 1.5 V but longer lasting, higher power,
and more stable current and voltage.
Slide 85
Batteries
Mercury Battery 0.9 V
Anode:
Cathode:
Zn(Hg) + 2OH- (aq)
HgO (s) + H2O (l) + 2eZn(Hg) + HgO (s)
ZnO (s) + H2O (l) + 2eHg (l) + 2OH- (aq)
ZnO (s) + Hg (l)
Non-rechargeable button cells for watches, hearing aids, and calculators
Slide 86
For the reaction given below, what
substance is oxidized and what is
reduced?
3 NO2- +
Cr2O72-
+
8 H+
2 Cr3+ + 3 NO3- + 4 H2O
4.
N of NO2- is reduced, Cr of Cr2O72- is
oxidized
N of NO2- is oxidized, Cr of Cr2O72- is
reduced
O of NO2- is oxidized, Cr of Cr2O72- is
reduced
Cr3+ is reduced, N of NO2- is oxidized
5.
N of NO3- is oxidized, Cr3+ is reduced
1.
2.
3.
For the reaction given below, what
substance is oxidized and what is
reduced?
3 NO2- +
Cr2O72-
+
8 H+
2 Cr3+ + 3 NO3- + 4 H2O
4.
N of NO2- is reduced, Cr of Cr2O72- is
oxidized
N of NO2- is oxidized, Cr of Cr2O72- is
reduced
O of NO2- is oxidized, Cr of Cr2O72- is
reduced
Cr3+ is reduced, N of NO2- is oxidized
5.
N of NO3- is oxidized, Cr3+ is reduced
1.
2.
3.
When the following reaction is balanced,
what are the coefficients for each
substance?
__Ag + __O2
1.
1, 1, 2, 1, 1
2.
1, 1, 4, 1, 2
3.
1, 1, 2, 1, 2
4.
4, 1, 2, 1, 2
5.
4, 1, 4, 4, 2
+ __H+
__Ag+
+
__H2O
When the following reaction is balanced,
what are the coefficients for each
substance?
__Ag + __O2
1.
1, 1, 2, 1, 1
2.
1, 1, 4, 1, 2
3.
1, 1, 2, 1, 2
4.
4, 1, 2, 1, 2
5.
4, 1, 4, 4, 2
+ __H+
__Ag+
+
__H2O
1.
2.
3.
4.
1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).
1 M solution for Cl2(g) and for Cl–(aq).
1 atm pressure for Cl2(g) and for Cl–(aq).
1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).
1.
2.
3.
4.
1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).
1 M solution for Cl2(g) and for Cl–(aq).
1 atm pressure for Cl2(g) and for Cl–(aq).
1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).
1Bar = .987 atm
Which substance is the stronger oxidizing
agent?
Br2
• O2
• NO3• H+
• Cl2
•
Which substance is the stronger oxidizing
agent?
Br2
• O2
• NO3• H+
• Cl2
•
Calculate the emf of the following cell:
Zn(s)|Zn2+(aq, 1 M)|| H+(aq, 1 M)|H2(g, 1 atm)|Pt
E° (Zn/Zn2+)= 0.76 V.
1.
2.
3.
4.
+0.76 V
+1.52 V
0.76 V
1.52 V
Correct Answer:
1.
2.
3.
4.
+0.76 V
+1.52 V
0.76 V
1.52 V
E°cell = E°red + E°ox
Zn is the anode, hydrogen at the
Pt wire is the cathode.
E°cell = 0.00 V + (0.76 V)
E°cell = +0.76 V
The End
Slide 97
Batteries
•
06
Nickel–Cadmium Battery: is rechargeable.
Anode: Cadmium metal.
Cd(s) + 2 OH–(aq)  Cd(OH)2(s) + 2 e–
Cathode: Nickel(III) compound on nickel metal.
NiO(OH) (s) + H2O(l) + e–  Ni(OH)2(s) + OH–(aq)
Electrolyte: Nickel oxyhydroxide, NiO(OH).
Cell Potential: 1.30 V
Slide 98
Batteries
1.5 V to about 3.7 V,
Slid Electrolyte:
Inorganic ceramic and
organic polymer
solid-electrolyte materials
are reviewed
Solid State Lithium Battery
Slide 99
Batteries
08
Cell Potential: 3.0 V
•
Lithium Ion (Li–ion): The newest rechargeable
battery is based on the migration of Li+ ions.
•Anode:
Li metal, or Li atom impregnated graphite.
Li(s)  Li+ + e–
Cathode: Metal oxide or sulfide that can accept Li+.
MnO2(s) + Li+(aq) + e–  LiMnO2(s)
Electrolyte: Lithium-containing salt such as LiClO4, in
organic solvent. Solid state polymers can also be used.
Slide 100
Batteries
•
•
07
Nickel–Metal–Hydride (NiMH):
Replaces toxic Cd anode with
a hydrogen atom impregnated ZrNi2
metal alloy.
Cathod:
Anode:
•
Applications of NiMH type batteries
includes hybrid vehicles such as the
Toyota Prius and consumer electronics.
1.2 V
Slide 101