John C. Kotz
Paul M. Treichel
John Townsend
http://academic.cengage.com/kotz
Chapter 20
Principles of Reactivity:
Electron Transfer Reactions
John C. Kotz • State University of New York, College at Oneonta
Important – Read Before Using Slides in Class
Instructor: This PowerPoint presentation contains photos and figures from the
text, as well as selected animations and videos. For animations and videos to
run properly, we recommend that you run this PowerPoint presentation from
the PowerLecture disc inserted in your computer. Also, for the mathematical
symbols to display properly, you must install the supplied font called
“Symb_chm,” supplied as a cross-platform TrueType font in the
“Font_for_Lectures” folder in the "Media" folder on this disc.
If you prefer to customize the presentation or run it without the
PowerLecture disc inserted, the animations and videos will only run properly if
you also copy the associated animation and video files for each chapter onto
your computer. Follow these steps:
1. Go to the disc drive directory containing the PowerLecture disc, and
then to the “Media” folder, and then to the
“PowerPoint_Lectures” folder.
2. In the “PowerPoint_Lectures” folder, copy the entire chapter
folder to your computer. Chapter folders are named “chapter1”,
“chapter2”, etc. Each chapter folder contains the PowerPoint Lecture
file as well as the animation and video files.
For assistance with installing the fonts or copying the animations and
video files, please visit our Technical Support at
http://academic.cengage.com/support or call (800) 423-0563. Thank you.
ELECTROCHEMISTRY
Chapter 20
© 2009 Brooks/Cole - Cengage
3
TRANSFER REACTIONS
Atom/Group transfer
HCl + H2O f Cl- + H3O+
Electron transfer
Cu(s) + 2 Ag+(aq) f Cu2+(aq) + 2 Ag(s)
© 2009 Brooks/Cole - Cengage
4
5
Electron Transfer Reactions
• Electron transfer reactions are oxidationreduction or redox reactions.
• Redox reactions can result in the
generation of an electric current or be
caused by imposing an electric current.
• Therefore, this field of chemistry is often
called ELECTROCHEMISTRY.
© 2009 Brooks/Cole - Cengage
Review of Terminology
for Redox Reactions
• OXIDATION—loss of electron(s) by a
species; increase in oxidation number.
• REDUCTION—gain of electron(s); decrease
in oxidation number.
• OXIDIZING AGENT—electron acceptor;
species is reduced.
• REDUCING AGENT—electron donor; species
is oxidized.
© 2009 Brooks/Cole - Cengage
6
7
OXIDATION-REDUCTION
REACTIONS
Direct Redox Reaction
Oxidizing and reducing agents in direct contact.
Cu(s) + 2 Ag+(aq) f Cu2+(aq) + 2 Ag(s)
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
8
OXIDATION-REDUCTION
REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons
through an external wire from the reducing
agent to the oxidizing agent.
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial
production of
chemicals such as
Cl2, NaOH, F2 and
Al
• Biological redox
reactions
© 2009 Brooks/Cole - Cengage
The heme group
9
10
Electrochemical Cells
• An apparatus that allows a
redox reaction to occur by
transferring electrons through
an external connector.
• Product favored reaction
f voltaic or galvanic cell f
chemical change produces
electric current
• Reactant favored reaction
f electrolytic cell
f electric current used to
cause chemical change.
© 2009 Brooks/Cole - Cengage
Batteries are voltaic
cells
Electrochemistry
Alessandro Volta,
1745-1827, Italian
scientist and inventor.
Luigi Galvani, 1737-1798,
Italian scientist and inventor.
© 2009 Brooks/Cole - Cengage
11
Balancing Equations
for Redox Reactions
Some redox reactions have equations that
must be balanced by special techniques
PLAY MOVIE
MnO4- + 5 Fe2+ + 8 H+ f Mn2+ + 5 Fe3+ + 4 H2O
Mn = +7
© 2009 Brooks/Cole - Cengage
Fe = +2
Mn = +2
Fe = +3
12
Balancing Equations
Cu + Ag+
© 2009 Brooks/Cole - Cengage
--give--> Cu2+ + Ag
13
Balancing Equations
Step 1:
Divide the reaction into half-reactions, one
for oxidation and the other for reduction.
Ox
Cu f Cu2+
Red
Ag+ f Ag
Step 2:
Balance each for mass. Already done in
this case.
Step 3:
Balance each half-reaction for charge by
adding electrons.
Ox
Cu f Cu2+ + 2eRed
Ag+ + e- f Ag
© 2009 Brooks/Cole - Cengage
14
Balancing Equations
Step 4:
Multiply each half-reaction by a factor so
that the reducing agent supplies as many electrons
as the oxidizing agent requires.
Reducing agent
Cu f Cu2+ + 2eOxidizing agent
2 Ag+ + 2 e- f 2 Ag
Step 5:
Add half-reactions to give the overall
equation.
Cu + 2 Ag+ f Cu2+ + 2Ag
The equation is now balanced for both
charge and mass.
© 2009 Brooks/Cole - Cengage
15
Reduction of VO2+ with Zn
© 2009 Brooks/Cole - Cengage
16
Balancing Equations
Balance the following in acid solution—
VO2+ + Zn f VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn f Zn2+
Red
VO2+ f VO2+
Step 2:
Balance each half-reaction for
mass.
Ox
Zn f Zn2+
Red
2 H+ + VO2+ f VO2+ + H2O
Add H2O on O-deficient side and add H+
on other side for H-balance.
© 2009 Brooks/Cole - Cengage
17
Balancing Equations
Step 3:
Ox
Red
Step 4:
Ox
Red 2e-
Balance half-reactions for charge.
Zn f Zn2+ + 2ee- + 2 H+ + VO2+ f VO2+ + H2O
Multiply by an appropriate factor.
Zn f Zn2+ + 2e+ 4 H+ + 2 VO2+
f 2 VO2+ + 2 H2O
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+
f Zn2+ + 2 VO2+ + 2 H2O
© 2009 Brooks/Cole - Cengage
18
19
Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end
to make sure mass and
charge are balanced.
• PRACTICE!
© 2009 Brooks/Cole - Cengage
CHEMICAL CHANGE f
ELECTRIC CURRENT
With time, Cu plates out onto Zn
metal strip, and Zn strip
“disappears.”
Electrons are transferred from Zn to
Cu2+, but there is no useful electric
current.
Oxidation: Zn(s) f Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- f Cu(s)
-------------------------------------------------------Cu2+(aq) + Zn(s) f Zn2+(aq) + Cu(s)
© 2009 Brooks/Cole - Cengage
20
CHEMICAL CHANGE f
ELECTRIC CURRENT
•To obtain a useful
current, we separate the
oxidizing and reducing
agents so that electron
transfer occurs thru an
external wire.
This is accomplished in a GALVANIC or
VOLTAIC cell.
A group of such cells is called a battery.
© 2009 Brooks/Cole - Cengage
21
22
Zn f Zn2+ + 2e-
Cu2+ + 2e- f Cu
Oxidation
Anode
Negative
Reduction
Cathode
Positive
rAnions
Cationsf
• Electrons travel thru external wire.
• Salt bridge allows anions and cations to move
between electrode compartments.
© 2009 Brooks/Cole - Cengage
The
© 2009 Brooks/Cole - Cengage
Cu|Cu2+
and
Ag|Ag+
Cell
23
24
Electrons move
from anode to
cathode in the wire.
Anions & cations
move thru the salt
bridge.
Electrochemical
Cell
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
25
Terms Used for Voltaic Cells
See Figure 20.6
© 2009 Brooks/Cole - Cengage
26
The Voltaic Pile
Drawing done by
Volta to show the
arrangement of
silver and zinc disks
to generate an
electric current.
What voltage
does a cell
generate?
© 2009 Brooks/Cole - Cengage
CELL POTENTIAL, E
27
1.10 V
Zn and Zn2+,
anode
Cu and Cu2+,
cathode
1.0 M
1.0 M
• Electrons are “driven” from anode to
cathode by an electromotive force or emf.
• For Zn/Cu cell, this is indicated by a voltage
of 1.10 V at 25 ˚C and when [Zn2+] and [Cu2+]
= 1.0 M.
© 2009 Brooks/Cole - Cengage
28
CELL POTENTIAL, E
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL
POTENTIAL, Eo
• —a quantitative measure of the tendency of
reactants to proceed to products when all
are in their standard states at 25 ˚C.
© 2009 Brooks/Cole - Cengage
29
Calculating Cell Voltage
• Balanced half-reactions can be added
together to get overall, balanced
equation.
Zn(s) f Zn2+(aq) + 2eCu2+(aq) + 2e- f Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) f Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we
could get Eo for net reaction.
© 2009 Brooks/Cole - Cengage
30
CELL POTENTIALS, Eo
Can’t measure 1/2 reaction Eo directly.
Therefore, measure it relative to a
STANDARD HYDROGEN CELL, SHE.
2 H+(aq, 1 M) + 2e- e H2(g, 1 atm)
Eo = 0.0 V
© 2009 Brooks/Cole - Cengage
31
Zn/Zn2+ half-cell hooked to a SHE.
Eo for the cell = +0.76 V
Negative
electrode
Supplier
of
electrons
Zn f Zn2+ + 2eOxidation
Anode
© 2009 Brooks/Cole - Cengage
Positive
electrode
Acceptor
of
electrons
2 H+ + 2e- f H2
Reduction
Cathode
Reduction of H+ by Zn
32
See Active
Figure
20.13
© 2009 Brooks/Cole - Cengage
33
Overall reaction is reduction of H+ by Zn metal.
Zn(s) + 2 H+ (aq) f Zn2+ + H2(g)
Eo = +0.76 V
Therefore, Eo for Zn f Zn2+ (aq) + 2e- is +0.76 V
Zn is a (better) (poorer) reducing agent than H2.
© 2009 Brooks/Cole - Cengage
Cu/Cu2+ and H2/H+ Cell
Eo = +0.34 V
Positive
Acceptor
of
electrons
Cu2+ + 2e- f Cu
Reduction
Cathode
© 2009 Brooks/Cole - Cengage
Negative
Supplier
of
electrons
H2 f 2 H+ + 2eOxidation
Anode
34
Cu/Cu2+
and
H2/H+
Cell
Overall reaction is reduction of Cu2+ by H2 gas.
Cu2+ (aq) + H2(g) f Cu(s) + 2 H+(aq)
Measured Eo = +0.34 V
Therefore, Eo for Cu2+ + 2e- f Cu is
+0.34 V
© 2009 Brooks/Cole - Cengage
35
Zn/Cu Electrochemical Cell
36
+
Anode,
negative,
source of
electrons
Zn(s) f Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- f Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) f Zn2+(aq) + Cu(s)
Eo (calc’d) = +1.10 V
© 2009 Brooks/Cole - Cengage
Cathode,
positive,
sink for
electrons
Uses of Eo Values
Organize halfreactions by
relative ability to
act as oxidizing
agents
Cu2+(aq) + 2e- f Cu(s)
Zn2+(aq) + 2e- f Zn(s)
Eo = +0.34 V
Eo = –0.76 V
Note that when a reaction is reversed the
sign of E˚ is reversed!
© 2009 Brooks/Cole - Cengage
37
Uses of Eo Values
• Organize halfreactions by
relative ability to
act as oxidizing
agents
• Table 20.1
• Use this to predict
direction of redox
reactions and cell
potentials.
© 2009 Brooks/Cole - Cengage
38
39
© 2009 Brooks/Cole - Cengage
40
Best
oxidizing
agents
See Figure 20.14
Potential Ladder for Reduction Half-Reactions
© 2009 Brooks/Cole - Cengage
Best
reducing
agents
Using Standard Potentials, Eo
Table 20.1
• Which is the best oxidizing agent:
O2, H2O2, or Cl2? _________________
• Which is the best reducing agent:
Hg, Al, or Sn? ____________________
© 2009 Brooks/Cole - Cengage
41
TABLE OF STANDARD
REDUCTION POTENTIALS
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
© 2009 Brooks/Cole - Cengage
42
Standard Redox Potentials, Eo
43
Any substance on the right
will reduce any substance
higher than it on the left.
• Zn can reduce H+ and
Cu2+.
• H2 can reduce Cu2+ but
not Zn2+
• Cu cannot reduce H+ or
Zn2+.
© 2009 Brooks/Cole - Cengage
44
Cu(s) | Cu2+(aq) || H+(aq) | H2(g)
Cathode
Positive
Anode
Negative
Electrons
r
Cu(s)
H2 (g)
salt bridge
KNO3
1 MCu(NO3 )2
Cu2+ + 2e- f Cu
Or
Cu f Cu2+ + 2 e© 2009 Brooks/Cole - Cengage
1 MH3 O+
H2 f 2 H+ + 2 eor
2 H+ + 2e- f H2
45
Cu(s) | Cu2+(aq) || H+(aq) | H2(g)
Cathode
Positive
Anode
Negative
Electrons
r
Cu(s)
H2 (g)
salt bridge
KNO3
1 MCu(NO3 )2
Cu2+ + 2e- f Cu
1 MH3 O+
H2 f 2 H+ + 2 e-
The sign of the electrode in Table 20.1 is the
polarity when hooked to the H+/H2 half-cell.
© 2009 Brooks/Cole - Cengage
Standard Redox Potentials,
Ox. agent Cu2+ + 2e- f Cu
2 H+ + 2e- f H2
Eo
46
+0.34
0.00
Zn2+ + 2e- f Zn
-0.76 Red. agent
Any substance on the right will reduce any
substance higher than it on the left.
Northwest-southeast rule: product-favored
reactions occur between
• reducing agent at southeast corner
• oxidizing agent at northwest corner
© 2009 Brooks/Cole - Cengage
Using Standard Potentials, Eo
Table 20.1
• In which direction do the following reactions
go?
• Cu(s) + 2 Ag+(aq) f Cu2+(aq) + 2 Ag(s)
–Goes right as written
• 2 Fe2+(aq) + Sn2+(aq) f 2 Fe3+(aq) + Sn(s)
–Goes LEFT opposite to direction written
• What is Eonet for the overall reaction?
© 2009 Brooks/Cole - Cengage
47
Standard Redox Potentials,
CATHODE Cu2+ + 2e- f Cu
+0.34
2 H+ + 2e- f H2
0.00
Zn2+ + 2e- f Zn
-0.76 ANODE
Northwest-southeast rule:
• reducing agent at southeast corner
= ANODE
• oxidizing agent at northwest corner
= CATHODE
© 2009 Brooks/Cole - Cengage
Eo
48
49
Standard Redox Potentials, Eo
E˚net = “distance” from “top” half-reaction
(cathode) to “bottom” half-reaction (anode)
E˚net = E˚cathode - E˚anode
Eonet for Cu/Ag+ reaction = +0.46 V
© 2009 Brooks/Cole - Cengage
Eo
for a Voltaic Cell
Cd f Cd2+ + 2eor
Cd2+ + 2e- f Cd
Fe f Fe2+ + 2eor
Fe2+ + 2e- f Fe
All ingredients are present. Which way does
reaction proceed?
© 2009 Brooks/Cole - Cengage
50
Eo for a Voltaic Cell
From the table, you see
• Fe is a better reducing
agent than Cd
• Cd2+ is a better
oxidizing agent than
Fe2+
Overall reaction
Fe + Cd2+ f Cd + Fe2+
Eo = E˚cathode - E˚anode
= (-0.40 V) - (-0.44 V)
= +0.04 V
© 2009 Brooks/Cole - Cengage
51
More About
Calculating Cell Voltage
52
Assume I- ion can reduce water.
2 H2O + 2e- f H2 + 2 OHCathode
2 I- f I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O f I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚net = E˚cathode - E˚anode
= (-0.828 V) - (+0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
© 2009 Brooks/Cole - Cengage
E at Nonstandard Conditions
0.0257 V
[Products]
E  EÞ ln
n
[Reactants]
• The NERNST EQUATION
• E = potential under nonstandard conditions
• n = no. of electrons exchanged
• ln = “natural log”
• If [P] and [R] = 1 mol/L, then E = E˚
• If [R] > [P], then E is ______________ than E˚
• If [R] < [P], then E is ______________ than E˚
© 2009 Brooks/Cole - Cengage
53
BATTERIES
Primary, Secondary, and Fuel Cells
© 2009 Brooks/Cole - Cengage
54
Dry Cell Battery
55
Primary battery — uses
redox reactions that cannot
be restored by recharge.
Anode (-)
Zn f Zn2+ + 2eCathode (+)
2 NH4+ + 2e- f
2 NH3 + H2
© 2009 Brooks/Cole - Cengage
Alkaline Battery
Nearly same reactions as in common dry
cell, but under basic conditions.
PLAY MOVIE
Anode (-): Zn + 2 OH- f ZnO + H2O + 2eCathode (+): 2 MnO2 + H2O + 2e- f
Mn2O3 + 2 OH© 2009 Brooks/Cole - Cengage
56
Lead Storage Battery
• Secondary battery
• Uses redox
reactions that can be
reversed.
• Can be restored by
recharging
© 2009 Brooks/Cole - Cengage
57
Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4- f PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2ef PbSO4 + 2 H2O
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
58
Ni-Cad Battery
Anode (-)
Cd + 2 OH- f Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- f Ni(OH)2 + OH-
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
59
Fuel Cells: H2 as a Fuel
60
•Fuel cell - reactants are
supplied continuously
from an external source.
•Cars can use electricity
generated by H2/O2 fuel
cells.
•H2 carried in tanks or
generated from
hydrocarbons.
© 2009 Brooks/Cole - Cengage
Hydrogen—Air Fuel Cell
See Figure 20.12
© 2009 Brooks/Cole - Cengage
61
H2 as a Fuel
Comparison of the volumes of substances
required to store 4 kg of hydrogen relative to
car size. (Energy, p. 264)
© 2009 Brooks/Cole - Cengage
62
Storing H2 as a Fuel
One way to store H2 is to adsorb the gas
onto a metal or metal alloy. (Energy, p. 264)
© 2009 Brooks/Cole - Cengage
63
Electrolysis
Using electrical energy to produce chemical change.
Sn2+(aq) + 2 Cl-(aq) f Sn(s) + Cl2(g)
© 2009 Brooks/Cole - Cengage
64
Electrolysis of Aqueous NaOH
Electric Energy f Chemical Change
Anode (+)
Anode
Cathode
4 OH- f
O2(g) + 2 H2O + 4e-
Cathode (-)
4 H2O + 4e- f
2 H2 + 4 OHEo for cell = -1.23 V
© 2009 Brooks/Cole - Cengage
PLAY MOVIE
65
66
Electrolysis
Electric Energy f Chemical Change
• Electrolysis of
molten NaCl.
• Here a battery
+
“pumps” electrons
Anode
from Cl- to Na+.
• NOTE: Polarity of
electrodes is
reversed from
batteries.
© 2009 Brooks/Cole - Cengage
electrons
BATTERY
Cathode
Cl-
Na +
Electrolysis of Molten NaCl
See Figure 20.18
© 2009 Brooks/Cole - Cengage
67
Electrolysis of Molten NaCl
electrons
Anode (+)
BATTERY
2 Cl- f Cl2(g) + 2e-
+
Anode
Cathode
Cl-
Na +
Cathode (-)
Na+ + e- f Na
Eo for cell (in water) = E˚c - E˚a
= - 2.71 V – (+1.36 V)
= - 4.07 V (in water)
External energy needed because Eo is (-).
© 2009 Brooks/Cole - Cengage
68
Electrolysis of Aqueous NaCl
69
Anode (+)
2 Cl- f Cl2(g) + 2e-
Cathode (-)
2 H2O + 2e- f H2 + 2 OHEo for cell = -2.19 V
Note that H2O is more
easily reduced
than Na+.
Also, Cl- is oxidized in
preference to H2O because of
kinetics.
© 2009 Brooks/Cole - Cengage
Electrolysis of Aqueous NaCl
70
Cells like these are the source of NaOH and Cl2.
In 1995: 25.1 x 109 lb Cl2 and 26.1 x 109 lb NaOH
Also the source of NaOCl for use in bleach.
© 2009 Brooks/Cole - Cengage
Electrolysis of Aqueous NaI
Anode (+):
2 I- f I2(g) + 2eCathode (-): 2 H2O + 2e- f H2 + 2 OHEo for cell = -1.36 V
© 2009 Brooks/Cole - Cengage
71
Electrolysis of Aqueous CuCl2
Anode (+)
2 Cl- f Cl2(g) + 2eCathode (-)
Cu2+ + 2e- f Cu
Eo for cell = -1.02 V
Note that Cu is more
easily reduced
than either H2O or Na+.
© 2009 Brooks/Cole - Cengage
72
Electrolytic Refining of Copper
See Figure 22.11
Impure copper is oxidized to Cu2+ at the anode.
The aqueous Cu2+ ions are reduced to Cu metal
at the cathode.
© 2009 Brooks/Cole - Cengage
73
Producing Aluminum
2 Al2O3 + 3 C f 4 Al + 3 CO2
Charles Hall (1863-1914) developed
electrolysis process. Founded Alcoa.
© 2009 Brooks/Cole - Cengage
74
Michael Faraday
1791-1867
Originated the terms anode,
cathode, anion, cation,
electrode.
Discoverer of
• electrolysis
• magnetic props. of matter
• electromagnetic induction
• benzene and other organic
chemicals
Was a popular lecturer.
© 2009 Brooks/Cole - Cengage
75
76
o
E
and Thermodynamics
• Eo is related to ∆Go, the free energy change for the
reaction.
• ∆G˚ proportional to –nE˚
∆Go = -nFEo
where F = Faraday constant
= 9.6485 x 104 J/V•mol of e(or 9.6485 x 104 coulombs/mol)
and n is the number of moles of electrons
transferred
© 2009 Brooks/Cole - Cengage
Eo and ∆Go
∆Go = - n F Eo
For a product-favored reaction
Reactants f Products
∆Go < 0 and so Eo > 0
Eo is positive
For a reactant-favored reaction
Reactants r Products
∆Go > 0 and so Eo < 0
Eo is negative
© 2009 Brooks/Cole - Cengage
77
Quantitative Aspects of
Electrochemistry
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- f Ag(s)
1 mol e- f 1 mol Ag
If we could measure the moles of e-, we
could know the quantity of Ag formed.
But how to measure moles of e-?
charge passing
Current =
time
coulombs
I (amps) =
seconds
© 2009 Brooks/Cole - Cengage
78
Quantitative Aspects of
Electrochemistry
coulombs
I (amps) =
seconds
charge passing
Current =
time
But how is charge related to moles of electrons?
=
=
96,500 C/mol e1 Faraday
© 2009 Brooks/Cole - Cengage
79
Quantitative Aspects of Electrochemistry
coulombs
I (amps) =
seconds
1.50 amps flow thru a Ag+(aq) solution for 15.0
min. What mass of Ag metal is deposited?
Solution
(a) Calc. charge
Charge (C) = current (A) x time (t)
= (1.5 amps)(15.0 min)(60 s/min) = 1350 C
© 2009 Brooks/Cole - Cengage
80
Quantitative Aspects of Electrochemistry
coulombs
I (amps) =
seconds
1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What
mass of Ag metal is deposited?
Solution
(a) Charge = 1350 C
(b) Calculate moles of e- used
(c)
Calc. quantity of Ag
© 2009 Brooks/Cole - Cengage
81
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) f 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-
c)
Calculate charge
4.38 mol e- • 96,500 C/mol e- = 423,000 C
© 2009 Brooks/Cole - Cengage
82
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) f 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) =
© 2009 Brooks/Cole - Cengage
Time (s) =
423,000 C
= 282,000 s
1.50 amp
Charge (C)
I (amps)
About 78 hours
83