Chapter 19. Electrochemistry
Chemistry 102(60) Summer 2004
19.1 Redox Reactions
19.2 Using Half-Reactions to Understand Redox Reactions
19.3 Electrochemical Cells
19.4 Electrochemical Cells and Voltage
19.5 Using Standard Cell Potentials
19.6 Eo and Gibbs Free Energy
19.7 Effect of Concentration on Cell Potential
19.8 Neuron Cells
19.9 Common Batteries
19.10 Fuel Cells
19.11 Electrolysis-Forcing Reactant-Favored Reactions to
Occur
19.12 Counting Electrons
19.13 Corrosion-Product-Favored Reactions
Instructor: Dr. Upali Siriwardane
• e-mail: upali@chem.latech.edu
• Office: CTH 311, Phone 257-4941
• Office Hours:
• 8:30-10:30 a.m, M, Tu, W,Th, F
•
•
•
•
July 15, 2004- Test 1: Ch.13
July 26, 2004- Test 2: Ch 14 & 15
Aug 3, 2004- Test 3: Ch. 16 & 17
Aug 12, 2004- Test 4: Ch. 18, 19
•
Aug 13, 2004- Make-up Test 13,14,15, 16, 17,18 , 19
Rules for Assigning
Oxidation States
More Definitions
oxidizing agent- substance that causes
oxidation while undergoing reduction;
attracts electrons away from a substance
reducing agent- substance that causes
reduction while undergoing oxidation;
looses electrons to another a substance
1. zero for uncombined element
2. charge on monatomic ion
3. F is always -1; other halogens -1 except when
combined with more electronegative halogen
or oxygenon ion or molecule
Rules for Assigning
Oxidation States
Balancing OxidationOxidation-Reduction
Equations
4. H is +1 except in metal hydrides, where H is 1
5. O is -2 except when combined with F (then +1
or +2) or in peroxides, -1.
6. sum of oxidation states equals charge on ion
or molecule
half-reaction technique
Fe+3 + Sn+2 => Fe+2 + Sn+4
ox 1/2 rx
Sn+2 => Sn+4 + 2ered 1/2 rx.
Fe+3 + 1e- => Fe+2
Page 1
Balancing OxidationOxidation-Reduction
Equations
Balancing OxidationOxidation-Reduction
Equations
half-reaction technique
Fe+3 + Sn+2 => Fe+2 + Sn+4
ox 1/2 rx
Sn+2 => Sn+4 + 2ered 1/2 rx.
1(
)
Fe+3 + 1e- => Fe+2
half-reaction technique
Fe+2 + MnO4- + H+ => Mn+2 + Fe+3 + H2O
2(
ox 1/2 rx
Fe+2 => Fe+3 + 1ered 1/2 rx
MnO4- + 8H+ + 5e- => Mn+2 + 4H2O
)
Sn+2 + 2Fe+3 => Sn+4 + 2Fe+2
Balancing OxidationOxidation-Reduction
Equations
Balancing Redox Reactions:
acid solution
H+ => H2O
base solution
OH- => H2O
half-reaction technique
Fe+2 + MnO4- + H+ => Mn+2 + Fe+3 + H2O
ox 1/2 rx
5(Fe+2 => Fe+3 + 1e-)
or
H2O => H+
or
H2O => OH-
red 1/2 rx
1 (MnO4- + 8H+ + 5e- => Mn+2 + 4H2O)
5Fe+2 + MnO4- + 8H+ => Mn+2 + 5Fe+3 + 4H2O
Electrochemical Cells
Daniel's Cell
• harnessed chemical reaction which produces an
electric current
• called Voltaic Cells or Batteries
• anode - electrode where oxidation occurs
• cathode - electrode where reduction occurs
• Cell Potential - the potential difference, in volts,
between the electrodes of an electrochemical cell
1.10 v
Zinc anode
Zn+2
Zn+2
Cu+2
Cu+2
Zn+2
Zn+2
Cu+2
Cu+2
Copper cathode
Page 2
• battery for telegraph
• copper electrode dipped
into a solution of
copper(II) sulfate
• zinc electrode dipped
into a solution of zinc
sulfate
Daniel's Cell
Simple Electrochemical Cell
Zn(s) + Cu+2(aq) => Zn+2(aq) + Cu(s)
oxidation half reaction
anodeZn(s) => Zn+2(aq) + 2 ereduction half reaction
cathode
Cu+2(aq) + 2 e- => Cu(s)
Apple Battery
Electrochemical Terminology
Copper and zinc
metal strips as
electrodes in an
apple to supply the
electrolytes.
Photo by Mike Condren
Standard Cell Potential
Electrical Work
• the potential difference, in volts, between the
electrodes of an electrochemical cell when the all
concentrations of all solutes is 1 molar, all the
partial pressures of any gases are 1 atm, and the
temperature at 25oC
• direction of Oxidation-Reduction Reactions
• positive value indicates a spontaneous reaction
electrical work = charge × potential energy difference
coulombs = amperes × seconds
1 volt = 1 joule/coulomb
thus
1 joule = 1 volt × 1 coulomb
Page 3
Standard Hydrogen Electrode
Zn vs. Hydrogen Electrode
• consists of a platinum
electrode covered with a
fine powder of platinum
around which H2(g) is
bubbled. Its potential is
defined as zero volts.
H2(g) = 2 H+(aq) + 2 ereversible reaction
Cu vs. Hydrogen Electrode
Cell Diagram
• the shorthand representation of an
electrochemical cell showing the two halfcells connected by a salt bridge or porous
barrier, such as:
Zn(s)/ZnSO4(aq)//CuSO4(aq)/Cu(s)
anode
cathode
Redox Reaction
Standard
Reduction
Potentials
• the reduction half reaction is the half reaction
with the more positive reduction potential
• the half reaction with the lower reduction
potential is reverse and the sign of the potential
is changed
• the cell potential is the sum of the half reaction
potentials
Page 4
Metal Displacement Reactions
Dental Voltaic Cell
• solid of more reactive metals will displace
ions of a less reactive metal from solution
• relative reactivity based on potentials of half
reactions
• metals with very different potentials react
most vigorously
Cell Potentials & ∆G
Cell Potentials, ∆G, and Q
∆G = -nFE or
where
n×
∆Go = -nFEo
number of electrons
involved in reaction
F×
Faraday's constant,
1 mole e- = 96,500 coulombs
thus
∆G = ∆Go + RT ln Q
Cell Potentials,
∆G = -nFE or ∆Go = -nFEo
∆G = ∆Go + RT ln Q
-nFE = -nFEo + RT ln Q
or
E = Eo - (RT/nF) ln Q
Effect of Concentration on Cell Voltage:
∆G, and K
The Nernst Equation
Ecell = Eocell - (RT/nF)ln Q
Ecell = Eocell - (0.0592/n)log Q
where Q × reaction quotient
E = Eo - (RT/nF) ln Q
at equilibrium
0 = Eo - (RT/nF) ln Keq
and
(nFEo)/(RT) = ln Keq
thus
Keq = e(nFE /RT)
Q = [products]/[reactants]
Page 5
Relationships Linking:
∆Go, Ecello, and Ko
EXAMPLE: What is the cell potential for the Daniel's
cell when the [Zn+2] = 10 [Cu+2]?
Ecell = Eocell - (0.0257/n)ln Q
Q = ([Zn+2]/[Cu+2 ] = (10 [Cu+2])/[Cu+2]) = 10
Eo = (0.34 V)Cu couple + (-(-0.76 V)Zn couple
n = 2, 2 electron change
Ecell = (1.10 - (0.0257/2)ln 10) V
Ecell = (1.10 - (0.0257/2)2.303) V
Ecell = (1.10 - 0.0296) V = 1.07 V
Ionic Concentrations and
Mammalian Neuron Cell
pH Meter
Eglass electrode
= 0.0592pH + constant
Batteries
Equilibrium Potential for K+ ions
Across a Neuronal Cell Membrane
Primary Cells
non-reversible, non-rechargeable electrochemical
cell
"dry" cell & alkaline cell 1.5 v/cell
mercury cell 1.34 v/cell
fuel cell 1.23v/cell
Page 6
Batteries
“Flash Light” Batteries
Secondary Cells
reversible, rechargeable electrochemical cell
lead-acid (automobile battery) 2 v/cell
NiCad 1.25 v/cell
Primary Cells
"Dry" Cell
Zn(s) + 2 MnO2(s) + 2 NH4+ ×
Zn+2(aq) + 2 MnO(OH)(s) + 2 NH3
Alkaline Cell
Zn(s) + 2 MnO2(s) × ZnO(s) + Mn2O3(s)
Mercury Battery
Leclanche
“Dry” Cell
Primary Cells
Zn(s) + HgO(s) × ZnO(aq) + Hg(l)
Lead-Acid
(Automobile Battery)
Secondary Cell
LeadLead-Acid
(Automobile Battery)
charging battery
requires electricity
Pb(s) + PbO2(s) + 2 H2SO4
2 PbSO4(s) + 2 H2O
discharging battery
requires electricity
2 v/cell thus 12 volt battery = 6-2 volt cells
Page 7
Lithium Battery
NickelNickel-Cadmium (Ni(Ni-Cad)
Anode reaction
Li(s) (in polymer) => Li+ (in polymer) + eCathode reaction
Li+(in CoO2) + e- => LiCoO2
Overall
Li(s) + CoO2(s) => LiCoO2(s)
Ecell = 3.4 volt
Secondary Cell
Cd(s) + 2 Ni(OH)3(s)
Cd(OH)2(s) + Ni(OH)2(s)
Electrolysis
Fuel Cells
• non-spontaneous reaction is caused by the passage
of an electric current through a solution
Quantitative Aspects of Electrolysis
1 coulomb = 1 amp sec
1 mole e- = 96,500 coulombs
anode
2( H2(g) + 2 OH-(aq) => 2 H2O(l) + 2 e-)
cathode
O2(g) + 2 H2O(l) + 4 e- => 4 OH-(aq)
Electrolysis of
Sodium chloride
Hall Process
aluminum oxide partially breaks into ions in molten cryolite,
Na3AlF6
Al+3 + 3e- => Al(l)
molten NaCl => liquid sodium and chlorine gas
aqueous NaCl => caustic soda (sodium
hydroxide) and chlorine gas
cathode
2O-2(melt) + C(s) => CO2(g)
anode
At the anode, graphite is consumed, and carbon dioxide is
formed. At the cathode, liquid aluminum is formed.
Page 8
Electrolytic Refining of Copper
EXAMPLE: How many grams of chromium
can be electroplated from a Cr+6 solution in
45 minutes at a 25 amp current?
Cu(s) + Cu+2(aq) => Cu+2(aq) + Cu(s)
impure
pure
anode
cathode
#g Cr =
impurities: anode mud; Ag, Au, Pb
(45 min)
(60 sec) (1 C) (25 amp)
(1 min) (1 amp sec)
(52 g Cr)
(1 mol e-)
(96,500 C) (6 mole e- )
= 58 g Cr
Corroding Iron Nails
Corrosion
O2(g) + 4 H+(aq) + 4 e- => 2 H2O(l)
Eo = 1.23 V
Rusting
Fe(s) ⌫ Fe+2(aq) + 2 eO2(g) + 4
H+(aq)
+ 4
e-
=> 2 H2O(l)
Eo = 0.44 V
Eo = 1.23 V
2 Fe(s) + O2(g) + 4 H+(aq) =>
2 H2O(l) + Fe+2(aq)
Preventing Corrosion
Galvanized Iron
painting
sacrificial cathode
galvanizing
Page 9
Eo = 1.67 V
TRANSFER
TRANSFER REACTIONS
REACTIONS
ELECTROCHEMISTRY
ELECTROCHEMISTRY
Chapter
Chapter 21
21
Atom transfer
A- + HOH- -- >HA + OH-
Electron transfer
Cu(s) + 2 Ag+(aq)
(aq)- -- >Cu2+(aq) + 2 Ag(s)
Electric automobile
Electron
Electron Transfer
Transfer Reactions
Reactions
Review
Review of
of Terminology
Terminology for
for
Redox
Redox Reactions
Reactions
• Electron transfer reactions are oxidation
reduction or redox reactions.
• Can generate electric current or be
caused by imposing an electric current.
•• OXIDATION—
—loss of
OXIDATION
OXIDATION—loss
of electron(s)
electron(s) by
by aa
species;
species; increase
increase in
in oxidation
oxidation number.
number.
•• REDUCTION—
—gain of
REDUCTION
REDUCTION—gain
of electron(s);
electron(s);
decrease
decrease in
in oxidation
oxidation number.
number.
•• OXIDIZING
—electron acceptor;
AGENT
OXIDIZING AGENT—
AGENT—electron
acceptor;
species
species is
is reduced.
reduced.
•• REDUCING
—electron donor;
AGENT
REDUCING AGENT—
AGENT—electron
donor;
species
species is
is oxidized.
oxidized.
• ELECTROCHEMISTRY.
OXIDATION
-REDUCTION
OXIDATION-REDUCTION
REACTIONS
REACTIONS
OXIDATION
-REDUCTION
OXIDATION-REDUCTION
REACTIONS
REACTIONS
Direct Redox Reaction
Oxidizing and reducing agents in direct contact.
Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq)
(aq)
Indirect Redox Reaction
A battery functions by transferring electrons
through an external wire from the reducing
agent to the oxidizing agent.
Page 10
Why
Why Study
Study Electrochemistry?
Electrochemistry?
Electrochemical
Electrochemical Cells
Cells
• Batteries
• Corrosion
• Industrial
production of
chemicals such as
Cl2, NaOH,
NaOH, F2 and
Al
• Biological redox
reactions
• An apparatus that allows a redox reaction to
occur by transferring electrons through an
external connector.
• Product favored reaction
voltaic or
galvanic cell (batteries)
current is used to perform work
• Reactant favored reaction
electrolytic cell
current used to cause chemical change.
The heme group
CHEMICAL
CHEMICAL CHANGE
CHANGE
ELECTRIC
ELECTRIC CURRENT
CURRENT
Zn
Zn metal
metal
2+ ions
Cu
Cu2+
ions
Adding
-reactions
half
Adding halfhalf-reactions
Oxidation: Zn(s)
Zn2+(aq) + 2e
Cu(s)
Reduction: Cu2+(aq) + 2e
-------------------------------------------------------Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
(already balanced)
With
With time,
time, Cu
Cu plates
plates out
out
onto
onto Zn
Zn metal
metal strip,
strip, and
and
Zn
Zn strip
strip “disappears.”
“disappears.”
•Zn is oxidized and is the reducing agent
Zn(s)
Zn2+(aq) + 2e
2+
•Cu is reduced and is the oxidizing agent
Cu2+(aq) + 2e
Cu(s)
•Separate the oxidizing and reducing
agents : electron transfer occurs thru an
external wire.
•This is accomplished in a GALVANIC or
VOLTAIC cell.
wire
-Oxidation
-Anode
(negative)
Electron
source
•A group of such cells is called a battery.
e le c t ro ns
Zn
Zn2+ ions
salt
bridge
Cu
Cu2+ ions
-reduction
-Cathode
(positive)
electron
sink
wire
e le c t ro ns
Zn
Zn2+ ions
salt
bridge
Salt
Salt bridge
bridge allows
allows anions
anions and
and cations
cations to
to
move
move between
between electrode
electrode compartments.
compartments.
(This
(This maintains
maintains electrical
electrical neutrality).
neutrality).
Cu
Cu2+ ions
Page 11
Electrons move
from anode to
cathode in the wire.
Anions & cations
move thru the salt
bridge.
Balancing
Balancing Equations
Equations
Consider the reduction of Ag+ ions with
copper metal.
Electrochemical
Cell
Cu + Ag+
Balancing
Balancing Equations
Equations
Step 1:
Ox
Red
Step 2:
Step 3:
Ox
Red
Balancing
Balancing Equations
Equations
Step 4:
Multiply each half- reaction to
balance electrons
Reducing agent
Cu
Cu2+ + 2e
+
Oxidizing agent
2 Ag + 2 -e
2 Ag
Step 5:
Add half- reactions to give the
overall equation.
Cu + 2 Ag+
Cu2+ + 2Ag
Divide the reaction into halfhalf-reactions
Cu
Cu2+
Ag+ -> Ag
Balance each for mass.
Already done in this case.
Balance each halfhalf-reaction for charge by
adding electrons.
Cu
Cu2+ + 2e2eAg+ + eAg
The equation is now balanced for
both charge and mass.
Balancing
Balancing Equations:
Equations:
Acidic
Solution
Acidic Solution
Balance the following in acid solution—
solution—
VO2+
Step 1:
Ox
Red
Step 2:
mass.
Ox
Red
Balancing
Balancing Equations
Equations
Step 3:
Ox
Red
Step 4:
Ox
Red 2-e
+ Zn
VO2+ + Zn2+
Write the half- reactions
Zn
Zn2+
VO2+
VO2+
Balance each half- reaction for
Zn
Zn2+
+
2 H + VO2+
VO2+
Cu2+ + Ag
Balance half- reactions for charge.
Zn
Zn2+ + 2e
-e + 2 H+ + VO2+
VO2+ + H2O
Multiply by an appropriate factor.
Zn
Zn2+ + 2e
+ 4 H+ + 2 VO2+
2 VO2+ + 2 H2O
Step 5:
Add half- reactions
Zn + 4 H+ + 2 VO2+
Zn2+ + 2 VO2+ + 2 H2O
+ H2O
Add H2O on OO-deficient side and
add H+ on other side for HH-balance.
Page 12
CELL
CELL POTENTIAL,
POTENTIAL, E
E
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.
Zn and Zn2+,
anode
Cu and Cu2+,
cathode
• Electrons are “driven” from anode to cathode by
an electromotive force or emf.
• Voltage = 1.10 V
• at 25 °C
C and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL POTENTIAL, Eo
Calculating
Calculating Cell
Cell Voltage
Voltage
CELL
CELL
POTENTIAL,
POTENTIAL, E
E
• Balanced half- reactions can be added
together to get overall, balanced
equation.
• a quantitative measure of the tendency of
reactants to proceed to products when all
are at unit concentration (1.00M)
22II- - --->
---> II22 ++ 2e2e+ 2e- ---> 2 OH- - + H
22HH2O
2O + 2e- ---> 2 OH + H22
--------------------------------------------------------------------------------------------------> I + 2 OH- - + H
22II- - ++ 22 HH2O
2O --> I22 + 2 OH + H22
• If we know Eo for each half- reaction, we
could get Eo for net reaction.
2+ half
Zn/Zn
-- cceellll hooked
Zn/Zn2+
half
hooked to
to aa SHE.
SHE.
EEoo for
for the
the cell
cell == +0.76
+0.76 VV
CELL
CELL POTENTIALS,
POTENTIALS, E
Eoo
• Can’t measure 1/2 reaction Eo directly.
Therefore, measure it relative to a
STANDARD HALF CELL, SHE.
Volts
Zn
22 H
(aq
2e
H++(aq,
(aq,, 11 M)
M) ++ 2e2e-
-
+
Salt Bridge
H
)
atm
H22(g,
(g, 11 atm)
atm)
Zn2+
Eo = 0.0 V
Zn
Zn2+ + 2eOXIDATION
ANODE
Page 13
H+
2 H+ + 2eH2
REDUCTION
CATHODE
H2
Volts
Volts
+
-
Zn
Zn2+
Zn(s)
Zn(s) + 2
Therefore,
(aq)
aq)- -
Eo
>Zn2+
Zn2+
for Zn - --
+ H2(g)
> Zn2+
Eo
= +0.76 V
(aq)
- is
aq) + 2e
+0.76 V.
V.
H2
H+
Zn
Zn2+ + 2eOXIDATION
ANODE
2 H+ + 2eH2
REDUCTION
CATHODE
Overall reaction is reduction of H+ by Zn metal.
H+
Salt Bridge
H2
H+
Zn
Zn2+ + 2eOXIDATION
ANODE
+
-
Zn
Salt Bridge
2 H+ + 2eH2
REDUCTION
CATHODE
Overall reaction is reduction of H+ by Zn metal.
Zn(s)
Zn(s) + 2 H+ (aq)
aq)- - >Zn2+ + H2(g)
Therefore,
Eo
for Zn - --
> Zn2+
Eo = +0.76 V
(aq)
- is +0.76 V.
aq) + 2e
V.
Zn is a (better) (poorer) reducing agent than H2.
Better!
2+ and
Cu/Cu
Cu/Cu2+
and H
H22/H
/H++ Cell
Cell
2+ and
Cu/Cu
Cu/Cu2+
and H
H22/H
/H++ Cell
Cell
Volts
Cu
-
+
Salt Bridge
Cu 2+
Cu 2+ + 2eCu
REDUCTION
CATHODE
Eo = +0.34 V
Volts
+
Cu
H2
H+
Salt Bridge
Cu2+
H2
2 H + + 2eOXIDATION
ANODE
Cu2+ + 2eCu
REDUCTION
CATHODE
Zn/Cu
Zn/Cu Electrochemical
Electrochemical Cell
Cell
H2
H+
H2
2 H+ + 2eOXIDATION
ANODE
Standard
Standard Redox
Redox Potentials,
Potentials, E
Eoo
Zn(s) - -- > Zn2+(aq) + 2e
Eo = +0.76 V
2+
- - -- >Cu(s)
Eo = +0.34 V
Cu (aq) + 2e
--------------------------------------------------------------Cu2+(aq) + Zn(s)- -- >Zn2+(aq) + Cu(s)
oxidizing
ability of ion
Eo = +1.10 V
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
Page 14
• 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+.
E
Eoo for
for aa Voltaic
Voltaic Cell
Cell
E
Eoo and
and ∆
∆G
Goo
Volts
Cd
Fe
Salt Bridge
Eo is related to ∆G
Go, the free
energy change for the reaction.
Fe2+
Cd2+
∆Go = - n F Eo
From table data :
• Fe is a better reducing agent than Cd
• Cd2+ is a better oxidizing agent than Fe2+
Overall reaction
Fe + Cd2+- -- >Cd + Fe2+
Eo = +0.04 V
F = Faraday constant
= 9.6485 x 104 J/V•mol
n = number of moles of electrons
Michael Faraday
transferred
17911791-1867
E
Eoo and
and ∆
∆G
Goo
∆ Go
Quantitative
Quantitative Aspects
Aspects of
of
Electrochemistry
Electrochemistry
Eo
= -nF
For a productproduct-favored reaction
Reactants ---->
----> Products
o
∆G < 0 and so Eo > 0
Eo is positive
For a reactantreactant-favored reaction
Reactants <---<---- Products
o
∆G > 0 and so Eo < 0
Eo is negative
Consider electrolysis of aqueous silver ion.
Ag+ (aq)
aq) + -e --- > Ag(s)
1 mol -e
--- > 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 ?
Quantitative
Quantitative Aspects
Aspects of
of
Electrochemistry
Electrochemistry
Current =
charge passing
time
I (amps)
=
Quantitative
Quantitative Aspects
Aspects of
of Electrochemistry
Electrochemistry
I (amps) =
1.50 amps flow thru a Ag+(aq)
(aq) solution for 15.0
min. What mass of Ag metal is deposited?
Solution
(a) Calc. charge
Coulombs = amps x time
= (1.5 amps)(15.0
amps)(15.0 min)(60 sec/min) = 1350 C
coulombs
seconds
But how is charge related to moles of
electrons?
Charge on 1 mol of -e =
(1.60 x 10-19 C/e
- )(6.02 x 1023 -e /mol)
= 96,500
coulombs
seconds
C/mol ee- = 1 Faraday
Page 15
Quantitative
Quantitative Aspects
Aspects of
of Electrochemistry
Electrochemistry
Quantitative
Quantitative Aspects
Aspects of
of Electrochemistry
Electrochemistry
The anode reaction in a lead storage battery
is
Pb(s)
Pb(s) + HSO4-(aq)- -- >PbSO4(s) + H+(aq)
(aq) + 2e
If a battery delivers 1.50 amp, and you have
454 g of Pb,
Pb, how long will the battery last?
coulombs
I (amps) =
seconds
1.50 amps flow thru a Ag+(aq)
(aq) solution for 15.0 min. What mass
of Ag metal is deposited?
Solution
(a) Charge = 1350 C
(b) Calculate moles of e
- used
1350 C •
(c)
1 mol e = 0.0140 mol e 96, 500 C
Calc. quantity of Ag
0.0140 mol e - •
1 mol Ag
= 0.0140 mol Ag or 1.51 g Ag
1 mol e -
Solution
a)
454 g Pb = 2.19 mol Pb
b)
moles of -e = 4.38
c)
Calculate charge
4.38 mol -e • 96,500 C/mol -e = 423,000 C
Quantitative
Quantitative Aspects
Aspects of
of Electrochemistry
Electrochemistry
Solution
a)
454 g Pb = 2.19 mol Pb
b)
Calculate moles of e
2.19 mol Pb •
a)
b)
c)
d)
2 mol e = 4.38 mol e 1 mol Pb
454 g Pb = 2.19 mol Pb
Mol of -e = 4.38 mol
Charge = 423,000 C
Calculate time
Time (s) =
Time (s) =
Charge (C)
I (amps)
423, 000 C
= 282, 000 s
1.50 amp
About 78 hours
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