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Slide 1
ElectrochemistryChapter 17
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Slide 2
Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?
• BatteriesBatteries• CorrosionCorrosion• Industrial production of Industrial production of chemicalschemicals such as Clsuch as Cl22, ,
NaOH, FNaOH, F22 and Al and Al• Biological redox Biological redox reactions,photosynthesisreactions,photosynthesis
6CO2 + 6H2O --> C6H12O6 + 6O2
C6H12O6 + O2 --> 6CO2 + 6H2O +Energy
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Slide 3
Redox Reactions 01Redox Reactions 01
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Slide 4
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.
Redox Reactions 01Redox Reactions 01
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Slide 5
• Oxidation Half-Reaction: Zn(s) Zn2+(aq) + 2 e–. • The Zn loses two electrons to form Zn2+.
Redox Reactions 02Redox Reactions 02
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Slide 6
Redox Reactions 03Redox Reactions 03
• Reduction Half-Reaction: Cu2+(aq) + 2 e– Cu(s)• The Cu2+ gains two electrons to form copper.
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Slide 7
Electrochemical Cells
spontaneousredox reaction
anodeoxidation
cathodereduction
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Slide 8
• Overall: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Redox Reactions 04Redox Reactions 04
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Slide 9
Electromotive Force (emf)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.
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Slide 10
Electromotive Force (emf)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.
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Slide 11
Cell PotentialCell Potential
Cell potential is measured in volts (V).
C (Coulomb):The amount of charge transferred when a current of 1 ampere (A) Flows for one second.
Volt, a potential difference between two points which results to a current of one
ampere through a resistance of one ohm
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Slide 12
Cell Potentials and Free-Energy Changes for Cell ReactionsCell Potentials and Free-Energy Changes for Cell Reactions
1 J = 1 C x 1 V
voltSI unit of electric potential
jouleSI unit of energy
coulombElectric charge
1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.
1 V = 1 JC
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Slide 13
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Slide 14
Electrochemical Cells 01Electrochemical 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
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Slide 15
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Slide 16
Cells Notation 02Cells Notation 02
Anode Half-Cell || Cathode Half-Cell
Electrode | Anode Soln || Cathode Soln | Electrode
Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)
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Slide 17
Electrochemical Cells 02Electrochemical Cells 02
Anode Half-Cell || Cathode Half-Cell
Electrode | Anode Soln || Cathode Soln | Electrode
Fe(s) | Fe2+(aq) || Fe3+(aq), | Pt(s)
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Slide 18
Write the cell notation for:Write the cell notation for:
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
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Slide 19
Electrochemical Cell PotentialsElectrochemical 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.
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Slide 20
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Slide 21
Cu2+ (aq) + 2e- Cu (s) E0 = 0.34 V∆G˚ = –nFE˚
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Slide 22
Electrochemical Cells 06Electrochemical Cells 06---
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Slide 23
• E0 is for the reaction as written
• The more positive E0 the greater the tendency for the substance to be reduced
• 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
∆G˚ = –nFE˚`n = number of moles of electrons in reaction
F = 96,500 C/mole
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Slide 24
Electrochemical Cells 03Electrochemical 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.
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Slide 25
Electrochemical Cells 07Electrochemical Cells 07
• 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) E° = 0.80 V
Zn2+(aq) + 2 e– Zn(s) E° = – 0.76 V
• Therefore, zinc forms the oxidation half-cell:Zn(s) Zn2+(aq) + 2 e– E° = – (–0.76 V)
E0 = E°Ox + E°Redcell0 0 E0 = 0.76 V+ 0.80 V = 1.56 V cell
0 0
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Slide 26
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
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V
Cd is the stronger oxidizer
Cd will oxidize Cr
2e- + Cd2+ (1 M) Cd (s)
Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):
Cathode (reduction):
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)
x 2
x 3
E°cell = E°oxidation + E°reduction
E0 = 0.74 V + (-0.4 V) cell
E0 = 0.34 V cell
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Slide 27
Spontaneity of a Reaction 01Spontaneity 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
• The value of K is related to ∆G˚ by:
∆G˚ = –RT ln K
F = 96,500 C/mole
G0 = -RT ln K = -nFEcell0
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Slide 28
Spontaneity of Redox Reactions
ΔG = -nFEcell
G0 = -nFEcell0
n = number of moles of electrons in reaction
F = 96,500J
V • mol = 96,500 C/mol
G0 = -RT ln K = -nFEcell0
Ecell0 =
RTnF
ln K(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)ln K=
=0.0257 V
nln KEcell
0
=0.0592 V
nlog KEcell
0
0.0257 Vx nE0 cellexpK =
ln K = 2.3035 Log K
º
( see slide 4)
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Slide 29
Spontaneity of Redox Reactions
∆G˚ = –RT ln K =0.0257 V
nln KEcell
0
---
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Slide 30
2e- + Fe2+ Fe
Oxidation:
Reduction:
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)
=0.0257 V
nln KEcell
0
E0 = (-0.80) +( -0.45)
E0 = -1.25 V
0.0257 Vx nE0 cellexpK =
n = 2
0.0257 Vx 2-1.25 V
= exp
K = 5.67 x 10-43
E°cell = E°oxidation + E°reduction
E0 = -0.80 V
E0 = -0.45V
2Ag 2Ag+ + 2e-
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Slide 31
Cell Emf Under Nonstandard Conditions
G = G0 + RT ln Q G = -nFE G0 = -nFE 0
-nFE = -nFE0 + RT ln Q
E = E0 - ln QRTnF
Nernst equation
At 298 K
-0.0257 V
nln QE0E = -
0.0592 Vn
log QE0E =
If we divide both sides by “-nF”
ln Q = 2.3035 Log Q
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Slide 32
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)
2e- + Fe2+ Fe
Cd Cd2+ + 2e-Oxidation:
Reduction:n = 2
E0 = 0.40 V + (-0.45 V)
E0 = -0.05 V -
0.0257 Vn
ln QE0E =
-0.0257 V
2ln -0.05 VE =
0.0100.60
E = 0.0026
E > 0Spontaneous
E0 = 0.40 V
E0 = -0.45 V
E°cell = E°oxidation + E°reduction
G = -nFEcell
ΔG < 0
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Slide 33
Write Cell Notation and Calculate the Cell Potential for the Following CellWrite Cell Notation and Calculate the Cell Potential for the Following Cell
E = 0.088 V
-0.0592 V
nlog QE0E =
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Slide 34
Concentration CellsConcentration Cells
• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
• For such a cell, would be 0, but Q would not.Ecell
• Therefore, as long as the concentrations are different, E will not be 0.
-0.0592V
nlog QE0E =
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Slide 35
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Slide 36
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Slide 37
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Slide 38
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Slide 39
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Slide 40
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Slide 41
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Slide 42
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Slide 43
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
-0.0592 V
nlog QE0E = Nernst equation
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Slide 44
Nernst Equation Could be Applied to Half Cell PotentialNernst 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–
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Slide 45
The Nernst Equation 05The Nernst Equation 05
• For the half-reaction: H2(g) 2 H+(aq) + 2 e–
• E° = 0 V for this reaction (standard hydrogen electrode). According to the problem, PH2
is 1 atm.
EH2 H+= EH2 H+o – 0.0592V
n logH+2
PH2
EH2 H+=0 V – 0.0592V
n log H+2
-0.0592 V
nlog QE0E =
refcell pHV 0592.0 EE
pHV 0592.0
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Slide 46
The Nernst Equation 07The Nernst Equation 07
• The overall potential is given by:
• Which rearranges to give an equation for the determination of pH:
pH V0.0592refcell EE
refcell pH V0592.0 EE
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Slide 47
Replacing Standard Hydrogen Electrode With Glass Electrode (Ag/AgCl wire in dilute HCl)
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Slide 48
pH meterpH 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
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Slide 49
pH Electrode 09pH 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.
pH V0.0592refcell EE
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Slide 50
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?
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);
6.9 V 0.0592
)13.0(28.0
VVpH
V 0.0592refcell EE
pH
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Slide 51
Batteries
Lead storagebattery
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BatteriesBatteries
Lead Storage Battery
2PbSO4(s) + 2H2O(l)Pb(s) + PbO2(s) + 2H1+(aq) + 2HSO41-(aq)
PbSO4(s) + 2H2O(l)PbO2(s) + 3H1+(aq) + HSO41-(aq) + 2e-
PbSO4(s) + H1+(aq) + 2e-Pb(s) + HSO41-(aq)
Overall:
Anode:
Cathode:
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Slide 53
Batteries
Leclanché cell
Dry cell
Zn (s) Zn2+ (aq) + 2e-Anode:
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
1.5 V but deteriorates to 0.8 V with use
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
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Slide 54
Batteries
A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-
2H2 (g) + O2 (g) 2H2O (l)
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Slide 55Figure 7.13: The essentials of a typical fuel cell.Figure 7.13: The essentials of a typical fuel cell.
© 2003 John Wiley and Sons Publishers
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Slide 56
Replacing KOH with Proton ExchangeMembrane
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Slide 57
Proton Exchange Membrane (PEM) Fuel CellProton Exchange Membrane (PEM) Fuel Cell
Anode side:2H2 => 4H+ + 4e-
Cathode side:O2 + 4H+ + 4e- => 2H2O
Net reaction:2H2 + O2 => 2H2O
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Slide 58
Corrosion 01Corrosion 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.
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Slide 59
CorrosionCorrosion
Corrosion: The oxidative deterioration of a metal.
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Slide 60
Corrosion Corrosion
4Fe+2(aq) + O2(g)+ 4H+(aq) 4Fe+3(aq) + 2H2O2Fe+3(aq) + 4H2O(l) Fe2O3.H2O + 6H+
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Slide 61
…Corrosion Prevention: Galvanizing…Corrosion Prevention: Galvanizing
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Slide 62
Corrosion Prevention 03Corrosion 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
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Slide 63
Electrolysis 01Electrolysis 01
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Slide 64
Electrolysis of Water
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Slide 65
• 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.
Electrolysis 01Electrolysis 01
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Slide 66
Electrolysis 05Electrolysis 05
• 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.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)
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Slide 67
Electrolysis Applications 01Electrolysis Applications 01
• Manufacture of Sodium (Downs Cell):
Bp (NaCl) = 801o CBp (NaCl-CaCl2) = 580oC
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Slide 68
Electrolysis 07Electrolysis 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.
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Slide 69
Electrolysis 08Electrolysis 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:
Molesofe– =Charge(C) 1 molee–
96,500C
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Slide 70
Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
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Slide 71
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: Ca2+ (l) + 2e- Ca (s)
2Cl- (l) Cl2 (g) + 2e-
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
0.452Cs
x 1.5 hr x 3600shr 96,500 C
1 mol e-
x2 mol e-
1 mol Cax
= 0.0126 mol Ca
= 0.50 g Ca
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Slide 72
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery isPb(s) + HSOPb(s) + HSO44
--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?
SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Calculate moles of e-Calculate moles of e-
2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb
= 4.38 mol e -
c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C
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Slide 73
Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery isPb(s) + HSOPb(s) + HSO44
--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-
If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?battery last?
SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Mol of e- = 4.38 molMol of e- = 4.38 molc)c) Charge = 423,000 CCharge = 423,000 C
Time (s) = Charge (C)
I (amps)Time (s) =
Charge (C)I (amps)
Time (s) = 423, 000 C1.50 amp
= 282,000 sTime (s) = 423, 000 C1.50 amp
= 282,000 s About 78 hoursAbout 78 hours
d)d) Calculate timeCalculate time
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Slide 74
Electrolysis Applications 01Electrolysis Applications 01
• Manufacture of Sodium (Downs Cell):
Bp (NaCl) = 801o CBp (NaCl-CaCl2) = 580oC
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Slide 75
Electrolysis Applications 02Electrolysis Applications 02
• Manufacture of Cl2 and NaOH (Chlor–Alkali):
Chlorine bleach: Cl2 + 2 NaOH → NaCl + NaClO + H2O
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Slide 76
• Manufacture of Aluminum (Hall–Heroult):
Anode: (positive electrode)C(s) + 2O2-
(l) ---> CO2(g) + 4e-Cathode: (negative electrode)Al3+
(l) + 3e- ---> Al(l)
Overall Reaction:2Al2O3(l) + 3C(s) ---> 4Al(l) + 3CO2(g)
Bauxite:Al2O3 + SiO2 + TiO2 + Fe2O3Hot NaOH used to dissolve alumnum Comounds and other materials separated by filration
Mixture mp. 1000 CAl2O3 mp. 2045 C
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Slide 77
• Electrorefining and Electroplating:
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Given the following reaction, which is true? Given the following reaction, which is true?
1. Plating Ag onto Cu is a spontaneous process.2. Plating Cu onto Ag is a spontaneous process.3. Plating Ag onto Cu is a nonspontaneous process.4. Plating Cu onto Ag is a nonspontaneous process.5. Energy will have to be put in for the reaction to
proceed.
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° = +0.46 V
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Given the following reaction, which is true? Given the following reaction, which is true?
1. Plating Ag onto Cu is a spontaneous process.2. Plating Cu onto Ag is a spontaneous process.3. Plating Ag onto Cu is a nonspontaneous process.4. Plating Cu onto Ag is a nonspontaneous process.5. Energy will have to be put in for the reaction to
proceed.
Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° = +0.46 V
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Based on the standard reduction potentials, which metal would not provide cathodic protection to iron?
1. Magnesium2. Nickel3. Sodium4. Aluminum
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Slide 81
Electrochemical Cells 06Electrochemical Cells 06---
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Correct Answer:
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.
1. Magnesium2. Nickel3. Sodium4. Aluminum
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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. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g
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Correct Answer:
Fn
FWti
mass
C/mol) 96,500(2
g/mol) (58.7s) (600. A2.43mass
g 0.444mass
1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g
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Slide 85
Alkaline Dry-Cell 04Alkaline 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.
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Slide 86
Batteries
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-Anode:
Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
Mercury Battery 0.9 V
Non-rechargeable button cells for watches, hearing aids, and calculators
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1. N of NO2- is reduced, Cr of Cr2O7
2- is oxidized
2. N of NO2- is oxidized, Cr of Cr2O7
2- is reduced
3. O of NO2- is oxidized, Cr of Cr2O7
2- is reduced
4. Cr3+ is reduced, N of NO2- is oxidized
5. N of NO3- is oxidized, Cr3+ is reduced
For the reaction given below, what substance is oxidized and what is reduced?
For the reaction given below, what substance is oxidized and what is reduced? 3 NO2
- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3
- + 4 H2O
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1. N of NO2- is reduced, Cr of Cr2O7
2- is oxidized
2. N of NO2- is oxidized, Cr of Cr2O7
2- is reduced
3. O of NO2- is oxidized, Cr of Cr2O7
2- is reduced
4. Cr3+ is reduced, N of NO2- is oxidized
5. N of NO3- is oxidized, Cr3+ is reduced
For the reaction given below, what substance is oxidized and what is reduced?
For the reaction given below, what substance is oxidized and what is reduced? 3 NO2
- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3
- + 4 H2O
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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
When the following reaction is balanced, what are the coefficients for each substance?
When the following reaction is balanced, what are the coefficients for each substance?
__Ag + __O2 + __H+ __Ag+ + __H2O
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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
When the following reaction is balanced, what are the coefficients for each substance?
When the following reaction is balanced, what are the coefficients for each substance?
__Ag + __O2 + __H+ __Ag+ + __H2O
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1. 1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).
2. 1 M solution for Cl2(g) and for Cl–(aq).
3. 1 atm pressure for Cl2(g) and for Cl–(aq).
4. 1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).
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1. 1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).
2. 1 M solution for Cl2(g) and for Cl–(aq).
3. 1 atm pressure for Cl2(g) and for Cl–(aq).
4. 1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).
1Bar = .987 atm
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Which substance is the stronger oxidizing agent? Which substance is the stronger oxidizing agent?
• Br2
• O2
• NO3-
• H+
• Cl2
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Which substance is the stronger oxidizing agent? Which substance is the stronger oxidizing agent?
• Br2
• O2
• NO3-
• H+
• Cl2
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1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V
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.
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Correct Answer:
Zn is the anode, hydrogen at the Pt wire is the cathode.
1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V
E°cell = E°red + E°ox
E°cell = 0.00 V + (0.76 V)
E°cell = +0.76 V
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Slide 97
The End
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Slide 98
Batteries 06Batteries 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
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Slide 99
Batteries
Solid State Lithium Battery
1.5 V to about 3.7 V,
Slid Electrolyte: Inorganic ceramic and organic polymer solid-electrolyte materials are reviewed
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Slide 100
Batteries 08Batteries 08
• 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.
Cell Potential: 3.0 V
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Slide 101
Batteries 07Batteries 07• Nickel–Metal–Hydride (NiMH):
• Replaces toxic Cd anode witha hydrogen atom impregnated ZrNi2 metal alloy.
• Applications of NiMH type batteries includes hybrid vehicles such as the Toyota Prius and consumer electronics. 1.2 V
Anode:
Cathod: