brdy 6ed ch20 electrochemistry
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Chapter 20:Electrochemistry
Chemistry: The Molecular Natureof Matter, 6E
Jespersen/Brady/Hyslop
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Electrochemistry
Study of chemical reactions that can produceelectricity or use electricity to produce desired
product. Study of interchange of chemical and electrical
energy
Electrochemical reaction always involvesoxidation-reduction reactions
Electron transfer reactions
Electrons transferred from one substance toanother
Also called redox reactions
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Oxidation-Reduction Reactions
Oxidation and reduction reactions occur inmany chemical and biochemical systems
Combustion reactions
Photosynthesis
6CO2 + 6H2O C6H12O6 + 6O2
Mitochondrial Respiration NADH NAD+
Methane monoxygenase
CH4 + NADH + H+ + O2 CH3OH + NAD
+ + H2O
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 4
Oxidation Reduction Reactions
Involves two processes:
Oxidation loss of electrons from one reactant
Na (s) Na+ + e Oxidation Half Reaction
Reduction gain of electrons from another
reactantCl2 (g) + 2 e
2 Cl Reduction Half Reaction
Net reaction
2 Na (s) + Cl2 (g) 2 Na+ + 2 Cl
Oxidation and reductionALWAYS occur together
Can't have one without the other
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 5
Oxidation-Reduction Reaction
Oxidizing Agent Substance that accepts the e's
Takes e's from another substance
is itself reduced
e acceptor
Cl2 (g) + 2 e
2 Cl
Reducing Agent
Substance that donates e's
Releases e's to another substance is itself oxidized
e donor
Na (s) Na+ + e
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 6
Oxidation Numbers Way to keep track of e'sRules for Assigning Oxidation Numbers
1. Sum of all oxidation numbers of atoms in moleculeor polyatomic ion must equal charge on particle
2. Oxidation number:
1. of any free element is zero (0)2. of any simple, monoatomic ion is equal to charge on ion
3. of fluorine in its compounds is 1
4. of hydrogen in its compounds is +15. of oxygen in its compounds is 2
3. If there is a conflict between 2 rules, apply the
rule with lower number and ignore conflicting rule.
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 7
Define Oxidation-Reduction in Terms
of Oxidation Number Oxidation
in oxidation number or more + oxidationnumber
Leo (Loss of es)
Oil (oxidation is e loss) Reduction
in oxidation number or more oxidationnumber
Ger (Gain of es)
Rig (Reduction is e gain)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Your Turn!
What is the oxidation number of Cl in HClO4?
A. +1B. +3
C. +5
D. +7
H is +1 and O is -2. There are 4 O atoms. +1 +(4)(-2) + x = 0
x = +78
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 9
Spontaneous Redox Reactions When a copper wire is placed in a solution of
silver nitrate Silver metal spontaneously precipitates
Copper ion spontaneously forms as evidenced byblue color of solution
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 10
Electrochemistry Possible to separate oxidation and reduction
processes and cause to occur in two differentlocations
Can use spontaneous redox reaction to produceelectricity
Can use electricity to make nonspontaneous redoxreactions happen
Biology does this by coupling nonspontaneous
redox reactions with spontaneous reactions thatprovide the driving force.
Can harness electrical energy to do work
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 11
Galvanic (or Voltaic) Cell
Electrochemical cell in which spontaneousredox reactions occur
Produce electricity spontaneously Energy released by this spontaneous redox
reaction can be used to perform electrical
work Transfer of electrons takes place through an
external pathway (wire) rather than directly
between reactants Ex. Batteries used to power laptops, cell phones,
camera, etc.
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 12
Spontaneous Redox Reactions
Ex 1. Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
Ex 2. Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
Lets look at Ex.1 in detail Imagine dividing up the reaction in Ex. 1 into
individual oxidation and reduction half reactions
Cu(s) Cu2+(aq) + 2e 2Ag+(aq) + 2e 2Ag(s)
Physically this can be accomplished by having a
strip of the given metal (called an electrode) ina solution of the corresponding ion
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 13
Example of Spontaneous Redox
Silver metal placed in a solution of AgNO3 Copper metal placed in solution of Cu(NO3)2 Each compartment is called a half-cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 14
Half-Cells
When metal ion collides withelectrode and gainselectrons, ion becomesreduced
If metal atom on surface ofelectrode loseselectrons,becomes oxidized
Left on their own, each individual cell quickly establishes an
equilibrium between metal and ions in solution.
M(s) Mn+(aq) + ne
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 15
Electrode Names By combining 2 different half-cells we can
cause es to flow from 1 cell to the other
One half-cell undergoes oxidation = anode
Other half-cell undergoes reduction = cathode
Anode = electrode at which oxidation (e
loss) occurs Cathode = electrode at which reduction (e
gain) occurs An Ox and a Red Cat
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 16
Learning Check:
Identify cathode and anode in these acidicsolutions
CrO3 (s) + MnO2 (s) MnO4 (aq) + Cr3+ (aq)
CrO3
(s) + 6H+ + 3 e Cr3+ (aq) + 3H2
O (redn)
MnO2(s) + 2H2O MnO4(aq) + 4H+ + 3e (oxidn)
Cathode = reduction = CrO3 Cr3+ (aq) half rxn
Anode = oxidation = MnO2(s) MnO4(aq) half rxn
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 17
Learning Check:
Identify the cathode and anode in these acidicsolutions
H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)
1H2
O2
(aq) O2
(g) + 2H+ (aq) + 2e (oxidn)
2CO2 (g) + 2H+ + 2e H2C2O4 (aq) (redn)
Cathode = reduction =
2CO2 (g) H2C2O4 (aq) half rxn
Anode = oxidation = H2O2 (aq) O2 (g) half rxn
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Your Turn!
Identify the half-reaction that occurs at the anode forthe reaction:
4H+(aq) + 3S(s) + 4NO3-(aq) 3SO2(g) + 4NO(g) + 2H2O
A. 3S(s) + 6H2O 3SO2(g) + 12H+(aq) + 12e-
B. 12e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2OC. 3S(s) + 6H2O 3SO2(g) + 12H
+(aq) + 8e-
D. 8e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2O
Oxidation occurs at the anode and oxidation is the loss ofelectrons. Answer C is not a balanced equation.
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 19
How can we get electrons to migrate intothe other solution?
Connect the two solutions with a metal wire that willallow es to pass from one cell to another
Now es can flow, but the reaction still will not initiate.
Why? Consider what would happen to solution if es did flow
from one cell to the other
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 20
At copper end, 2 es are given up Cu(s) Cu2+(aq) + 2e
Travel across with wire to the silver half-cell
2Ag+
(aq) + 2e
2Ag(s) What happens to the charge in each half-cell?
Each time the reaction occurs Net charge at copper end by 2
Net charge at silver end by 2
Violates principle of electroneutrality Cannot have solution with a net charge
We need to balance the charge in
order for the es to flow
Electron Flow Between Half-Cells
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 21
Two Types of Charge Conduction
Metallic Conduction External to the cell
Electrical charge is transported from one electrodeto the other by movement of electrons through thewires
How metals conduct electricity Electrolytic Conduction
Inside electrochemical cells
Electrical charge is carried through the liquid bymovement of ions
Transport of electrical charge by ions
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 22
Needed to complete circuit Tube filled with solution of an electrolyte
Salt composed of ions not involved in cell reaction
KNO3 and KCl often used
Porous plugs at each end of tube Prevent solution from pouring out
Enable ions from salt bridge to migrate betweenhalf-cells to neutralize charges in cell compartments Anions always migrate toward anode
Cations always migrate toward cathode
Salt Bridge
Salt bridge
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 24
Summary of a Galvanic Cell
Half-cells (compartments containing reactantsfor each half-reaction)
Electrodes to conduct current through solution Wire connecting two half-cells
Salt bridge to offset ionmovement
Resistance to electrical flow
Supporting electrolyte
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Your Turn!
Which of the following species would beappropriate for a salt bridge solution?
A. AgCl
B. C12H22O11 sucrose (sugar)
C. NaCl
D. C6H6
The solution needs to be an electrolyte. AgCl isnot soluble and the organic compounds are not
ionic.25
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 26
Cell Reaction
Net overall reaction in the cell To get add individual half-reactions, after make
sure that number of es gained in reduction = number of es
lost in oxidation
Cu(s) Cu2+(aq) + 2e
(oxidation)2Ag+(aq) + 2e 2Ag(s) (reduction)
2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
Use ion electron method to balance half-reactions (see Ch 6)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 27
Standard Cell Notation
Drawing sketches of electrochemical cells canbe cumbersome
Simpler representation called standard cellnotation used instead
Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s) Anode electrode where oxidation occurs is
placed on left
Cathode electrode where reduction occurs inplaced on right
anode cathode
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 28
Cell Notation
Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)
Single slash = boundary between phases (solidelectrode and aqueous solution of ions)
Double slash represents salt bridge Separates cell reactions
In each half (half-cell)
Electrodes appear at outsides Reaction electrolytes in inner section
Species in same state separated with ;
Concentrations shown in ( )
anode cathode
anode
electrode
anode
electrolyte
cathode
electrolyte
cathode
electrode
Salt Bridge
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 29
Learning Check
Write the standard cell notation for thefollowing electrochemical cells:
Fe (s) + Cd2+
(aq)
Cd(s) + Fe2+
(aq)Anode = ox = Fe(s)
Cathode = red = Cd2+(aq)
Fe(s)|Fe2+(aq)||Cd2+(aq)|Cd(s)
Al(s) + Au3+(aq) Al3+(aq) + Au(s)
Anode = ox = Al(s)
Cathode = red = Au3+(aq)
Al(s)|Al3+
(aq)||Au3+
(aq)|Au(s)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E30
Learning Check
Where there are no conductive metals involvedin a process, an inert electrode is used. C(gr)or Pt(s) are often used. Write the Standard
Cell notation for the reactions
H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)
Anode = ox = H2O2(aq) | O2(g)Cathode = an = CO2(g) | H2C2O4(aq)
C(gr)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H+(aq)|C(gr)
Pt(s)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H
+(aq)|Pt(s)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Your Turn!
Write the standard cell notation (Pt electrodes) for thefollowing reaction:
2Mn3+(aq) + 2I-(aq) Mn2+(aq) + I2(s)
A. Pt(s)|Mn3+(aq); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)
B. Pt(s)|I-(aq)|I2(s)||Mn3+(aq); Mn2+(aq)|Pt(s)
C. Mn3+(aq)|Pt(s); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)
D. Pt(s)|Mn3+
(aq); I-
(aq)||Mn2+
(aq)|I2(s)|Pt(s)
Oxidation reaction is on the right and reduction
reaction is on the left of the salt bridge (||).31
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 32
Measuring Cell Potential
Instead of just a wire we can connect avoltmeter between 2 half-cells
Voltmeter Measures potential difference between 2 half-cells
Unit of Potential Volt (V)
Measure of amount of energy(Joules, J) that can bedelivered per SI unit ofcharge (coulomb, C)
1 V = 1 J/C
l
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 33
Voltage
Voltage is driving force for reactions Greater voltage = greater driving force
Voltage or potential of galvanic cell varies withamount of current flowing through the circuit
Voltage is always positive in direction of aspontaneous reaction
Cell Potential (Ecell)
Maximum voltage that a given cell can generate
Depends on Composition of electrodes
[Ions] in half-cells
Temperature
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 34
Standard Cell Potential, Ecell
To compare potentials for different cells, weneed a set of standard conditions
E
cell means at standard conditions T = 25C = 298 K
All gases at P = 1 atm
All [Ions] = 1 M
Rarely larger than a few volts Ex. 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
If you need batteries with higher voltages,arrange several cells in series Ex. car batteries
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 35
Electrical Potential Cell potential
Competition between two half-cells
Every half-cell has tendency to gain electrons
and proceed as reduction half-reaction Reduction Potential, Ered
Relative ease of gaining electrons
Standard Reduction Potential, Ered Reduction potential measured under standard
conditions All [solutes] = 1M
All gases at P = 1 atm
T = 25C
St d d C ll P t ti l E
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 36
Standard Cell Potential, Ecell When two half-cells are connected The one with larger (more +) Ered acquires electrons
Undergoes reduction
The one with lower (more ) Ered donates electrons Undergoes oxidation
Measure cell potential, Ecell
Represents magnitude of difference between Ered of
one half-cell and Ered of other half-cell Is always taken as positive number for spontaneous
redox reactions Ecell = E
red E
red
Of substance
reduced
Of substance
oxidized
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 37
Ex: Calculating Ecell Look at reaction in zinc-copper cell Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)
Zinc is oxidized
Copper is reduced
Half reduction reactions
Zn2+
(aq) + 2 e
Zn (s) Cu2+ (aq) + 2 e Cu (s)
Reaction for Cu2+ must have greatertendency to proceed than Zn2+
as Cu2+ is the one reduced
Ecell = E
Cu2+ E
Zn2+
Refe ence Elect ode
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 38
Reference Electrode
No way to measure reduction potential of isolatedhalf-reaction Can only measure difference in potential between 2 half-cells
To assign Ered to half reactions
Must arbitrarily choose a reference electrode (half-reaction)
Set its potential to exactly 0.00V
Standard Hydrogen Electrode H2 gas at 1 atm bubbled over
Pt electrode coated with finely divided Pt to provide large surface area
T = 25C
1.00 M [H+]
Standard Hydrogen Electrode
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 39
Standard Hydrogen Electrode
2H+(aq, 1.00M) +2e H2(g, 1 atm) EH+ = 0.00 V
Double arrow indicates only that reaction is
reversible Not that there is true equilibrium
Whether oxidation or reduction occurs depends on
what half-reaction this is paired with How to list in cell notation
Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu (s)
Zn(s)|Zn2+(aq)|| H+(aq)|Pt(s), H2(g)
Ex. Galvanic cell of Cu/Cu2+ with Hydrogen
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 40
Ex. Galvanic cell of Cu/Cu with HydrogenElectrode
Need to know which half-cell undergoes reduction andwhich undergoes oxidation
When measuring potential in galvanic cell Cathode (reduction) carries + charge Anode (oxidation carries charge
Use voltmeter to determine potential
Shows that Cu carries + charge (cathode)
Ex Galvanic cell of Cu/Cu2+ with
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 41
Ex. Galvanic cell of Cu/Cu2+ with
Hydrogen ElectrodeCu2+(aq) + 2e Cu(s) (cathode)
H2(g) 2H+(aq) + 2e (anode)
Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) (cell reaction)
E
cell = E
reduced E
oxidized
Ecell = E
cathode E
anode
Ecell = ECu2+ E
H+
0.34 V = ECu2+ 0V
ECu2+ = 0.34 V
Ex Galvanic cell of Zn/Zn2+ with
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 42
Ex. Galvanic cell of Zn/Zn2+ with
Hydrogen Electrode In this case the hydrogen electrode is positive so it
is the cathode
Zn(s) Zn2+(aq) + 2e (anode)2H+(aq) + 2e H2(g) (cathode)
Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) (cell reaction)
Ecell = E
reduced E
oxidized
E
cell = E
cathode E
anode Ecell = E
H+ E
Zn2+
0.76 V = 0.00V EZn2+
EZn2+ = 0.76 V
Standard Reduction Potentials (E )
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 43
Standard Reduction Potentials (E red)
See Table 20.1 in the Jespersen textAll tabulated as reduction potentials
oxidized form + electrons reduced form
Can be ions, elements or compounds
Arranged from top to bottom by Ered
High values of E
red (E
red > 0) meanssubstance is easily reduced
More positive value of Ered = morelikely to
undergo reduction
F2(g, 1atm) + 2 e 2 F(aq, 1M) E = 2.87 V
Most easily reduced
Standard Reduction Potentials (E )
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Standard Reduction Potentials (E red)
Low value of Ered (Ered < 0) means substanceis easily oxidized More negative value of Ered = less likely to
undergo reduction
Li+ (aq, 1M) + e Li(s) E = 3.05 V More likely the reverse reaction occurs
Li(s) Li+ (aq, 1M) + e E = +3.05 V
When reaction reversed, sign of E reversed
All substances are compared to H+
, which has aEred of 0.00 V. Sign of Ered is sign of electrode when attached to
H+
/H2
Using E to Calculate E
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 45
Using E cell to Calculate E red Ex. Go back to galvanic cell with
2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
Ecell = 0.46V Cu(s) undergoes oxidation
ECu
= + 0.34 V
Ag+(aq) undergoes reduction
What is Ered for the half-reaction of Ag+?
Ecell = EAg+ ECu +0.46 V = EAg+ (+0.34 V)
E
Ag+ = 0.46 V + 0.34 V = +0.80 V
Can we Use E to Predict
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 46
Can we Use E red to Predict
Spontaneous Reactions?Yes!!
For substances in two half-reactions
Substance with more positive Eredalways occurs as written
Reduction Other half-reaction with less positive Ered
always forces to run in reverse
Oxidation
Predicting Reaction Spontaneity
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 47
Predicting Reaction Spontaneity
When have list of Ered as given in Table 20.1 Half-reaction with more positive potentials
(higher up in table) and half-reaction occurs as
written Other half-reaction (lower in table) is reversed and
occurs as oxidation
For spontaneous reaction Reactants found in left side on higher half-reaction
and on right side of lower half-reaction
Predicting Reaction Spontaneity
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 48
Predicting Reaction Spontaneity
Ex. What spontaneous reaction occurs when Crand Au are added to a solution of Au3+ andCr3+ ?
Cr3+(aq) + 3e Cr(s) E = 0.74 VAu3+(aq) + 3e Au(s) E = +1.50 V
Au is more + so reduction
Cr (oxidations) so half reaction is reversed
Au3+(aq) + 3e Au(s)
Cr(s) Cr3+(aq) + 3e
Au3+(aq) + Cr(s)Au(s) + Cr3+(aq)
Is this Answer Reasonable?
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 49
Is this Answer Reasonable?
Use reduction potentials to predict Ecell for thereaction.
Cr3+(aq) + 3e Cr(s) E = 0.74 V
Au3+(aq) + 3e Au(s) E = +1.50 V
Net Reaction =
Au3+(aq) + Cr(s)Au(s) + Cr3+(aq)
Ecell = E
reduced E
oxidized
Ecell = E
Au3+ E
Cr3+
Ecell = +1.50 V (0.74 V)
Ecell = +2.24 V
Calculating Cell Potentials E
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Calculating Cell Potentials, E cell Predict the reaction that will occur when Mg
and Cd are added to a solution containingMg2+ and Cd2+ ion.
Cd2+(aq) + 2e Cd(s) E = 0.40 V
Mg2+
(aq) + 2e
Mg(s) E
= 2.37 V Since Cd2+ higher in Table (more positive), it
is reduced and Mg is oxidized
Cd2+(aq) + 2e Cd(s) (reduction)Mg(s) Mg2+(aq) + 2e (oxidation)
Cd2+(aq) + Mg(s) Cd(s) + Mg2+(aq)
Calculating Cell Potentials E ll
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 51
Calculating Cell Potentials, E cell
What would be the cell reaction and cellpotential for the galvanic cell employing the
following half-reactions? Co2+(aq) + 2e Co(s) E = 0.28 V
Au3+
(aq) + 3e
Au(s) E
= +1.50 VAu3+ has more positive Ered so
Au3+(aq) + 3e Au(s) reduction
Co(s) Co2+(aq) + 2e oxidation
2Au3+(aq) + 3 Co(s) 2Au(s) + 3Co2+(aq) net
2[
3[ ]
]
Calculating Cell Potentials, E ll
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 52
Calculating Cell Potentials, E cellEcell = Ereduced EoxidizedEcell = E
Au3+ E
Co2+
Ecell
= +1.50 V (0.28 V)
Ecell = +1.78 V
Note:Although multiplying half-reaction by factors
to make electrons cancel, Do NOT multiply Ered by these factors
Ered are intensive quantities, do not depend on
amount Units = V
Same number of joules available for each coulomb of
charge regardless of total number of electrons shown
Predicting if Reaction is Spontaneous
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 53
Predicting if Reaction is Spontaneous
Given any redox reaction
Calculate Ecell for reaction as written
If E
cell > 0 V (positive), then reaction isspontaneous
Galvanic cell = spontaneous
Ecell always positive
If Ecell < 0 V (negative), then reaction is
nonspontaneous Electrolytic cell = nonspontaneous
Ecell always negative
Using Ecell to Predict Spontaneity
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 54
Using E cell to Predict Spontaneity
Determine whether the following reaction isspontaneous as written
6I
(aq) + BrO3
(aq) + 6H+
(aq) 3I2(s) + Br
(aq) + 6H2O I2(s) + 2e
2I(aq) Ered = + 0.54 V
BrO3
(aq) + 6H+(aq) + 6e Br(aq) + 6H2O
Ered = + 1.44 V BrO3
(aq) + 6H+(aq) + 6e Br(aq) + 6H2O redn
3 x [2I(aq) I2(s) + 2e ] oxidn
Ecell = EBrO3 E
I2
Ecell = 1.44 V (+0.54 V)
E
cell = +0.90 V spontaneous
Using Ecell to Predict Spontaneity
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 55
Using E cell to Predict Spontaneity
Determine whether the following reaction isspontaneous as written
Au(s) + Al3+(aq) Au3+(aq) + Al(s)
Al3+(aq) + 3e Al(s) E = 1.66 V
Au3+(aq) + 3e Au(s) E = +1.50 V
Au(s) Au3+(aq) + 3e oxidation
Al3+(aq) + 3e Al(s) reduction
Ecell = EAl3+ EAu3+
Ecell = 1.66 V (+1.50 V)
Ecell
= 3.16 V nonspontaneous
Your Turn!
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Is the following reaction spontaneous and what is itsstandard cell potential?
2H2O2(aq) + S(s) SO2(g) + 2H2O
H2O2(aq) + 2H+(aq) + 2e- 2H2O Eo = +1.77 VSO2(g) + 4H
+(aq) +4e- S(s) + 2H2O Eo = +.45 V
A. spontaneous; +2.22V
B. spontaneous; +1.32V
C. spontaneous; +3.09V
D. non-spontaneous; -1.32V Eocell = (+1.77 0.45) V = +1.32 V
spontaneous since Eocell > 0
56
Are Ecell and G Related?
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 57
cell
Yes, both let us predict spontaneity of reaction For chemical reaction
G = maximum work system can do In electrical system
Maximum work= nFEcell Where n= number of moles of electrons transferred
F= Faradays constant = number of Coulombs of
charge equivalent to 1 mole of e
1 F= 96,486 C
Ecell = potential of cell in Volts
Relationship between Ecell and G
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 58
p cell
( ) Je
e =
=
coulomboules
molcoulombsmolWorkMax
Equating the 2 expressions for maximum work gives
G = nF
Ecell
If using standard potentials, you can calculate
standard free energy changes
G = nFEcell
n F Ecell
Calculating Cell Potentials, Ecell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
cell
VC
JV
e
CeG
11
1)16.3(
mol1
485,96)mol3(o
59
Ex. Calculate G for the following reaction
Au3+(aq) + Al(s) Au(s) + Al3+(aq)
Ecell = +3.16 V
Solving
Std conditions so T = 25C = 298 K
G = 914,678 J/mol = -915 kJ/mol
Applications of Electrochemistry
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 60
Determination of Equilibrium Constant, K If we combine
G = RTlnKC
with
G = nFEcell
We get nFEcell = RTlnKC
Rearranging gives
Ccell KnRTE lnF
=o
Determine KC from E
cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 61
C cell
Ex. Calculate KC for the redox reaction6I(aq) + BrO3
(aq) + 6H+(aq) 3I2(s) + Br
(aq) + 6H2O
E
cell = +0.90 V Determine n
BrO3
(aq) + 6H+(aq) + 6e Br(aq) + 6H
2O redn
6I(aq) 3I2(s) + 6e oxidn
So n= 6 T = 298 K
R= 8.3145 J/(molK)
Determine KC from E
cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 62
C cell
Next Rearrange eqn to get lnKc
Plug values into equation
lnKc = 210.3
Ecell = +0.90 V, G = 521 kJ/mol, Kc = 2.1 10
90
Reaction spontaneous; goes to completion
RTnEK cellC
F
o
=ln
KKmolJ
molCCJKC 298/3145.8
485,96690.0ln
=
903.210
101.2 == eKC
Ccell
Kn
RTE ln
F
=o
Your Turn!
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Determine the equilibrium constant, KC , for areaction that has a standard cell potential of
+2.55 V and transfers four electrons in the
reaction process.
A. infiniteB. 1.00
C. 1.6 x 10397
D. 3.2 x 10172
63
Your Turn! - Solution
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 64
172
2.55 / 4 96,485 /ln
8.3145 / 298ln 397.2
3.2 x 10
C
C
C
J C C mol K
J mol K K K
K
=
=
=
Relationship between Ecell and Ecell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 65
Ecell = cell potential at standard conditions All gases at 1 atm pressure
All ion concentrations = 1.00M
25C
When concentrations, pressures or temperature
changes, Ecell changes Just as G changes
Ex. In operating battery
Potential gradually as reactants are used up andcell reaction approaches equilibrium
When reaches equilibrium, Ecell = 0
Relationship between Ecell and Ecell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 66
To determine effect of [ion] on Ecell look athow G affected by [ion]
G =
G
+ RT lnQ Since G = nFEcell and G = nFE
cell
Then
nFEcell = nFEcell + RT lnQ
Dividing both sides by nF gives
Nernst equation
QnRTEE cellcell lnF
= o
Using the Nernst Eqn
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 67
Need to construct mass action expression forredox reaction in either
[ ] in M for solutions
P in atm for gases
Ex. For the reaction,
Zn(s) + 2 H+(aq)
Zn2+(aq) + H2 (g)
2
2
]H[
]Zn[
ln2
+
+
=
H
cellcell
P
n
RTEE
F
o
2
2
]H[
]Zn[
ln2
+
+
=
HP
Q
Ex. Using the Nernst Eqn Calculate the cell potential for a galvanic cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 68
Calculate the cell potential for a galvanic cellemploying the following half-reactions if[Fe3+]=0.0100 M, [Fe2+]=0.250 M,[Co2+]=0.050 M,
Fe3+(aq) + e Fe2+(aq) Ered = +0.77 V
Co2+(aq) + 2e Co(s) Ered = 0.28 V
Solution Fe3+ has more + Ered , so reduction
Fe3+(aq) + e Fe2+(aq) reduction
Co(s) Co2+(aq) + 2e oxidation
2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq) net rxn
]2 [
Ex. Using the Nernst Eqn n = total number of e transferred = 2
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 69
n= total number of e transferred = 2
2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq)
Ecell = EFe
3+ ECo2+
Ecell = 0.77 V (0.28V) = +1.05 V
Plug in R, T, nand concentrations
E
cell = +1.05 V 0.0738 V = 0.98 V
23
222
]Fe[
]Co[]Fe[ln
2 +++
=F
RTEE cellcell
o
2
2
]0100.0[
]050.0[]250.0[ln
/485,962
298)/(3145.8V1.05
molC
KKmolJEcell
=
)100.1(
)03125.0(ln)V8012.0(V1.05
4=cellE
Ex. Using the Nernst Eqn
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 70
In order to determine H+ concentration, agalvanic cell was constructed by connecting aCu electrode dipped in 0.50 M CuSO4 to a H2electrode at gas pressure of 1 atm dipped inthe H+ solution of unknown concentration. A
value of 0.49 V is recorded for Ecell at 298K.What is the pH of hydrogen half-cell?
Step 1: First write balanced redox reaction
and determine Ecell. H2 (g) + Cu
2+ (aq) 2H+ (aq) + Cu (s)
E
cell = ECu2+ EH+ = 0.34 V 0.00 V = 0.34 V
Ex. Determination of pH from Ecell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 71
Step 2: Determine the form of the Nernstequation for this reaction
n = 2 as 2 electrons are transferred
Step 3: Solve for [H+]
][
][ln
2 2
2
+
+
=Cu
HRTEE cellcell
F
o
)050.0(
][
ln)/500,962
298)/(314.8
34.049.0
2+
=
H
molC
KKmolJ
VV
)050.0(
][lnV01284.0V34.0V49.0
2+
= H
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Your Turn!
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
What is the pOH of the following reaction when thepressure of O2 is 1 atm, the temperature is 25
oC, the
concentration of Al3+ is 1.00M, and the cell potential
reads 1.341V?3O2(g) + 6H2O + 4Al(s) 4Al
3+(aq) + 12OH-(aq)
O2(g) + 2H2O + 4e- 4OH-(aq) Eo = +0.401V
Al3+(aq) Al(s) + 3e- Eo = -1.66V
A. 12.99
B. 7.00C. 1.01
D. Cannot be determined from the data given
73
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Electrolytic Cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 75
Nonspontaneous Use electrical energy to force nonspontaneous
reaction to occur
Electrodes switch relative to galvanic cells What was cathode becomes anode where oxidation occurs
What was anode becomes cathode where reduction occurs
Type of reaction that occurs when you rechargebatteries
Substance undergoing electrolysis must be molten or
in solution so ions can move freely and conduction canoccur
Example Electrolysis Cell
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 76
Molten NaCl cell (801C) Inert electrodes placed into
molten liquid and connected
to electrical source Cathode: Na+() + e Na()
Anode: 2Cl() Cl2 (g) + 2e
Multiply first half-cell by 2to balance electrons
2Na+() + 2e 2Na()
2Na+() + 2Cl() 2Na() + Cl2(g) (Net) Since nonspontaneous
2Na+() + 2Cl() 2Na() + Cl2
(g)
cathode anode
electrolysis
Electrolytic vs. Galvanic Cells
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 77
Electrolysis means application of electricity Electrolytic means that a particular reaction is not
spontaneous
Electrolytic Cell Galvanic CellCathode is positive (reduction)
Anode is negative (oxidation)
Non-spontaneous
Requires a battery
Cathode is positive (reduction)
Anode is negative (oxidation)
Spontaneous
Is a Battery
In both galvanic and electrolytic cells, positive ionsmove to cathode and negative ions move to anode Attracted there in electrolytic cell by charges on electrodes
Diffuse there in galvanic cell to balance developing charges
Electrical Conduction
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 78
Only occurs because reactions take place at surface ofelectrode
Consider molten NaCl electrolytic cell When charged anode placed in molten NaCl Positive charge attracts Cl ions positive charge on electrode pulls e away from Cl to form Cl
atoms that combine to form Cl2
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Electrolysis Reactions in Aq. Solution
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 80
More complex May have competingreactions Must consider possible
oxidation and reduction ofsolute
Must also consider redox
reactions of water Ex. K2SO4 in H2O Expected products
K and S2O82 Actual products H2 and O2 gases
cathode anode
SO42
Electrolysis of Aqueous K2SO4
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 81
Why this difference in products? Must examine reduction potential data
Cathode
K+ (aq) + e K(s) E = 2.92 V
2H2O + 2e H2(g) + 2OH
(aq) E = 0.83 V
H2O much easier to reduce than K+
Anode
S2O82(aq) + 2e 2SO4
2 E = +2.01 V
O2(g) + 4H+(aq) + 4e 2H2OE = +1.23 V
S2O82 easier to reduce than O2
So H2O easier to oxidize than SO42
Electrolysis of Aqueous K2SO4 Alternatively calculate E
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 82
ycell
1st for K+ and S2O82 reaction
Then for H2O to H2 and O2
K+
(aq) + e
K(s) 2SO4
2 S2O82(aq) + 2e
Ecell = EK+ ES2O82 = 2.92 V (+2.01 V)
Ecell = 4.93 V 2H2O + 2e
H2(g) + 2OH(aq)
2H2O O2(g) + 4H+(aq) + 4e
Ecell = EH2O EO2= 0.83 V (+1.23 V)
Ecell = = 2.06 V
So water reaction costs much less energy
Electrolysis of Aqueous K2SO4
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 83
Net reaction
2H2O + 2e H2(g) + 2OH
(aq)
2H2O O2(g) + 4H+(aq) + 4e
6H2O O2(g) + 2H2(g) + 4OH(aq) + 4H+(aq)
Net = 2H2O
H2(g) + O2(g) So what is the role of K2SO4?
If not present, no electrolysis occurs
K2SO4 is the electrolyte K+ and SO4
2 carry charges through solution
Without it no current can pass through solution
2 [ ]
4H2O
2
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Using Reduction Potentials to Predict
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 85
Electrolysis Products Possible anode reactions Oxidation of Br or H2O
Br2(aq) + 2e 2Br(aq) E = +1.07 V O2(g) + 4H
+(aq) + 4e 2H2OE = +1.23 V
EO2
more + so
O2 easier to reduce and Br easier to oxidize
Expected reaction
Cu2+(aq) + 2e Cu(s) (cathode) 2Br(aq) Br2(aq) + 2e
(anode)
Cu2+(aq) + 2Br(aq) Cu(s) + Br2(aq) (net)
Electrolysis of CuBr2 in H2)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 86
Anode Cathode
Blue color is Cu2+
Black bits on electrode
are Cu metal
Yellow-orange
color is Br2
Learning Check
h h d d f h
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 87
What are the expected products for theelectrolysis of an aqueous solution of Na2S withtwo platinum electrodes connected to anexternal electrical source?
Possible cathode reactions
Reduction of Na+ or H2O
Na+(aq) + e Na(s) E = 2.71 V
2H2O + 2e H2(g) + 2OH
(aq) E = 0.83 V
H2O easier to reduce
Using Reduction Potentials to Predict
El t l i P d t
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 88
Electrolysis Products Possible anode reactions Oxidation of S2 or H2O
S(s) + 2e S2(aq) E = 0.48 V O2(g) + 4H
+(aq) + 4e 2H2OE = +1.23 V
EO2
more + so
O2 easier to reduce and S2 easier to oxidize
Expected reaction
2H2O + 2e H2(g) + 2OH(aq) (cathode)S2(aq) S(s) + 2e (anode)
2H2O + S2(aq) H2(g) + S(s) + 2OH
(aq) (net)
Your Turn!
Whi h f th f ll i ti i t lik l
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E
Which of the following reactions is most likelyto occur at the anode?
A. Cl2 + 2e- Cl- Eo = +1.36V
B. I2 + 2e-I- Eo = +0.54
C. Br2 + 2e-
Br- Eo = +1.07D. Li+ + e- Li Eo = -3.05V
Lithium is the easiest to oxidize
89
Michael Faraday
D t i d th t t f h i l h
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 90
Determined that amount of chemical changeoccurring during electrolysis is directlyproportional to amount of electrical charge
passed through cellAmpere (A)
SI unit of electrical current
Coulomb (C) SI unit of charge
1 coulomb = 1 ampere 1 second
1 C = 1A s
Faraday (F) 1F = 96,485 C/mol e
Kinetics of Electrolysis
F d E ti It
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 91
Faradays Equation q = charge (coulombs, C)
I = current (Amperes, A, or C/s)
t = time (s) n = moles of electrons transferred in process
F= Faradays constant (96,485 C/mol)
Units tell us how these quantities are related
This equation addresses kinetic aspects of
electrochemical cells
FnItq ==
F)()(mole
)(time)(current)(coulombs
Faradaysne
sAC
=
=
Electroplating
In Faradays equation use number of moles
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 92
In Faradays equation, use number of molesof electrons transferred, ne Because we cant see electrons, we gauge
this by the amount of metal deposited or lost Using the half-reaction and stoichiometry, we
can relate the metal to the number of molesof electrons
entcoefficien
tcoefficien
MM
molesmass
m
e
metal
metalmetal =
Calculations Related to Electrolysis
In the electrolysis of a solution containing Ni2+
( ) t lli Ni ( ) d it th th d
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 93
(aq), metallic Ni (s) deposits on the cathode.Using a current of 0.150 A for 12.2 min, what
mass of nickel will form?Faradays
(mole e)
Current
& time
Charge (C)
Mass (g) reactant
(or product)
A s = C
F
mol en
mol reactant1
Mol reactant
(or product)
MM
Solution
1 Calculate charge passing through cell in
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 94
1. Calculate charge passing through cell in12.2 min
Coulombs (C) = current (A) time (s)
C = 0.150 A 12.2 min 60 s/min
C = 109.8 C
2. Calculate moles of electrons
== emol
C
emolCemol 10138.1
485,96
18.109 3
Solution (cont.)
3 Calculate amount of Ni in g
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 95
3. Calculate amount of Ni in g
= 0.033 g Ni
Could also do the entire calculation in one
step:
Ni1
Ni69.58
2
Ni110138.1 3
mol
g
emol
molemol
Ni033.0Ni1
Ni69.58
2
Ni1
1
485,96
1min/60min2.12150.0
gmol
g
emol
mol
sA
C
C
emolsA
=
Calculations Related to Electrolysis How long must a current of 0.800 A flow to
form 2 50 g of silver metal in an electroplating
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 96
form 2.50 g of silver metal in an electroplatingexperiment? The cathode reaction is Ag+(aq) +e Ag(s)
Faradays
(mole e)
Current
& time
Charge (C)
Mass (g) reactant
(or product)
A s = C
Fmol en
mol
reactant1
Mol reactant
(or product)
MM
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Solution3. Calculate time required
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 98
Could also do the entire calculation in onestep:
min6.4660
min1
1
1
0.800
2236
)(current
)(Charge=
==
sC
sA
A
C
A
Ctime
min6.4660
min1
0.800
/1
1
485,96
Ag1
1
Ag07.91
Ag1Ag500.2
=
sA
Cs
emol
C
mol
emol
g
molg
Your Turn!60.3 minutes are required to deposit 1.00 g of
ld th th d f l ti f 1 00M
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gold on the cathode from a solution of 1.00M
Au2+ solution. How many amps are used in the
electroplating process?
A. 0.250 AB. 0.125 A
C. 15.0 A
D. 0.0625 A
99
Your Turn! - Solution
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1 Au 21.00 Au
196.97 Au 1 Au96,485 1 1min
i65.3min 601
i = 0.250 amps
mol mol e g
g mol C
smol e
=
Applications of Electrochemistry Batteries
G l i ll i ti
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Galvanic cells in action
Electroplating
Electrolysis in action
101
Batteries Galvanic cells that generate portable electrical
energy
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energy Usually a collection of several linked in series to
get higher voltages
Ex. Car batteries,usually 6 cellseach capable ofproducing 2 Vso Net = 12 V
Two Major Classes of BatteriesPrimary Cell Secondary Cell
Non rechargeable Rechargeable
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Non-rechargeable
Ex. Alkaline dry cell
Rechargeable
Pb storage Battery
Alkaline Battery Zn/MnO
2battery 1.5 V
Uses basic or alkaline electrolyte
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Uses basic or alkaline electrolyte Not rechargeable Compared to dry cell
Longer shelf life Delivers higher currents Less expensive
Reactions: Anode
Zn (s) + 2OH (aq) ZnO (s) + H2O + 2e
Cathode2MnO2 (s) + H2O + 2e
Mn2O3 (s) + 2OH (aq)
NetZn (s) + 2MnO2 (s) ZnO (s) + Mn2O3 (s)
Nicad Battery 1.4 V
Nickel Cadmium storage cellRecha geable
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g Rechargeable
High energy density
Release energy quickly
Rapidly recharged
Used in portable power tools,
CD players, electric cars Reactions:
Anode: Cd(s) + 2OH(aq) Cd(OH)2(s) + 2e
Cathode: NiO2(s) + 2H2O + 2e Ni(OH)2(s) + 2OH (aq)
Net: Cd(s) + NiO2(s) Cd(OH)2(s) + Ni(OH)2(s)
Disadvantage = Cd toxicity, so disposal a problem
Properties that modern battery designers consider for
applications such as
Important Properties of Batteries
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pp Cell phones, laptops, digital cameras, cordless tools and
pacemakers
Shelf-life How long do batteries hold their charge when not in use?
Rate of energy output
High currents Energy density
Ratio of available energy to battery volume
Specific energy Ratio of available energy to weight
Ni-MH Batteries Nickel Metal Hydride
Rechargeable
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a g ab
1.35V
Some metals and alloys can
absorb H2 gas and effectively
store it, release to redox reaction
Advantages: 50% more power/volume than NiCad
Useful longer
Reactions:Anode: MH(s) + OH(aq) M(s) + H2O + e
Cathode: NiO(OH)(s) + H2O + e Ni(OH)2(s) + OH
(aq)
Net: MH(s) + NiO(OH)(s) Ni(OH)2(s) + M(s)
Lithium Ion Batteries
Rechargeable high specific energy (due tolow mass)
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g g p gy (low mass)
High energy density (due to very () E0red)
Dont actually involve true oxidation andreduction
Li+
ions can slip between layers of atoms insolids such as graphite or LiCoO2 Process called intercalation
Li+ ion battery is based on transport of Li+
ions
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Fuel Cells
Galvanic cells in which reactants are continuouslysupplied with reagents
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supplied with reagents
Able to operate as long as supply of reactants is
maintained Attractive power source for long-term generation of
electricity
Major advantages: Clean burning
No electrode loss
Easily replenished High operational temperature
Highly efficient
Hydrogen-Oxygen Fuel Cell
Early design Cathode: O2(g) + 2H2O + 4e
4OH(aq)
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Cathode: O2(g) 2H2O 4e 4OH (aq) Anode: H2(g) + 2OH
(aq) 2H2O + 2e
Net: 2H2(g) + O2(g) 2H2O Electrolyte
Hot (~200C) concentratedKOH in center compartment
In contact with 2 porouselectrodes containing Ptcatalyst that facilitatesreactions
Advances led to loweroperating temperatures anduse of H+ transfer membranes
Application of Electrolysis
Electroplating Put fork and bar of silver in AgCN bath
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Set potential so that cell runs electrolytically
Ag dissolves from Ag electrode
Ag deposited on metal utensil
cathode anode
Bar of Silver
Uses of Electrolysis Separating metals in aqueous solution
If you have several metals in solution each with
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ydifferent reduction potentials, you can oftenseparate by electrolysis
Start at very low voltage and gradually turn up
Ag+ + e Ag E = 0.80 V
Cu2+ + 2e Cu E = 0.34 V Zn2+ + 2e Zn E = 0.76 V
Order of oxidizing ability: Ag+ > Cu2+ > Zn2+
So Ag plates out first, then Cu, then Zn as voltage
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Application of Electrolysis
Electrorefining ofMetals
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Purifying metals such as
Cu Ultrapure Cu sheets =
cathodes
Lower between slabs ofimpure Cu = anodes
Aqueous CuSO4 electrolytesolution
4 weeks for anodes todissolve and pure Cu toplate onto cathodes
115
Refining Copper
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Corrosion of Iron Steel mostly iron
Fe2O3 coating NOT impervious to further corrosion
Ph i l i l i i l
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Physical stress points leave strain in metal More susceptible to corrosion
Creates anodic and cathodic regions
Electrochemical process
Anode region: Fe Fe2+ + 2e
Electrons flow through steel to cathode region Cathode region: O2 + 2H2O + 4e
4OH
117
Prevention of Corrosion
How to conserve our natural resources Primaryapply coating
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y pp y g
Paint
Metal plating Use chromium and tin to plate steel
Galvanizingusing zinc to coat steel
Zinc more active metal so oxidized first
Zn(s) = sacrificial coating
Alloyingmixture of metals
Stainless steel contains chromium and nickel
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Your Turn!Which metal would provide the best cathodic
protection for an iron tank?
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A. CrB. Pb
C. CuD. Sn
Cr is a better reducing agent than the otherchoices.
120
Extra Practice
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121
Learning CheckCalculate E
cell. Which are spontaneous?
Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s)
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 122
Cr2O72(aq) + MnO2 (s) MnO4
(aq) + Cr(s)
Cu2+/Cu 0.337V Cr 2O72/Cr
1.33V
Ag+/Ag 0.799V MnO4/MnO2 1.695V
Pb2+/Pb 0.126V
[0.799 0.337]V = 0.462V
[1.33 1.695]V = 0.365V
Learning Check Calculate G0 in kJ.. Which reactions are spontaneous
under standard conditions? Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
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Cr2O72(aq) + MnO2 (s) MnO4
(aq) + Cr(s)
Cu2+/Cu 0.337V Cr 2O72/Cr 1.33V
Ag+
/Ag 0.799V MnO4
/MnO2 1.695VPb2+/Pb -0.126V
G = 2 mol x 96,485 C/mol x 0.462 J/C
G = 89.1 kJ
G = 6 mol x 96,485 C/mol x 0.365 J/CG = 423 kJ
Learning Check: Calculate Kfor the following cells:
Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s) E= 0.462 V
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 124
K= e35.967 = 4.18 1015
Cr2O72(aq) + MnO2 (s) MnO4(aq) + Cr(s)
E= 0.365 V
K= e170.486 = 9.10 10-75
967.3515298)(3148
4620485962ln =
= = K.KmolJ/.
J/C.C/mol,
RT
EnK
oF
486.17015298)(314836504859612ln =
==
K.KmolJ/.J/C.C/mol,
RTEnK
o
F
Learning Check: Calculate Ecell
Al|Al3+ (aq, 0.5M)||Zn2+(aq, 0.2M)|Zn
Al3+/Al 1.662V Zn2+/Zn 0.763V
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 125
3Zn2+(aq) + 2Al(s) 3Zn(s) + 2Al3+(aq)
E = [0.763 (1.662)]V = 0.899V
E = 0.899 V 0.015 V = 0.884 V
Al|Al3+ (aq, 0.5M)||Zn2+ (aq, 1M)|Zn
E
= 0.899 V + 0.006 V = 0.905 V
3
2
]2.0[
]5.0[ln
6
02569.0899.0
VVE =
3
2
]0.1[]5.0[ln
602569.0899.0 VVE =
Learning Check: What current must be supplied to deposit
3.00 g Au from a solution of AuCl3 in 200.0 s?
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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 126
How much time (in s) does it take to plate10.2 g of Ag+ using a 0.1mA power source?
22.0 A
9 107 s
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