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1
CHAPTER 20
� Electron Transfer Reactions
Why Study Electrochemistry?Why Study Why Study Electrochemistry?Electrochemistry?
�� BatteriesBatteries
�� CorrosionCorrosion
�� Industrial production Industrial production
of chemicals such as of chemicals such as
ClCl22, , NaOHNaOH, F, F22 and Aland Al
�� Biological Biological redoxredox
reactionsreactions
The The hemeheme groupgroup
TRANSFER REACTIONSTRANSFER REACTIONS
Atom/Group transferAtom/Group transfer
HClHCl + H+ H22O O ------> > ClCl-- + H+ H33OO
++
Electron transferElectron transfer
Cu(s) + 2 Cu(s) + 2 AgAg++(aq(aq) ) ------> Cu> Cu2+2+(aq) + 2 Ag(s) (aq) + 2 Ag(s)
4
Oxidation-Reduction Reactions
� Rules for assigning oxidation numbers
� Alkali (+1), Alkaline Earth (+2), Al family (+3)
� Halides (-1) w/out oxygen
� Oxygen family (-2) (except peroxide,
superoxide)
� H (+1) (except with metal -1, hydride)
� Transition metals (by difference)
Oxidation-Reduction Reactions
� Oxidation – increase in oxidation #
� Loss of electron
� Fe(III) � Fe(II) + e-
� Reduction – decrease in oxidation #
� Gain of electron
� MnO4- + 5 e- � Mn(II)
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Review of Terminology for Redox ReactionsReview of Terminology Review of Terminology for for RedoxRedox ReactionsReactions
� OXIDATION—loss of electron(s) by a species;
increase in oxidation number.
� REDUCTION—gain of electron(s); decrease in
oxidation number.
� OXIDIZING AGENT—electron acceptor;
species is reduced.
� REDUCING AGENT—electron donor; species is
oxidized.
�� OXIDATIONOXIDATION——loss of electron(s) by a species; loss of electron(s) by a species;
increase in oxidation number.increase in oxidation number.
�� REDUCTIONREDUCTION——gain of electron(s); decrease in gain of electron(s); decrease in
oxidation number.oxidation number.
�� OXIDIZING AGENTOXIDIZING AGENT——electron acceptor; electron acceptor;
species is reduced.species is reduced.
�� REDUCING AGENTREDUCING AGENT——electron donor; species is electron donor; species is
oxidized.oxidized.
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The Half-Reaction Method
� Half reaction method rules:
1. Write the unbalanced reaction.
2. Break the reaction into 2 half reactions:
One oxidation half-reaction and
One reduction half-reaction
Each reaction must have complete formulas for
molecules and ions.
3. Mass balance each half reaction by adding
appropriate stoichiometric coefficients. To balance H and O we can add:
� H+ or H2O in acidic solutions.
� OH- or H2O in basic solutions.8
The Half-Reaction Method
4. Charge balance the half reactions by adding appropriate numbers of electrons.� Electrons will be products in the oxidation half-
reaction.
� Electrons will be reactants in the reduction half-reaction.
5. Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction.
6. Add the two half reactions.
7. Eliminate any common terms and reduce coefficients to smallest whole numbers.
Tips on Balancing EquationsTips on Balancing Equations
�� Never add ONever add O22, O atoms, or , O atoms, or
OO22-- to balance oxygen.to balance oxygen.
�� Never add HNever add H22 or H atoms to or H atoms to
balance hydrogen.balance hydrogen.
�� Be sure to write the correct Be sure to write the correct
charges on all the ions.charges on all the ions.
�� Check your work at the end to Check your work at the end to
make sure mass and charge make sure mass and charge
are balanced.are balanced.
�� PRACTICE!PRACTICE!
Balancing EquationsBalancing Equations
Cu + AgCu + Ag++ --------> Cu> Cu2+2+ + Ag+ Ag
Balancing EquationsBalancing Equations
Step 1:Step 1: Divide the reaction into halfDivide the reaction into half--reactions, reactions,
one for oxidation and the other for reduction.one for oxidation and the other for reduction.
OxOx Cu Cu ------> Cu> Cu2+2+
RedRed AgAg++ ------> Ag> Ag
Step 2:Step 2: Balance each for mass. Already done Balance each for mass. Already done
in this case.in this case.
Step 3:Step 3: Balance each halfBalance each half--reaction for charge reaction for charge
by adding electrons.by adding electrons.
OxOx Cu Cu ------> Cu> Cu2+2+ + + 2e2e--
RedRed AgAg++ + + ee-- ------> Ag> Ag
Balancing EquationsBalancing Equations
Step 4:Step 4: Multiply each halfMultiply each half--reaction by a factor so that reaction by a factor so that
the reducing agent supplies as many electrons as the the reducing agent supplies as many electrons as the
oxidizing agent requires.oxidizing agent requires.
Reducing agentReducing agent Cu Cu ------> Cu> Cu2+2+ + 2e+ 2e--
Oxidizing agentOxidizing agent 22 AgAg++ + + 22 ee-- ------> > 22 AgAg
Step 5:Step 5: Add halfAdd half--reactions to give the overall reactions to give the overall
equation.equation.
Cu + 2 AgCu + 2 Ag++ ------> Cu> Cu2+2+ + 2Ag+ 2Ag
The equation is now balanced for both charge and mass.The equation is now balanced for both charge and mass.
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The Half-Reaction Method
� Example: Tin (II) ions are oxidized to tin (IV)
by bromine.
-+4
2
+2 BrSnBrSn
Reaction Starting
+→+
Reduction of VOReduction of VO22++ with Znwith Zn
Balance the following in Balance the following in acidacid solutionsolution——
VOVO22++ + Zn + Zn ------> VO> VO2+ 2+ + Zn+ Zn2+2+
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The Half-Reaction Method
� Example: Dichromate ions oxidize iron (II) ions to
iron (III) ions and are reduced to chromium (III)
ions in acidic solution. Write and balance the net
ionic equation for the reaction.
reaction starting FeCrFeOCr +3+3+22
72 +→+−
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The Half-Reaction Method
� Example: In basic solution hydrogen
peroxide oxidizes chromite ions, Cr(OH)4-, to
chromate ions, CrO42-. The hydrogen
peroxide is reduced to hydroxide ions. Write
and balance the net ionic equation for this
reaction.
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Stoichiometry of Redox Reactions
� Just as we have done stoichiometry with
acid-base reactions, it can also be done with
redox reactions.
� Example: What volume of 0.200 M KMnO4 is
required to oxidize 35.0 mL of 0.150 M HCl?
The balanced reaction is:
OH 8Cl 5MnCl 2+KCl 2HCl 16 KMnO 2 2224 ++→+
You do it!You do it!
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Stoichiometry of Redox Reactions
� Example: A volume of 40.0 mL of iron
(II) sulfate is oxidized to iron (III) by
20.0 mL of 0.100 M potassium
dichromate solution. What is the
concentration of the iron (II) sulfate
solution? The balanced equation is:
OH 7Cr 2Fe 6H 14OCrFe 6 2
+3+3+2
72
+2++→++
−
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Electrochemistry
� Electrochemical reactions are oxidation-reduction
reactions.
� The two parts of the reaction are physically separated.
� The oxidation reaction occurs in one cell.
� The reduction reaction occurs in the other cell.
� There are two kinds electrochemical cells.
1. Electrochemical cells containing in nonspontaneous
chemical reactions are called electrolytic cells.
2. Electrochemical cells containing spontaneous chemical
reactions are called voltaic or galvanic cells.
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Electrical Conduction
� In ionic or electrolytic conduction ionic
motion transports the electrons.
� Positively charged ions, cations, move toward
the negative electrode.
� Negatively charged ions, anions, move toward
the positive electrode.
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Electrodes
� The following convention for electrodes is
correct for either electrolytic or voltaic cells:
� The cathode is the electrode at which
reduction occurs.� The cathode is negative in electrolytic cells and
positive in voltaic cells.
� The anode is the electrode at which
oxidation occurs.� The anode is positive in electrolytic cells and negative
in voltaic cells.22
Electrolytic Cells
� Electrical energy is used to force
nonspontaneous chemical reactions to
occur.
� The process is called electrolysis.
� Two examples of commercial electrolytic
reactions are:
1. The electroplating of jewelry and auto parts.
2. The electrolysis of chemical compounds.
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Electrolytic Cells
� Electrolytic cells consist of:
1. A container for the reaction mixture.
2. Two electrodes immersed in the reaction
mixture.
3. A source of direct current.
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The Electrolysis of Molten Potassium Chloride
� Liquid potassium is produced at one electrode.
� Indicates that the reaction K+(l) + e
- → K(l) occurs at this electrode.
� Is this electrode the anode or cathode?
� Gaseous chlorine is produced at the other electrode.
� Indicates that the reaction 2 Cl- → Cl2(g) + 2 e-
occurs at this electrode.
� Is this electrode the anode or cathode?
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The Electrolysis of Molten Potassium Chloride
Diagram of this electrolytic cell.
Porous barrier
e-
K+ + e- → K(l)
cathode reaction
2Cl- → Cl2 (g) + 2e-
anode reaction
Generator-source
of DC
- electrode + electrode
e-
molten KCl
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The Electrolysis of Aqueous Potassium Chloride
2 H2O + 2e- → H2 (g) + 2 OH
-
cathode reaction
2Cl- → Cl2 (g) + 2e-
anode reaction
Cell diagram Battery, a source
of direct currente- flow
- electrode + electrode
e- flow
aqueous KCl
Cl2 gasH2 gas
+ pole of battery- pole of battery
Electrolysis of Molten Electrolysis of Molten NaClNaCl
Figure 20.18Figure 20.18
Electrolysis of Molten Electrolysis of Molten NaClNaCl
Anode (+) Anode (+)
2 2 ClCl-- ------> Cl> Cl22(g) + 2e(g) + 2e--
Cathode (Cathode (--) )
NaNa++ + e+ e-- ------> Na> Na
BATTERY
+
Na+Cl-
Anode Cathode
electrons
EEoo for cell (in water) = Efor cell (in water) = E˚̊cc -- EE˚̊aa
= = -- 2.71 V 2.71 V –– (+1.36 V)(+1.36 V)
= = -- 4.07 V (in water)4.07 V (in water)
External energy needed because External energy needed because EEoo is (is (--). ).
Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl
Anode (+) Anode (+)
2 2 ClCl-- ------> >
ClCl22(g) + 2e(g) + 2e--
Cathode (Cathode (--) )
2 H2 H22O + 2eO + 2e-- ------> >
HH22 + 2 OH+ 2 OH--
EEoo for cell = for cell = --2.19 V2.19 V
Note that HNote that H22O is more O is more
easily reduced easily reduced
than Nathan Na++. .
BATTERY
+
Na+Cl -
Anode Cathode
H2O
electrons
Also, Cl- is oxidized in preference to H2O because of kinetics.
Also, Also, ClCl-- is oxidized in is oxidized in
preference to Hpreference to H22O because of O because of
kinetics.kinetics.
Michael FaradayMichael Faraday17911791--18671867
Originated the terms anode, Originated the terms anode,
cathode, anion, cathode, anion, cationcation, ,
electrode.electrode.
Discoverer of Discoverer of
�� electrolysiselectrolysis
�� magnetic props. of mattermagnetic props. of matter
�� electromagnetic inductionelectromagnetic induction
�� benzene and other organic benzene and other organic
chemicalschemicals
Was a popular lecturer.Was a popular lecturer.
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Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis
� Faraday’s Law - The amount of substance
undergoing chemical reaction at each electrode
during electrolysis is directly proportional to the
amount of electricity that passes through the
electrolytic cell.
� A faraday is the amount of electricity that reduces
one equivalent of a species at the cathode and
oxidizes one equivalent of a species at the anode.
-23 e 106.022y electricit offaraday 1 ×≡
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Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis
� A coulomb is the amount of charge that passes a
given point when a current of one ampere (A) flows
for one second.
� 1 amp = 1 coulomb/second
coulombs 487,96e 106.022faraday 1 -23 ≡×≡
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.
AgAg++ ((aqaq) + e) + e-- ------> Ag(s)> Ag(s)
1 mol e1 mol e-- ------> 1 mol Ag> 1 mol Ag
If we could measure the moles of eIf we could measure the moles of e--, we could , we could
know the quantity of Ag formed.know the quantity of Ag formed.
But how to measure moles of eBut how to measure moles of e--??
Current = charge passing
time
I (amps) = coulombs
seconds
But how is charge related to moles of But how is charge related to moles of electrons?electrons?
Current = charge passing
timeI (amps) =
coulombs
seconds
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
= = 96,500 C/mol e96,500 C/mol e--= = 1 Faraday1 Faraday
Michael FaradayMichael Faraday
17911791--18671867
Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a 1.50 amps flow thru a AgAg++(aq(aq) solution for 15.0 ) solution for 15.0
min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?
SolutionSolution
(a)(a) Calc. chargeCalc. charge
Charge (C) = current (A) x time (t)Charge (C) = current (A) x time (t)
= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C
I (amps) = coulombs
seconds
Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C
(b)(b) Calculate moles of eCalculate moles of e-- usedused
I (amps) = coulombs
seconds
1350 C • 1 mol e -
96,500 C==== 0.0140 mol e -
0.0140 mol e - • 1 mol Ag
1 mol e -==== 0.0140 mol Ag or 1.51 g Ag
1.50 amps flow thru a 1.50 amps flow thru a AgAg++(aq(aq) solution for 15.0 min. What ) solution for 15.0 min. What
mass of Ag metal is deposited?mass of Ag metal is deposited?
(c)(c) Calc. quantity of AgCalc. quantity of Ag
Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(sPb(s) + HSO) + HSO44--(aq) (aq) ------> PbSO> PbSO44(s) + (s) + HH++(aq(aq) + 2e) + 2e--
If a battery delivers 1.50 amp, and you have 454 g of If a battery delivers 1.50 amp, and you have 454 g of PbPb, ,
how long will the battery last?how long will the battery last?
SolutionSolution
a)a) 454 g 454 g PbPb = 2.19 mol = 2.19 mol PbPb
b)b) Calculate moles of eCalculate moles of e--
2.19 mol Pb • 2 mol e -
1 mol Pb = 4.38 mol e -
c)c) Calculate chargeCalculate charge
4.38 mol e4.38 mol e-- •• 96,500 C/mol e96,500 C/mol e-- = 423,000 C= 423,000 C
Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(sPb(s) + HSO) + HSO44--(aq) (aq) ------> PbSO> PbSO44(s) + (s) + HH++(aq(aq) + 2e) + 2e--
If a battery delivers 1.50 amp, and you have 454 g of If a battery delivers 1.50 amp, and you have 454 g of PbPb, how long will the battery last?, how long will the battery last?
SolutionSolution
a)a) 454 g 454 g PbPb = 2.19 mol = 2.19 mol PbPb
b)b) Mol of eMol of e-- = 4.38 mol= 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
Time (s) = Charge (C)
I (amps)
Time (s) = 423, 000 C
1.50 amp = 282, 000 s About 78 hoursAbout 78 hours
d)d) Calculate timeCalculate time
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Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the
reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
40
Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the
reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
( )
Cathode: Pd + 2e Pd
1 mol 2 mol 1 mol
106 g 2(96,500) 106 g
3.20 amp = 3.20
g = 30.0 min60 s
min
C
s
g Pd
2 96,500 C g Pd
2+ - 0
Cs
→
× × × =?.
.320 106
316
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Commercial Applications of Electrolytic Cells
Electrolytic Refining and Electroplating of Metals
� Impure metallic copper can be purified
electrolytically to ≈ 100% pure Cu.
� The impurities commonly include some active metals
plus less active metals such as: Ag, Au, and Pt.
� The cathode is a thin sheet of copper metal
connected to the negative terminal of a direct
current source.
� The anode is large impure bars of copper.
Oxidation:Oxidation: Zn(s) Zn(s) ------> Zn> Zn2+2+(aq) + (aq) + 2e2e--
Reduction:Reduction: CuCu2+2+(aq) + (aq) + 2e2e-- ------> Cu(s)> Cu(s)
----------------------------------------------------------------------------------------------------------------
CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)
Zn metal
Cu2+ ions
Zn metal
Cu2+ ionsElectrons are transferred from Zn to
Cu2+, but there is no useful electric
current.
Electrons are transferred from Zn to
Cu2+, but there is no useful electric
current.
With time, Cu plates out onto Zn metal strip, and Zn strip
“disappears.”
With time, Cu plates out onto Zn metal strip, and Zn strip
“disappears.”
Chemical Change���� Electrical
Current
��To obtain a useful current, To obtain a useful current,
we separate the oxidizing we separate the oxidizing
and reducing agents so that and reducing agents so that
electron transfer occurs thru electron transfer occurs thru
an external wire. an external wire.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
elect rons
This is accomplished in a This is accomplished in a GALVANICGALVANIC or or
VOLTAICVOLTAIC cell.cell.
A group of such cells is called a A group of such cells is called a batterybattery..
Chemical Change���� Electrical
Current
44
Voltaic or Galvanic Cells
� Electrochemical cells in which a spontaneous
chemical reaction produces electrical energy.
� Cell halves are physically separated so that
electrons (from redox reaction) are forced to travel
through wires and creating a potential difference.
� Examples of voltaic cells include:
batteries calculator andComputer
batteries Flashlight
batteries Auto
45
The Construction of Simple Voltaic Cells
� Voltaic cells consist of two half-cells which
contain the oxidized and reduced forms of an
element (or other chemical species) in
contact with each other.
� A simple half-cell consists of:
� A piece of metal immersed in a solution of its
ions.
� A wire to connect the two half-cells.
� And a salt bridge to complete the circuit, maintain
neutrality, and prevent solution mixing.
Terms Used for Voltaic CellsTerms Used for Voltaic Cells
Figure 20.6Figure 20.6
47
Standard Electrode Potential
� To measure relative
electrode potentials, we
must establish an arbitrary
standard.
� That standard is the
Standard Hydrogen
Electrode (SHE).
� The SHE is assigned an
arbitrary voltage of
0.000000… V
48
The Zinc-SHE Cell
� For this cell the
components are:
1. A Zn strip
immersed in 1.0
M zinc (II) sulfate.
2. The other
electrode is the
Standard
Hydrogen
Electrode.
3. A wire and a salt
bridge to
complete the
circuit.
� The initial cell
voltage is 0.763
volts.
49
The Zinc-SHE Cell
( )
( ) V 763.0E H+Zn2H+Znreaction Cell
V 0.000 He 2+H 2reaction Cathode
V 0.763 e 2+Zn Znreaction Anode
E
0
cellg2
+2+0
g2
-+
-+20
0
=→
→
→
� The cathode is the Standard Hydrogen Electrode.
� In other words Zn reduces H+ to H2.
� The anode is Zn metal.
� Zn metal is oxidized to Zn2+ ions.
Volts
Zn
H2
Salt Bridge
Zn2+ H+
Zn Zn2+ + 2e- OXIDATION ANODE
2 H+ + 2e- H2REDUCTIONCATHODE
-+
Overall reaction is reduction of HOverall reaction is reduction of H++ by Zn metal.by Zn metal.
Zn(sZn(s) + 2 H) + 2 H++ ((aqaq) ) ----> Zn> Zn2+2+ + H+ H22(g)(g) EEoo = +0.76 V= +0.76 V
Therefore, Therefore, EEoo for for Zn Zn ------> Zn> Zn2+2+ ((aqaq) + 2e) + 2e-- is is +0.76 V+0.76 V
Zn is a Zn is a (better) (poorer)(better) (poorer) reducing agent than Hreducing agent than H22..
Volts
Cu
H2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-
OXIDATION ANODE
-+
Cu/CuCu/Cu2+2+ and Hand H22/H/H++ CellCell
EEoo = +0.34 V= +0.34 V
Acceptor Acceptor
of of
electronselectrons
Supplier Supplier
of of
electronselectrons
CuCu2+2+ + 2e+ 2e-- ----> Cu> Cu
ReductionReduction
CathodeCathode
HH22 ----> 2 H> 2 H++ + 2e+ 2e--
OxidationOxidation
AnodeAnode
PositivePositive NegativeNegative
Cu/CuCu/Cu2+2+ and Hand H22/H/H++ CellCell
Overall reaction is reduction of CuOverall reaction is reduction of Cu2+2+ by Hby H22 gas.gas.
CuCu2+2+ ((aqaq) + H) + H22(g) (g) ------> Cu(s) + 2 > Cu(s) + 2 HH++(aq(aq))
Measured Measured EEoo = +0.34 V= +0.34 V
Therefore, Therefore, EEoo for Cufor Cu2+2+ + 2e+ 2e-- ------> Cu> Cu isis
Volts
Cu
H2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-
OXIDATION ANODE
-+
+0.34 V+0.34 V
53
Uses of Standard Electrode Potentials
� Electrodes that force the SHE to act as an anode are assigned
positive standard reduction potentials.
� Electrodes that force the SHE to act as the cathode are assigned
negative standard reduction potentials.
� Standard electrode (reduction) potentials tell us the tendencies of
half-reactions to occur as written.
� For example, the half-reaction for the standard potassium
electrode is:
V -2.925=E K e K 00→+ −+
The large negative value tells us that this reaction
will occur only under extreme conditions.
54
Uses of Standard Electrode Potentials
� Compare the potassium half-reaction to fluorine’s half-
reaction:
� The large positive value denotes that this reaction
occurs readily as written.
� Positive E0 values denote that the reaction tends to
occur to the right.
� The larger the value, the greater the tendency to occur
to the right.
� It is the opposite for negative values of Eo.
F + 2 e 2 F E = +2.87 V20 - - 0→
Using Standard Potentials, Using Standard Potentials, EEoo
Table 20.1Table 20.1
� Which is the best oxidizing agent:
O2, H2O2, or Cl2? _________________
� Which is the best reducing agent:
Hg, Al, or Sn? ____________________
Standard Standard RedoxRedox Potentials, Potentials, EEoo
Any substance on the right will Any substance on the right will
reduce any substance higher reduce any substance higher
than it on the left.than it on the left.
�� Zn can reduce HZn can reduce H++ and Cuand Cu2+2+..
�� HH22 can reduce Cucan reduce Cu2+2+ but not Znbut not Zn2+2+
�� Cu cannot reduce HCu cannot reduce H++ or Znor Zn2+2+..
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
Standard Standard RedoxRedox Potentials, Potentials, EEoo
Cu2+ + 2e- --> Cu +0.34
+2 H + 2e- --> H2 0.00
Zn2+ + 2e- --> Zn -0.76
NorthwestNorthwest--southeast rule:southeast rule: productproduct--favored favored
reactions occur between reactions occur between
•• reducing agent at southeast corner reducing agent at southeast corner
•• oxidizing agent at northwest corneroxidizing agent at northwest corner
Any substance on the right will reduce any substance Any substance on the right will reduce any substance
higher than it on the left.higher than it on the left.
Ox. agentOx. agent
Red. agentRed. agent
Using Standard Potentials, Using Standard Potentials, EEooTable 20.1Table 20.1
�� In which direction do the following reactions go?In which direction do the following reactions go?
�� Cu(s) + 2 Cu(s) + 2 AgAg++(aq(aq) ) ------> Cu> Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)
�� Goes right as writtenGoes right as written
�� 2 Fe2 Fe2+2+(aq) +(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s)
�� Goes LEFT opposite to direction writtenGoes LEFT opposite to direction written
� What is Eonet for the overall reaction?
Standard Standard RedoxRedox Potentials, Potentials, EEoo
EE˚̊netnet = = ““distancedistance”” from from ““toptop”” halfhalf--reaction reaction
(cathode)(cathode) to to ““bottombottom”” halfhalf--reaction reaction (anode)(anode)
EE˚̊netnet = E= E˚̊cathodecathode -- EE˚̊anodeanode
EEoonetnet for Cu/Agfor Cu/Ag++ reaction = +0.46 Vreaction = +0.46 V
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
elect rons
•Electrons travel thru external wire.
�Salt bridge allows anions and cations to move
between electrode compartments.
••Electrons travel thru external wire.Electrons travel thru external wire.
��Salt bridge Salt bridge allows anions and allows anions and cationscations to move to move
between electrode compartments.between electrode compartments.
Zn Zn ----> Zn> Zn2+2+ + 2e+ 2e-- CuCu2+2+ + 2e+ 2e-- ----> Cu> Cu
<<----AnionsAnions
CationsCations---->>
Oxidation
Anode
Negative
OxidationOxidation
AnodeAnode
NegativeNegative
Reduction
CathodePositive
ReductionReduction
CathodeCathode
PositivePositive
CELL POTENTIAL, ECELL POTENTIAL, E
�� Electrons are Electrons are ““drivendriven”” from anode to cathode by an from anode to cathode by an electromotive forceelectromotive force or or
emfemf..
�� For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚̊C and when C and when
[Zn[Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.
�� STANDARD CELL POTENTIAL, STANDARD CELL POTENTIAL, EEoo
�� ——a quantitative measure of the tendency of reactants to proceed ta quantitative measure of the tendency of reactants to proceed to o
products when all are in their standard states at 25 products when all are in their standard states at 25 ˚̊C. C.
Zn and ZnZn and Zn2+2+,,
anodeanodeCu and CuCu and Cu2+2+,,
cathodecathode
Zn
Zn2+ ions
Cu
Cu 2+ ions
wire
saltbridge
e lect rons
1.10 V1.10 V
1.0 M1.0 M 1.0 M1.0 M
Calculating Cell VoltageCalculating Cell Voltage
�� Balanced halfBalanced half--reactions can be reactions can be
added together to get overall, added together to get overall,
balanced equation. balanced equation.
Zn(s) ---> Zn2+(aq) + 2e-Cu2+(aq) + 2e- ---> Cu(s)--------------------------------------------
Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Zn(s) Zn(s) ------> Zn> Zn2+2+(aq) + 2e(aq) + 2e--
CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s)
----------------------------------------------------------------------------------------
CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)
If we know If we know EEoo for each halffor each half--reaction, we reaction, we
could get could get EEoo for net reaction.for net reaction.
Uses of Eo ValuesUses of Uses of EEoo ValuesValues
Organize halfOrganize half--reactions by reactions by relative ability to act as relative ability to act as oxidizing agentsoxidizing agents
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
elect rons
CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s) EEoo = +0.34 V= +0.34 V
ZnZn2+2+(aq) + 2e(aq) + 2e-- ------> Zn(s) > Zn(s) EEoo = = ––0.76 V0.76 V
Zn(s) Zn(s) ------> Zn> Zn2+2+(aq) + 2e(aq) + 2e-- EEoo = +0.76 V= +0.76 V
CuCu2+2+(aq) + 2e(aq) + 2e-- ------> Cu(s)> Cu(s) EEoo = +0.34 V= +0.34 V
------------------------------------------------------------------------------------------------------------------------------
CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) ------> Zn> Zn2+2+(aq) + Cu(s) (aq) + Cu(s)
EEoo ((calccalc’’dd) = +1.10 V) = +1.10 V
� In all voltaic cells, electrons flow spontaneously from the
negative electrode (anode) to the positive electrode (cathode).
65
The Zinc-Copper Cell
� There is a commonly used short hand notation for voltaic cells.
� The Zn-Cu cell provides a good example.
Zn/Zn2+(1.0 M ) || Cu2+(1.0 M )/Cu
species (and concentrations)in contact with electrode surfaces
electrode surfaces
salt bridge
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
CdCd ----> Cd> Cd2+2+ + 2e+ 2e--
oror
CdCd2+2+ + 2e+ 2e-- ----> > CdCd
Fe Fe ----> Fe> Fe2+2+ + 2e+ 2e--
oror
FeFe2+2+ + 2e+ 2e-- ----> Fe> Fe
EEoo for a Voltaic Cellfor a Voltaic Cell
All ingredients are present. Which way does All ingredients are present. Which way does
reaction proceed? What is the value of Ereaction proceed? What is the value of E00cellcell??
67
Effect of Concentrations (or Partial Pressures) on Electrode Potentials
The Nernst Equation
� Standard electrode potentials, those
compiled in appendices, are determined at
thermodynamic standard conditions.
� Reminder of standard conditions. 1.00 M solution concentrations
1.00 atm of pressure for gases
All liquids and solids in their standard thermodynamic
states.
Temperature of 250 C.68
The Nernst Equation
� The value of the cell potentials change if conditions are nonstandard.
� The Nernst equation describes the electrode potentials at nonstandard conditions.
� The Nernst equation is:
( )
quotientreaction =Q
e mol J/V 487,96
VCJ 1)e C/mol (96,487=faraday the=F
ed transferrelectrons ofnumber =n
Kin etemperatur=T
K molJ 8.314=constant gas universal=R
conditions standardunder potentialE
interest ofcondition under potential=E
Q log nF
2.303RT-E=E
-
.-
0
0
=
×
=
69
The Nernst Equation
� Substitution of the values of the constants into the
Nernst equation at 25o C gives:
Q log n
0.0592-E=E Thus
0592.0e mol J/V 96,487
K 298314.8303.2
F
RT 2.303
0
-
K molJ
=××
=
70
The Nernst Equation
� For this half-reaction:
� The corresponding Nernst equation
is:
V +0.153=ECueCu 0-+2 ←→+ +
[ ][ ]
E = E -0.0592
1 log
Cu
Cu
0
+
2+
71
The Nernst Equation
� Substituting E0 into the above expression gives:
� If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard
conditions, then E = E0 because the concentration
term equals zero.
[ ][ ]
E = 0.153 V -0.0592
1 log
Cu
Cu
+
2+
E = 0.153 V -0.0592
1 log
1
172
The Nernst Equation
� The Nernst equation can also be used to calculate
the potential for a cell that consists of two
nonstandard electrodes.
� Example: Calculate the initial potential of a cell that
consists of an Fe3+/Fe2+ electrode in which
[Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to
a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and
[Sn2+]=0.10 M . A wire and salt bridge complete
the circuit.
73
Relationship of E0cell to
∆∆∆∆G0 and K� From previous chapters we know the
relationship of ∆G0 and K for a reaction.
K log RT 303.2G
or
lnK -RTG
0
0
−=∆
=∆
e ofnumber n
e mol J/V 96,487 F where
E F-n G
-
-
0
cell
0
=
=
=∆
EEoo and and ∆∆GGoo
∆∆GGoo = = -- n F n F EEoo
For a For a productproduct--favoredfavored reactionreaction
Reactants Reactants --------> Products> Products
∆∆GGo o < 0 and so < 0 and so EEoo > 0> 0
EEoo is positiveis positive
For a For a reactantreactant--favoredfavored reactionreaction
Reactants <Reactants <-------- ProductsProducts
∆∆GGo o > 0 and so > 0 and so EEoo < 0< 0
EEoo is negativeis negative
75
Relationship of E0cell to ∆∆∆∆G
0
and K
� Example: Calculate the standard Gibbs free
energy change, ∆G0 , at 250C for the
following reaction.
Cu PbCuPb 22 +→+ ++
76
Relationship of E0cell to ∆∆∆∆G
0
and K
� Example: Calculate the Gibbs Free Energy
change, ∆G and the equilibrium constant at
250C for the following reaction with the
indicated concentrations.
( ) ( )MM 0.50 Zn Ag2 0.30 Ag 2 Zn 2++ +→+
Dry Cell BatteryDry Cell Battery
Anode (Anode (--))
Zn Zn ------> Zn> Zn2+2+ + 2e+ 2e--
Cathode (+)Cathode (+)
2 NH2 NH44++ + 2e+ 2e-- ------> >
2 NH2 NH33 + H+ H22
Primary batteryPrimary battery —— uses uses
redoxredox reactions that cannot reactions that cannot
be restored by recharge.be restored by recharge.
Nearly same reactions as in common dry Nearly same reactions as in common dry
cell, but under basic conditions.cell, but under basic conditions.
Alkaline BatteryAlkaline Battery
Anode (Anode (--): ): Zn + 2 OHZn + 2 OH-- ------> > ZnOZnO + H+ H22O + 2eO + 2e--
Cathode (+): Cathode (+): 2 MnO2 MnO22 + H+ H22O + 2eO + 2e-- ------> >
MnMn22OO33 + 2 OH+ 2 OH--
79
The Lead Storage Battery
� Diagram of the lead storage battery.
Lead Storage BatteryLead Storage Battery
Anode (Anode (--) ) EEoo = +0.36 V= +0.36 V
PbPb + HSO+ HSO44-- ------> PbSO> PbSO44 + H+ H++ + 2e+ 2e--
Cathode (+) Cathode (+) EEoo = +1.68 V= +1.68 V
PbOPbO22 + HSO+ HSO44-- + 3 H+ 3 H++ + 2e+ 2e--
------> PbSO> PbSO44 + 2 H+ 2 H22OO
NiNi--Cad BatteryCad BatteryAnode (Anode (--))
CdCd + 2 OH+ 2 OH-- ------> Cd(OH)> Cd(OH)22 + 2e+ 2e--
Cathode (+) Cathode (+)
NiO(OHNiO(OH) + H) + H22O + eO + e-- ------> Ni(OH)> Ni(OH)22 + OH+ OH--
Fuel Cells: HFuel Cells: H22 as a Fuelas a Fuel
••Fuel cellFuel cell -- reactants are reactants are
supplied continuously supplied continuously
from an external source.from an external source.
••Cars can use electricity Cars can use electricity
generated by Hgenerated by H22/O/O22 fuel fuel
cells.cells.
••HH22 carried in tanks or carried in tanks or
generated from generated from
hydrocarbons.hydrocarbons.
HydrogenHydrogen——Air Fuel CellAir Fuel Cell
Figure 20.12Figure 20.12
HH22 as a Fuelas a Fuel
Comparison of the volumes of substances Comparison of the volumes of substances
required to store 4 kg of hydrogen relative to required to store 4 kg of hydrogen relative to
car size. car size. (Energy, p. 290)(Energy, p. 290)