L1 364 General Electrochemistry
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ElectrochemistryElectrochemistryCHCH 364364
4/13/2011 1
Prof. Dr. Saeed REFAEYChemistry DepartmentFaculty of ScienceMinia University
Welcome
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Redox Reactions andRedox Reactions and
Electrochemical CellsElectrochemical Cells
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OxidationOxidation--Reduction Concepts ReviewReduction Concepts ReviewOxidationOxidation ±± Loss of ElectronsLoss of Electrons
ReductionReduction ±± Gain of ElectronsGain of Electrons
Oxidizing Agent Oxidizing Agent ±± Species that causes anotherSpecies that causes anotherspecies to be oxidized (lose electrons)species to be oxidized (lose electrons)
Oxidizing agent is reduced (gains eOxidizing agent is reduced (gains e--))Reducing Agent Reducing Agent ±± Species that cause anotherSpecies that cause another
species to be reduced (gain electrons)species to be reduced (gain electrons)
Reducing agent is oxidized (loses eReducing agent is oxidized (loses e--))Oxidation (eOxidation (e-- loss) always accompaniesloss) always accompanies
Reduction (eReduction (e-- gain)gain)Total number of electrons gained by the atoms/ionsTotal number of electrons gained by the atoms/ions
of the oxidizing agent always equals the totalof the oxidizing agent always equals the totalnumber of electrons lost by the reducing agent number of electrons lost by the reducing agent
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Oxidation NumberOxidation NumberA number = the magnitude of theA number = the magnitude of the
charge an atomcharge an atom
The oxidation number in aThe oxidation number in a binary ionicbinary ioniccompoundcompound equals the ionic chargeequals the ionic charge
The oxidation number for each element inThe oxidation number for each element inpolyatomic ionpolyatomic ion are assigned according to theare assigned according to the
relativerelative attraction of an atom for electronsattraction of an atom for electrons
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Balancing Redox ReactionsBalancing Redox ReactionsOxidation Number MethodOxidation Number Method
Half Half--Reaction MethodReaction Method
The balancing process must insure that:The balancing process must insure that:
the number of electrons lost =the number of electrons lost =
the number of electrons gainedthe number of electrons gained
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Oxidation Number MethodOxidation Number Method
Assign oxidation numbers to all elements inAssign oxidation numbers to all elements inthethereaction.reaction.
compute the number of compute the number of electrons lost electrons lost inin thetheoxidation andoxidation and electron gainedelectron gained in thein the
reductionreductionfrom the oxidation number change .from the oxidation number change .
Multiply one or both these number by factorsMultiply one or both these number by factorsto maketo make the electrons lost = to thethe electrons lost = to the
electronselectrons4/13/2011 8
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Practice
Problem
Practice
Problem 11
Balance equation with Oxidation NumberBalance equation with Oxidation Number
method:method:
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3 3 2 2 2Cu(s) + HNO (aq) Cu(NO ) + NO (g) + H O(l)p
3 3 2 2 2Cu(s) + HNO (aq) Cu(NO ) + NO (g) + H O(l)p
-2+5 +5
0 +1 -2 -2-2+2 +4 +1
3 3 2 2 2(s) (a ) ( ) (g) (l)p
Loses 2e-
Gains 1e-
3 3 2 2 2Cu(s) + 4HNO (aq) Cu(NO ) + 2NO (g) + 2H O(l)p
Balance the Equation
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Reaction MethodReaction Method--Half Half
Divide the overall reaction into:Divide the overall reaction into:Oxidation Half Oxidation Half--ReactionReaction
Reduction Half Reduction Half--ReactionReaction
Balance each half Balance each half--reaction for atoms &reaction for atoms &chargecharge
Multiply one or both reactions by someMultiply one or both reactions by some
electrons gained equal toelectrons gained equal tointeger to makeinteger to makeelectrons lost electrons lost
Recombine to given balanced redoxRecombine to given balanced redoxe ati ne ati n
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Redox Half Redox Half--Reaction MethodReaction Method ±±ExampleExample
Divide steps into Half Divide steps into Half--ReactionsReactions
Balance Atoms & Charges for CrBalance Atoms & Charges for Cr22OO7722-- / Cr/ Cr33++
Balance Atoms & Charges for IBalance Atoms & Charges for I-- / I/ I22
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2- - +32 7 2Cr O (aq) + I (aq) Cr (aq) + I (s)p
2- 3+ -2 7Cr O Cr (e gain - reduction)p - -
2I I (e loss - oxidation)p
2- 3+2 7Cr O 2Crp
2- 3+2 7 2Cr O 2Cr + 7 Op
+ 2- 3+2 7 214 + Cr O 2Cr + 7 Op
AddWater Molecules
Add 14 H+ ions on left to
balance 14 H on right
Add 6 electrons (e-) on left
to balance charge on right
- + 2- 3+2 7 26e + 14H + Cr O 2Cr + 7H Op
-22I Ip
- -22I I + 2ep
No need to add H2O or H+
Add 2 electrons (e-) on right
to balance charge on left
(6 electrons gained @ this is the reduction reaction
(2 electrons lost @ this is the oxidation reaction
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Half Half--Reaction Method inReaction Method in ³³BasicBasic´́solutionsolution
Sodium Permanganate & Sodium OxalateSodium Permanganate & Sodium OxalateNaMnONaMnO44 NaNa22CC22OO44
Half Half--ReactionsReactions
Multiply each reaction by appropriate integerMultiply each reaction by appropriate integer
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- 2- 2-4 2 4 2 3MnO (aq) + C O MnO (s) + CO (aq) (basic solution)p
-
4 2MnO MnOp 2- 2-
2 4 3p
-4 2 22p
-4 2 24 2p
- ¡ -
4 2 23e 4 2(re ctio )
p
2- 2-2 4 32p
2- 2-2 2 4 32 2p
2- 2-
2 2 4 32 2 4p
2- 2- + -
2 2 4 32 H O + C O 2CO + 4H + 2e(oxidation)
p
- + -4 2 26e + 8H + 2MnO 2MnO + 4H Op
2- 2- + -2 2 4 36H O + 3 C O 6CO + 12H + 6ep
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Sodium Permanganate & Sodium OxalateSodium Permanganate & Sodium Oxalate
(con(con¶¶t)t)Add reactionsAdd reactions
Basic SolutionBasic Solution -- Add OHAdd OH--
to neutralize Hto neutralize H++ and cancel Hand cancel H22OO
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- + -4 2 26e + 8H + 2MnO 2MnO + 4H Op
2- 2- + -2 2 4 36H O + 3 C O 6CO + 12H + 6ep
- 2- 2- +4 2 2 4 2 32 O + 2H O + 3 C O 2 O + 6CO + 4Hp
- 2- - 2- + -4 2 2 4 2 32 O + 2H O + 3 C O + 4OH 2 O + 6CO + 4H + 4OHp
- 2- - 2-4 2 2 4 2 3 22 O + 2H O + 3 C O + 4OH 2 O + 6CO + 4H Op
- 2- - 2-4 2 4 2 3 22 O + 3 C O + 4OH 2 O + 6CO + 2H Op
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Electrochemical CellsElectrochemical CellsGalvanic CellsGalvanic Cells
Use spontaneous reaction (Use spontaneous reaction (((G <G < 00) to) togenerate electrical energygenerate electrical energy
Difference in Chemical Potential energyDifference in Chemical Potential energy
betweenbetween higherhigher energy reactantsenergy reactants andandlower energy productslower energy products is converted theis converted the
electrical energy to power electricalelectrical energy to power electrical
devices (as car)devices (as car)
doesdoessystemsystemTheThe--ThermodynamicallyThermodynamically
on the surroundingson the surroundingsworkwork
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Electrochemical CellsElectrochemical CellsElectrolytic CellsElectrolytic Cells
Uses electrical energy to driveUses electrical energy to drive
nonspontaneousnonspontaneous reaction (reaction (((G >G > 00))
Electrical energy from an external powerElectrical energy from an external power
supply convertssupply converts lower energy reactantslower energy reactants toto
higherhigher energy productsenergy products
ThermodynamicallyThermodynamically ±± TheThe surroundings do worksurroundings do workon the systemon the system
ExamplesExamples ±± Electroplating and recovering metalsElectroplating and recovering metalsfrom oresfrom ores
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Electrochemical CellsElectrochemical CellsCell diagram is used to describe structure of Cell diagram is used to describe structure of galvanic cellgalvanic cell
is:is:cell diagramcell diagramFor the Zn/Cu cell, theFor the Zn/Cu cell, the
Zn(s)Zn(s) ZnZn22++(aq)(aq) CuCu22++(aq)(aq) Cu(s)Cu(s)= phase boundary (solid Zn vs. Aqueous= phase boundary (solid Zn vs. AqueousZnZn22++))= salt bridge= salt bridge
of the salt bridgeof the salt bridgeleft left Anode reaction (oxidation) isAnode reaction (oxidation) is
Cathode reaction (reduction) isCathode reaction (reduction) is right right of the salt of the salt bridgebridge
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Galvanic CellsGalvanic Cells
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a van c e sa van c e sZinc metal (Zn) in solution of CuZinc metal (Zn) in solution of Cu++++ ionsions
Construction of a Galvanic CellConstruction of a Galvanic CellThe oxidizing agent (Zn) and reducing agent The oxidizing agent (Zn) and reducing agent
(Cu(Cu22++) in) in
the same beakerthe same beaker will not generate electricalwill not generate electricalenergy.energy.
Separate the half Separate the half--reactions byreactions by a barriera barrier andandconnect connect
them via an external circuit them via an external circuit (wire).(wire).
Set upSet up salt bridgesalt bridge between chambers to maintainbetween chambers to maintain
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2 -(a ) 2e (s) [re ctio ]p
2 -Z (s) Z (a ) 2e [o i atio ]p
2 2Z (s) (a ) Z (s)p
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Oxidation Half Oxidation Half--CellCell
AnodeAnode Compartment Compartment ±± Oxidation of ZincOxidation of Zinc (An Ox)(An Ox)
Zinc metal in solution of ZnZinc metal in solution of Zn22++ electrolyte (ZnSOelectrolyte (ZnSO44))
Zn is reactant in oxidation half Zn is reactant in oxidation half--reactionreaction
Conducts released electrons (eConducts released electrons (e--)) out ofout of its half its half--cellcell
Reduction Half Reduction Half--CellCell
CathodeCathode Compartment Compartment ±± Reduction of CopperReduction of Copper (Red(RedCat)Cat)
Copper bar in solution of CuCopper bar in solution of Cu22++
electrolyte (CuSOelectrolyte (CuSO44))Copper metal is product in reduction half Copper metal is product in reduction half--cell reactioncell reaction
Conducts electronsConducts electrons intointo its half its half--cellcell
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Zn Cu Galvanic Cell
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Relative Charges on theRelative Charges on theAnode/CathodeAnode/Cathode
electrodeselectrodesare determined by theare determined by thechargeschargesElectrodeElectrodesourcesource
of the electrons and theof the electrons and the directiondirection of electronof electronflowflow
Zinc atoms oxidized to ZnZinc atoms oxidized to Zn22++ at at anodeanodeAnodeAnode ±± negative charge (enegative charge (e-- rich)rich)
Released electrons flow to right towardReleased electrons flow to right towardcathodecathode
to be accepted by Cuto be accepted by Cu22++ to form Cu(s)to form Cu(s)
--4/13/2011 23
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Active vs Inactive ElectrodesActive vs Inactive Electrodes
Active ElectrodesActive ElectrodesElectrodes in Zn/CuElectrodes in Zn/Cu22++ cell are activecell are active
Zn & Cu bars are components of Zn & Cu bars are components of
the cell reactionsthe cell reactions
Mass of Zn bar decreases as ZnMass of Zn bar decreases as Zn22++ ions inions in
cell solution increasecell solution increase
Mass of Cu bar increases as CuMass of Cu bar increases as Cu22++
ions accept electron to form more Cuions accept electron to form more Cu
metalmetal4/13/2011 24
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Active vs Inactive ElectrodesActive vs Inactive ElectrodesInactive ElectrodesInactive Electrodes
Inactive electrodesInactive electrodes -- graphite or Pt:graphite or Pt:
Can conduct electrons into and out of Can conduct electrons into and out of
half half--cellscells
Cannot take part in the half Cannot take part in the half--reactionsreactions
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Galvanic Cell withInactive Graphite
Electrodes
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Practice ProblemPractice Problem 11A mercury battery, used for hearing aids andA mercury battery, used for hearing aids and
electric watches, delivers a constant voltageelectric watches, delivers a constant voltage((11..35 35 V) for long periods. The half reactions areV) for long periods. The half reactions aregiven below. Which half reaction occurs at thegiven below. Which half reaction occurs at theanode and which occurs at the cathode? Whatanode and which occurs at the cathode? What
is the overall cell reaction?is the overall cell reaction?HgO(s) + HHgO(s) + H
22O(l) +O(l) + 22ee--
pp Hg(l) +Hg(l) + 22 OHOH--(aq)(aq)
n(s) +n(s) + 22 OHOH--(aq)(aq) pp n(OH)n(OH)22(s) +(s) + 22ee--
Ans:Ans:Reduction occurs at Cathode (Red Cat)Reduction occurs at Cathode (Red Cat)to form Hgto form Hgreduced)reduced)((--ee22gainsgains++22HgHg
Oxidation occurs at the Anode (An
Oxidation occurs at the Anode (An Ox)Ox)
++22to form Znto form Znoxidized)oxidized)((--ee22Zn losesZn loses
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2 2HgO(s) + n(s) + H O n(OH) + Hg(l)p
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Practice Problem 2
Write the cell diagrame for a Galvanic cell with the following cellreaction:
Ni(s) + Pb
2+
(aq)p
Ni
2+
(aq) + Pb(s)
Ans:
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2 -i(s) i 2e ( i atio - A o e eactio )p
2 -(a ) 2e (s) ( e ctio - at o e eactio )p
2 2i(s) i (a ) (a ) (s)PI I
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Practice ProblemPractice Problem 33Write the cell reaction for the following galvanicWrite the cell reaction for the following galvanic
cellcellPt|HPt|H22(g) | H(g) | H++(aq) Br (aq) Br 22(l) | Br (l) | Br --(aq)|Pt(aq)|PtNote: Platinum (Pt) serves as a reaction site, butNote: Platinum (Pt) serves as a reaction site, but
doesdoesnot participate in the reactionnot participate in the reaction
Ans:Ans:
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+ - -2H (g) 2H (aq) + 2e (lose 2 e - oxidation (cathode) reaction)p
- - -2r (l) 2e 2 r (a ) (gai 2 e - re ctio (a o e) reactio ) p
+ -2 2Pt H (g) + Br (l) 2H (aq) + 2Br (aq) PtpI I
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Cell PotentialsCell Potentials
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Cell PotentialCell PotentialTheThe movement of electronsmovement of electrons isisanalogous to the pumping of water fromanalogous to the pumping of water from
one point to anotherone point to another
Water moves from a point of highWater moves from a point of highpressure to a point of lower pressure.pressure to a point of lower pressure.
Thus,Thus, aa pressure differencepressure difference is requiredis required
TheThe workwork expended in moving the water expended in moving the water volume of volume of through a pipe depends on thethrough a pipe depends on the
pressure differencepressure differenceand theand thewater water
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Cell PotentialCell PotentialMovement of ElectronsMovement of Electrons
An electric charge moves from a point of highAn electric charge moves from a point of high
electrical potential (electrical potential (high electrical pressurehigh electrical pressure) to) to
one of lower electrical potentialone of lower electrical potential
TheThe workwork expended in moving the electrical chargeexpended in moving the electrical charge
potentialpotentialthrough a conductor depends on thethrough a conductor depends on the
difference and the amount of chargedifference and the amount of charge
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cell
Work(w) = Potential difference (E) Charge
w = E charge
v
v
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Cell PotentialCell Potential
Purpose of a galvanic cell is to convert Purpose of a galvanic cell is to convert thethe
free energy of a spontaneous reactionfree energy of a spontaneous reactionintointo
thethe kinetic energykinetic energy of electrons movingof electrons movingthrough an external circuit (electricalthrough an external circuit (electrical
energy)energy)
Electrical energy is proportional to theElectrical energy is proportional to thedifference in the electrical potentialdifference in the electrical potential
between the two cell electrodesbetween the two cell electrodes ±± CellCellPotentialPotential
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Cell PotentialCell PotentialElectrons flowElectrons flow±±Cell PotentialCell PotentialPositivePositive
spontaneouslyspontaneously from the negative electrodefrom the negative electrode((anodeanode) to the positive electrode () to the positive electrode (cathodecathode))
NegativeNegative cell potentialcell potential ±± is associated with ais associated with a³³nonspontaneousnonspontaneous´́ cell reactioncell reaction
Cell potential for an cell reaction at equilibriumCell potential for an cell reaction at equilibrium
would bewould be³³
00´́
As with Entropy, there is a clear relationshipAs with Entropy, there is a clear relationshipbetween Ebetween Ecellcell , K, and, K, and ((GG
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cellE < 0 for a nonspontaneous process
cellE = 0 for an equilibrium process
cellE > 0 for a spontaneous process
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Units of Cell PotentialUnits of Cell PotentialThe SI (metric) unit of electrical charge is theThe SI (metric) unit of electrical charge is the
³³Coulomb (C)Coulomb (C)´́The SI (metric) unit of current is theThe SI (metric) unit of current is the ³³Ampere (A)Ampere (A)´́
The SI (metric) unit of electrical potential isThe SI (metric) unit of electrical potential isthethe ³³Volt (V)Volt (V)´́
By definition, the energy released by a potentialBy definition, the energy released by a potential
difference of difference of one volt one volt moving between the anode andmoving between the anode and
cathode of a galvanic cell releasescathode of a galvanic cell releases1
1
joulejoule of work perof work percoulomb of chargecoulomb of charge
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1 coulomb1 ampere = 1A = 1 C/ s
second
1 J 1 J1 volt = 1 C =
C V
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TheThe charge (F)charge (F) that flows through a cell equals thethat flows through a cell equals thenumber of moles of electrons (n) transferred timesnumber of moles of electrons (n) transferred times
the charge of the charge of 11 mol of electronsmol of electrons
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4
-
JF 9.65 10
�mol e
v-
-
chargeCharge = moles of e
mol e
Charge = n - araday Constant
-
96, 485 C J= C - Coulom b , SI unit of charge
mol e V
¨ ¸© ¹ª º
-Moles e = n
-
Charge 96,485 C= = arday Constant =
Mole mole
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Standard Cell PotentialStandard Cell PotentialEEoocellcell ±±The potential measured at aThe potential measured at aspecificspecific
temperature (temperature (298298 K)K) with no current with no current flowing and all concentrations in theirflowing and all concentrations in their
³³standard statesstandard states´́
11 atm for gasesatm for gases
11 M for solutionsM for solutions
Pure solids for electrodesPure solids for electrodes
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2 2 ocellZ (s) (a ; 1 ) Z (a ; 1 ) (s) 1.10p
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Standard Electrode Half Standard Electrode Half--CellCellPotentialsPotentials
EEoo
half half--cellcell ±± Potential associated with a given half Potential associated with a given half--cellcellreaction (electrode compartment) when allreaction (electrode compartment) when allcomponents are incomponents are in ³³standard statesstandard states´́
Standard Electrode Potential for a half Standard Electrode Potential for a half--cell reaction,cell reaction,whether anode (oxidation) or cathode (reduction)whether anode (oxidation) or cathode (reduction)
´́reductionreduction³³written as awritten as aisisEx.Ex.
would be written:would be written:
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2 -(a ) 2e (s) [re ctio - cat o e]p
2 -Z (s) Z (a ) 2e [o i atio - a o e]p
2+ - o oCopper cathodeCu (aq) + 2e Cu(s) E (E ) [reduction]p
2+ - oinc anoden (aq) + 2e n(s) E (E ) [reduction]p
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Standard Electrode Half Standard Electrode Half--Cell PotentialsCell PotentialsElectrons flow spontaneously fromElectrons flow spontaneously from anodeanode (negative)(negative)
toto cathodecathode (positive)(positive)Cathode must have a moreCathode must have a more ³³PositivePositive´́ EEoo
half half--cellcell thanthanthe Anodethe Anode
cellcellooEE´́positivepositive³³For aFor a
betweenbetweendifferencedifferenceThe standard cell potential is theThe standard cell potential is thethe standard electrode potential of thethe standard electrode potential of the ³³CathodeCathode´́(reduction) half (reduction) half--cell and the standard electrodecell and the standard electrode
potential of thepotential of the ³³AnodeAnode´́ (oxidation) half (oxidation) half--cellcell
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o o ocell cat o e (re ctio ) a o e (o i atio )( - ) > 0
o o ocell copper i c-
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Practice ProblemPractice Problem 11Write out the overall equation for the cellWrite out the overall equation for the cell
reaction and determine the standard cellreaction and determine the standard cellpotential for the following galvanic cell :potential for the following galvanic cell :[E[Eoo (Ag(Ag++/Ag) =/Ag) = 00..8080; E; Eoo (Ni(Ni22++/Ni) =/Ni) = -- 00..2626]]
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+ -
2Ag (a ) + 2e 2Ag(s) [re ctio (cat o e)]p 2+ -i(s) i (a ) + 2e [o i atio (a o e)]p
+ 2+Ni(s) + 2Ag (a ) Ni + 2Ag(s)p
2+ +Ni(s) Ni (a ) Ag (a ) Ag(s)PI I
+ -2Ag (aq) + 2e 2Ag(s) [reduction (cathode)]p
2 + -Ni (a ) + 2e Ni(s) [re ctio (a o e)]p
o o ocell Silver NickelE = E - E
ocellE = 0.80 - (-0.26) = 1.06 V
Writ t r cti s i "r cti " f r
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The Standard Hydrogen ElectrodeThe Standard Hydrogen Electrode
The values found in tables (other elements) areThe values found in tables (other elements) aredetermined relative to adetermined relative to a ³³StandardStandard´́
The Standard Electrode potential is defined as zeroThe Standard Electrode potential is defined as zero
(E(E
ooreferencereference) =) = 00..0000
TheThe ³³standard reference half standard reference half--cellcell´́ is a standardis a standard³³HydrogenHydrogen´́ electrodeelectrode
Specially prepared Platinum electrode immersed inSpecially prepared Platinum electrode immersed inaa
11 M aqueous solution of a strong acid throughM aqueous solution of a strong acid throughwhich Hwhich H22 gas at gas at 11 atm is bubbledatm is bubbled
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+ - o2 refere ce2 (a ; 1 ) + 2e (g); 1 atm) 0.00
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Reference Half Reference Half--Cell and UnknownCell and UnknownHalf Half--CellCell
cellcell--) at anode half ) at anode half --(lose e(lose e22of Hof HOxidationOxidation andandreduction of unknown at cathode half reduction of unknown at cathode half--cellcell
--) at cathode half ) at cathode half --(gain e(gain e++of Hof HReductionReduction
cellcelland oxidation of unknown at anode half and oxidation of unknown at anode half--cellcell
4/13/2011 42
o o o o ocell cathode anode unknown referenceE = E - E E - E
o o ocell unknown unknownE = E - 0.00 V = E
o o o o o
cell cathode anode reference unknownE = E - E E - E
o o ocell unknown unknownE = 0.00 V - E = - E
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Practice ProblemPractice Problem 11Determine the standard electrode potential,Determine the standard electrode potential,
EEoo
zinczinc, using a galvanic cell consisting of the, using a galvanic cell consisting of theZn/ZnZn/Zn22++ half half--reaction and the Hreaction and the H++/H/H22 half half--reaction.reaction.
Zinc electrode is negativeZinc electrode is negativeEEoo
cellcell = += +00..7676
Ans: Zinc electrode is anodeAns: Zinc electrode is anode @@Zinc is beingZinc is beingoxidizedoxidized
4/13/2011 43
+ - o2 refere ce2 (a ) + 2e (g) 0.00p
2+ - ozi cZ (s) Z (a ) + 2e ?p
+ 2+ o2Z (s) + 2 (a ) Z (a ) + (g) cell 0.76p
o o o o ocell cathode anode reference incE = E - E E - E
o o oZi c refere ce cell- 0.00 - 0.76 - 0.76
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Free Energy and ElectricalFree Energy and ElectricalWorkWork
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Free Energy and Electrical WorkFree Energy and Electrical WorkElectrical WorkElectrical Work
Potential (EPotential (Ecellcell, in volts) times the charge, in volts) times the charge
EEcellcell measured withmeasured with no current flowingno current flowingEEcellcell voltage isvoltage is maximummaximum possible for cellpossible for cell
Work isWork is maximummaximum possiblepossible
OnlyOnly reversible processreversible process can do maximum workcan do maximum work
Reversible process with no current flow:Reversible process with no current flow:
Forward reaction if opposing potential is smallerForward reaction if opposing potential is smallerReverse reaction if opposing potential is largerReverse reaction if opposing potential is larger
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cell
Work(w) = Potential difference (E) Charge
w = E charge
v
v
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Spontaneous ReactionSpontaneous Reaction ±± ((G <G < 00
Spontaneous ReactionSpontaneous Reaction ±± EEcellcell >> 00
The galvanic cell loses energy as it does work onThe galvanic cell loses energy as it does work onthe surroundings; thus the work termthe surroundings; thus the work term (w(wmaxmax)) isis
negativenegative
4/13/2011 46
ma cell- c argev
max cellw = - E charge = Gv (
( vcell
G = - E charge
o o
cellG = - E F (com o e ts i sta ar states)v
cell- F( v
4 4
-
o lom s JF 9.65 10 9.65 10
mole � mol e
v v
moles
maxw
Recall :
wcell
G - E
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4/13/2011 47
Electrical Work (cont)
Relate standard cell potential to equilibrium constant (K) of the redox reaction
oG = - RT lnK
o
cell-E n = - RT lnK v
)
o
cell -
4
-
J8.314 298.15
mol rl 2.303(log )mol e JF
(9.65 10mol r �mole
vy
v
v
vocell 0.0592 log
o
celllog (at 298.15 )0.0592
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4/13/2011 48
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Effect of Concentration on CellEffect of Concentration on CellPotentialPotential
Most cells do not start withMost cells do not start with
concentrations in theirconcentrations in their ³³standardstandard´́ statesstatesRecall:Recall:
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oG = G + RTlnQ
o o
cellG = - n × E (Standard State)
cell
G = - n E( v
ocell cell- F - F + lv v
ocell cell
l- (Ner st atio )
F
)
o
cell cell -
4
-
J8.314 298.15 K
mol rxn K E = E - 2.303× log Qn mol e J
(9.65 10mol rxn V � mole
vy
v
o
cell cell
0.0592- log (at 298.15 )
ecall : co tai s o ly s ecies it co ce tratio s ( ress res) t at c a ge
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Practice ProblemPractice Problem 11Calculate ECalculate Ecellcell for a galvanic cell containing thefor a galvanic cell containing the
following half following half--cells:cells:[Zn[Zn22++] =] = 00..010010 M [HM [H++] =] = 22..55 M PM PHH22==
00..3030 atmatmConstruct EConstruct Eoo
cellcell
Calculate QCalculate Q
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2+ + 2n(s) n (aq) H H (g)I II I
+ - o22H (aq) + 2e H (g) ( 0.00 ) (re ; cat o e) p
ocell 0.00 - (-0.76 ) 0.76
2+
-4H2+ 2 2P [ n ] 0.30 0.010Q 4.8 10
[H ] 2.5
vv v
o ocell cell cell
0.0592- 2.303 log Q - log Q
nF n
-4cell
0.05920.76 - log(4.8 10 ) 0.76 - (-0.0982 0.86
2
« »v v¬ ¼½
2+ o
n(s) n + 2e - ( - 0.76 ) (o ; ano e) p+ 2+
22 (a ) + Z (s) (g) + Z (a ) p
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Practice ProblemPractice Problem 22What is the equilibrium constant for the followingWhat is the equilibrium constant for the following
reactionreaction[E[Eoo(Ce(Ce44++/Ce/Ce33++) =) = 11..72 72 EEoo(Cl(Cl22/Cl/Cl--) =) = 11..3636))2 2 ClCl--(aq) + (aq) + 2 2 CeCe44++(aq)(aq) mm ClCl22(g) + (g) + 2 2 CeCe33++(aq)(aq)
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- 4 + 3 +
2l (a ) l (g) e (a ) e (a )I II I
4+ - 3+ o2 e (a ) + 2e 2 e (a ) ( 1.72 ) (re ; cat o e)p
ocell 1.72 - 1.36 0.36
vo
cellnE 2 0.36 V
log K = = = 12.2 (at 298.15 K)0.0592 V 0.0592 V
12.2 12K = 10 = 1.45 10v
- o22Cl (aq) Cl + 2e - (E = 1.36 V) (ox; anode) p
4+ - 3+
22 e (a ) + 2 l l (g) + 2 e (a ) p
vo
cell0.0592 VE = log K
n
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Practice ProblemPractice Problem 33Write out the overall equation for the cellWrite out the overall equation for the cell
reaction and determine the standard cellreaction and determine the standard cellpotential for the following galvanic cell.potential for the following galvanic cell.[E[Eoo (Ag(Ag++/Ag) =/Ag) = 00..80 80 EEoo (Ni(Ni22++/Ni) =/Ni) = --00..2626]]
Ni(s)|NiNi(s)|Ni22++(aq)Ag(aq)Ag++(aq)|Ag(s)(aq)|Ag(s)
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+ - + o
2Ag (aq) + 2e 2Ag (aq) (E = 0.80 V) (red; cathode)p
ocellE = 0.80V - (- 0.26) = 1.06 V
+ 2+ +2Ag (a ) + Ni(a ) Ni (a ) + 2Ag (a )p
2+ oNi(a ) Ni + 2e - ( - 0.26 ) (o ; a o e)p
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Practice ProblemPractice Problem 44What is the maximum work you can obtain fromWhat is the maximum work you can obtain from
1515..0
0
g of Ni in the galvanic cell shown in theg of Ni in the galvanic cell shown in theprevious problem when the Eprevious problem when the Ecellcell isis 00..9797 V?V?[E[Eoo (Ag(Ag++/Ag) =/Ag) = 00..80 80 EEoo (Ni(Ni22++/Ni) =/Ni) = --00..2626]]
Ni(s)|NiNi(s)|Ni22++(aq)Ag(aq)Ag++(aq)|Ag(s)(aq)|Ag(s)
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vma cell
- c arge c arge F-
96,485 C JF 96,485
mol e V � mol e -
ass 15.0gmol(Ni) 0.256 molol gt 58.69 g / mol
+ 2 + +2Ag (a ) + Ni(a ) Ni (a ) + 2Ag (a )p
5 4ma 0.256 mol (-1.87 10 J / mol -4.78 10 J
4ma - 4.78 10 J - 4.78 kJ
-5
ma cell -Ni
mol e J-E n -0.97 V 2 96,485 - 1.87 10 J / mol
mol V � mol ev v v
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Practice ProblemPractice Problem 55What is the cell voltage (EWhat is the cell voltage (Ecellcell) for the following) for the following
galvanic cell?galvanic cell?Cd(s)|CdCd(s)|Cd22++((00..026 026 M)NiM)Ni22++((00..00420 00420 M)|Ni(s)M)|Ni(s)
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2+ - oNi (aq) + 2e Ni(s) E = - 0.25 V (Reduction)p
2+ - oCd(s) Cd + 2e E = - 0.40 V (Oxidation)p
2+ 2+Cd(s) + Ni (aq) Cd + Ni(s)p
ocell - 0.25 - (-0.40 ) 0.15
2+
2+
[Cd ] 0.026MQ 6.19
0.00420M[Ni ]
o
cell cell0.0592 VE = E - × log Q (at 298.15 K)
n
vcell
0.05920.15 - log 6.19 0.15 - (0.0296 0.79) 0.13
2
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Practice ProblemPractice Problem 66Construct a Galvanic Cell to determine the pH of Construct a Galvanic Cell to determine the pH of
an UnknownSolutionan UnknownSolutionCompartment #Compartment #11 ±± Cathode consisting of StandardCathode consisting of Standard
Hydrogen Electrode based onHydrogen Electrode based onHH22/H/H
++
half half--cell reaction at standardcell reaction at standardconditions (Hconditions (H++ ±± 11 M; HM; H22 ±± 11
atm)atm)Compartment #Compartment #22 ±± Anode consisting of sameAnode consisting of same
apparatusapparatusbut dipping into a solution of but dipping into a solution of
unknown Hunknown H++(pH)(pH)
Although EAlthough Eoocellcell == 00, the individual half , the individual half--cells differ incells differ in[H[H++]] and Eand Ecellcell is not =is not = 00
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+ -22H (aq; 1 M) + 2e H (g; 1 atm) [cathode; reduction]p
+ -2H (g; 1atm) 2H (aq; unknown) + 2e [anode; oxidation]p
+ +2H (aq; 1M) 2H (aq; unknown)p cell ?Con¶t
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Practice ProblemPractice Problem 77What is the pH of the test solution when EWhat is the pH of the test solution when Ecellcell ==
00
..612
612
V atV at25
25
oo
C?C?PtPt||HH22(g)((g)(1 1 atm)|Hatm)|H++(test sol¶n)AgCl(s),Ag(s)|Cl(test sol¶n)AgCl(s),Ag(s)|Cl--
((22..80 80 M)M)
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+ - o2 (g; 1atm) 2 (a ; u nk no n) + 2e E = 0.0 [ano e; o i ation]p
- - oAgCl(s) + e Ag(s) + Cl (aq) E = 0.22V [cathode; reduction]p
- +2 (g;1atm) + 2AgCl(s) 2Ag(s) + 2Cl (aq) + 2 (aq, unkno n)p
ocellE = 0.22 V - 0.0 V = 0.22 V
- 2 + 2
o o
cell cell cell 2
0.0592 V 0.0592 V [Cl ] [ ]E = E - × log Q = E - × log
n 2 (g;1atm)
- 2 + 2
2
[Cl ] [ ]Q =
(g;1atm)
2 + 2
+
0.0592 V (2.80) [ ]0.612 V = 0.22 V - × log = 0.22 V - 0.0296(log(7.84) + 2log[ ])
2 1
+-log[ ] = p = 7.07
+-0.0296 2log[ ] = 0.612 - 0.22 + 0.0296 0.894v v
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Summary EquationsSummary Equations 11
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v-
-
chargeCharge = moles of e
mol e
Charge = nF F - Fara ay Constant
-
96,485F = ( - coulomb, SI unit of c arge)
mol e
o o ocell cat ode (reduction) anode (oxidation)E = E - E
p + - oAg(s) Ag (a ) + e E = 0.80 (ano e - o i ation)
p+ - o
22H (a ) + 2e H (g) E = 0.00 (catho e - re uction)
p+ +22Ag(s) + 2H (aq) H (g) + 2Ag (aq)
o
cellE = 0.00 V - 0.80 V = - 0.80 V
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Summary EquationsSummary Equations 22
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( vcell
G = - E charge
cellG = - E n( v
o o
cell = - E n F (com onents in standard states)v
o = - lnK
o
cell-E n F = - lnK v
oG = G + RTlnQ
ocell cell-n E = - n E + RTlnQv v
ocell cell
RTlnQE = E - (Nernst E uation)
n
o
cell cell
0.0592E = E - log (at 298.15 K )
n
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Goodbye