L1 364 General Electrochemistry

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ElectrochemistryElectrochemistryCHCH 364364

4/13/2011 1

Prof. Dr. Saeed REFAEYChemistry DepartmentFaculty of ScienceMinia University

Welcome



Redox Reactions andRedox Reactions and

Electrochemical CellsElectrochemical Cells



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

4/13/2011 3



4/13/2011 4



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

4/13/2011 5



4/13/2011 6



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

4/13/2011 7



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



Practice

Problem

Practice

Problem 11

Balance equation with Oxidation NumberBalance equation with Oxidation Number

method:method:

4/13/2011 9

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



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

4/13/2011 10



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

4/13/2011 11

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





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

4/13/2011 13

- 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



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

4/13/2011 14

- + -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



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

4/13/2011 15



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

4/13/2011 18



Galvanic CellsGalvanic Cells



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

4/13/2011 20

2 -(a ) 2e (s) [re ctio ]p

2 -Z (s) Z (a ) 2e [o i atio ]p

2 2Z (s) (a ) Z (s)p



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

4/13/2011 21



4/13/2011 22

Zn Cu Galvanic Cell



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



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



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

4/13/2011 25



4/13/2011 26

Galvanic Cell withInactive Graphite

Electrodes



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

4/13/2011 27

2 2HgO(s) + n(s) + H O n(OH) + Hg(l)p



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:

4/13/2011 28

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



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:

4/13/2011 29

+ - -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



Cell PotentialsCell Potentials



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

4/13/2011 31



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

4/13/2011 32

cell

Work(w) = Potential difference (E) Charge

w = E charge

v

v



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

4/13/2011 33



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

4/13/2011 34

cellE < 0 for a nonspontaneous process

cellE = 0 for an equilibrium process

cellE > 0 for a spontaneous process



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

4/13/2011 35

1 coulomb1 ampere = 1A = 1 C/ s

second

1 J 1 J1 volt = 1 C =

C V



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

4/13/201136

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



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

4/13/201137

2 2 ocellZ (s) (a ; 1 ) Z (a ; 1 ) (s) 1.10p



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:

4/13/201138

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



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

4/13/2011 39

o o ocell cat o e (re ctio ) a o e (o i atio )( - ) > 0

o o ocell copper i c-



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]]

4/13/2011 40

+ -

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



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

4/13/2011 41

+ - o2 refere ce2 (a ; 1 ) + 2e (g); 1 atm) 0.00



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



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



Free Energy and ElectricalFree Energy and ElectricalWorkWork



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

4/13/2011 45

cell

Work(w) = Potential difference (E) Charge

w = E charge

v

v



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



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



4/13/2011 48



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:

4/13/2011 49

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



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

4/13/2011 50

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



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)

4/13/2011 51

- 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



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)

4/13/2011 52

+ - + 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



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)

4/13/2011 53

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



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)

4/13/2011 54

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



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

4/13/2011 55

+ -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





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)

4/13/2011 57

+ - 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



Summary EquationsSummary Equations 11

4/13/2011 58

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



Summary EquationsSummary Equations 22

4/13/2011 59

( 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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L1 364 General Electrochemistry

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