electrochemistry

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ElectrochemistryElectrochemistry

Half-reactionsHalf-reactions

ElectrodesElectrodes

CellsCells

Reduction PotentialsReduction Potentials

The Electrochemical SeriesThe Electrochemical Series

Cell Potentials and Cell Potentials and Thermodynamic FunctionsThermodynamic Functions

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Some TermsSome Terms

GalvanicGalvanic or or voltaic cellvoltaic cell Electrolytic cellElectrolytic cell OxidationOxidation (what happens to the (what happens to the reducing agentreducing agent)) ReductionReduction (what happens to the (what happens to the oxidizing oxidizing

agentagent)) Half ReactionsHalf Reactions Redox CoupleRedox Couple AnodeAnode CathodeCathode Standard Electrode PotentialStandard Electrode Potential

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Exercise E6.1Exercise E6.1

Identify the species that have undergone Identify the species that have undergone oxidation and reduction in the reaction oxidation and reduction in the reaction CuS(s) + O CuS(s) + O22(g) (g) Cu(s) + SO Cu(s) + SO22(g) (g)

Solution: Solution: The The Cu(II) on the left is Cu(II) on the left is reducedreduced from the +2 to the 0 oxidation from the +2 to the 0 oxidation state. The state. The SS2-2- on the left has been on the left has been oxidizedoxidized from the -2 to the +4 oxidation from the -2 to the +4 oxidation state. The state. The O on the left has been O on the left has been reducedreduced from the 0 to the -2 oxidation from the 0 to the -2 oxidation state. state.

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Exercise E6.2Exercise E6.2 Express the formation of HExpress the formation of H22O from HO from H22 and and

OO22 in acid solution as the difference of two in acid solution as the difference of two reduction half-reactions.reduction half-reactions.

Solution:Solution: The overall reaction is The overall reaction is 2 H 2 H22(g) + O(g) + O22(g)(g) 2 H2 H22O(l)O(l)

Oxygen is reduced according to OOxygen is reduced according to O22 2 H2 H22O and O and in acid solution this is balanced by adding Hin acid solution this is balanced by adding H++ and and ee to getto get OO22(g) + 4 (g) + 4 HH++(aq) + 4e(aq) + 4e 2 H2 H22O(l)O(l)

The reaction to be subtracted from this is The reaction to be subtracted from this is 4 H4 H++(aq) + 4 e(aq) + 4 e 2 H2 H22(g)(g)

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Exercise E6.3Exercise E6.3 Express the oxidation of NADH (nicotinamide Express the oxidation of NADH (nicotinamide

adenine dinucleotide, which participates in the adenine dinucleotide, which participates in the chain of oxidations that constitutes respiration) to chain of oxidations that constitutes respiration) to NADNAD++ by oxygen, when the latter is reduced to by oxygen, when the latter is reduced to HH22OO2 2 in aqueous solution, as the difference of in aqueous solution, as the difference of two reduction half-reactions.two reduction half-reactions.

Solution:Solution: O O22 is the oxidizing agent; is the oxidizing agent; therefore Otherefore O22 is reduced: is reduced: OO22 + 2 H+ 2 H++ + 2 e + 2 e H H22OO22

The reverse oxidation reaction to be The reverse oxidation reaction to be subtracted from this is then subtracted from this is then NADNAD++ + H + H++ + 2 e + 2 e NADHNADH

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Reactions at electrodesReactions at electrodes Left: Galvanic Left: Galvanic

cell. Electrons cell. Electrons are deposited on are deposited on the anode (so it the anode (so it is neg) and is neg) and collected from collected from the cathode (so it the cathode (so it is positive)is positive)

Right: Electro-Right: Electro-lytic cell. Elec-lytic cell. Elec-trons are forced trons are forced out of the anode out of the anode (positive) and (positive) and into the cathode into the cathode (negative)(negative)

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Types of ElectrodesTypes of Electrodes

(a) metal/metal (a) metal/metal ion electrodeion electrode

(b) metal/ (b) metal/ insoluble salt insoluble salt electrodeelectrode

(c) gas electrode(c) gas electrode (d) redox (d) redox

electrodeelectrode

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Write the half-reaction and the reaction Write the half-reaction and the reaction quotient for a chlorine-gas electrode.quotient for a chlorine-gas electrode.

Solution:Solution: Written as a reduction, the half- Written as a reduction, the half-reaction at the chlorine electrode is reaction at the chlorine electrode is ½½ ClCl22(g) + e(g) + e Cl Cl(aq)(aq)

And the reaction quotient then becomes And the reaction quotient then becomes Q = a(ClQ = a(Cl) / a(Cl) / a(Cl2 2 ))½½

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Types of Electrode Types of Electrode (continued)(continued)

EElleeccttrrooddee TTyyppee DDeessiiggnnaattiioonn

MMeettaall//mmeettaall iioonn MM((ss))||MM++((aaqq))

GGaass PPtt((ss))||XX22((gg))||XX++((aaqq)) oorrPPtt((ss))||XX22((gg))||XX--((aaqq))

MMeettaall//iinnssoolluubbllee ssaalltt MM((ss))||MMXX((ss))||XX--((aaqq))

RReeddooxx PPtt((ss))||MM++((aaqq)),,MM22++((aaqq))

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Types of Electrode Types of Electrode (continued)(continued)

RReeddooxx CCoouuppllee HHaallff rreeaaccttiioonn

MM++//MM MM++((aaqq)) ++ ee-- MM((ss))

XX++//XX22

XX22//XX--XX++((aaqq)) ++ ee-- ½½ XX22((gg))½½ XX22((gg)) ++ ee-- XX--((aaqq))

MMXX//MM//XX-- MMXX((ss)) ++ ee-- MM((ss)) ++ XX--((aaqq))

MM22++//MM++ MM22++((aaqq)) ++ ee-- MM++ ((ss))

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Write the half-reaction and the reaction Write the half-reaction and the reaction quotient for the calomel electrode, Hg(l)|quotient for the calomel electrode, Hg(l)|HgHg22ClCl22(s)|Cl(s)|Cl(aq), in which mercury(I) chloride (aq), in which mercury(I) chloride (calomel) is reduced to mercury metal in the (calomel) is reduced to mercury metal in the presence of chloride ions. This electrode is a presence of chloride ions. This electrode is a component of instruments used to measure component of instruments used to measure pH, as explained later.pH, as explained later.

Solution: The reduction of HgSolution: The reduction of Hg22ClCl22 to Hg is to Hg is given by given by HgHg22ClCl22((ss)) + 2 e + 2 e 2 Hg2 Hg((ll)) + 2 Cl + 2 Cl((aqaq))

Notice that everything except the chloride ion Notice that everything except the chloride ion is a pure solid or liquid. Therefore only the is a pure solid or liquid. Therefore only the chloride appears in the reaction quotient. chloride appears in the reaction quotient. Q = aQ = a((ClCl22

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Varieties of CellVarieties of Cell

The two basic types are The two basic types are concentration concentration cellscells and and chemical cellschemical cells. .

Concentration cells are either Concentration cells are either electrolyte electrolyte concentration cellsconcentration cells, where the electrode , where the electrode compartments are identical except for the compartments are identical except for the concentrations of the electrolytes, or concentrations of the electrolytes, or electrode concentration cellselectrode concentration cells, in which , in which the electrodes themselves have different the electrodes themselves have different concentrations, such as amalgams or concentrations, such as amalgams or gas electrodes at different pressures.gas electrodes at different pressures.

Most cells are chemical cells.Most cells are chemical cells.

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Two Versions of the Daniell Two Versions of the Daniell CellCell

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Constructing a Daniell Cell Constructing a Daniell Cell

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Two Practical CellsTwo Practical Cells

At left is At left is a a pri-pri-mary mary cell cell (used (used once once only). only).

At right At right is a is a secon-secon-dary dary cell cell (may be (may be re-re-charged)charged)

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Cell NotationCell Notation

In the version of the Daniell cell with the In the version of the Daniell cell with the porous pot there is a liquid junction. This is porous pot there is a liquid junction. This is denoted as Zn(s)|ZnSOdenoted as Zn(s)|ZnSO44(aq):CuSO(aq):CuSO44(aq)|(aq)|Cu(s)Cu(s)

When the liquid junction potential has been When the liquid junction potential has been essentially eliminated by use of a salt bridge essentially eliminated by use of a salt bridge the Daniell cell is denoted the Daniell cell is denoted Zn(s)|ZnSO Zn(s)|ZnSO44(aq)||CuSO(aq)||CuSO44(aq)|Cu(s)(aq)|Cu(s)

Other punctuation in cell notations includes a Other punctuation in cell notations includes a comma to separate two species present in comma to separate two species present in the same phase.the same phase.

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Cells with a Common Cells with a Common ElectrolyteElectrolyte

A cell in which the A cell in which the anode is a anode is a hydrogen electrode hydrogen electrode and the cathode is a and the cathode is a silver-silver chloride silver-silver chloride electrode is denoted electrode is denoted Pt|HPt|H22((gg)|H)|H++((aqaq), ), ClCl((aqaq) | ) | . . AgCl( AgCl(ss)|Ag()|Ag(ss))

or, without the comma, or, without the comma, Pt|HPt|H22((gg))|HCl|HCl((aqaq))|AgCl |AgCl ((ss))|Ag|Ag((ss))

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Give the notation for a cell in which the oxidation Give the notation for a cell in which the oxidation of NADH by oxygen could by studied.of NADH by oxygen could by studied.

Solution:Solution: Refer to Exercise E6.3, where the half- Refer to Exercise E6.3, where the half-reactions are Oreactions are O22 + 2 H + 2 H++ + 2 e + 2 e HH22OO22 and NADand NAD++ + H + H++ + 2 e + 2 e NADH NADH

Putting the oxidation half-reaction into conven-Putting the oxidation half-reaction into conven-tional cell notation, as written on the left, it tional cell notation, as written on the left, it becomes Pt|NADH(aqbecomes Pt|NADH(aq)), NAD, NAD++(aq(aq)), H, H++(aq(aq))

The reduction half-reaction is written The reduction half-reaction is written Pt|O Pt|O22((gg))|H|H++((aqaq),),HH22OO22((aqaq))

Putting these together, Putting these together, Pt|NADH(aq), NADPt|NADH(aq), NAD++(aq), (aq), HH++(aq)|| H(aq)|| H22OO22(aq), H(aq), H++(aq)|O(aq)|O22(g) | Pt(g) | Pt

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Write the chemical equation for the cell in Write the chemical equation for the cell in Exercise E6.6.Exercise E6.6.

Solution:Solution: We can again refer to Exercise E6.3, We can again refer to Exercise E6.3, where the half-reactions are where the half-reactions are O O22 + 2 H + 2 H++ + 2 e + 2 e HH22OO22 and NADand NAD++ + H + H++ + 2 e + 2 e NADH NADH

Subtracting the oxidation half-reaction from the Subtracting the oxidation half-reaction from the reduction half-reaction gives reduction half-reaction gives NADHNADH((aqaq)) + O + O22((gg)) + H + H++((aqaq)) NADNAD+ +

((aqaq)) + + . . HH22OO2 2 ((aqaq))

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The Cell Potential The Cell Potential

Since w = Since w = G G = = work work output, and output, and since electrical since electrical work output = work output = (charge) x (vol-(charge) x (vol-tage) = tage) = FE ,FE ,

G =G = FEFE A spontaneous A spontaneous

rxn has a neg.rxn has a neg. G and a pos. EG and a pos. E• E is E is intensiveintensive

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The Nernst EquationThe Nernst Equation

SubstitutingSubstituting G = G = FE intoFE into GG == GGoo + RT ln Q+ RT ln Q givesgives

FE FE == FEFEoo + RT ln Q + RT ln Q or or

E = EE = Eoo RT/ RT/FF ln Qln Q At 25At 25ooC, RT/F = 0.02569 v = 25.69 mVC, RT/F = 0.02569 v = 25.69 mV A practical form of the NerA practical form of the Ner nst equation isnst equation is E = EE = Eoo (25.69 mV/ (25.69 mV/ln Qln Q At equilibrium, E = 0 and Q = K, so At equilibrium, E = 0 and Q = K, so

ln K =ln K = FEFEoo/RT/RT = = EEoo/(25.69 mV)/(25.69 mV)

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Concentration CellsConcentration Cells

A concentration cell derives its potential from A concentration cell derives its potential from the difference in concentration between the the difference in concentration between the right and left sides.right and left sides.

M|MM|M++(aq, L)||M(aq, L)||M++(aq, R)|M(aq, R)|M The cell reaction is MThe cell reaction is M++(aq, R) (aq, R) MM++(aq, L)(aq, L) Using the Nernst equation, E = EUsing the Nernst equation, E = Eoo - -

(RT/nF) ln Q(RT/nF) ln Q But EBut Eoo = 0 ! = 0 ! (Do you see why?) (Do you see why?) ln Q = ln Q = aaLL//aaRR

So for a conc. cell,So for a conc. cell, E = - (RT/ E = - (RT/F) ln (F) ln (aaLL//aaRR))

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Standard Electrode Standard Electrode PotentialsPotentials

EEoocellcell can be found from can be found from rrGGoo using the using the

equationequation rrGGoo = - = -FEFEoo

• (or in general,(or in general, rrG = -G = -FEFE))

But EBut Eoocellcell can also be found from values of E can also be found from values of Eoo

for the two electrodes involved.for the two electrodes involved. Standard electrode potentials are given in Standard electrode potentials are given in

Table 6.1, p. 216.Table 6.1, p. 216.• Since it is impossible to measure the potential of Since it is impossible to measure the potential of

one electrode alone, these are all relative to H.one electrode alone, these are all relative to H.

Using Table 6.1,Using Table 6.1, EEoocellcell = E = Eoo

R R - E - EooLL

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E and SpontaneityE and Spontaneity EEoo > 0 goes with K > 1, which indicates a > 0 goes with K > 1, which indicates a

spontaneous reactionspontaneous reaction from reactants in from reactants in their standard state to products in their standard state to products in their standard statetheir standard state..

However, the direction of a reaction can However, the direction of a reaction can sometimes be reversed by judicious sometimes be reversed by judicious manipulation of the concentrations of manipulation of the concentrations of product and reactant species. (That is, product and reactant species. (That is, by altering Q.)by altering Q.)

Any given reaction proceeds left to Any given reaction proceeds left to right when E right when E (not E(not Eoo)) > 0 > 0

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Is the equilibrium constant for the Is the equilibrium constant for the displacement of copper by zinc greater or displacement of copper by zinc greater or smaller than 1?smaller than 1?

Solution:Solution: For this cell reaction For this cell reaction, , R (red’n):R (red’n): Cu Cu2+ 2+ + 2e + 2e Cu Cu E Eoo

RR = +0.34 v = +0.34 v L (ox’n):L (ox’n): Zn Zn2+ 2+ + 2e + 2e Zn Zn E Eoo

LL = = 0.76 v0.76 v

EEoocellcell = E = Eoo

R R EEooLL = +0.34 v = +0.34 v 0.76 v0.76 v= =

+1.10 v+1.10 v Since ESince Eoo is positive, the reaction is is positive, the reaction is

spontaneous, and spontaneous, and K>1K>1. .

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Calculate the equilibrium constant for the rxn SnCalculate the equilibrium constant for the rxn Sn2+2+(g) (g) + Pb(s)+ Pb(s) Sn(s) + PbSn(s) + Pb2+2+(s) (s) at 25at 25ooCC

Solution:Solution: For this cell reaction, For this cell reaction, R (red’n): R (red’n): Sn Sn2+ 2+ + 2e + 2e Sn Sn E Eoo

RR = = 0.14 V 0.14 V L (ox’n):L (ox’n): Pb Pb2+ 2+ + 2e + 2e Pb Pb E Eoo

LL = = 0.13 V0.13 V


R R EEooLL = = 0.14 V 0.14 V 0.13 V0.13 V= =

0.01 V0.01 V ln K =ln K = EEoo/(25.69 mV) = 2 (/(25.69 mV) = 2 (10 mV)/(25.69 mV) = 10 mV)/(25.69 mV) =

0.780.78

K = K = 0.460.46

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Exercise E6.10Exercise E6.10 What is the equilibrium constant for the reduction What is the equilibrium constant for the reduction

of riboflavin with rubredoxin in the reaction of riboflavin with rubredoxin in the reaction rib riboxox + rub + rubredred rib ribredred + rub + ruboxox given that at pH = 7 the reduction potential for given that at pH = 7 the reduction potential for rub-redoxin is rub-redoxin is 0.06 V and that for riboflavin is 0.06 V and that for riboflavin is 0.21 V 0.21 V

Solution:Solution: For this cell reaction, For this cell reaction, R (red’n): rib R (red’n): riboxox+ 2 H+ 2 H+ + + 2e + 2e ribribredred + H + H22OO L (ox’n): rub L (ox’n): ruboxox + 2e + 2e ribribredred


R R EEooLL = = 0.21 V 0.21 V 0.06 V0.06 V= = 0.15 V0.15 V

ln K = ln K = EEoo//25.69 mV25.69 mV= 2 = 2 150 mV150 mV//25.69 25.69 mVmV= = 11.6811.68

K = K = 8.5 8.5 xx 10 10-6-6 , making reactants strongly favored , making reactants strongly favored

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The Hydrogen Electrode The Hydrogen Electrode and pHand pH

The potential of a hydrogen electrode is directly The potential of a hydrogen electrode is directly proportional to the pH of the solution. Consider proportional to the pH of the solution. Consider the calomel-hydrogen cell Hgthe calomel-hydrogen cell Hgll| Hg| Hg22ClCl22ss| |

ClClaqaq|| H|| H++aqaq|H|H22gg|Pt , for which the cell |Pt , for which the cell

reaction is Hgreaction is Hg22ClCl22ss + H + H22gg 2 Hg 2 Hgll+ 2 + 2 ClClaqaq+ 2 H+ 2 H++aqaq

If the HIf the H22gg is at standard pressure and the is at standard pressure and the chloride ion activity is constant and incorporated chloride ion activity is constant and incorporated into Einto Eo†o†, the Nernst equation becomes E = E, the Nernst equation becomes E = Eo†o† - - RT/FRT/F ln a ln aHH++= E= Eo†o† + + RT ln 10/FRT ln 10/FxxpH = pH = EEo†o† + + 59.15 mV59.15 mVxxpH pH

So the pH can be determined from the cell So the pH can be determined from the cell potential.potential.

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Exercise E6.11Exercise E6.11 Calculate the biological standard potential of Calculate the biological standard potential of

the half-reaction Othe half-reaction O22gg + 4 H + 4 H++aqaq + 4 e + 4 e HH22OOll at 25 at 25ooC given its value of +1.23 V under C given its value of +1.23 V under thermo-dynamic standard conditions.thermo-dynamic standard conditions.

Solution:Solution: Just as biochemical values of Just as biochemical values of GGoo can can be calculated from the thermodynamic be calculated from the thermodynamic GGoo values, it is also possible to calculate values, it is also possible to calculate biochemical Ebiochemical Eoo values from E values from Eoo values. values.

The relationship is EThe relationship is Eoo = E = Eoo RT /RT /FFln 10ln 10n n xx pHpH= E= Eo o 0.05915 V /0.05915 V /7n7n• = the number of electrons; n = the number of H= the number of electrons; n = the number of H++

ionsions EEoo = 1.23 V = 1.23 V 0.05915 V / 40.05915 V / 42828 = 1.23 V - = 1.23 V -

0.41 V0.41 V = = 0.82 V0.82 V

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Exercise E6.12Exercise E6.12 What range should a voltmeter have to display What range should a voltmeter have to display

changes of pH from 1 to 14 at 25changes of pH from 1 to 14 at 25ooC if it is C if it is arranged to give a reading of 0 when pH = 7 ?arranged to give a reading of 0 when pH = 7 ?

Solution:Solution: Each pH unit changes the cell Each pH unit changes the cell potential by potential by RT ln 10/FRT ln 10/F= 0.05916 V. An = 0.05916 V. An increase in pH makes the measured potential increase in pH makes the measured potential more negative and a decrease in pH makes more negative and a decrease in pH makes the measured potential more positive. the measured potential more positive.

To go from 7 to 14 changes the value of E by To go from 7 to 14 changes the value of E by 7 7 xx 0.05916 V = 0.05916 V = 0.414 V0.414 V

To go from 7 to 1 changes the value of E by To go from 7 to 1 changes the value of E by 6 6 xx 0.05916 V = 0.05916 V = + 0.355 V+ 0.355 V

Therefore, a range of Therefore, a range of 0.77 V0.77 V would be needed. would be needed.

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The Electrochemical SeriesThe Electrochemical Series

A species with a low standard reduction A species with a low standard reduction potential has a thermodynamic tendency to potential has a thermodynamic tendency to reduce a species with a high standard reduce a species with a high standard reduction potential.reduction potential.• More briefly, More briefly, low reduces highlow reduces high..• Equivalently, Equivalently, high oxidizes lowhigh oxidizes low..

This is the basis for the This is the basis for the activity seriesactivity series of metals.of metals.

Other couples can also be fitted into the activity Other couples can also be fitted into the activity series.series.

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Activity Series of MetalsActivity Series of Metals

potassiumsodium

potassiumsodium

calciumcalcium

magnesiumaluminum

zincchromium

magnesiumaluminum

zincchromium

ironnickel

tinlead

ironnickel

tinlead

coppersilver

platinumgold

coppersilver

platinumgold

incr

easi

ng

rea

ctiv

ity

React violently with cold water

React slowly with cold water

React very slowly with steambut quite reactive in acid

React moderately with highlevels of acid

< HYDROGEN comes here

Unreactive in acid

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Does acidified dichromate (CrDoes acidified dichromate (Cr22OO772-2-) have ) have

a thermodynamic tendency to oxidize a thermodynamic tendency to oxidize mercury to mercury(I)?mercury to mercury(I)?

Solutions:Solutions: The relevant E The relevant Eoo’s are +1.33 V ’s are +1.33 V for Crfor Cr22OO77

2-2-|Cr|Cr3+3+ and +0.79 V for Hg and +0.79 V for Hg222+2+|Hg|Hg

Since high oxidizes low, the answer isSince high oxidizes low, the answer is YesYes, Cr, Cr22OO77

2-2- will oxidize Hg to Hg will oxidize Hg to Hg222+2+

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Can lead displace (a) iron(II) ions, Can lead displace (a) iron(II) ions, (b) copper(II) ions from solutions at 298 K?(b) copper(II) ions from solutions at 298 K?

Solutions:Solutions: The relevant E The relevant Eoo’s are:’s are: -0.13 V for Pb-0.13 V for Pb2+2+|Pb|Pb -0.44 V for Fe-0.44 V for Fe2+2+|Fe|Fe +0.34 V for Cu+0.34 V for Cu2+2+|Cu|Cu (a) Lead is NOT lower than iron(II), so the (a) Lead is NOT lower than iron(II), so the

answer is answer is NONO.. (b) Lead IS lower than copper(II), so the (b) Lead IS lower than copper(II), so the

answer is answer is YESYES

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Thermodynamic Functions Thermodynamic Functions and E’s and E’s

Because of the relationshipBecause of the relationship• rrG = - G = - FEFE

• rrGGoo = - = - FEFEoo

It is possible to obtain the thermodynamic It is possible to obtain the thermodynamic value of the standard reaction Gibbs value of the standard reaction Gibbs energy by measuring cell potentials.energy by measuring cell potentials.

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Estimate the standard reaction Gibbs Estimate the standard reaction Gibbs energy of Agenergy of Ag++

(aq) + ½ H(aq) + ½ H22(g)(g) HH++(aq) + Ag(s) given that (aq) + Ag(s) given that

the standard potential of the cell Hthe standard potential of the cell H22| |

HH++||Ag||Ag++|Ag is E|Ag is Eoo = +0.7996 V = +0.7996 V Solution:Solution: rrGGoo = - = - FEFEoo

rrGGoo = - (1)(96.485 kC)(+0.7996 V) = - (1)(96.485 kC)(+0.7996 V)

= = -77.15 kJ-77.15 kJ

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Finding EFinding Eoo via via GGoo

EEoo’s for two half-reactions can be combined ’s for two half-reactions can be combined directly as long as the number of electrons is the directly as long as the number of electrons is the same in each half-reaction and the electrons same in each half-reaction and the electrons cancel out when the half-reactions are cancel out when the half-reactions are combined.combined.

EEoo’s cannot be directly combined for half-’s cannot be directly combined for half-reactions in which the electrons do not cancel reactions in which the electrons do not cancel out.out.• For instance, EFor instance, Eoo for Cu for Cu2+2+|Cu|Cu++ cannot be found by cannot be found by

directly combining Edirectly combining Eoo’s for Cu’s for Cu2+2+|Cu and Cu|Cu and Cu++|Cu .|Cu . In these cases, EIn these cases, Eoo’s may be converted to ’s may be converted to GGoo’s ’s

for and the for and the GGoo’s for then directly combined.’s for then directly combined.

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Other Thermodynamic Other Thermodynamic Values - Values - rrHHoo

The van’t Hoff equation (Eq. 3.25) may The van’t Hoff equation (Eq. 3.25) may be modified to give be modified to give rrHHo o if Eif Eoo is measured is measured

at two different temperatures.at two different temperatures.• Substitute -Substitute -rrGGoo/RT for ln K/RT for ln K

• Substitute -Substitute -FEFEoo for for rrGGoo

– or do it in one step by replacing or do it in one step by replacing FEFEoo /RT for ln K /RT for ln K

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Other Thermodynamic Other Thermodynamic Values - Values - rrSSoo

With expressions for both With expressions for both rrGGo o and and rrHHo o

the entropy can also be obtained.the entropy can also be obtained. rrSSo o is given by the relationis given by the relation

EEoo(T(T) - E) - Eoo(T) (T) rrSSoo = = F xF x . . . . TT- T - T

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Predict the standard potential of the Harned cellPredict the standard potential of the Harned cell Pt|Pt|HH22(g)|HCl(aq)|AgCl(s)|Ag(s)(g)|HCl(aq)|AgCl(s)|Ag(s) at 303 K from at 303 K from

tables of thermodynamic data at 298 K.tables of thermodynamic data at 298 K. Solution:Solution: The cell reaction is AgCl(s) The cell reaction is AgCl(s)

+ ½ H+ ½ H22(g)(g) HH++(aq) + Cl(aq) + Cl(aq) + Ag(s) (aq) + Ag(s)

From tables of thermodynamic data, From tables of thermodynamic data, rrGGoo == (131.23 131.23

+ 109.79) kJ = -21.44 kJ and + 109.79) kJ = -21.44 kJ and rrSSoo == 0 + 56.5 + 42.55 0 + 56.5 + 42.55

96.2 96.2 ½(130.684) J/K = -62.5 J/K; at 298 K, E½(130.684) J/K = -62.5 J/K; at 298 K, Eo o ==

rrGGooF F = = 0.2222 v. 0.2222 v.

EEoo(303) = E(303) = Eoo(303) + (303) + rrSSo o (T(T- T)- T)F = 0.2222 v + (-62.5 F = 0.2222 v + (-62.5

J/K)(303-298)K/(96,485 C) = J/K)(303-298)K/(96,485 C) = +0.2190 v+0.2190 v

electrochemistry

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m2 aq e

reaction quotient

cl2g e

calomel electrode

chlorine electrode

overall reaction

h2olthe reaction

reverse oxidation reaction