Constructing Electrochemical Cells

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http:\\aliasadipour.kmu.ac.ir91100222Electrochemistryhttp:\\aliasadipour.kmu.ac.ir91100233ElectrochemistryAll of Chemical reactins are related to ELECTRONS

Redox reactions

Voltaic or Galvanic cellsElectrochemical cells

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Electric power conversion in electrochemistry Chemical ReactionsElectric PowerPower consumptionPower generationElectrolysisGalvanic cellshttp:\\aliasadipour.kmu.ac.ir91100255ElectrochemistryConduction1)Metalic2)Electrolytic

Temprature Motion of ions Resistance

1C=1AS /// 1J=1CV

------------------------------------------http:\\aliasadipour.kmu.ac.ir9110026battery+-powersourcee-e-IonsChemical change

(-)(+)Aqueous NaClInterionic attractions................................Ions Solvation .Solvent viscosity ..

Na+Cl-H2OElectrolytic conductionIon-Ion Attr.Ion- Solvent Attr.SolventSolvent Attr.

TempratureAttractions & Kinetic energyConduction

http:\\aliasadipour.kmu.ac.ir911002Conduction Ions mobility7battery+-inertelectrodespowersourcevessele-e-conductivemediumElectrolytic CellConstructionhttp:\\aliasadipour.kmu.ac.ir9110028+-batteryNa (l)electrode half-cellelectrode half-cellMolten NaClNa+Cl-Cl-Na+Na+Na+ + e- Na2Cl- Cl2 + 2e-Cl2 (g) escapesObserve the reactions at the electrodesNaCl (l)(-)Cl-(+)http:\\aliasadipour.kmu.ac.ir9110029+-batterye-e-NaCl (l)(-)(+)cathodeanodeMolten NaClNa+Cl-Cl-Cl-Na+Na+Na+ + e- Na2Cl- Cl2 + 2e-cationsmigrate toward (-) electrodeanionsmigrate toward (+) electrodeAt the microscopic levelhttp:\\aliasadipour.kmu.ac.ir91100210Molten NaCl Electrolytic Cellcathode half-cell (-)REDUCTION Na+ + e- Na

anode half-cell (+)OXIDATION2Cl- Cl2 + 2e-

overall cell reaction2Na+ + 2Cl- 2Na + Cl2

X 2Non-spontaneous reaction!http:\\aliasadipour.kmu.ac.ir91100211What chemical species would be present in a vessel of aqueous sodium chloride, NaCl (aq)?Na+Cl-H2OWill the half-cell reactions be the same or different?http:\\aliasadipour.kmu.ac.ir911002Water Complications in ElectrolysisIn an electrolysis, the most easily oxidized and most easily reduced reaction occurs.

When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown.

ElectrodeIons ...Anode RxnCathode Rxn EPt (inert)H2O H2O(l)+ 2e- g H2(g)+ 2OH-(aq) -0.83 VH2O 2 H2O(l) g 4e- + 4H+(g) + O2(g) -1.23 VNet Rxn Occurring:2 H2O g 2 H2(g)+ O2 (g) E = - 2.06 V

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http:\\aliasadipour.kmu.ac.ir911002CHM 102Sinex1314battery+-powersourcee-e-NaCl (aq)(-)(+)cathodedifferent half-cellAqueous NaClanode2Cl- Cl2 + 2e-Na+Cl-H2OWhat could be reduced at the cathode?http:\\aliasadipour.kmu.ac.ir2H2O + 2e- H2 + 2OH-91100215Aqueous NaCl Electrolysispossible cathode half-cells (-)REDUCTION Na+ + e- Na

2H2O + 2e- H2 + 2OH-

possible anode half-cells (+)OXIDATION2Cl- Cl2 + 2e-

2H2O O2 + 4H+ + 4e-

overall cell reaction2Cl- + 2H2O H2 + Cl2 + 2OH-

http:\\aliasadipour.kmu.ac.ir91100216Aqueous CuCl2 Electrolysispossible cathode half-cells (-)REDUCTION Cu2+ + 2e- Cu

2H2O + 2e- H2 + 2OH-

possible anode half-cells (+)OXIDATION2Cl- Cl2 + 2e-

2H2O O2 + 4H+ + 4e-

overall cell reaction Cu2+ + 2Cl- Cu(s) + Cl2(g)

http:\\aliasadipour.kmu.ac.ir91100217Aqueous Na2SO4 Electrolysispossible cathode half-cells (-)REDUCTION Na+ + e- Na

[2H2O + 2e- H2 + 2OH- ]

possible anode half-cells (+)OXIDATION SO42- S4O82_ + 2e-

2H2O O2 + 4H+ + 4e-

overall cell reaction 6H2O 2H2 + O2 +4H+ + 4OH-

http:\\aliasadipour.kmu.ac.ir911002218Faradays LawThe mass deposited or eroded from an electrode depends on the quantity of electricity.Quantity of electricity = coulomb (Q)Q = Itcoulombcurrent in amperes (amp)time in secondshttp:\\aliasadipour.kmu.ac.ir91100219e-Ag+AgFor every electron, an atom of silver is plated on the electrode.Ag+ + e- AgElectrical current is expressed in terms of the ampere, which is defined as that strength of current which, when passed thru a solution of AgNO3 (aq) under standard conditions, will deposit silver at the rate of 0.001118 g Ag/sec 1 amp = 0.001118 g Ag/sechttp:\\aliasadipour.kmu.ac.ir9110021 coulomb = 1 amp-sec = 0.001118 g Ag20Ag+ + e- Ag1.00 mole e- = 1.00 mole Ag = 107.87 g Ag 107.87 g Ag/mole e-0.001118 g Ag/coul= 96,485 coul/mole e-1 Faraday (F )mole e- = Q/F http:\\aliasadipour.kmu.ac.ir91100221A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode.battery-++++---1.0 M Au+31.0 M Zn+21.0 M Ag+Au+3 + 3e- AuZn+2 + 2e- ZnAg+ + e- Age-e-e-e-http:\\aliasadipour.kmu.ac.ir91100222Examples using Faradays Law1)How many grams of Cu will be deposited in 3.00 hours by a current of 4.00 amps?(Cu=64) Cu+2 + 2e- Cu

2)The charge on a single electron is 1.6021 x 10-19 coulomb. Calculate Avogadros number from the fact that 1 F = 96,487 coulombs/mole e-.

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http:\\aliasadipour.kmu.ac.ir9110022521-8 Industrial Electrolysis Processes

Slide 25 of 52http:\\aliasadipour.kmu.ac.ir91100226

http:\\aliasadipour.kmu.ac.ir911002CHM 102Sinex2627

Voltas battery (1800)

Alessandro Volta 1745 - 1827Paper moisturized with NaCl solution

Cu

Znhttp:\\aliasadipour.kmu.ac.ir91100228

Galvanic Cells19.2spontaneousredox reactionanodeoxidationcathodereductionhttp:\\aliasadipour.kmu.ac.ir91100229Cu1.0 M CuSO4Zn1.0 M ZnSO4Salt bridge KCl in agarProvides conduction between half-cellsGalvanic CellConstructionObserve the electrodes to see what is occurring.http:\\aliasadipour.kmu.ac.ir91100230Cu1.0 M CuSO4Zn1.0 M ZnSO4Cu plates out or deposits on electrodeZn electrode erodes or dissolvesCu+2 + 2e- Cucathode half-cellZn Zn+2 + 2e-anode half-cell

Anod-Cathod+What about half-cell reactions?What about the sign of the electrodes?What happened at each electrode?Why?http:\\aliasadipour.kmu.ac.ir911002Compare with Electrolytic cells31+-batterye-e-NaCl (l)(-)(+)Cathode -Anode +Electrolytic cells sign of the electrodes? Na+Cl-Cl-Cl-Na+Na+Na+ + e- Na2Cl- Cl2 + 2e-http:\\aliasadipour.kmu.ac.ir91100232

Electrodes are passive (not involved in the reaction)Olmsted Williamshttp:\\aliasadipour.kmu.ac.ir91100233H2 input1.00 atminert metalHow do we calculate Standard Redox Potentials?We need a standard electrode to make measurements against!The Standard Hydrogen Electrode (SHE)Pt1.00 M H+25oC1.00 M H+1.00 atm H2Half-cell2H+ + 2e- H2EoSHE = 0.0 voltshttp:\\aliasadipour.kmu.ac.ir9110023419.3E0 is for the reaction as written E0red // E0oxThe more positive E0 the greater the tendency for the substance to be reducedThe half-cell reactions are reversibleThe sign of E0 changes when the reaction is reversedChanging the stoichiometric coefficients of a half-cell reaction does not change the value of E0

http:\\aliasadipour.kmu.ac.ir9110023535Cell EMFOxidizing and Reducing Agents

http:\\aliasadipour.kmu.ac.ir91100236Measuring E0red Cu2+& Zn2+

Slide 36 of 52cathodecathodeanodeanodehttp:\\aliasadipour.kmu.ac.ir911002Cu+2 + 2e- CuE=E0redZn Zn+2 + 2e-E=E0ox

-E=E0red

37Cu1.0 M CuSO4Zn1.0 M ZnSO4cathode half-cellCu+2 + 2e- Cuanode half-cellZn Zn+2 + 2e--+Measuring E0 of a cell1.1 voltshttp:\\aliasadipour.kmu.ac.ir91100238What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V Cd is the stronger oxidizerCd will oxidize Cr2e- + Cd2+ (1 M) Cd (s)Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):Cathode (reduction):2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)x 2x 3E0 cell = -0.40 +0.74=0.34 cellE0 = 0.34 V cell19.3http:\\aliasadipour.kmu.ac.irE0 = -0.40 V E0 = 0.74 V 91100239Calculating the cell potential, Eocell, at standard conditionsFe+2 + 2e- Fe Eo = -0.44 vO2 (g) + 2H2O + 4e- 4 OH-Eo = +0.40 v This is spontaneoues corrosion or the oxidation of a metal.Consider a drop of oxygenated water on an iron object

FeH2O with O2Fe Fe+2 + 2e- -Eo = +0.44 v2x2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 vreversehttp:\\aliasadipour.kmu.ac.irFe + O2 (g) + H2O Fe(OH)2 (s)91100240

http:\\aliasadipour.kmu.ac.ir91100241DGo = -nFEocellFree Energy and the Cell PotentialCu Cu+2 + 2e- Eo = - 0.34Ag+ + e- Ag Eo = + 0.80 v2xCu + 2Ag+ Cu+2 + 2AgEocell= +0.46 vwhere n is the number of electrons for the balanced reactionWhat is the free energy for the cell?1F = 96,500 J/vhttp:\\aliasadipour.kmu.ac.irCu + 2Ag+ Cu+2 + 2AgDGo = -2965000.46=-88780 J91100242

-E depends on: -Related half reaction -Concentration

-kinetic------------------------------------------------------ 2e- +2H+ H2 E0 = 0.000 Fe 3e- +Fe3+ E0 = 0.036 ------------------------------------------ Fe +H+ Fe3+ +H2 E0 = 0.036 Spontaneous redox reaction ????? !!!!!!!No===========================================================================================

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0.036 V91100243

0.337 Vhttp:\\aliasadipour.kmu.ac.ir

e- +Cu+ Cu E0 = 0.521 VCu+ Cu2++e- E0 = -0.153 V-------------------------------------------2Cu+ Cu2++Cu E0 = 0.368V

2Cu+ Cu2++CuAuto redox=Dis proportionation

91100244

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0.036 VAuto redox=Dis proportionation ?????? 2e- +Fe2+ Fe E0 = -0.440 VFe2+ Fe3++e- E0 = -0.771 V

9110022 -------------------------------------------3Fe2+ 2Fe3++Fe E0 = -1.221V

NO45

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-0.036 V911002

-------------------------------------------------------3e +Fe3+ Fe E0=+0.331 ?

No e isnt a function state

1) e +Fe3+ Fe2+ E0= 0.7712) 2e +Fe2+ Fe E0=-0.4402e- +Fe2+ Fe E0 = -0.440 VFe2+ Fe3++e- E0 = -0.771 V-------------------------------------------3Fe2+ 2Fe3++Fe E0 = -1.221V

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G0 =-nE0f= -3E0f

http:\\aliasadipour.kmu.ac.ir911002G0 =-nE0f2) 2e +Fe2+ Fe E0=-0.4401) e +Fe3+ Fe2+ E0= 0.771G0=-1(+0.771) F=-0.771f

G0=-2(-0.440) F=+0.880f3e +Fe3+ Fe G0=+0.109f

------------------------------------------------------=+0.109f3E0=-0.109 E0=-0.036 v

Free Energy and Chemical Reactions47WqGHTSSpontaneous reactionIdeal reverse cellOperating cell911002http:\\aliasadipour.kmu.ac.irG = H - TSW = H - q4748http:\\aliasadipour.kmu.ac.irNi(s) | Ni2+(XM) || Sn2+(YM) | Sn(s) A cell2 e- + Sn2+ Sn(s)Ni(s) 2 e- + Ni2+Ni(s) + Sn2+ Ni2+ + Sn(s) Redox reaction CathodeAnodeRepresentation of a cell91100249http:\\aliasadipour.kmu.ac.irNi(s) | Ni2+(1M)|| Sn2+(1M) | Sn(s)Ni(s) 2 e- + Ni2+ E =0.230 V

Ni(s) + Sn2+(1M) Ni2+(1M) + Sn(s) CathodeAnodeEmf of a standard cellE =0.230 -0.140 =0.090V 2 e- + Sn2+ Sn(s) E=-0.140V 911002------------------------------------5050Effect of Concentration on Cell EMFA voltaic cell is functional until E = 0 at which point equilibrium has been reached.The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction.The Nernst Equation

http:\\aliasadipour.kmu.ac.ir911002E = Eo RT ln Q n/-nfE = Eo - 0.0591 log Q n 51Effect of Concentration on Cell EMFat 25oC: E = Eo - 0.0591 log Ni2+/ Sn2+ n Calculate the Ered for the hydrogen electrode where 0.50 M H+ and 0.95 atm H2.http:\\aliasadipour.kmu.ac.irNi(s) | Ni2+ (XM) || Sn2+ (YM) | Sn(s)Ni(s) + Sn2+ (YM) Ni2+ (XM) + Sn(s) E= 0.090 V 911002Q= Ni2+/ Sn2+E=0.090-0.059/2logx/yE=0.000-0.059/2logpH2/[H+]22H++2e H2Q=X/Y-------------------------------------------------------52Ni(s) + Sn2+ Ni2+ + Sn(s) E= 0.090 V

http:\\aliasadipour.kmu.ac.irNi(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)Emf of a cell91100253Emf of a cellSn(s) | Sn2+(1.0M)|| Pb2+(0.0010M) | Pb(s)2 e- + Pb2+ Pb(s) E=-0.126 V Sn(s) 2 e- + Sn2+ E=0.136Vhttp:\\aliasadipour.kmu.ac.irE=E-0.059/2log[Sn2+]/[Pb2+]E=-0.079 !!!=Reversed cell

Sn(s) + Pb2+(0.0010M) Sn2+(1.0M) + Pb(s)Ecell=0.010 Vpb(s) | pb2+(1.0M)|| sn2+(0.0010M) | sn(s)911002E=+0.079(Electrolytic cell)(Galvanic cell)54equilibrium constant of a cell at equilibrium E = 0 Nernst Equation:

http:\\aliasadipour.kmu.ac.irE = Eo - 0.0591 log B n A911002A B55

http:\\aliasadipour.kmu.ac.irNi(s) | Ni2+(0.600M)|| Sn2+(0.300M) | Sn(s)

Ni(s) + Sn2+ Ni2+ + Sn(s) E= 0.090 V

equilibrium constant of a cell91100256

http:\\aliasadipour.kmu.ac.ir91100257Electrod potential and electrolysisTheoritical emf of a Voltaic cell is maximum voltage.(Practically is less)Theoritical emf of an electrolysis cell is minimum voltage.(Practically is more)

Emf is related to:ResistanceConcentrationOvervoltagehttp:\\aliasadipour.kmu.ac.ir91100258Electrod potential and electrolysis2H2O + 2e- H2 + 2OH- (E = - 0.83 V)In aqueous salts electrolysis [OH-] =1 10-7Mhttp:\\aliasadipour.kmu.ac.irE = E - log Qn0.059 VE = E - log [OH-] 2pH220.059 VE = -0.83 -0.0295 log [1*10-7] 2*1=-0.41791100259Electrod potential and electrolysis2H2O O2 + 4H+ + 4e- (E = - 1.23V)

In aqueous salts electrolysis [H+] =1 10-7Mhttp:\\aliasadipour.kmu.ac.irE = E - log Qn0.059 VE = E - log [H+] 4pO240.059 VE = -1.23 -0.01475 log [1*10-7] 4*1=-0.81791100260Effect of concentration in aqueous Na2SO4 electrolysispossible cathode half-cells (-)REDUCTION Na+ + e- Na (E = - 2.71 V)

[2H2O + 2e- H2 + 2OH- ] (E = - 0.83 V) (E = - 0.417 V)

possible anode half-cells (+)OXIDATION SO42- S4O82_ + 2e- (E = - 2.01 V)

2H2O O2 + 4H+ + 4e- (E = - 1.23V) (E = - 0.817 V) overall cell reaction 6H2O 2H2 + O2 + 4H+ + 4OH- E=-0.417-0.817=-1.234

http:\\aliasadipour.kmu.ac.ir9110022 61Electrod potential and electrolysisOvervoltage(OV): (Because of slow rate of reaction)OV of deposition of metals are lowOV of liberation of gases are appreciable (O2 & H2 >Cl2)

http:\\aliasadipour.kmu.ac.ir91100262Effect of overvoltage & concentrationin aqueous NaCl Electrolysispossible cathode half-cells (-)REDUCTION Na+ + e- Na (E = - 2.71 V)

[2H2O + 2e- H2 + 2OH- ] (E = - 0.83 V) (E = - 0.417 V)

possible anode half-cells (+)OXIDATION2Cl- Cl2 + 2e- (E = - 1.36 V)

2H2O O2 + 4H+ + 4e- (E = - 1.23V) (E = - 0.817 V)

OVERVOLTAGE H2 & O2 > Cl2overall cell reaction2Cl- + 2H2O H2 + Cl2 + 2OH-

http:\\aliasadipour.kmu.ac.ir91100263Effect of overvoltage & concentrationin aqueous CuCl2 Electrolysispossible cathode half-cells (-)REDUCTION Cu2+ + 2e- Cu (E = +0.337V)

2H2O + 2e- H2 + 2OH- (E = - 0.83 V) (E = - 0.417 V)

possible anode half-cells (+)OXIDATION2Cl- Cl2 + 2e- (E = - 1.36 V)

2H2O O2 + 4H+ + 4e- (E = - 1.23V) (E = -0.817V) OVERVOLTAGE H2 & O2 > Cl2overall cell reaction Cu2+ + 2Cl- Cu(s) + Cl2(g)

http:\\aliasadipour.kmu.ac.ir91100264CuCuCuSO4CuCu+2 + 2e- CuCu Cu+2 + 2e-What happened at each electrode?http:\\aliasadipour.kmu.ac.irbatteryImpure Cupure CuAnode+Cathode-Pure Cu deposit on cathode =(Pure cathodic Cu)What happens if aqueous CuSO4 electrolyze between 2 Cu electrodes ?=purification of Cu91100265What happens if aqueous CuSO4 electrolyze between 2 Cu electrodes ?=purification of Cupossible anode half-cells (+) (Impure Cu)OXIDATION Cu Cu2+ + 2e- (E = -0.337 V)

2H2O O2 + 4H+ + 4e- (E = - 1.23V) (E = - 0.817 V) 2SO42- S2O82- + 2e- (E = -2.01V)

possible cathode half-cells (-) (Purified Cu)REDUCTION Cu2+ + 2e- Cu (E = +0.337V)

2H2O + 2e- H2 + 2OH- (E = - 0.83 V) (E = - 0.417 V)

http:\\aliasadipour.kmu.ac.ir(((Purified cathodic Cu)))overall cell reaction Cu2+ + Cu(s) Anod Cu2+ + Cu(s) Cathode91100266Cu1.0 M CuSO4Cu1.0 M CuSO4A cell with the similar electrods and electrolytes0.0http:\\aliasadipour.kmu.ac.irvolts91100267Cu1.0 M CuSO4Cu1.0 M CuSO4A cell with the similar electrods but different concentration electrolyteshttp:\\aliasadipour.kmu.ac.irvolts91100268Electrolysis of CopperConcentration CellsA concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions.

http:\\aliasadipour.kmu.ac.ir911002http:\\aliasadipour.kmu.ac.ir69CuCu2+ (0.1M) Cu2+ (1.0 M)Cu Anod cathod

E=E0-0.059/2Log(0.1/1.0) =+0.0296

Concentration CellsCu+Cu2+ (1.0 M)Cu2+ (0.1M)+Cu91100270

pH meter, A concentration Cell

911002http:\\aliasadipour.kmu.ac.irSlide 70 of 522 H+(1 M) 2 H+(x M)Pt | H2 (1 atm)|H+(x M) ||H+(1.0 M) |H2(1 atm) | Pt(s)2 H+(1 M) + 2 e- H2(g, 1 atm)H2(g, 1 atm) 2 H+(x M) + 2 e- H2(g, 1 atm) +2 H+(1 M) 2 H+(x M) + H2(g, 1 atm)Chemistry 140 Fall 20027071

Slide 71 of 52Ecell = Ecell - logn0.059 Vx212Ecell = 0 - log20.059 Vx21Ecell = - 0.059 V log xEcell = (0.059 V) pH2 H+(1 M) 2 H+(x M)Ecell = Ecell - log Qn0.059 Vhttp:\\aliasadipour.kmu.ac.irpH = Ecell /(0.059) 91100272The pH MeterIn practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!

A stable reference electrode and a glass-membrane electrode are contained within a combination pH electrode.The electrode is merely dipped into a solution, and the potential difference between the electrodes is displayed as pH.http:\\aliasadipour.kmu.ac.ir91100273galvanicelectrolyticneedpowersourcetwoelectrodesproduces electrical currentanode (-)cathode (+)anode (+)cathode (-)salt bridgevessel conductive mediumComparison of Electrochemical CellsDG < 0DG > 0http:\\aliasadipour.kmu.ac.ir91100274

Corrosion of Fe: Unwanted Voltaic CellsO2+2H2O+4e-4OH-Rust formation: Fe2+Fe3+ +e E0=-0.771 VO2(g) + 4 H+(aq) + 4 e- 4 H2O (aq) E0 = 1.229 V---------------------------------------------------------------- 4Fe2+(aq) + O2(g) + 4H+(aq) 4Fe3+(aq) + 2H2O(l) E0 =0.458V 2Fe3+(aq) + 4H2O(l) Fe2O3H2O(s) + 6H+(aq)http:\\aliasadipour.kmu.ac.irE0=0.440 VE0=1.229 VE0=0.401 V91100275Prevention of CorrosionCover the Fe surface with a protective coatingPaintTin (Tin plate)Zn (Galvanized iron)http:\\aliasadipour.kmu.ac.ir91100276Corrosion Protection

Slide 76 of 52http:\\aliasadipour.kmu.ac.ir FeFe2+ +2e E0=0.440 V Cu Cu2+ +2e E0=0.337 V

Fe Fe2+ +2e E0=0.440 V Zn Zn2+ +2e E0=0.763 V

91100277

Corrrosion Protection(cathode)(electrolyte)(anode)http:\\aliasadipour.kmu.ac.ir Fe Fe2+ +2e E0=0.440 V MgMg2+ +2e E0=2.363 V

Steel pipe dont rust911002Fe Fe2+ +2e E0=0.440 V 78Cathodic ProtectionIn cathodic protection, an iron object to be protected is connected to a chunk of an active metal.The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.Water heaters often employ a magnesium anode for cathodic protection.

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Fig 18-26 Pg 901
The electroplating of silver.

Pg 901
Courtesy International Silver Plating, Inc.

Fig 18-9 Pg 872
Diagram of a galvanic cell containing passive electrodes. the two platinum electrodes do not take part in the redoxchemistry of this cell. Theyonly conduct electrons to andfrom the interfaces.