5_Elektrokimia.ppt

Chapter 18Electrochemistry

2007, Prentice HallChemistry: A Molecular Approach, 1st Ed.Nivaldo TroRoy KennedyMassachusetts Bay Community CollegeWellesley Hills, MA

Tro, Chemistry: A Molecular Approach

Redox Reactionone or more elements change oxidation number

all single displacement, and combustion,some synthesis and decompositionalways have both oxidation and reduction

split reaction into oxidation half-reaction and a reduction half-reactionaka electron transfer reactions

half-reactions include electronsoxidizing agent is reactant molecule that causes oxidation

contains element reducedreducing agent is reactant molecule that causes reduction

contains the element oxidized


Oxidation & Reductionoxidation is the process that occurs when

oxidation number of an element increaseselement loses electronscompound adds oxygencompound loses hydrogenhalf-reaction has electrons as productsreduction is the process that occurs when

oxidation number of an element decreaseselement gains electronscompound loses oxygencompound gains hydrogenhalf-reactions have electrons as reactants


Rules for Assigning Oxidation Statesrules are in order of priority

free elements have an oxidation state = 0Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)monatomic ions have an oxidation state equal to their chargeNa = +1 and Cl = -1 in NaCl (a) the sum of the oxidation states of all the atoms in a compound is 0Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0


Rules for Assigning Oxidation States(b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ionN = +5 and O = -2 in NO3, (+5) + 3(-2) = -1(a) Group I metals have an oxidation state of +1 in all their compoundsNa = +1 in NaCl

(b) Group II metals have an oxidation state of +2 in all their compoundsMg = +2 in MgCl2


Rules for Assigning Oxidation Statesin their compounds, nonmetals have oxidation states according to the table belownonmetals higher on the table take priority

NonmetalOxidation StateExampleF-1CF4H+1CH4O-2CO2Group 7A-1CCl4Group 6A-2CS2Group 5A-3NH3


Oxidation and Reductionoxidation occurs when an atoms oxidation state increases during a reactionreduction occurs when an atoms oxidation state decreases during a reaction

CH4 + 2 O2 CO2 + 2 H2O-4 +1 0 +4 2 +1 -2


OxidationReductionoxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent

the reducing agent contains the element that is oxidizedthe reactant that oxidizes an element in another reactant is called the oxidizing agent

the oxidizing agent contains the element that is reduced2 Na(s) + Cl2(g) 2 Na+Cl(s)Na is oxidized, Cl is reducedNa is the reducing agent, Cl2 is the oxidizing agent


Identify the Oxidizing and Reducing Agents in Each of the Following3 H2S + 2 NO3 + 2 H+ 3 S + 2 NO + 4 H2OMnO2 + 4 HBr MnBr2 + Br2 + 2 H2O


Identify the Oxidizing and Reducing Agents in Each of the Following3 H2S + 2 NO3 + 2 H+ 3 S + 2 NO + 4 H2OMnO2 + 4 HBr MnBr2 + Br2 + 2 H2O+1 -2 +5 -2 +1 0 +2 -2 +1 -2ox agred ag+4 -2 +1 -1 +2 -1 0 +1 -2red agox ag


Common Oxidizing Agents

Oxidizing Agent

Product when Reduced

O2

O-2

H2O2

H2O

F2, Cl2, Br2, I2

F-1, Cl-1, Br-1, I-1

ClO3-1 (BrO3-1, IO3-1)

Cl-1, (Br-1, I-1)

H2SO4 (conc)

SO2 or S or H2S

SO3-2

S2O3-2, or S or H2S

HNO3 (conc) or NO3-1

NO2, or NO, or N2O, or N2, or NH3

MnO4-1 (base)

MnO2

MnO4-1 (acid)

Mn+2

CrO4-2 (base)

Cr(OH)3

Cr2O7-2 (acid)

Cr+3

Common Reducing Agents

Reducing AgentProduct when OxidizedH2H+1H2O2O2I-1I2NH3, N2H4N2S-2, H2SSSO3-2SO4-2NO2-1NO3-1C (as coke or charcoal)CO or CO2Fe+2 (acid)Fe+3Cr+2 Cr+3Sn+2Sn+4metalsmetal ions

Balancing Redox Reactionsassign oxidation numbersdetermine element oxidized and element reducedwrite ox. & red. half-reactions, including electronsox. electrons on right, red. electrons on left of arrowbalance half-reactions by massfirst balance elements other than H and Oadd H2O where need Oadd H+1 where need Hneutralize H+ with OH- in basebalance half-reactions by chargebalance charge by adjusting electronsbalance electrons between half-reactionsadd half-reactionscheck


Ex 18.3 Balance the equation:I(aq) + MnO4(aq) I2(aq) + MnO2(s) in basic solution

Assign Oxidation StatesI(aq) + MnO4(aq) I2(aq) + MnO2(s)Separate into half-reactions ox:red:

Assign Oxidation StatesSeparate into half-reactions ox: I(aq) I2(aq)red: MnO4(aq) MnO2(s)



Balance half-reactions by mass ox: I(aq) I2(aq)red: MnO4(aq) MnO2(s)

Balance half-reactions by mass ox: 2 I(aq) I2(aq)red: MnO4(aq) MnO2(s)

Balance half-reactions by mass then O by adding H2Oox: 2 I(aq) I2(aq)red: MnO4(aq) MnO2(s) + 2 H2O(l)

Balance half-reactions by mass then H by adding H+ox: 2 I(aq) I2(aq)red: 4 H+(aq) + MnO4(aq) MnO2(s) + 2 H2O(l)

Balance half-reactions by mass in base, neutralize the H+ with OH-ox: 2 I(aq) I2(aq)red: 4 H+(aq) + MnO4(aq) MnO2(s) + 2 H2O(l)

4 H+(aq) + 4 OH(aq) + MnO4(aq) MnO2(s) + 2 H2O(l) + 4 OH(aq)

4 H2O(aq) + MnO4(aq) MnO2(s) + 2 H2O(l) + 4 OH(aq)

MnO4(aq) + 2 H2O(l) MnO2(s) + 4 OH(aq)



Balance Half-reactions by chargeox: 2 I(aq) I2(aq) + 2 e red: MnO4(aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH(aq)Balance electrons between half-reactions ox: 2 I(aq) I2(aq) + 2 e } x3red: MnO4(aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH(aq) }x2ox: 6 I(aq) 3 I2(aq) + 6 ered: 2 MnO4(aq) + 4 H2O(l) + 6 e 2 MnO2(s) + 8 OH(aq)



Add the Half-reactionsox: 6 I(aq) 3 I2(aq) + 6 ered: 2 MnO4(aq) + 4 H2O(l) + 6 e 2 MnO2(s) + 8 OH(aq)tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l) 3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)Check

ReactantCountElementProductCount6I62Mn212O128H82charge2


Practice - Balance the Equation H2O2 + KI + H2SO4 K2SO4 + I2 + H2O


Practice - Balance the Equation H2O2 + KI + H2SO4 K2SO4 + I2 + H2O

+1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2oxidationreductionox:2 I-1 I2 + 2e-1red:H2O2 + 2e-1 + 2 H+ 2 H2Otot2 I-1 + H2O2 + 2 H+ I2 + 2 H2O1 H2O2 + 2 KI + H2SO4 K2SO4 + 1 I2 + 2 H2O


Practice - Balance the EquationClO3-1 + Cl-1 Cl2 (in acid)


Practice - Balance the EquationClO3-1 + Cl-1 Cl2 (in acid)+5 -2 -1 0oxidationreductionox:2 Cl-1 Cl2 + 2 e-1 } x5red:2 ClO3-1 + 10 e-1 + 12 H+ Cl2 + 6 H2O} x1tot10 Cl-1 + 2 ClO3-1 + 12 H+ 6 Cl2 + 6 H2O1 ClO3-1 + 5 Cl-1 + 6 H+1 3 Cl2 + 3 H2O


Electrical Currentwhen we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of timewhen we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time

whether as electrons flowing through a wire or ions flowing through a solution


Redox Reactions & Currentredox reactions involve the transfer of electrons from one substance to anothertherefore, redox reactions have the potential to generate an electric currentin order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring


Electric Current Flowing Directly Between Atoms


Electric Current Flowing Indirectly Between Atoms


Electrochemical Cellselectrochemistry is the study of redox reactions that produce or require an electric currentthe conversion between chemical energy and electrical energy is carried out in an electrochemical cellspontaneous redox reactions take place in a voltaic cell

aka galvanic cellsnonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy


Electrochemical Cellsoxidation and reduction reactions kept separate

half-cellselectron flow through a wire along with ion flow through a solution constitutes an electric circuitrequires a conductive solid (metal or graphite) electrode to allow the transfer of electrons

through external circuition exchange between the two halves of the system

electrolyte


ElectrodesAnode

electrode where oxidation occursanions attracted to itconnected to positive end of battery in electrolytic cellloses weight in electrolytic cellCathode

electrode where reduction occurscations attracted to itconnected to negative end of battery in electrolytic cellgains weight in electrolytic cellelectrode where plating takes place in electroplating


Voltaic Cellthe salt bridge is required to complete the circuit and maintain charge balance


Current and Voltagethe number of electrons that flow through the system per second is the current

unit = Ampere1 A of current = 1 Coulomb of charge flowing by each second1 A = 6.242 x 1018 electrons/secondElectrode surface area dictates the number of electrons that can flowthe difference in potential energy between the reactants and products is the potential difference

unit = Volt1 V of force = 1 J of energy/Coulomb of chargethe voltage needed to drive electrons through the external circuitamount of force pushing the electrons through the wire is called the electromotive force, emf


Cell Potentialthe difference in potential energy between the anode the cathode in a voltaic cell is called the cell potentialthe cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anodethe cell potential under standard conditions is called the standard emf, Ecell

25C, 1 atm for gases, 1 M concentration of solutionsum of the cell potentials for the half-reactions


Cell Notationshorthand description of Voltaic cellelectrode | electrolyte || electrolyte | electrodeoxidation half-cell on left, reduction half-cell on the rightsingle | = phase barrier

if multiple electrolytes in same phase, a comma is used rather than |often use an inert electrodedouble line || = salt bridge


Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)


Standard Reduction Potentiala half-reaction with a strong tendency to occur has a large + half-cell potentialwhen two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occurwe cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reactionwe select as a standard half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 v

standard hydrogen electrode, SHE


Half-Cell PotentialsSHE reduction potential is defined to be exactly 0 vhalf-reactions with a stronger tendency toward reduction than the SHE have a + value for Eredhalf-reactions with a stronger tendency toward oxidation than the SHE have a value for EredEcell = Eoxidation + Ereduction

Eoxidation = Ereductionwhen adding E values for the half-cells, do not multiply the half-cell E values, even if you need to multiply the half-reactions to balance the equation


Ex 18.4 Calculate Ecell for the reaction at 25CAl(s) + NO3(aq) + 4 H+(aq) Al3+(aq) + NO(g) + 2 H2O(l)

Separate the reaction into the oxidation and reduction half-reactionsox:Al(s) Al3+(aq) + 3 e

red:NO3(aq) + 4 H+(aq) + 3 e NO(g) + 2 H2O(l)find the E for each half-reaction and sum to get EcellEox = Ered = +1.66 vEred = +0.96 vEcell = (+1.66 v) + (+0.96 v) = +2.62 v


Ex 18.4a Predict if the following reaction is spontaneous under standard conditionsFe(s) + Mg2+(aq) Fe2+(aq) + Mg(s)

Separate the reaction into the oxidation and reduction half-reactionsox:Fe(s) Fe2+(aq) + 2 e

red: Mg2+(aq) + 2 e Mg(s) look up the relative positions of the reduction half-reactionsred: Mg2+(aq) + 2 e Mg(s) red: Fe2+(aq) + 2 e Fe(s) since Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written


the reaction is spontaneous in the reverse directionMg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)ox:Mg(s) Mg2+(aq) + 2 ered:Fe2+(aq) + 2 e Fe(s)

sketch the cell and label the parts oxidation occurs at the anode; electrons flow from anode to cathode


Practice - Sketch and Label the Voltaic CellFe(s) Fe2+(aq) Pb2+(aq) Pb(s) , Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.


ox: Fe(s) Fe2+(aq) + 2 e E = +0.45 V red: Pb2+(aq) + 2 e Pb(s) E = 0.13 V tot: Pb2+(aq) + Fe(s) Fe2+(aq) + Pb(s) E = +0.32 V


Predicting Whether a Metal Will Dissolve in an Acidacids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+(aq)metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid


Ecell, DG and Kfor a spontaneous reaction

one the proceeds in the forward direction with the chemicals in their standard statesDG < 1 (negative)E > 1 (positive)K > 1 DG = RTlnK = nFEcell

n is the number of electronsF = Faradays Constant = 96,485 C/mol e


Example 18.6- Calculate DG for the reactionI2(s) + 2 Br(aq) Br2(l) + 2 I(aq) since DG is +, the reaction is not spontaneous in the forward direction under standard conditionsAnswer:Solve:Concept Plan:

Relationships: I2(s) + 2 Br(aq) Br2(l) + 2 I(aq) DG, (J)Given:Find:ox: 2 Br(aq) Br2(l) + 2 e E = 1.09 vred: I2(l) + 2 e 2 I(aq) E = +0.54 vtot: I2(l) + 2Br(aq) 2I(aq) + Br2(l) E = 0.55 v


Example 18.7- Calculate K at 25C for the reactionCu(s) + 2 H+(aq) H2(g) + Cu2+(aq) since K < 1, the position of equilibrium lies far to the left under standard conditionsAnswer:Solve:Concept Plan:

Relationships: Cu(s) + 2 H+(aq) H2(g) + Cu2+(aq) KGiven:Find:ox: Cu(s) Cu2+(aq) + 2 e E = 0.34 vred: 2 H+(aq) + 2 e H2(aq) E = +0.00 vtot: Cu(s) + 2H+(aq) Cu2+(aq) + H2(g) E = 0.34 v


Nonstandard Conditions - the Nernst Equation DG = DG + RT ln Q E = E - (0.0592/n) log Q at 25Cwhen Q = K, E = 0use to calculate E when concentrations not 1 M


E at Nonstandard Conditions


Example 18.8- Calculate Ecell at 25C for the reaction3 Cu(s) + 2 MnO4(aq) + 8 H+(aq) 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)units are correct, Ecell > Ecell as expected because [MnO4] > 1 M and [Cu2+] < 1 MCheck:Solve:Concept Plan:Relationships:3 Cu(s) + 2 MnO4(aq) + 8 H+(aq) 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)[Cu2+] = 0.010 M, [MnO4] = 2.0 M, [H+] = 1.0 M EcellGiven:

Find:ox: Cu(s) Cu2+(aq) + 2 e }x3E = 0.34 vred: MnO4(aq) + 4 H+(aq) + 3 e MnO2(s) + 2 H2O(l) }x2 E = +1.68 vtot: 3 Cu(s) + 2 MnO4(aq) + 8 H+(aq) 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)) E = +1.34 v


Concentration Cellsit is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are differentthe difference in energy is due to the entropic difference in the solutions

the more concentrated solution has lower entropy than the less concentratedelectrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution

oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution the less concentrated solution has the anodereduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration the more concentrated solution has the cathode


Concentration CellCu(s) Cu2+(aq) (0.010 M) Cu2+(aq) (2.0 M) Cu(s)


LeClanche Acidic Dry Cellelectrolyte in paste form

ZnCl2 + NH4Clor MgBr2anode = Zn (or Mg)

Zn(s) Zn2+(aq) + 2 e-cathode = graphite rodMnO2 is reduced

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e- 2 NH4OH(aq) + 2 Mn(O)OH(s)cell voltage = 1.5 vexpensive, nonrechargeable, heavy, easily corroded


Alkaline Dry Cellsame basic cell as acidic dry cell, except electrolyte is alkaline KOH pasteanode = Zn (or Mg)

Zn(s) Zn2+(aq) + 2 e-cathode = brass rodMnO2 is reduced

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e- 2 NH4OH(aq) + 2 Mn(O)OH(s)cell voltage = 1.54 vlonger shelf life than acidic dry cells and rechargeable, little corrosion of zinc


Lead Storage Battery6 cells in serieselectrolyte = 30% H2SO4anode = Pb

Pb(s) + SO42-(aq) PbSO4(s) + 2 e-cathode = Pb coated with PbO2PbO2 is reduced

PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- PbSO4(s) + 2 H2O(l) cell voltage = 2.09 vrechargeable, heavy


NiCad Batteryelectrolyte is concentrated KOH solutionanode = Cd

Cd(s) + 2 OH-1(aq) Cd(OH)2(s) + 2 e-1E0 = 0.81 vcathode = Ni coated with NiO2NiO2 is reduced

NiO2(s) + 2 H2O(l) + 2 e-1 Ni(OH)2(s) + 2OH-1 E0 = 0.49 vcell voltage = 1.30 vrechargeable, long life, light however recharging incorrectly can lead to battery breakdown


Ni-MH Batteryelectrolyte is concentrated KOH solutionanode = metal alloy with dissolved hydrogen

oxidation of H from H0 to H+1MH(s) + OH-1(aq) M(s) + H2O(l) + e-1E = 0.89 vcathode = Ni coated with NiO2NiO2 is reduced

NiO2(s) + 2 H2O(l) + 2 e-1 Ni(OH)2(s) + 2OH-1 E0 = 0.49 vcell voltage = 1.30 vrechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad


Lithium Ion Batteryelectrolyte is concentrated KOH solutionanode = graphite impregnated with Li ionscathode = Li - transition metal oxide

reduction of transition metalwork on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathoderechargeable, long life, very light, more environmentally friendly, greater energy density


Fuel Cellslike batteries in which reactants are constantly being added

so it never runs down!Anode and Cathode both Pt coated metalElectrolyte is OH solutionAnode Reaction:

2 H2 + 4 OH 4 H2O(l) + 4 e-Cathode Reaction:

O2 + 4 H2O + 4 e- 4 OH


Electrolytic Celluses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction

must be DC sourcethe + terminal of the battery = anodethe - terminal of the battery = cathodecations attracted to the cathode, anions to the anodecations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidizedsome electrolysis reactions require more voltage than Etot, called the overvoltage


electroplating In electroplating, the work piece is the cathode.Cations are reduced at cathode and plate to the surface of the work piece.The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution


Electrochemical Cellsin all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathodein voltaic cells,

anode is the source of electrons and has a () chargecathode draws electrons and has a (+) chargein electrolytic cells

electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the batteryelectrons are forced toward the anode, so it must have a source of electrons, the terminal of the battery


Electrolysis electrolysis is the process of using electricity to break a compound apartelectrolysis is done in an electrolytic cellelectrolytic cells can be used to separate elements from their compounds

generate H2 from water for fuel cellsrecover metals from their ores


Electrolysis of Water


Electrolysis of Pure Compoundsmust be in molten (liquid) stateelectrodes normally graphitecations are reduced at the cathode to metal elementanions oxidized at anode to nonmetal element


Electrolysis of NaCl(l)


Mixtures of Ionswhen more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode

least negative or most positive Ered when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode

least negative or most positive Eox


Electrolysis of Aqueous SolutionsComplicated by more than one possible oxidation and reductionpossible cathode reactions

reduction of cation to metalreduction of water to H22 H2O + 2 e-1 H2 + 2 OH-1E = -0.83 v @ stand. cond.E = -0.41 v @ pH 7possible anode reactions

oxidation of anion to elementoxidation of H2O to O22 H2O O2 + 4e-1 + 4H+1E = -1.23 v @ stand. cond.E = -0.82 v @ pH 7oxidation of electrodeparticularly Cugraphite doesnt oxidizehalf-reactions that lead to least negative Etot will occur

unless overvoltage changes the conditions


Electrolysis of NaI(aq) with Inert Electrodespossible oxidations2 I-1 I2 + 2 e-1E = 0.54 v2 H2O O2 + 4e-1 + 4H+1E = 0.82 vpossible reductionsNa+1 + 1e-1 Na0E = 2.71 v2 H2O + 2 e-1 H2 + 2 OH-1E = 0.41 vpossible oxidations2 I-1 I2 + 2 e-1E = 0.54 v2 H2O O2 + 4e-1 + 4H+1E = 0.82 vpossible reductionsNa+1 + 1e-1 Na0E = 2.71 v2 H2O + 2 e-1 H2 + 2 OH-1E = 0.41 voverall reaction2 I(aq) + 2 H2O(l) I2(aq) + H2(g) + 2 OH-1(aq)


Faradays Lawthe amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs

charge that flows through the cell = current x time


Example 18.10- Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e Au(s)units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol eCheck:Solve:Concept Plan:Relationships:3 mol e : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, gGiven:Find:


Corrosioncorrosion is the spontaneous oxidation of a metal by chemicals in the environmentsince many materials we use are active metals, corrosion can be a very big problem


Rustingrust is hydrated iron(III) oxidemoisture must be present

water is a reactantrequired for flow between cathode and anodeelectrolytes promote rusting

enhances current flowacids promote rusting

lower pH = lower Ered


Preventing Corrosionone way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment

paintsome metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corrodinganother method to protect one metal is to attach it to a more reactive metal that is cheap

sacrificial electrodegalvanized nails


Sacrificial Anode


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5_Elektrokimia.ppt