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AP Chemistry Wolpa AP Chemistry Wolpa CDO CDO Chapter 21 Chapter 21 Electrochemistry Electrochemistry
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Chapter 21. Electrochemistry. Oxidation-Reduction Reactions. oxidation: loss of electrons LEO reduction: gain of electrons GER Can be determined from change in oxidation numbers. Oxidation Number. - PowerPoint PPT Presentation
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Page 1: Chapter 21

AP Chemistry Wolpa CDO AP Chemistry Wolpa CDO

Chapter 21Chapter 21

ElectrochemistryElectrochemistry

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Oxidation-Reduction ReactionsOxidation-Reduction Reactions

oxidation: loss of electrons LEOoxidation: loss of electrons LEO reduction: gain of electrons GERreduction: gain of electrons GER Can be determined from change in Can be determined from change in

oxidation numbers.oxidation numbers.

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Oxidation NumberOxidation Number

Charge an atom has or would have if Charge an atom has or would have if all the bonding electrons were all the bonding electrons were assigned to the most electronegative assigned to the most electronegative element.element.

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Rules for assigning oxidation Rules for assigning oxidation numbersnumbers

11). Oxidation number pure neutral element = 0 ). Oxidation number pure neutral element = 0 (e.g. Li, H(e.g. Li, H22, C etc) , C etc)

2). Oxidation number monatomic ion = ionic 2). Oxidation number monatomic ion = ionic charge (Nacharge (Na++ = +1, Mg = +1, Mg2+2+ = +2, etc) = +2, etc)

3). Oxidation number of O = -2 in all its 3). Oxidation number of O = -2 in all its compounds compounds exceptexcept: : those with O-O bond (peroxides e.g. Na-O-O-Na those with O-O bond (peroxides e.g. Na-O-O-Na then ox. number = -1) superoxides, (e.g. KOthen ox. number = -1) superoxides, (e.g. KO22 then ox. number = -1/2) then ox. number = -1/2)

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Rules cont.Rules cont.

4). H is +1 in all its compounds (4). H is +1 in all its compounds (exceptexcept metal hydrides then = -1) metal hydrides then = -1)

5). F is - 1 in all its compounds (5). F is - 1 in all its compounds (exceptexcept, of , of course, Fcourse, F22) )

6). Alkali metals are always +1; alkaline 6). Alkali metals are always +1; alkaline earths are always +2. earths are always +2.

7). Sum of oxidation number is zero for 7). Sum of oxidation number is zero for neutral compounds, and equal to the neutral compounds, and equal to the overall charge for polyatomic ionic overall charge for polyatomic ionic species. species.

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Oxidation and Reduction must occur Oxidation and Reduction must occur simultaneouslysimultaneously

Zn(s) + CuZn(s) + Cu2+2+(aq) -------> Zn(aq) -------> Zn2+2+(aq) + (aq) + Cu(s)Cu(s)

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Example 1Example 1

Indicate which of the reactants is Indicate which of the reactants is reducing and which is oxidizingreducing and which is oxidizing

2Ce2Ce4+4+(aq) + Sn(aq) + Sn2+2+(aq) (aq) 2Ce 2Ce3+3+(aq) + Sn(aq) + Sn4+4+(aq)(aq)

8H8H++(aq) + MnO(aq) + MnO44-1-1(aq) + 5Fe(aq) + 5Fe2+2+(aq) (aq) 5Fe 5Fe3+3+(aq) + (aq) +

MnMn2+2+(aq) + 4H(aq) + 4H22O(l)O(l)

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Electrochemical CellsElectrochemical Cells

System consisting of electrodes that System consisting of electrodes that dip into an electrolyte and in which an dip into an electrolyte and in which an oxidation-reduction reaction either oxidation-reduction reaction either uses or generates an electric current.uses or generates an electric current.

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Types of Electrochemical Types of Electrochemical CellCell

Voltaic or GalvanicVoltaic or Galvanic A spontaneous reaction generates an A spontaneous reaction generates an

electric currentelectric current

ElectrolyticElectrolytic An electric current is used to drive a An electric current is used to drive a

nonspontaneous reactionnonspontaneous reaction

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ElectrodesElectrodes

Cathode: electrode where reduction Cathode: electrode where reduction reaction occursreaction occurs

Anode: electrode where oxidation Anode: electrode where oxidation reaction occurs.reaction occurs.

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Balancing Redox ReactionsBalancing Redox Reactions

Neutral SolutionNeutral Solution

1. Split into two half-reactions1. Split into two half-reactions

2. Balance each half-reaction2. Balance each half-reaction

a. Balance the elementsa. Balance the elements

b. Balance the chargeb. Balance the charge

3. Combine half-reactions so that 3. Combine half-reactions so that electrons cancel.electrons cancel.

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Example 2Example 2

ClCl22(g) + Br(g) + Br--(aq) --> Br(aq) --> Br22(l) + Cl(l) + Cl--(aq)(aq)

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Balancing Redox ReactionsBalancing Redox Reactions

Acidic SolutionAcidic Solution1. Split into two half-reactions1. Split into two half-reactions2. Balance each half-reaction2. Balance each half-reaction

a. Balance the elementsa. Balance the elementsb. Balance oxygens by adding waterb. Balance oxygens by adding waterc. Balance hydrogens by adding hydrogen c. Balance hydrogens by adding hydrogen ionsionsd. Balance the charged. Balance the charge

3. Combine half-reactions so that electrons 3. Combine half-reactions so that electrons cancel.cancel.

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Example 3Example 3

Zn(s) + VOZn(s) + VO2+2+(aq) --> V(aq) --> V2+2+(aq) + Zn(aq) + Zn2+2+(aq)(aq)

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Balancing Redox ReactionsBalancing Redox Reactions

Basic SolutionBasic Solution1. Split into two half-reactions1. Split into two half-reactions2. Balance each half-reaction2. Balance each half-reaction

a. Balance the elementsa. Balance the elementsb. Balance oxygens by adding waterb. Balance oxygens by adding waterc. Balance hydrogens by adding hydrogen c. Balance hydrogens by adding hydrogen ionsionsd. Neutralize the H+ ions by adding OH- ionsd. Neutralize the H+ ions by adding OH- ionse. Balance the chargee. Balance the charge

3. Combine half-reactions so that electrons 3. Combine half-reactions so that electrons cancel.cancel.

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Example 4Example 4

Zn(s) + ClZn(s) + Cl22 (g) --> Zn(OH)(g) --> Zn(OH)22(s) + Cl(s) + Cl- -

(aq) (aq)

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Voltaic CellVoltaic Cell

DescriptionDescription anode: Zn ----------> Znanode: Zn ----------> Zn2+2+ + 2e- + 2e- cathode: Cucathode: Cu2+2+ + 2e- ----------> Cu + 2e- ----------> Cu Consists of two half-cells that are electrically Consists of two half-cells that are electrically

connected. The salt bridge consists of a connected. The salt bridge consists of a tube of an electrolyte that is connected to tube of an electrolyte that is connected to two half-cells of the voltaic cell. Allows the two half-cells of the voltaic cell. Allows the flow of charge ions but prevents diffusional flow of charge ions but prevents diffusional mixing of the different solutions.mixing of the different solutions.

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Voltaic Cell Key PointsVoltaic Cell Key Points

1. There are always two separate half-cells 1. There are always two separate half-cells connected by a wire and a salt bridgeconnected by a wire and a salt bridge

2. One of the compartments is the anode and the 2. One of the compartments is the anode and the other is the cathodeother is the cathode

3. If a metal participates in a cell reaction it is 3. If a metal participates in a cell reaction it is ordinarily chosen as the electrode. If no metal is ordinarily chosen as the electrode. If no metal is involved in the half-reaction an electrically involved in the half-reaction an electrically conduction solid like platinum or graphite is usedconduction solid like platinum or graphite is used

4. Electrons flow from the anode to the cathode.4. Electrons flow from the anode to the cathode.5. Cations move to the cation and anions to the 5. Cations move to the cation and anions to the

anode via the salt bridge.anode via the salt bridge.

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Cell NotationCell Notation

Zn/ZnZn/Zn2+2+ ll Cu ll Cu2+2+/Cu/Cu

anode reaction is shown at leftanode reaction is shown at left

salt bridge is indicated by llsalt bridge is indicated by ll

cathode reaction is shown at rightcathode reaction is shown at right

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Example 5Example 5

Draw cell diagram and write the cell Draw cell diagram and write the cell notation notation

FeFe3+3+(aq) + H(aq) + H22(g) (g) Fe Fe2+2+(aq) + 2H(aq) + 2H++(aq)(aq)

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Electrochemical Cells and Electrochemical Cells and PotentialsPotentials

EmfEmf

Electrons generated at the site of Electrons generated at the site of oxidation of a cell are thought to be oxidation of a cell are thought to be "driven" or "pushed" toward the cathode "driven" or "pushed" toward the cathode by an electromotive force, or emf. This by an electromotive force, or emf. This force is due to the difference in electric force is due to the difference in electric potential of an electron at the two potential of an electron at the two electrodes.electrodes.

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Standard PotentialStandard Potential

Measure of the driving force of the Measure of the driving force of the cell reaction when all ions and cell reaction when all ions and molecules in the solution are at a molecules in the solution are at a concentration of 1 M and all gases concentration of 1 M and all gases are at a pressure of one atm.are at a pressure of one atm.

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Calculating the Potential ECalculating the Potential Eoo of of an Electrochemical Cellan Electrochemical Cell

Eotot = EEotot = Eooox + Eox + Eooredred

EEooox: standard voltage for the ox: standard voltage for the oxidation half-reactionoxidation half-reaction

EEoored: standard voltage for the red: standard voltage for the reduction half-reactionreduction half-reaction

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Standard Hydrogen Half-CellStandard Hydrogen Half-Cell

Cannot determine an individual half-reaction Cannot determine an individual half-reaction voltage therefore we arbitrarily take the voltage therefore we arbitrarily take the standard voltage for the reduction of H+ ions standard voltage for the reduction of H+ ions to H2(g) to be zero (standard hydrogen half-to H2(g) to be zero (standard hydrogen half-cell):cell):

2H2H++(aq,1M) + 2e- ----------> H(aq,1M) + 2e- ----------> H22(g, 1atm) (g, 1atm) EEoored red = 0= 0

With this assigned voltage others can be With this assigned voltage others can be determined from measurement by hook-up determined from measurement by hook-up with knowns.with knowns.

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Standard Hydrogen Half-Standard Hydrogen Half-CellsCells

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Standard PotentialsStandard Potentials

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Using Standard PotentialsUsing Standard Potentials

1. The E1. The Eoo values are for reactions written in the form values are for reactions written in the form "oxidized form + electrons --------> reduced form." "oxidized form + electrons --------> reduced form." The species on the left side of the reaction is an The species on the left side of the reaction is an oxidizing agent, and the species on the right is a oxidizing agent, and the species on the right is a reducing agent. All potentials are therefore for reducing agent. All potentials are therefore for reduction reactions.reduction reactions.

2. When writing the reaction "reduced form --------> 2. When writing the reaction "reduced form --------> oxidized form + electrons," the sign of Eoxidized form + electrons," the sign of Eo o is reversed, is reversed, but the value of Ebut the value of Eoo is unaffected. is unaffected.

3. All the half-reactions are reversible.3. All the half-reactions are reversible.4. The more positive the value of E4. The more positive the value of Eoo for the reactions for the reactions

the better the oxidizing ability of the ion or compound.the better the oxidizing ability of the ion or compound.

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5. The more negative the value of the reduction 5. The more negative the value of the reduction potential Epotential Eoo the less likely the reaction occurs as a the less likely the reaction occurs as a reduction, and the more likely the reverse reaction reduction, and the more likely the reverse reaction occurs.occurs.

6. The reaction between any substance on the left in 6. The reaction between any substance on the left in this table with any substance lower than it on the this table with any substance lower than it on the right is product-favored under standard conditions.right is product-favored under standard conditions.

7. The algebraic sign of the half-reaction potential is 7. The algebraic sign of the half-reaction potential is the sign of the electron when it is attached to the sign of the electron when it is attached to HH22/H/H33OO++ standard cell. standard cell.

8. Electrochemical potential depends on the nature of 8. Electrochemical potential depends on the nature of the reactants and products and their the reactants and products and their concentrations, not on the quantities of material concentrations, not on the quantities of material used.used.

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LargeLarge negativenegative value means value means oxidationoxidation strongly favored; strongly favored; strong strong reducingreducing agent. agent.

LargeLarge positivepositive value means value means reductionreduction strongly favored; strongly favored; strong strong oxidizingoxidizing agent. agent.

Relative values in table give an indication that one half-Relative values in table give an indication that one half-reaction favored over other. Summing half-cell reactions reaction favored over other. Summing half-cell reactions allow determination of standard cell potential.allow determination of standard cell potential.

Half-cell potential intensive property Half-cell potential intensive property independent of independent of amount of material amount of material we we don’t use stoichiometric don’t use stoichiometric coefficients for determining standard cell potentialscoefficients for determining standard cell potentials..

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Example 6Example 6

Determine the cell potential of Determine the cell potential of

BrBr22(l) + 2I(l) + 2I(aq) (aq) I I22(l) + 2Br(l) + 2Br(aq)(aq)

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Example 7Example 7

Determine the cell potential of Determine the cell potential of

2Ag2Ag++(aq) + Cu(s) (aq) + Cu(s) 2Ag(s) + Cu 2Ag(s) + Cu2+2+(aq)(aq)

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Example 8Example 8

Determine cell potential:Determine cell potential:

MnOMnO44--(aq) + Fe(s) (aq) + Fe(s) Fe Fe2+2+(aq) + Mn(aq) + Mn2+2+

(aq) (balanced?) (aq) (balanced?)

when it is operated galvanically. when it is operated galvanically. Which is the oxidizing agent? Which is the oxidizing agent? reducing agent?reducing agent?

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Example 9Example 9

Determine if the reaction below is Determine if the reaction below is spontaneous in the direction written. spontaneous in the direction written.

FeFe3+3+(aq) + Ag(s) (aq) + Ag(s) ? ?

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EEootot and tot and GGoo

1. 1. G = free energy changeG = free energy change

= maximum amount of useful work= maximum amount of useful work

= -w= -wmaxmax

2. Electric work2. Electric work

= charge x potential energy difference = charge x potential energy difference

= coulomb x volt= coulomb x volt

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EEootot and tot and GGoo

wwmax max = nFE = coulombs x volts = electrical = nFE = coulombs x volts = electrical energy in joules where F is the Faraday energy in joules where F is the Faraday constant, 9.65 x 104 J/V mol.constant, 9.65 x 104 J/V mol.

GGoo = -nFE = -nFEoo tot = -96.5 nE tot = -96.5 nEoo tot (in KJ) tot (in KJ)

Because spontaneous reactions have a Because spontaneous reactions have a negative free energy change, negative free energy change, G, the G, the negative sign in the equation above negative sign in the equation above confirms that all product-favored electron confirms that all product-favored electron transfer reactions have a positive Etransfer reactions have a positive Eoo..

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Electrochemical Cells at Electrochemical Cells at Nonstandard ConditionsNonstandard Conditions

Voltage is a measure of reaction Voltage is a measure of reaction spontaneity.spontaneity.

Hence: Hence: voltage is increased by increasing voltage is increased by increasing

concentrations of reactants or decreasing concentrations of reactants or decreasing concentrations of products.concentrations of products.

voltage is decreased by decreasing voltage is decreased by decreasing concentrations of reactants or increasing concentrations of reactants or increasing concentrations of products.concentrations of products.

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Quantitative: Nernst Quantitative: Nernst EquationEquation

aA + bB ----------> cC + dDaA + bB ----------> cC + dD

Etot = EEtot = Eootot -RT/nF ln [products]/[reactants] tot -RT/nF ln [products]/[reactants] raised to the power of their coefficientsraised to the power of their coefficients

where R = 8.314510 J/K mol, F = 9.645309 where R = 8.314510 J/K mol, F = 9.645309 x 10x 1044 J/v mol and number of electrons J/v mol and number of electrons transferred at 25 Ctransferred at 25 C

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Nernst EquationNernst Equation

Etot = EEtot = Eootot - 0.0257/n tot - 0.0257/n ln[products]/[reactants] raised to the ln[products]/[reactants] raised to the power of their coefficients power of their coefficients

n = number of electrons transferredn = number of electrons transferred

[ ] use molarities for aqueous [ ] use molarities for aqueous solutions, atm for gasessolutions, atm for gases

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K: Equilibrium ConstantK: Equilibrium Constant

Etot = EEtot = Eootot - - 0.0257/n ln [products]/[reactants] tot - - 0.0257/n ln [products]/[reactants] raised to the power of their coefficientsraised to the power of their coefficients

at equilibrium Etot = zeroat equilibrium Etot = zero

andand

K = [products]/[reactants] raised to the power of K = [products]/[reactants] raised to the power of their coefficientstheir coefficients

thereforetherefore

0 = E0 = Eootot - 0.0257/n ln Ktot - 0.0257/n ln K

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EEootot and Ktot and K

ln K = nEln K = nEootot/0.0257tot/0.0257

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Example 10Example 10

Determine free energy and equilibrium Determine free energy and equilibrium constant for reaction below constant for reaction below (unbalanced). (unbalanced).

MnOMnO44--(aq) + Fe(s) (aq) + Fe(s) Fe Fe2+2+(aq) + Mn(aq) + Mn2+2+

(aq)(aq)

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Example 11Example 11

Determine cell potential and Determine cell potential and equilibrium constant of Clequilibrium constant of Cl22/Br/Br22 cell. cell.

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Example 12Example 12

The following cell has a potential of The following cell has a potential of 0.578 V at 25°C; determine K0.578 V at 25°C; determine Kspsp..

Ag(s)|AgCl(s)|ClAg(s)|AgCl(s)|Cl(1.0 M)||Ag(1.0 M)||Ag++(1.0 M)|(1.0 M)|Ag(s).Ag(s).

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BatteriesBatteries

A A batterybattery is a package of one or more is a package of one or more galvanic cells used for the production and galvanic cells used for the production and storage of electric energy by chemical storage of electric energy by chemical means. A means. A galvanic cellgalvanic cell consists of at least consists of at least two half cells, a reduction cell and an two half cells, a reduction cell and an oxidation cell. Chemical reactions in the two oxidation cell. Chemical reactions in the two half cells provide the energy for the galvanic half cells provide the energy for the galvanic cell operations. cell operations.

Each Each half cellhalf cell consists of an electrode and consists of an electrode and an electrolyte solution. Usually the solution an electrolyte solution. Usually the solution contains ions derived from the electrode by contains ions derived from the electrode by oxidation or reduction reaction. oxidation or reduction reaction.

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Types of BatteriesTypes of Batteries

Batteries can be divided into two Batteries can be divided into two types: primary or disposable types: primary or disposable batteries and secondary or batteries and secondary or rechargeable batteries. rechargeable batteries.

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Fuel cells are different from batteries Fuel cells are different from batteries in that they consume reactant, which in that they consume reactant, which must be replenished, whereas must be replenished, whereas batteries store electrical energy batteries store electrical energy chemically in a closed system. chemically in a closed system.

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Corrosion of Metals/Cathodic Corrosion of Metals/Cathodic ProtectionProtection

Cathodic protectionCathodic protection ( (CPCP) is a ) is a technique to control the corrosion of technique to control the corrosion of a metal surface by making that a metal surface by making that surface the cathode of an surface the cathode of an electrochemical cell. electrochemical cell.

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Cathodic ProtectionCathodic Protection

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Electrolytic CellElectrolytic Cell

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Aqueous ElectrolysisAqueous Electrolysis

Cathode(reduction)Cathode(reduction)

a. cation reduced to metal; usually a. cation reduced to metal; usually occurs with transition metal cationsoccurs with transition metal cations

b. Hb. H++ ions reduced to H ions reduced to H22; occurs with ; occurs with strong acidsstrong acids

c. Hc. H22O molecules reduced to HO molecules reduced to H22; occurs ; occurs with Group 1, Group 2 metals and Alwith Group 1, Group 2 metals and Al

2H2H22O(l) + 2e- ----------> HO(l) + 2e- ----------> H22(g) + 2OH(g) + 2OH--(aq)(aq)

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Aqueous ElectrolysisAqueous ElectrolysisAnode(oxidation)Anode(oxidation)

a. anion oxidized to nonmetala. anion oxidized to nonmetal

b. OH- ions oxidized to Ob. OH- ions oxidized to O22; occurs ; occurs with strong baseswith strong bases

c. Hc. H22O molecules oxidized to OO molecules oxidized to O22; with ; with NONO33

--, SO, SO442-2- and F and F--

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Overall resultOverall result

NiClNiCl22: Ni + Cl: Ni + Cl22

NaCl: HNaCl: H22, OH, OH--, Cl, Cl22

CuSOCuSO44: Cu, O: Cu, O22, H, H++

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Example 13Example 13

What reactions take place at the What reactions take place at the anode and cathode when each of the anode and cathode when each of the following is electrolyzed?following is electrolyzed?

1.0 M NiBr1.0 M NiBr22

1.0 M AlF1.0 M AlF33

1.0 M MnI1.0 M MnI22

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Stoichiometry of ElectrolysisStoichiometry of Electrolysis

UnitsUnits

Faraday = charge = 9.65 x 104 C = 1 mole of Faraday = charge = 9.65 x 104 C = 1 mole of e-e-

Coulomb = chargeCoulomb = charge

Ampere = current = charge/time = C/sAmpere = current = charge/time = C/s

Joule = electrical work = C x voltsJoule = electrical work = C x volts

watt = energy/time = 1 joule/secondwatt = energy/time = 1 joule/second

volt = potential differencevolt = potential difference

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CalculationsCalculations

Electric Charge = electric current x Electric Charge = electric current x time elapsedtime elapsed

Electric Energy = Coulombs x voltsElectric Energy = Coulombs x volts

watt = volts x ampereswatt = volts x amperes

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Example 14Example 14

Determine amount of CuDetermine amount of Cu2+2+ electrolyzed electrolyzed from solution at constant current of from solution at constant current of 6.00 A for period of 1.00 hour. 6.00 A for period of 1.00 hour.

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Example 15Example 15

What volumes of HWhat volumes of H22(g) and O(g) and O22(g) rat (g) rat STP are produced from the STP are produced from the electrolysis of water by a current of electrolysis of water by a current of 2.50 A in 15.0 minutes?2.50 A in 15.0 minutes?

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Example 16Example 16

Electrolysis of a molten metal Electrolysis of a molten metal chloride (MClchloride (MCl33) using a current of ) using a current of 6.50 amp for 1397 seconds deposits 6.50 amp for 1397 seconds deposits 0.471 g of metal at the cathode. 0.471 g of metal at the cathode. What is the identity of the alkaline What is the identity of the alkaline earth metal?earth metal?