1 chapter 20 principles of reactivity: electron transfer reactions 1-11 + all bold numbered problems
TRANSCRIPT
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CHAPTER 20CHAPTER 20PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:
ELECTRON TRANSFER ELECTRON TRANSFER REACTIONSREACTIONS
CHAPTER 20CHAPTER 20PRINCIPLES OF REACTIVITY: PRINCIPLES OF REACTIVITY:
ELECTRON TRANSFER ELECTRON TRANSFER REACTIONSREACTIONS
1-11 + all bold numbered problems1-11 + all bold numbered problems
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ELECTROCHEMISTRYELECTROCHEMISTRYELECTROCHEMISTRYELECTROCHEMISTRY
Electric automobile
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Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?
• BatteriesBatteries• CorrosionCorrosion• Industrial Industrial
production of production of chemicalschemicals such such as Clas Cl22, NaOH, , NaOH, FF22 and Al and Al
• Biological Biological redox reactionsredox reactions
The heme group
A rusted car.
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20.1 OXIDATION-REDUCTION 20.1 OXIDATION-REDUCTION REACTIONSREACTIONS
• Redox reactions make up Redox reactions make up
electrochemical cells. electrochemical cells.
• Those that Those that produceproduce energy are called energy are called
batteries or batteries or galvanic cellsgalvanic cells (voltaic). (voltaic).
• Those that Those that requirerequire energy to operate are energy to operate are
called called electrolysis cellselectrolysis cells..
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TRANSFER REACTIONSTRANSFER REACTIONSTRANSFER REACTIONSTRANSFER REACTIONSAtom transferAtom transfer
HOAc + HHOAc + H22O ---> OAcO ---> OAc-- + H + H33OO++
no change in oxidation state of any of the elementsno change in oxidation state of any of the elements
Electron transferElectron transfer
Cu(s) + 2 AgCu(s) + 2 Ag++(aq) ---> Cu(aq) ---> Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)
Cu looses 2 eCu looses 2 e-- and and AuAu++ gains 1 e gains 1 e--
OIL OIL RIGRIG
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Electron Transfer ReactionsElectron Transfer ReactionsElectron Transfer ReactionsElectron Transfer Reactions
• Electron transfer reactions are Electron transfer reactions are oxidation-reduction or oxidation-reduction or redoxredox reactions.reactions.
• Redox reactions can result in the Redox reactions can result in the generation of an electric current generation of an electric current or be caused by imposing an or be caused by imposing an electric current. electric current.
• Therefore, this field of chemistry Therefore, this field of chemistry is often called is often called ELECTROCHEMISTRY.ELECTROCHEMISTRY.
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CHAPTER OVERVIEWCHAPTER OVERVIEW
• This chapter examines electron This chapter examines electron transfer reactions, transfer reactions,
redoxredoxReduction/OxidationReduction/Oxidation
Note: Oil Rig = Note: Oil Rig =
Oxidation is Loss of an electronOxidation is Loss of an electron
Reduction is Gain of an electronReduction is Gain of an electron
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Review of Terminology for Review of Terminology for Redox ReactionsRedox Reactions
Review of Terminology for Review of Terminology for Redox ReactionsRedox Reactions
• OXIDATIONOXIDATION—loss of electron(s) by —loss of electron(s) by a species; increase in oxidation a species; increase in oxidation number.number.
• REDUCTIONREDUCTION—gain of electron(s); —gain of electron(s); decrease in oxidation number.decrease in oxidation number.
• OXIDIZING AGENTOXIDIZING AGENT—electron —electron acceptor; species is reduced.acceptor; species is reduced.
• REDUCING AGENTREDUCING AGENT—electron —electron donor; species is oxidized.donor; species is oxidized.
• OXIDATIONOXIDATION—loss of electron(s) by —loss of electron(s) by a species; increase in oxidation a species; increase in oxidation number.number.
• REDUCTIONREDUCTION—gain of electron(s); —gain of electron(s); decrease in oxidation number.decrease in oxidation number.
• OXIDIZING AGENTOXIDIZING AGENT—electron —electron acceptor; species is reduced.acceptor; species is reduced.
• REDUCING AGENTREDUCING AGENT—electron —electron donor; species is oxidized.donor; species is oxidized.
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Solid interacting with aqueous ions
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OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS
OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS
Direct Redox ReactionDirect Redox ReactionOxidizing and reducing agents in direct Oxidizing and reducing agents in direct
contact.contact.Cu(s) + 2 AgCu(s) + 2 Ag++(aq) ---> Cu(aq) ---> Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)
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Electrochemical CellsElectrochemical CellsElectrochemical CellsElectrochemical Cells• An apparatus that allows An apparatus that allows
a redox reaction to occur a redox reaction to occur by transferring electrons by transferring electrons through an external through an external connector.connector.
• Product favored reaction Product favored reaction ---> voltaic or galvanic cell ---> voltaic or galvanic cell ----> electric current----> electric current
• Reactant favored reaction Reactant favored reaction ---> electrolytic cell ---> ---> electrolytic cell ---> electric current used to electric current used to cause chemical change.cause chemical change.
Batteries are voltaic Batteries are voltaic cellscells
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OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS
OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS
Indirect Redox ReactionIndirect Redox ReactionA battery functions by transferring A battery functions by transferring electrons through an external wire electrons through an external wire
from the reducing agent to the from the reducing agent to the oxidizing agent.oxidizing agent.
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CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
Cu2+ ions
Zn metal
Cu2+ ions
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
Zn is oxidized Zn is oxidized and is the reducing agent and is the reducing agent Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e(aq) + 2e--
CuCu2+2+ is reduced is reduced and is the oxidizing agentand is the oxidizing agentCuCu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s)
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CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
Oxidation: Zn(s) ---> ZnOxidation: Zn(s) ---> Zn2+2+(aq) + 2e(aq) + 2e--
Reduction: CuReduction: Cu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s)----------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s)(aq) + Cu(s)
Electrons are transferred from Zn to Cu2+, but there is no useful electric current.
Electrons are transferred from Zn to Cu2+, but there is no useful electric current.
Cu2+ ions
Zn metal
Cu2+ ions
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CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
CHEMICAL CHANGE -->CHEMICAL CHANGE -->ELECTRIC CURRENTELECTRIC CURRENT
•To obtain a useful current, To obtain a useful current, we separate the oxidizing we separate the oxidizing and reducing agents so and reducing agents so that electron transfer that electron transfer occurs thru an external occurs thru an external wire. wire.
•This is accomplished in a This is accomplished in a GALVANICGALVANIC or or VOLTAICVOLTAIC cell.cell.
•A group of such cells is A group of such cells is
called a called a batterybattery..
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
e le c t ro ns
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••Electrons travel thru external wire.Electrons travel thru external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.•This maintains electrical This maintains electrical neutralityneutrality..
••Electrons travel thru external wire.Electrons travel thru external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.•This maintains electrical This maintains electrical neutralityneutrality..
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Electrochemical Electrochemical Cell or BatteryCell or Battery
Electrochemical Electrochemical Cell or BatteryCell or Battery
Electrons move from anode to cathode in the wire.Anions & cations move thru the salt bridge.
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Balancing Redox EquationsBalancing Redox EquationsOXIDATION-REDUCTION REACTIONSOXIDATION-REDUCTION REACTIONS
• To be useful, the reactions for these To be useful, the reactions for these cells must be balanced. cells must be balanced.
• The simplest method of balancing is The simplest method of balancing is the ion-electron or half-reaction the ion-electron or half-reaction method. method.
• We learned this method in the first We learned this method in the first half of General Chemistry, but a quick half of General Chemistry, but a quick review is in order. review is in order.
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OXIDATION-REDUCTION REACTIONSOXIDATION-REDUCTION REACTIONS
• Oxidation is the loss of electron, Oxidation is the loss of electron, and reducing agents are oxidized. and reducing agents are oxidized.
• Reduction is the gain of electrons, Reduction is the gain of electrons, and oxidizing agents are reduced.and oxidizing agents are reduced.
• The simple steps in balancing a The simple steps in balancing a redox equation are:redox equation are:
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BALANCING REDOX REACTIONSBALANCING REDOX REACTIONS1. Separate the reaction into half reactions.1. Separate the reaction into half reactions.
2. Perform a mass balance for each element 2. Perform a mass balance for each element beginning with the element being oxidized beginning with the element being oxidized or reduced. Next balance the oxygen or reduced. Next balance the oxygen atoms with water and the hydrogen atoms atoms with water and the hydrogen atoms with hydrogen ions.with hydrogen ions.
3. Perform a charge balance by adding 3. Perform a charge balance by adding electrons to balance the charge.electrons to balance the charge.
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BALANCING REDOX REACTIONSBALANCING REDOX REACTIONS
4. If the reaction is in basic 4. If the reaction is in basic solution, add as many hydroxide solution, add as many hydroxide ions to both sides as there are ions to both sides as there are hydrogen ions and simplify by hydrogen ions and simplify by forming water. Eliminate the forming water. Eliminate the excess water, if any occurs on excess water, if any occurs on both sides of the reaction.both sides of the reaction.
5. Multiply each half-reaction by 5. Multiply each half-reaction by the appropriate factor so the the appropriate factor so the electrons are eliminated when the electrons are eliminated when the half-reactions are added.half-reactions are added.
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Balancing Equations for Balancing Equations for Redox ReactionsRedox Reactions
Balancing Equations for Balancing Equations for Redox ReactionsRedox Reactions
Some redox reactions have equations Some redox reactions have equations that must be balanced by special that must be balanced by special techniques.techniques.
MnOMnO44-- + 5 Fe + 5 Fe2+2+ + 8 H + 8 H++
---> Mn---> Mn2+2++ 5 Fe+ 5 Fe3+3+ + 4 H + 4 H22OO
22
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing Equations Consider the reduction of AgConsider the reduction of Ag++
ions with copper metal.ions with copper metal.
Cu + AgCu + Ag++ -- give --> Cu -- give --> Cu2+2+ + Ag + Ag
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 1:Step 1: Divide the reaction into half-Divide the reaction into half-
reactions, one for oxidation and the other reactions, one for oxidation and the other for reduction.for reduction.
OxOx Cu ---> CuCu ---> Cu2+2+
RedRed Ag Ag++ ---> Ag ---> AgStep 2:Step 2: Balance each for mass. Already Balance each for mass. Already
done in this case.done in this case.Step 3:Step 3: Balance each half-reaction for Balance each half-reaction for
charge by adding electrons.charge by adding electrons.OxOx Cu ---> Cu Cu ---> Cu2+2+ + + 2e2e--
RedRed Ag Ag++ + + ee-- ---> Ag---> Ag
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 4:Step 4: Multiply each half-reaction by a Multiply each half-reaction by a
factor that means the reducing agent factor that means the reducing agent supplies as many electrons as the supplies as many electrons as the oxidizing agent requires.oxidizing agent requires.
Reducing agentReducing agent Cu ---> CuCu ---> Cu2+2+ + 2e + 2e--
Oxidizing agentOxidizing agent 22 Ag Ag++ + + 22 e e-- ---> ---> 22 AgAg
Step 5:Step 5: Add half-reactions to give the Add half-reactions to give the overall equation.overall equation.
Cu + 2 AgCu + 2 Ag++ ---> Cu ---> Cu2+2+ + 2Ag + 2AgThe equation is now balanced for The equation is now balanced for
both charge and mass.both charge and mass.
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsBalance the following in acid Balance the following in acid
solution—solution—
VOVO22++ + Zn ---> VO + Zn ---> VO2+ 2+ + Zn+ Zn2+2+
Step 1:Step 1: Write the half-reactionsWrite the half-reactions
OxOx Zn ---> ZnZn ---> Zn2+2+
RedRed VOVO22++ ---> VO ---> VO2+2+
Step 2:Step 2: Balance each half-reaction Balance each half-reaction for mass.for mass.
OxOx Zn ---> ZnZn ---> Zn2+2+
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsBalance the following in acid solution—Balance the following in acid solution— VOVO22
++ + Zn ---> VO + Zn ---> VO2+ 2+ + Zn+ Zn2+2+
Step 1:Step 1: Write the half-reactionsWrite the half-reactionsOxOx Zn ---> ZnZn ---> Zn2+2+
RedRed VOVO22++ ---> VO ---> VO2+2+
Step 2:Step 2: Balance each half-reaction for Balance each half-reaction for mass.mass.
OxOx Zn ---> ZnZn ---> Zn2+2+
RedRed VO VO22++ ---> VO ---> VO2+2+ + + HH22OO2 H2 H++
++ Add HAdd H22O on O-deficient side and add HO on O-deficient side and add H++ on other side on other side for H-balance.for H-balance.
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Balancing EquationsBalancing EquationsBalancing EquationsBalancing EquationsStep 3:Step 3: Balance half-reactions for charge.Balance half-reactions for charge.
OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e 2e--
RedRed ee-- + 2 H+ 2 H++ + VO + VO22++ ---> VO ---> VO2+2+ + H + H22OO
Step 4:Step 4: Multiply by an appropriate factor.Multiply by an appropriate factor.
OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e2e--
RedRed 22ee-- + + 44 H H++ + + 22 VO VO22++
---> ---> 22 VO VO2+2+ + + 22 H H22OO
Step 5:Step 5: Add half-reactionsAdd half-reactions
Zn + 4 HZn + 4 H++ + 2 VO + 2 VO22++
---> Zn ---> Zn2+2+ + 2 VO + 2 VO2+2+ + 2 H + 2 H22OO
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Tips on Balancing EquationsTips on Balancing Equations• Never add ONever add O22, O atoms, or , O atoms, or
OO2-2- to balance oxygen. to balance oxygen.• Never add HNever add H22 or H atoms or H atoms
to balance hydrogen.to balance hydrogen.• Be sure to write the Be sure to write the
correct charges on all correct charges on all the ions.the ions.
• Check your work at the Check your work at the end to make sure mass end to make sure mass and charge are balanced.and charge are balanced.
MORE PRACTICE!!
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20.2 CHEMICAL CHANGES LEADING 20.2 CHEMICAL CHANGES LEADING TO ELECTRIC CURRENTTO ELECTRIC CURRENT
• An electrochemical cell has two An electrochemical cell has two electrodes, the electrodes, the anodeanode where where oxidation oxidation occursoccurs, and the , and the cathodecathode where where reductionreduction occurs. occurs.
• The compartment containing the anode The compartment containing the anode is called the is called the anodic compartmentanodic compartment
• The compartment containing the The compartment containing the cathode is called the cathode is called the cathodic cathodic compartmentcompartment. .
• These two half-cells are frequently These two half-cells are frequently connected by a porous barrier or a connected by a porous barrier or a salt salt
bridgebridge. .
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CHEMICAL CHANGES LEADING CHEMICAL CHANGES LEADING TO ELECTRIC CURRENT TO ELECTRIC CURRENT
• The electrical current is carried by The electrical current is carried by electrons in the metals present, and electrons in the metals present, and by ions in the solution. by ions in the solution.
• Ions carrying charge are called Ions carrying charge are called electrolytes.electrolytes.
• Electrons travel from the anode (-) to Electrons travel from the anode (-) to the cathode (+)the cathode (+)
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Reduction
Oxidation
Cathodiccompartment Anodic
compartment
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Illustrates a general cell
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Abbreviated Cell NotationAbbreviated Cell NotationAnode | Anodic sol’n || Cathodic sol’n | CathodeAnode | Anodic sol’n || Cathodic sol’n | CathodeThis simple notation is very efficient
2Fe3+(aq) + H2(g) 2Fe2+(aq) + 2H+(aq)Pt|H2(1atm)|H+(1M)||Fe2+(1M),Fe3+(1M)|Pt
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Cu|Cu2+(1.0 M)||Ag+(1.0 M)|Ag
Anode | Anodic sol’n || Cathodic sol’n | CathodeAnode | Anodic sol’n || Cathodic sol’n | Cathode
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20.3 ELECTROCHEMICAL 20.3 ELECTROCHEMICAL CELLS AND POTENTIALSCELLS AND POTENTIALS
• The electrical potential is called the The electrical potential is called the electromotive force or emf, and is given electromotive force or emf, and is given the symbol E. the symbol E.
• The unit of charge is called the The unit of charge is called the coulomb coulomb
• emf is measured in joules/coulomb or emf is measured in joules/coulomb or volts, Vvolts, V
• Under standard conditions, 1.0 M Under standard conditions, 1.0 M for solutes, 1.0 bar for gases, and for solutes, 1.0 bar for gases, and 298.15 K, the emf is given the symbol 298.15 K, the emf is given the symbol EEoo, and called the standard potential., and called the standard potential.
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EEoo vs. vs. GGoo
• The electrical work is nFE, where F is The electrical work is nFE, where F is Faraday's constant, 96, 485 coulombs/ Faraday's constant, 96, 485 coulombs/ mole electrons or J/V mole, and n is mole electrons or J/V mole, and n is the moles of electrons in the balanced the moles of electrons in the balanced equation. equation.
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3838
EEoo vs. vs. GGoo
This leads to the following equation This leads to the following equation between E and between E and G: G: G = - nFE.G = - nFE.
• This can be derived from: This can be derived from:
G = H -TS = E + PG = H -TS = E + PV - TV - TSS
• Since Since E = q + w and w = - PE = q + w and w = - PV – wV – wextext
• If the process is done reversibly at If the process is done reversibly at constant temperature and pressure, q constant temperature and pressure, q = T= TS.S.
G = q - PG = q - PV – wV – wext ext + P+ PV - TV - TSS
• This leads toThis leads toG = - wG = - wextext. .
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3939
EEoo vs. vs. GGoo
In the standard state, In the standard state, GGoo = - nFE = - nFEoo..• Now we can use E and ENow we can use E and Eoo to predict to predict
reaction spontaneity, i.e. whether a reaction spontaneity, i.e. whether a reaction is product-favored. reaction is product-favored.
• Calculations of Calculations of GGoo from E from Eoo and visa and visa versa are algebraic.versa are algebraic.
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Calculating the Potential ECalculating the Potential Eoo of of an Electrochemical Cellan Electrochemical Cell
• All standard potentials are bases All standard potentials are bases on a relative scale where the on a relative scale where the hydrogen half-cell is assigned a hydrogen half-cell is assigned a value to 0.00 V at all temperature, value to 0.00 V at all temperature, and is called the standard and is called the standard hydrogen electrode, SHE. hydrogen electrode, SHE.
• All electrode potentials are written All electrode potentials are written as reductions and derive their as reductions and derive their values from direct or indirect values from direct or indirect measurements vs. the SHE. measurements vs. the SHE.
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4141
Calculating the Potential ECalculating the Potential Eoo of of an Electrochemical Cellan Electrochemical Cell
• Demonstrates this measure for the zinc Demonstrates this measure for the zinc
electrode potential.electrode potential.
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4242
Demonstrates
this measure for
the copper
electrode
potential.
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4343
CELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, ECELL POTENTIAL, E
• Electrons are “driven” from anode to cathode by Electrons are “driven” from anode to cathode by an electromotive force or an electromotive force or emfemf..
• For Zn/Cu cell, this is indicated by a voltage of For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 1.10 V at 25 C and when [ZnC and when [Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.
Zn and ZnZn and Zn2+2+,,anodeanode
Cu and CuCu and Cu2+2+,,cathodecathode
43
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4444
CELL CELL POTENTIAL, EPOTENTIAL, E
CELL CELL POTENTIAL, EPOTENTIAL, E
• For Zn/Cu cell, voltage is 1.10 V at 25 For Zn/Cu cell, voltage is 1.10 V at 25 C C and when [Znand when [Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.
• This is the This is the STANDARD CELL STANDARD CELL POTENTIAL, EPOTENTIAL, Eoo
• -- a quantitative measure of the tendency -- a quantitative measure of the tendency of reactants to proceed to products when of reactants to proceed to products when all are in their standard states at 25 all are in their standard states at 25 C. C.
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4545
Calculating Cell VoltageCalculating Cell VoltageCalculating Cell VoltageCalculating Cell Voltage• Balanced half-reactions can be added Balanced half-reactions can be added
together to get overall, balanced equation. together to get overall, balanced equation.
2 I- ---> I2 + 2e-
2 H2O + 2e- ---> 2 OH- + H2
-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
2 I- ---> I2 + 2e-
2 H2O + 2e- ---> 2 OH- + H2
-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
If we know Eo for each half-reaction, we could get Eo for net reaction.
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4646
CELL POTENTIALS, ECELL POTENTIALS, EooCELL POTENTIALS, ECELL POTENTIALS, Eoo
• Can’t measure 1/2 reaction ECan’t measure 1/2 reaction Eoo directly. directly. Therefore, measure it relative to a STANDARD Therefore, measure it relative to a STANDARD HALF CELL, HALF CELL, SHESHE..
2 H2 H++(aq, 1 M) + 2e(aq, 1 M) + 2e-- -->--> H H22(g, 1 atm)(g, 1 atm)2 H2 H++(aq, 1 M) + 2e(aq, 1 M) + 2e-- -->--> H H22(g, 1 atm)(g, 1 atm)
EEoo = 0.0 V = 0.0 V
Typically made of Pt or Au, are inert, and importantTypically made of Pt or Au, are inert, and importantbecause SHE’s can act as either an anode orbecause SHE’s can act as either an anode orcathode, difficult to maintain, not used often cathode, difficult to maintain, not used often
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4747
Other Reference Electrodes, know EOther Reference Electrodes, know Eoo, easy to make, easy to makeSelf-contained, only make once, sealed, last a longtimeSelf-contained, only make once, sealed, last a longtime
||Hg||Hg22ClCl22(sat),KCl(xM)|Hg(sat),KCl(xM)|Hg
||AgCl(sat), KCl(xM)|Ag||AgCl(sat), KCl(xM)|Ag||TlCl(sat),KCl(xM)|Tl||TlCl(sat),KCl(xM)|Tl
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4848
Volts
ZnH2
Salt Bridge
Zn 2+ H+
Zn Zn 2+ + 2e-
OXIDATION ANODE
2 H+ + 2e- H2
REDUCTIONCATHODE
- +
Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V
Zn/Zn2+ half-cell hooked to a SHE. Eo for the cell = +0.76 V
Remember ERemember Eoo for SHE = 0.0 Vfor SHE = 0.0 V
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4949
Overall reaction is reduction of H+ by Zn metal.
Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = + 0.76 V
Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is
+ 0.76 V+ 0.76 V. Reduction potential is - 0.76 V- 0.76 V.
Zn is a betterbetter reducing agent (Zn gets oxidized)
than) H2.
Volts
ZnH2
Salt Bridge
Zn2+ H+
Zn Zn2+
+ 2 e-
OXIDATION ANODE
2 H+ + 2 e- H2
REDUCTIONCATHODE
-+
Remember ERemember Eoo for SHE = 0.0 Vfor SHE = 0.0 V
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5050
Cu/CuCu/Cu2+2+ and H and H22/H/H++ Cell CellCu/CuCu/Cu2+2+ and H and H22/H/H++ Cell Cell
Volts
CuH2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-OXIDATION ANODE
-+
Volts
CuH2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-OXIDATION ANODE
-+
EEoo = +0.34 V = +0.34 V
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5151
Cu/CuCu/Cu2+2+ and H and H22/H/H++ Cell CellCu/CuCu/Cu2+2+ and H and H22/H/H++ Cell Cell
• Overall reaction is reduction of CuOverall reaction is reduction of Cu2+2+ by H by H22 gas. gas.• CuCu2+2+ (aq) + H (aq) + H22(g) ---> Cu(s) + 2 H(g) ---> Cu(s) + 2 H++(aq)(aq)• Measured EMeasured Eoo = + 0.34 V = + 0.34 V• Therefore, ETherefore, Eoo for Cu for Cu2+2+ + 2e + 2e-- ---> Cu is ---> Cu is
+ 0.34 V + 0.34 V so Cu is a better Oxidizing agent (Cu gets so Cu is a better Oxidizing agent (Cu gets reduced) the Hreduced) the H++
Volts
CuH2
Salt Bridge
Cu2+ H+
Cu2+ + 2e- Cu REDUCTION CATHODE
H2 2 H+ + 2e-OXIDATION ANODE
-+
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5252
• Lets Combine the work we just did Lets Combine the work we just did with the Zinc and Copper Cells with the Zinc and Copper Cells and combine them into ONE cell and combine them into ONE cell and see the net effectand see the net effect
COMBINING Voltaic CELLSCOMBINING Voltaic CELLS
Zn2+(aq) + 2e- ---> Zn(s) Eo = - 0.76 VCu2+(aq) + 2e- ---> Cu(s) Eo = + 0.34 V
Cu is a better Oxidizing agent than HCu is a better Oxidizing agent than H++
Zn is a better better reducing agent than H+
From this info we know Zn From this info we know Zn MUSTMUST be oxidized be oxidized and that Cu and that Cu MUSTMUST be reduced be reduced
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Zn/Cu Electrochemical CellZn/Cu Electrochemical CellZn/Cu Electrochemical CellZn/Cu Electrochemical Cell
Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e(aq) + 2e-- EEoo = + 0.76 V = + 0.76 VCuCu2+2+(aq) + 2e(aq) + 2e-- ---> Cu(s) ---> Cu(s) EEoo = + 0.34 V = + 0.34 V------------------------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s)
EEoo (calculated) = + 1.10 V (calculated) = + 1.10 V
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Cathode, positive, sink for electrons(reduction)
Anode, negative, source of electrons(oxidation)
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5454
Uses of EUses of Eoo Values ValuesUses of EUses of Eoo Values ValuesThese experiments show we canThese experiments show we can
a) decide on relative ability of a) decide on relative ability of elements to act as reducing elements to act as reducing agents (or oxidizing agents)agents (or oxidizing agents)
b) assign a voltage to a half-b) assign a voltage to a half-reaction that reflects this ability.reaction that reflects this ability.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
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5555
USING STANDARD POTENTIALSUSING STANDARD POTENTIALS
EEoocellcell = E = Eoo
cathodiccathodic - E - Eooanodicanodic
EEoocellcell = E = Eoo
reductionreduction + E + Eoooxidationoxidation
• The first equation uses the values right The first equation uses the values right out of the table of standard electrode out of the table of standard electrode potentials. potentials.
• The table can also give us information The table can also give us information about which reactions will be about which reactions will be product-product-favoredfavored (produce current) in the (produce current) in the standard state since these reactions standard state since these reactions with have a positive Ewith have a positive Eoo
cellcell..
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5656
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5757
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USING STANDARD POTENTIALSUSING STANDARD POTENTIALS
• The following is an The following is an abbreviated table form the abbreviated table form the previous tables to give previous tables to give practice at using them.practice at using them.
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TABLE OF STANDARD TABLE OF STANDARD POTENTIALSPOTENTIALS
TABLE OF STANDARD TABLE OF STANDARD POTENTIALSPOTENTIALS
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
OxidizingOxidizingability of Ion ability of Ion
reducing abilityreducing abilityof elementof element
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6161
Standard Redox Potentials, EStandard Redox Potentials, EooStandard Redox Potentials, EStandard Redox Potentials, Eoo
• Any substance on Any substance on the right will reduce the right will reduce any substance any substance higher than it on the higher than it on the left.left.
• Zn can reduce HZn can reduce H++ and Cuand Cu2+2+..
• HH22 can reduce Cu can reduce Cu2+2+ but not Znbut not Zn2+2+
• Cu cannot reduce Cu cannot reduce HH++ or Zn or Zn2+2+..
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
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6262
Standard Redox Potentials, EStandard Redox Potentials, EooStandard Redox Potentials, EStandard Redox Potentials, Eoo
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
Eo (V)
Cu2+ + 2e- Cu +0.34
2 H+ + 2e- H2 0.00
Zn2+ + 2e- Zn -0.76
oxidizingability of ion
reducing abilityof element
• Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e- EEoo = + 0.76 V = + 0.76 V• CuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s) EEoo = + 0.34 V = + 0.34 V• ------------------------------------------------------------------------------------------------------------------------------• CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s) • EEoo (calculated) = + 1.10 V (calculated) = + 1.10 V
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6363
Using Standard Potentials, EUsing Standard Potentials, EooUsing Standard Potentials, EUsing Standard Potentials, Eoo
Which is the best oxidizing agent:Which is the best oxidizing agent:OO22, H, H22OO22, or Cl, or Cl22? _________________? _________________
• Which is the best reducing agent:Which is the best reducing agent:
Hg, Al, or Sn? ____________________Hg, Al, or Sn? ____________________
• Which is the correct direction for the following Which is the correct direction for the following
reaction to produce a favorable potential?reaction to produce a favorable potential?
2Al(s) + 3 Cl2Al(s) + 3 Cl22(aq) ---> AlCl(aq) ---> AlCl33(aq) (aq)
AlClAlCl33(aq) ---> 2Al(s) + 3 Cl(aq) ---> 2Al(s) + 3 Cl22(aq) (aq)
See next slideSee next slide
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6565
EEoo for a Voltaic Cell for a Voltaic CellEEoo for a Voltaic Cell for a Voltaic Cell
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Cd --> Cd2+ + 2e-
orCd2+ + 2e- --> Cd
Fe --> Fe2+ + 2e-
orFe2+ + 2e- --> Fe
See previous slideSee previous slide
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6666
From the table, you see From the table, you see •• Fe is a better reducing Fe is a better reducing
agent than Cdagent than Cd•• CdCd2+2+ is a better is a better
oxidizing agent than oxidizing agent than FeFe2+2+
Overall reactionOverall reactionFe + CdFe + Cd2+2+ ------
> Cd + Fe> Cd + Fe2+2+
EEoo = + 0.04 V = + 0.04 V
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
Volts
Cd Salt Bridge
Cd2+
Fe
Fe2+
EEoo for a Voltaic Cell for a Voltaic CellEEoo for a Voltaic Cell for a Voltaic Cell
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Examples of Cells that NEED electricity Examples of Cells that NEED electricity (opposite of Voltaic or Galvanic cells(opposite of Voltaic or Galvanic cells
Production of Production of Oxygen and Oxygen and HydrogenHydrogen
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6868
Electrolysis of SnClElectrolysis of SnCl22(aq)(aq)
SnSn2+2+(aq) + 2Cl(aq) + 2Cl--(aq) (aq) Sn(s) and Cl Sn(s) and Cl22(g)(g)
Anode (+)Anode (+) Cathode (-)Cathode (-)
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6969
NaINaI
phenolphthaleinphenolphthalein
More on this later, industrialapplications
OH- detected
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7070
EEoo and and GGooEEoo and and GGoo
EEoo is related to is related to GGoo, the free , the free energy change for the energy change for the reaction.reaction.
GGoo = - n F E = - n F Eoo where F = Faraday constant where F = Faraday constant
= 9.6485 x 10= 9.6485 x 1044 J/V•molJ/V•mol
and n is the number of moles and n is the number of moles of electrons transferredof electrons transferred
Michael Faraday1791-1867
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Michael FaradayMichael Faraday1791-18671791-1867
Michael FaradayMichael Faraday1791-18671791-1867
Originated the terms anode, Originated the terms anode, cathode, anion, cation, cathode, anion, cation, electrode.electrode.
Discoverer of Discoverer of • electrolysiselectrolysis• magnetic props. of mattermagnetic props. of matter• electromagnetic inductionelectromagnetic induction• benzene and other organic benzene and other organic
chemicalschemicalsWas a popular lecturer.Was a popular lecturer.
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EEoo and and GGooEEoo and and GGoo
GGoo = - n F E = - n F Eoo
For a For a product-favoredproduct-favored reaction reaction Reactants ----> ProductsReactants ----> Products
GGo o < 0 and so E < 0 and so Eo o > 0 > 0EEoo is positive, is positive, results in results in
currentcurrentFor a For a reactant-favoredreactant-favored reaction reaction Reactants <---- ProductsReactants <---- Products
GGo o > 0 and so E > 0 and so Eo o < 0 < 0EEoo is negative is negative , needs current, needs current
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7373
ELECTROCHEMICAL CELLS AT ELECTROCHEMICAL CELLS AT NON STANDARD CONDITIONS NON STANDARD CONDITIONS
• We learned the relationship between We learned the relationship between G and G and GGoo in chapter 19. in chapter 19.
G = G = GGoo + RT + RT lnln Q. Q. • This yields: -nFE = -nFEThis yields: -nFE = -nFEoo + RT + RT lnln Q Q • At 298.15 K, we solve for EAt 298.15 K, we solve for E
E = EE = Eoo - [0.0592 V/ - [0.0592 V/nn] ] loglog Q or Q or
E = EE = Eoo - [0.0257 V/ - [0.0257 V/nn]] lnln Q Q
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20.5 E20.5 Eoo and the Equilibrium Constant and the Equilibrium ConstantNernst EquationsNernst Equations
• At At equilibriumequilibrium for the cell reaction, for the cell reaction, E = 0, (equivalent to saying the cell is E = 0, (equivalent to saying the cell is dead) Therefore:dead) Therefore:
EEoo = [0.0592 V/ = [0.0592 V/nn] ] loglog K or K or
EEoo = [0.0257 V/ = [0.0257 V/nn] ] lnln K K
• These equations can be solved for K These equations can be solved for K using the inverse functions.using the inverse functions.
• nn is the number of moles of e is the number of moles of e--
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7575
Calculating KCalculating Kspsp, K, Kff, and other K's , and other K's from cell potentialsfrom cell potentials
• In the laboratory we will be calculating KIn the laboratory we will be calculating Kspsp's and 's and KKff 's for reactions using electrochemical cells. 's for reactions using electrochemical cells.
• Determine the KDetermine the Kff for AlF for AlF66-3-3 from the cell below: E from the cell below: E
= 2.465 V= 2.465 V
Al |Al |10.0 mL 0.10 M10.0 mL 0.10 M Al(NOAl(NO33))33,,40.0 mL 1.0 M40.0 mL 1.0 M NaF NaF ||||0.10 M0.10 M CuSOCuSO44||CuCu
2 Al + 3 Cu2 Al + 3 Cu+2 +2 ===> 2 Al ===> 2 Al+3+3 + 3 Cu + 3 Cu AlAl+3+3 + 6F + 6F-- <===> AlF <===> AlF66
-3-3
Given: [FGiven: [F--] = 0.68 M, [AlF] = 0.68 M, [AlF66-3-3] = 0.020 M] = 0.020 M
Note: no reaction between AlNote: no reaction between Al3+3+ or Cu or Cu2+2+ and SO and SO44-2-2
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7676
2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu
Eo = 1.676 + .337 = 2.013 VEcell = 2.465 V (given), 0.10 M Cu+2
Ecell = Eo - [0.0257 V/n] ln K
2.465 = 2.013 - [0.0257/6] Ln (Al+3)2/(.1)3
solved for Al+3, note that as soon as Al+3 forms it immediately reacts with F-, so Al+3 is NOT .1M
Which gives: [Al+3] = 4.0x10-25 M
Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M
Sub these values into Kf
Kf = 5.0 x 1023
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21.6 BATTERIES AND FUEL CELLS21.6 BATTERIES AND FUEL CELLS
• There are two types of batteries: There are two types of batteries:
• Primary, which are not Primary, which are not rechargeable rechargeable
• Secondary, or storage batteries Secondary, or storage batteries which are easily recharged.which are easily recharged.
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7878
Primary BatteriesPrimary Batteries
• 3 types: Dry, Alkaline, Button (Hg, Li)3 types: Dry, Alkaline, Button (Hg, Li)• The The dry celldry cell produces H produces H22 gas gas• The battery has a zinc anode and The battery has a zinc anode and
reduces ammonium ion to ammonia and reduces ammonium ion to ammonia and hydrogen gas at a graphite cathode. hydrogen gas at a graphite cathode.
• The net-reaction does not contain the The net-reaction does not contain the hydrogen gas since MnOhydrogen gas since MnO22 is added to is added to oxidize the hydrogen gas to hydrogen oxidize the hydrogen gas to hydrogen ion.ion.
• This battery has a voltage of 1.5 Volts This battery has a voltage of 1.5 Volts when new. when new.
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7979
Dry Cell BatteryDry Cell BatteryDry Cell BatteryDry Cell Battery
Anode (-)Anode (-)
Zn ---> ZnZn ---> Zn2+2+ + 2e + 2e--
Cathode (+)Cathode (+)
2 NH2 NH44++ + 2e + 2e-- ---> --->
2 NH2 NH33 + H + H22
Common dry cell
79
1.5 Volts1.5 Volts
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8080
+ -4 3 2
2+ -
2NH + 2e 2NH ( ) + H ( )
Zn Zn + 2e
g g
Problem alkaline & button: gasses formedProblem alkaline & button: gasses formed
3 2The gasses NH ( ) + H ( ) are removed:g g
2+ -3 3 2 2
2 2 2 3 2
Zn + 2NH ( ) + 2Cl Zn(NH ) Cl ( )
2MnO ( ) + H ( ) Mn O ( ) + H O( )
g s
and
s g s l
anode
cathode
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8181
Primary Batteries: No HPrimary Batteries: No H22 Gas produced Gas produced
• The The alkaline batteryalkaline battery has a zinc anode has a zinc anode and a graphite cathode, the MnOand a graphite cathode, the MnO22 is is
reduced directly to Mnreduced directly to Mn22OO33 in a basic in a basic
environment.environment.
• It's voltage is 1.54 Volts. It's voltage is 1.54 Volts.
• The mercury and lithium batteries are The mercury and lithium batteries are buttonbutton shaped and have voltages of shaped and have voltages of 1.35 V and 3.0 V respectively.1.35 V and 3.0 V respectively.
• 50% stronger then a dry cell50% stronger then a dry cell
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Nearly same reactions as in Nearly same reactions as in common dry cell, but under basic common dry cell, but under basic conditions (1.54V)conditions (1.54V)
Alkaline BatteryAlkaline BatteryAlkaline BatteryAlkaline Battery
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8383
Mercury BatteryMercury BatteryMercury BatteryMercury BatteryAnode:Anode:
Zn is reducing agent under basic conditionsZn is reducing agent under basic conditionsCathode:Cathode:
HgO + HHgO + H22O + 2eO + 2e-- ---> Hg + 2 OH ---> Hg + 2 OH--
Primary, Hg (1.35V, Li (3.0V)Primary, Hg (1.35V, Li (3.0V)
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8484
- -2
- -2 2 2
Zn( ) + 2OH ZnO + 2H O( ) + 2e
2MnO + H O( ) + 2e MnO + 2OH
s l
l
Note OHNote OH--, either KOH or NaOH, either KOH or NaOH
anode
cathode
Also note that Zn here is the cathode, it is the anode in a dry cell
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8585
Primary
At anode
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8686Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)
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8787
Secondary BatteriesSecondary Batteries
• The net-reaction turns lead and The net-reaction turns lead and lead(IV)oxide, in the presence of lead(IV)oxide, in the presence of sulfuric acid, into lead(II)sulfate and sulfuric acid, into lead(II)sulfate and waterwater
• The nickel-cadmium battery is a The nickel-cadmium battery is a rechargeable battery. rechargeable battery.
Fuel CellsFuel Cells• Fuel cells convert hydrogen and Fuel cells convert hydrogen and
oxygen gasoxygen gas into water and electrical into water and electrical energy without combustion.energy without combustion.
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8989
Lead Storage BatteryLead Storage BatteryLead Storage BatteryLead Storage BatteryAnode (-) Anode (-) EEoo = +0.36 V = +0.36 VPb + HSOPb + HSO44
-- ---> PbSO ---> PbSO44 + H + H++ + 2e + 2e--
Cathode (+) Cathode (+) EEoo = +1.68 V = +1.68 VPbOPbO22 + HSO + HSO44
-- + 3 H + 3 H++ + 2e + 2e-- ---> PbSO ---> PbSO44 + 2 H + 2 H22OO
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9090
Ni-Cad BatteryNi-Cad BatteryNi-Cad BatteryNi-Cad BatteryAnode (-)Anode (-)
Cd + 2 OHCd + 2 OH-- ---> Cd(OH) ---> Cd(OH)22 + 2e + 2e--
Cathode (+) Cathode (+)
NiO(OH) + HNiO(OH) + H22O + eO + e-- ---> Ni(OH) ---> Ni(OH)22 + OH + OH--
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9191
CORROSION: REDOX REACTIONS CORROSION: REDOX REACTIONS IN THE ENVIRONMENT IN THE ENVIRONMENT
• Corrosion of iron produces Corrosion of iron produces iron(III)oxide and other iron oxides iron(III)oxide and other iron oxides depending on the conditions, depending on the conditions, moisture and oxygen levels. moisture and oxygen levels.
• Other active metals also corrode Other active metals also corrode and form oxides, but frequently this and form oxides, but frequently this is in the form of a stable oxide is in the form of a stable oxide coating on the metal surface. coating on the metal surface.
• The oxides of iron flake off easily, The oxides of iron flake off easily, exposing fresh iron surfaces for exposing fresh iron surfaces for additional oxidation. additional oxidation.
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9292
CORROSION: REDOX REACTIONS CORROSION: REDOX REACTIONS IN THE ENVIRONMENT IN THE ENVIRONMENT
• These processes are true These processes are true electrochemical processes and a cathode electrochemical processes and a cathode and an anode can be identified. and an anode can be identified.
• The presence of conducting ions The presence of conducting ions enhances the process. (Salt water spray, enhances the process. (Salt water spray, salt water from ice suppression, etc.)salt water from ice suppression, etc.)
• One protection process is called anodic One protection process is called anodic inhibition and involves changing the inhibition and involves changing the surface of the metal. surface of the metal.
• Paint and other coatings are examples. Paint and other coatings are examples.
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9393
CORROSIONCORROSION
• A more effective method is A more effective method is cathodic protection which places cathodic protection which places the metal in electrical contact with the metal in electrical contact with a more active metal like a more active metal like magnesium or zinc. magnesium or zinc.
• The latter with iron is called The latter with iron is called galvanization and the former with galvanization and the former with iron is called a sacrificial anode. iron is called a sacrificial anode.
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9494
Cathodic Protection
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9595
Fe corrosion to rust FeFe corrosion to rust Fe22OO33
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9696
Why Aluminum does not rustWhy Aluminum does not rust
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9797
20.8 ELECTROLYSIS: 20.8 ELECTROLYSIS: CHEMICAL CHANGE FROM CHEMICAL CHANGE FROM ELECTRIC ENERGYELECTRIC ENERGY• When a reaction is driven by an When a reaction is driven by an
external electrical energy source external electrical energy source the process is called electrolysis. the process is called electrolysis.
• The anode and cathode of such The anode and cathode of such cells are the sites of oxidation and cells are the sites of oxidation and reduction as in a galvanic cell, but reduction as in a galvanic cell, but the polarity of the electrodes is the polarity of the electrodes is reversed. reversed.
• The anode is now (+) and The anode is now (+) and the cathode (-) . the cathode (-) .
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9898
ELECTROLYSISELECTROLYSIS
• For the electrolysis of molten NaCl, NaFor the electrolysis of molten NaCl, Na++ is is reduced to Na, and Clreduced to Na, and Cl-- is oxidized to Cl is oxidized to Cl22. .
• In aqueous solution, HIn aqueous solution, H22O is reduced to HO is reduced to H2 2
and OHand OH-- and Cl and Cl-- is oxidized to Cl is oxidized to Cl22. .
• These decisions are made by consulting These decisions are made by consulting the standard electrode potential table to the standard electrode potential table to determine which species are most easily determine which species are most easily oxidized and reduced. oxidized and reduced.
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9999
ELECTROLYSISELECTROLYSIS• The oxidation and reduction of water The oxidation and reduction of water
can be a little tricky since the table is a can be a little tricky since the table is a Standard electrode potential table Standard electrode potential table which means the all reactants and which means the all reactants and products are present at standard products are present at standard concentrations, 1M and 1 atm. concentrations, 1M and 1 atm.
• Reduction of water to form HReduction of water to form H2(g)2(g) occurs occurs at - 0.414 V when concentration is at - 0.414 V when concentration is taken into account, and oxidation of taken into account, and oxidation of water to form Owater to form O2(g)2(g) occurs at - 0.816 V. occurs at - 0.816 V.
• Over-voltages for hydrogen and Over-voltages for hydrogen and oxygen gases also complicate matters. oxygen gases also complicate matters.
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100100
Electrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOHElectrolysis of Aqueous NaOH
Anode (+) Anode (+) EEoo = - 0.40 V = - 0.40 V
4 OH4 OH-- ---> O ---> O22(g) + 2 H(g) + 2 H22O + 2eO + 2e--
Cathode (-) Cathode (-) EEoo = - 0.83 V = - 0.83 V
4 H4 H22O + 4eO + 4e-- ---> 2 H ---> 2 H22 + 4 OH + 4 OH--
EEoo for cell = - 1.23 V for cell = - 1.23 V
Electric Energy ----> Chemical ChangeElectric Energy ----> Chemical Change
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101101
ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change
ElectrolysisElectrolysisElectric Energy ---> Chemical ChangeElectric Energy ---> Chemical Change
BATTERY
+
Na+Cl-
Anode Cathode
electrons
BATTERY
+
Na+Cl-
Anode Cathode
electrons
Electrolysis of molten NaCl.Here a battery “pumps” electrons from Cl- to Na+.
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102102
Electrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaClElectrolysis of Molten NaCl
Anode (+) Anode (+) EEoo = - 1.36 V = - 1.36 V
2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e(g) + 2e--
Cathode (-) Cathode (-) EEoo = - 2.71 V = - 2.71 V
NaNa++ + e + e-- ---> Na ---> Na
EEoo for cell = - 4.07 V for cell = - 4.07 V
External energy needed External energy needed because Ebecause Eoo is (-). is (-).
Note that signs of electrodes Note that signs of electrodes are reversed from batteries.are reversed from batteries.
BATTERY
+
Na+Cl-
Anode Cathode
electrons
BATTERY
+
Na+Cl-
Anode Cathode
electrons
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103103
Simpler process for making Na(s)?Simpler process for making Na(s)?
• We just saw the molten process of We just saw the molten process of taking NaCl and by applying a taking NaCl and by applying a voltage (electrolysis) we can make voltage (electrolysis) we can make both Na(s) and Clboth Na(s) and Cl22(g)(g)
• Wouldn’t it be easier to take sea-Wouldn’t it be easier to take sea-water, perform electrolysis on the water, perform electrolysis on the “brine” to get Na(s) and Cl“brine” to get Na(s) and Cl22(g)(g)
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104104
What are the possible reactionsWhat are the possible reactions
At the anodeNa+ + e- Na(s)2H2O(l) + 2e- H2(g) + 2OH-
At the cathode2Cl- Cl2 + 2e-
2H2O(l) O2(g) + 4H+ + 4e-
E = -2.714E = -0.8277
E = -1.360E = -1.229
Once a potential is applied the process with the lowest potential will start FIRST!
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105105
At the anodeNa+ + e- Na(s)2H2O(l) + 2e- H2(g) + 2OH-
At the cathode2Cl- Cl2 + 2e-
2H2O(l) O2(g) + 4H+ + 4e-
E = -2.714E = -0.8277
E = -1.360E = -1.229
Once a potential is applied the process with the lowest potential will start FIRST!
The process with the lowest potential:4H2O(l) + 4e- 2H2(g) + 4OH-
2H2O(l) O2(g) + 4H+ + 4e-
2H2O(l) O2(g) + 2H2
E = -0.8277E = -1.229Ecell = -2.057V
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106106
Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl
Electrolysis of Aqueous Electrolysis of Aqueous NaClNaCl
Since Na(s) can not be made from sea-water due to hydrogen and oxygen gas are made another technique is used on sea-water
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107107
“Brine Process” for making NaOH from sea-water
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108108
Electrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClElectrolysis of Aqueous NaClCells like these are the source of Cells like these are the source of
NaOH and ClNaOH and Cl22..
In 1995In 1995
25.1 x 1025.1 x 1099 lb Cl lb Cl22
26.1 x 1026.1 x 1099 lb NaOH lb NaOH
Also the source of NaOCl for use in bleach.
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109109
Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22
Anode (+) Anode (+) EEoo = - 1.36 V = - 1.36 V
2 Cl2 Cl-- ---> Cl ---> Cl22(g) + 2e(g) + 2e--
Cathode (-) Cathode (-) EEoo = + 0.34 V = + 0.34 V
CuCu2+2+ + 2e + 2e-- ---> Cu ---> Cu
EEoo for cell = - 1.02 V for cell = - 1.02 V
Note that CuNote that Cu+2+2 is more is more easily reduced than easily reduced than either Heither H22O or NaO or Na++..
BATTERY
+
Cu2+Cl-
Anode Cathode
electrons
H2O
BATTERY
+
Cu2+Cl-
Anode Cathode
electrons
H2OThe water process potential:4H2O(l) + 4e- 2H2(g) + 4OH-
2H2O(l) O2(g) + 4H+ + 4e-
2H2O(l) O2(g) + 2H2
E = -0.8277E = -1.229Ecell = -2.057V
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• AluminumAluminum• Chlorine, sodium hydroxide, Chlorine, sodium hydroxide,
and hydrogenand hydrogen• These chemicals are extremely These chemicals are extremely
important to the economy. important to the economy. • The power required for these The power required for these
processes can be enormous. processes can be enormous.
THE COMMERCIAL PRODUCTION THE COMMERCIAL PRODUCTION OF CHEMICALS BY OF CHEMICALS BY ELECTROCHEMICAL METHODSELECTROCHEMICAL METHODS
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Producing AluminumProducing AluminumProducing AluminumProducing Aluminum2 Al2 Al22OO33 + 3 C ---> 4 Al + 3 CO + 3 C ---> 4 Al + 3 CO22
Charles Hall (1863-1914) at age of 19 developed the Al electrolysis process. Founded Alcoa!
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COMMERCIAL PRODUCTION COMMERCIAL PRODUCTION OF CHEMICALSOF CHEMICALS
• Power is measured in watts and Power is measured in watts and has basic units of has basic units of joules/second. joules/second.
• Power is calculated from the Power is calculated from the equationequation
P = V I, (J/C) (C/S) = J/SP = V I, (J/C) (C/S) = J/S• A kilowatt-hour is an energy A kilowatt-hour is an energy
unit and is equal to unit and is equal to 3.60x103.60x1066J:J:3600 S/hr times 1000 W/kW3600 S/hr times 1000 W/kW
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20.8 COUNTING ELECTRONS20.8 COUNTING ELECTRONS• Current, I, is measured in amperes and Current, I, is measured in amperes and
has basic units of coulombs/second. has basic units of coulombs/second. • Faraday's Law tells us that the current Faraday's Law tells us that the current
times the time can be used to calculate the times the time can be used to calculate the mass of material oxidized or reduced. mass of material oxidized or reduced.
• FF, Faraday's constant, is 9.65x10, Faraday's constant, is 9.65x1044C/mol eC/mol e--. . • These unit factor problems change: These unit factor problems change:
current x time ==> coulombs ==> current x time ==> coulombs ==> moles emoles e- - ==> moles oxidized or reduced ==> moles oxidized or reduced species ==> massspecies ==> mass
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• The balanced half-reaction The balanced half-reaction provides the conversion from provides the conversion from moles electrons to moles of moles electrons to moles of species sought. species sought.
• Calculate the grams of aluminum Calculate the grams of aluminum formed at the cathode if a current formed at the cathode if a current of 4.40 A flows for 1.50 hours.of 4.40 A flows for 1.50 hours.
–Let’s do an example first.Let’s do an example first.
COUNTING ELECTRONSCOUNTING ELECTRONS
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Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous Consider electrolysis of aqueous silver ion.silver ion.
AgAg++ (aq) + e (aq) + e-- ---> Ag(s) ---> Ag(s)1 mol e1 mol e-- ---> 1 mol Ag---> 1 mol AgIf we could measure the moles of If we could measure the moles of
ee--, we could know the quantity of , we could know the quantity of Ag formed.Ag formed.
But how to measure moles of eBut how to measure moles of e--??
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Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.
AgAg++ (aq) + e (aq) + e-- ---> Ag(s) ---> Ag(s)1 mol e1 mol e-- ---> 1 mol Ag---> 1 mol AgIf we could measure the moles of eIf we could measure the moles of e--, we , we
could know the quantity of Ag formed.could know the quantity of Ag formed.But how to measure moles of eBut how to measure moles of e--??
Current = charge passingtime
I (amps) = coulombsseconds
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But how is charge related to moles of But how is charge related to moles of electrons?electrons?
Charge on 1 mol of eCharge on 1 mol of e-- = = (1.60 x 10(1.60 x 10-19 -19 C/eC/e--)(6.02 x 10)(6.02 x 102323 e e--/mol)/mol)
= = 96,500 C/mol e96,500 C/mol e- - = = 1 Faraday1 Faraday
Current = charge passing
time
I (amps) = coulombsseconds
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry
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Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) (aq) solution for 15.0 min. What mass solution for 15.0 min. What mass of Ag metal is deposited?of Ag metal is deposited?
SolutionSolution(a)(a) Calculate the chargeCalculate the charge
Coulombs = amps x timeCoulombs = amps x time= (1.5 amps)(15.0 min)(60 s/min) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C= 1350 C
I (amps) = coulombsseconds
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Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 (aq) solution for 15.0 min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?
SolutionSolution
(a)(a) Charge = 1350 CCharge = 1350 C(b)(b) Calculate moles of eCalculate moles of e-- used used
1350 C • 1 mol e-
96,500 C 0.0140 mol e-
(c)(c) Calculate quantity of AgCalculate quantity of Ag0.0140 mol e- •
1 mol Ag1 mol e-
0.0140 mol Ag or 1.51 g Ag
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Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is
Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e(aq) + 2e--
If a battery delivers 1.50 amp, and you have If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?454 g of Pb, how long will the battery last?
SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Calculate moles of eCalculate moles of e--
2.19 mol Pb • 2 mol e-
1 mol Pb = 4.38 mol e-
c)c) Calculate chargeCalculate charge
4.38 mol e4.38 mol e-- • 96,500 C/mol e • 96,500 C/mol e-- = 423,000 C = 423,000 C
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Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry
SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Mol of e- = 4.38 molMol of e- = 4.38 molc)c) Charge = 423,000 CCharge = 423,000 C
d)d) Calculate timeCalculate time
Time (s) = Charge (C)
I (amps)
Time (s) = 423,000 C1.50 amp
= 282,000 s
About 78 hoursAbout 78 hours
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• Remember:Remember:–The balanced half-reaction The balanced half-reaction provides the conversion provides the conversion from moles electrons to from moles electrons to moles of species sought.moles of species sought.
Now solve the problem:Now solve the problem:
• Calculate the grams of aluminum Calculate the grams of aluminum formed at the cathode if a current formed at the cathode if a current of 4.40 A flows for 1.50 hours.of 4.40 A flows for 1.50 hours.
COUNTING ELECTRONSCOUNTING ELECTRONS
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BALANCING REDOX BALANCING REDOX EQUATIONSEQUATIONS
The Half-Reaction MethodThe Half-Reaction Method• Separate the equation into half-Separate the equation into half-
reactions.reactions.
• Balance the half-reactions.Balance the half-reactions.
• Combine the half-reactions to form a Combine the half-reactions to form a balanced equation containing no balanced equation containing no electrons.electrons.
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Balancing Half-ReactionsBalancing Half-ReactionsBalancing Half-ReactionsBalancing Half-Reactions
• First balance the element changing First balance the element changing oxidation state.oxidation state.
• Balance the oxygen atoms with water.Balance the oxygen atoms with water.
• Balance the hydrogen atoms with HBalance the hydrogen atoms with H++..
• Balance theBalance the charge with electrons.charge with electrons.
After combining the half-reactions, After combining the half-reactions,
check for mass and charge balance.check for mass and charge balance.
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Practice ProblemsPractice ProblemsPractice ProblemsPractice ProblemsBalance the following equation:Balance the following equation:
MnOMnO44-- + H + H22SOSO3 3 ----> Mn ----> Mn+2+2 + SO + SO44
-2-2
MnOMnO44-- ----> Mn ----> Mn+2+2
HH22SOSO3 3 ----> SO ----> SO44-2-2
2. Balance atoms:
8 H8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22O O
HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++
1. Separate into half reactions:
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems3. Balance charges:
5 e5 e-- + 8 H + 8 H++ + MnO + MnO44------> Mn----> Mn+2+2 + 4 H + 4 H22O O
HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H++ + 2 e + 2 e--
4. Equal electrons gained and lost:
2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))
5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2+ 4 H+ 4 H+++ 2 e+ 2 e--))
10 16 2 2 8
5 5 5 20 10
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
5. Simplify and Add:
2(2(5 e5 e-- + 8 H + 8 H++ + MnO + MnO44-- ----> Mn ----> Mn+2+2 + 4 H + 4 H22OO))
5(5(HH22O + HO + H22SOSO3 3 ----> SO ----> SO44-2-2 + 4 H + 4 H+++2 e+2 e--))
10 16 2 2 8
5 5 5 20 10
2MnO2MnO44-- + 5 H + 5 H22SOSO3 3 ----> ---->
2 Mn2 Mn+2 +2 + 5 SO+ 5 SO44-2-2 + 4 H + 4 H++ + + 33 HH22O O
4
3
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Practice ProblemsPractice ProblemsPractice ProblemsPractice ProblemsBalance the following equation:Balance the following equation:
Al + NOAl + NO33-- ---> Al(OH) ---> Al(OH)44
-- + NH + NH33
Al ----> Al(OH)Al ----> Al(OH)44--
NONO33-- ----> NH ----> NH33
2. Balance atoms:
4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++
9 H9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 H+ 3 H22OO
1. Separate into half reactions:
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems3. Balance charges:
4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 + 4
HH++ + 3 e + 3 e--8 e8 e-- + 9 H + 9 H+ + + NO+ NO33-- ----> NH ----> NH3 3 + 3 + 3
HH22OO 4. Equal electrons gained and lost:
8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++
+ 3 e+ 3 e--)) 3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33
-- ----> NH ----> NH3 3 + 3 + 3
HH22OO))
32 8 8 32 24
24 27 3 3 9
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
5. Simplify and Add:
8(8(4 H4 H22O + Al ----> Al(OH)O + Al ----> Al(OH)44-- + 4 H + 4 H++
+ 3 e+ 3 e--))3(3(8 e8 e-- + 9 H + 9 H+ + + NO+ NO33
-- ----> NH ----> NH3 3 + 3 + 3
HH22OO))
32 8 8 32 24
24 27 3 3 9
23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ---->
8 8
Al(OH)Al(OH)44-- + 5 H + 5 H++ + 3 NH + 3 NH3 3
523
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
6. Change to basic solution:
23 H23 H22O + 8 Al + 3 NOO + 8 Al + 3 NO33-- ----> ----> 8 Al(OH)8 Al(OH)44
--
+ 5 H+ 5 H++ + 3 NH + 3 NH3 3 + 5 OH- + 5 OH -
5 H2O
18
18 H18 H22O + 5 OHO + 5 OH-- + 8 Al + 3 NO + 8 Al + 3 NO33-- ---->8 ---->8
Al(OH)Al(OH)44-- + 3 NH + 3 NH3 3
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
1. MnO2 + HBr --> Br2 + MnBr2
2e2e-- + 4 H + 4 H++ + MnO + MnO22 ----> Mn ----> Mn2+2+ + 2 + 2
HH22OO 2 Br2 Br-- ----> Br ----> Br2 2 + 2e+ 2e--
4 H4 H++ + MnO + MnO22 + 2 Br + 2 Br-- ----> Mn ----> Mn2+2+ + 2 + 2
HH22O + BrO + Br22
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2. Cl2 + NaBr --> NaCl + Br2
2e2e-- + Cl + Cl22 --> 2 Cl --> 2 Cl--
2 Br --> Br2 Br --> Br22 + 2e + 2e--
ClCl22 + 2 Br + 2 Br-- ----> 2 Cl ----> 2 Cl-- + +
BrBr2 2
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
3. H2S + HNO3 --> S + NO
3(3(HH22S ----> S + 2 HS ----> S + 2 H++ + 2 e + 2 e--))
2(2(3 e3 e-- + 4 H + 4 H+ + + NO+ NO33-- ----> NO ----> NO + 2 + 2
HH22OO))
3 3 6 6
6 8 2 2 42
3 H2S + 2 H+ + 2 NO3- ----> 3 S + 2 NO + 4 H2O
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Practice ProblemsPractice ProblemsPractice ProblemsPractice Problems
4. PbO2 + Sb --> PbO + NaSbO2
(base)
3(3(2 e- +2 e- + 2H2H++ + PbO + PbO22 ----> PbO + ----> PbO + HH22OO))
2(2(2 H2 H22O + Sb ----> SbOO + Sb ----> SbO22-- + 4 + 4
HH++ + 3 e + 3 e--))
6 6 3 3 3
4 2 2 8 6
3 PbO3 PbO22 + H + H22O + 2 Sb --->3 PbO + 2 O + 2 Sb --->3 PbO + 2
SbOSbO22-- + 2 H + 2 H++
21
+ 2 OH- + 2 OH-
2 H2O
3PbO3PbO22 + 2 OH- + 2Sb --->3PbO + + 2 OH- + 2Sb --->3PbO +
2SbO2SbO22-- + H + H22OO
1