1 CHAPTER 20 Electron Transfer Reactions Why Study Electrochemistry? Why Study Why Study Electrochemistry? Electrochemistry? Batteries Batteries Corrosion Corrosion Industrial production Industrial production of chemicals such as of chemicals such as Cl Cl 2 , , NaOH NaOH, F , F 2 and Al and Al Biological Biological redox redox reactions reactions The The heme heme group group TRANSFER REACTIONS TRANSFER REACTIONS Atom/Group transfer Atom/Group transfer HCl HCl + H + H 2 O O --- --- > > Cl Cl - + H + H 3 O + Electron transfer Electron transfer Cu(s) + 2 Cu(s) + 2 Ag Ag + (aq (aq) ) --- --- > Cu > Cu 2+ 2+ (aq) + 2 Ag(s) (aq) + 2 Ag(s) 4 Oxidation-Reduction Reactions Rules for assigning oxidation numbers Alkali (+1), Alkaline Earth (+2), Al family (+3) Halides (-1) w/out oxygen Oxygen family (-2) (except peroxide, superoxide) H (+1) (except with metal -1, hydride) Transition metals (by difference) Oxidation-Reduction Reactions Oxidation – increase in oxidation # Loss of electron Fe(III) Fe(II) + e - Reduction – decrease in oxidation # Gain of electron MnO 4 - + 5 e - Mn(II) 5 Review of Terminology for Redox Reactions Review of Terminology Review of Terminology for for Redox Redox Reactions Reactions OXIDATION—loss of electron(s) by a species; increase in oxidation number. REDUCTION—gain of electron(s); decrease in oxidation number. OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized. OXIDATION OXIDATION—loss of electron(s) by a species; loss of electron(s) by a species; increase in oxidation number. increase in oxidation number. REDUCTION REDUCTION—gain of electron(s); decrease in gain of electron(s); decrease in oxidation number. oxidation number. OXIDIZING AGENT OXIDIZING AGENT—electron acceptor; electron acceptor; species is reduced. species is reduced. REDUCING AGENT REDUCING AGENT—electron donor; species is electron donor; species is oxidized. oxidized.
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CHAPTER 20
� Electron Transfer Reactions
Why Study Electrochemistry?Why Study Why Study Electrochemistry?Electrochemistry?
� Alkali (+1), Alkaline Earth (+2), Al family (+3)
� Halides (-1) w/out oxygen
� Oxygen family (-2) (except peroxide,
superoxide)
� H (+1) (except with metal -1, hydride)
� Transition metals (by difference)
Oxidation-Reduction Reactions
� Oxidation – increase in oxidation #
� Loss of electron
� Fe(III) � Fe(II) + e-
� Reduction – decrease in oxidation #
� Gain of electron
� MnO4- + 5 e- � Mn(II)
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Review of Terminology for Redox ReactionsReview of Terminology Review of Terminology for for RedoxRedox ReactionsReactions
� OXIDATION—loss of electron(s) by a species;
increase in oxidation number.
� REDUCTION—gain of electron(s); decrease in
oxidation number.
� OXIDIZING AGENT—electron acceptor;
species is reduced.
� REDUCING AGENT—electron donor; species is
oxidized.
�� OXIDATIONOXIDATION——loss of electron(s) by a species; loss of electron(s) by a species;
increase in oxidation number.increase in oxidation number.
�� REDUCTIONREDUCTION——gain of electron(s); decrease in gain of electron(s); decrease in
oxidation number.oxidation number.
�� OXIDIZING AGENTOXIDIZING AGENT——electron acceptor; electron acceptor;
species is reduced.species is reduced.
�� REDUCING AGENTREDUCING AGENT——electron donor; species is electron donor; species is
oxidized.oxidized.
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The Half-Reaction Method
� Half reaction method rules:
1. Write the unbalanced reaction.
2. Break the reaction into 2 half reactions:
One oxidation half-reaction and
One reduction half-reaction
Each reaction must have complete formulas for
molecules and ions.
3. Mass balance each half reaction by adding
appropriate stoichiometric coefficients. To balance H and O we can add:
� H+ or H2O in acidic solutions.
� OH- or H2O in basic solutions.8
The Half-Reaction Method
4. Charge balance the half reactions by adding appropriate numbers of electrons.� Electrons will be products in the oxidation half-
reaction.
� Electrons will be reactants in the reduction half-reaction.
5. Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction.
6. Add the two half reactions.
7. Eliminate any common terms and reduce coefficients to smallest whole numbers.
Tips on Balancing EquationsTips on Balancing Equations
�� Never add ONever add O22, O atoms, or , O atoms, or
OO22-- to balance oxygen.to balance oxygen.
�� Never add HNever add H22 or H atoms to or H atoms to
balance hydrogen.balance hydrogen.
�� Be sure to write the correct Be sure to write the correct
charges on all the ions.charges on all the ions.
�� Check your work at the end to Check your work at the end to
make sure mass and charge make sure mass and charge
are balanced.are balanced.
�� PRACTICE!PRACTICE!
Balancing EquationsBalancing Equations
Cu + AgCu + Ag++ --------> Cu> Cu2+2+ + Ag+ Ag
Balancing EquationsBalancing Equations
Step 1:Step 1: Divide the reaction into halfDivide the reaction into half--reactions, reactions,
one for oxidation and the other for reduction.one for oxidation and the other for reduction.
OxOx Cu Cu ------> Cu> Cu2+2+
RedRed AgAg++ ------> Ag> Ag
Step 2:Step 2: Balance each for mass. Already done Balance each for mass. Already done
in this case.in this case.
Step 3:Step 3: Balance each halfBalance each half--reaction for charge reaction for charge
by adding electrons.by adding electrons.
OxOx Cu Cu ------> Cu> Cu2+2+ + + 2e2e--
RedRed AgAg++ + + ee-- ------> Ag> Ag
Balancing EquationsBalancing Equations
Step 4:Step 4: Multiply each halfMultiply each half--reaction by a factor so that reaction by a factor so that
the reducing agent supplies as many electrons as the the reducing agent supplies as many electrons as the
If a battery delivers 1.50 amp, and you have 454 g of If a battery delivers 1.50 amp, and you have 454 g of PbPb, how long will the battery last?, how long will the battery last?
SolutionSolution
a)a) 454 g 454 g PbPb = 2.19 mol = 2.19 mol PbPb
b)b) Mol of eMol of e-- = 4.38 mol= 4.38 mol
c)c) Charge = 423,000 CCharge = 423,000 C
Time (s) = Charge (C)
I (amps)
Time (s) = 423, 000 C
1.50 amp = 282, 000 s About 78 hoursAbout 78 hours
d)d) Calculate timeCalculate time
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Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the
reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
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Counting Electrons:
Coulometry and Faraday’s Law of Electrolysis� Example: Calculate the mass of palladium produced by the
reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.
( )
Cathode: Pd + 2e Pd
1 mol 2 mol 1 mol
106 g 2(96,500) 106 g
3.20 amp = 3.20
g = 30.0 min60 s
min
C
s
g Pd
2 96,500 C g Pd
2+ - 0
Cs
→
× × × =?.
.320 106
316
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Commercial Applications of Electrolytic Cells
Electrolytic Refining and Electroplating of Metals
� Impure metallic copper can be purified
electrolytically to ≈ 100% pure Cu.
� The impurities commonly include some active metals
�� Electrons are Electrons are ““drivendriven”” from anode to cathode by an from anode to cathode by an electromotive forceelectromotive force or or
emfemf..
�� For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚̊C and when C and when
[Zn[Zn2+2+] and [Cu] and [Cu2+2+] = 1.0 M.] = 1.0 M.
�� STANDARD CELL POTENTIAL, STANDARD CELL POTENTIAL, EEoo
�� ——a quantitative measure of the tendency of reactants to proceed ta quantitative measure of the tendency of reactants to proceed to o
products when all are in their standard states at 25 products when all are in their standard states at 25 ˚̊C. C.
Zn and ZnZn and Zn2+2+,,
anodeanodeCu and CuCu and Cu2+2+,,
cathodecathode
Zn
Zn2+ ions
Cu
Cu 2+ ions
wire
saltbridge
e lect rons
1.10 V1.10 V
1.0 M1.0 M 1.0 M1.0 M
Calculating Cell VoltageCalculating Cell Voltage
�� Balanced halfBalanced half--reactions can be reactions can be
added together to get overall, added together to get overall,