electrochemistry: oxidation-reduction reactions zn(s) + cu +2 (aq) zn 2+ (aq) + cu(s) loss of 2e -...
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Electrochemistry: Oxidation-Reduction Reactions
Zn(s) + Cu+2 (aq) Zn2+(aq) + Cu(s)
loss of 2e-
gaining to 2e-
Zinc is oxidized - it goes up in charge and is the reducing agent
Copper is reducedreduced because it goes down in charge and is the oxidizing agent
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Electrochemistry
Messing with electrons
What is the charge on each atom
KMnO4 Cr(OH)3 Fe(OH)2+1
-2
-8
+1
+1 +7
+7 -2 +1
-3+3
+3 -1
-2 = 1+3
+3
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Electrochemistry: Oxidation-Reduction Reactions
By writing the oxidation number of each element under the reaction equation, we can easily see the oxidation state
changes that occur
Zn(s) + 2H+(aq) Zn2+(aq) + H2(aq)0 +1 +2 0
In any oxidation-reduction reaction (redox), both oxidation and reduction must occur.
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Electrochemistry: Balancing Oxidation-Reduction Reactions
Oxidation Number Method
Al(l) + MnO2 Al2O3 + Mn 0 +4 -2 +3 -2 0
-3e’s
+4e’s
X 4 = -12
X 3 = +12
4 3 2 3
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Electrochemistry: Balancing Oxidation-Reduction Reactions
Sample problem:
I2O5(s) + CO(g) I2(s) + CO2(g)
-2e’s
+5e’s
+2 +4
+5 0X 2 = 10
X 5 = -10
5 5
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:
MnO4-(aq) + C2O4
2-(aq) Mn2+(aq) + CO2(g) (acid)
Half-Reaction Method
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: Half-Reaction Method (in acid)
MnO4- Mn2+
C2O42- CO2
MnO4-(aq) + C2O4
2-(aq) Mn2+(aq) + CO2(g)
Step 1: Divide into two half-reactions,
Step 2: Balance main element
MnO4- Mn2+
C2O42- 2CO2
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: : Step 3: Balance the O atoms by adding H2O
MnO4- Mn2+
C2O42- 2CO2
Step 5: Balance charge with electrons
8H++ MnO4- Mn2+ + 4H2O
C2O42- 2CO2
+ 4H2O
8H+ +
+7 +2
5e- + -2 0
+ 2e-
Step 4: Balance the H atoms by adding H+
MnO4- Mn2+ + 4H2O
C2O42- 2CO2
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: Step 6: Make the electrons balance
5e- + 8H++ MnO4- Mn2+ + 4H2O
C2O42- 2CO2 + 2e-
10e- + 16H++ 2MnO4- 2Mn2+ + 8H2O
5C2O42- 10CO2 +10e-
Step 7: Cancel and add
16H+ + 2MnO4- + 5C2O4
2- 2Mn2+ + 10CO2 + 8H2O
10 16 2 2 8
5 10 10
+
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Balancing in base
CN- + MnO4- CNO- + MnO2
CN- CNO-
MnO4- MnO2
H2O +
+ 2H2O
+ 2H+
4H+ +
-1 +1
+3 0
+ 2e-
3e-+
3 3 3 6 6
6 8 2 2 4
2H+ + 3CN- + 2MnO4- 3CNO- + 2MnO2 + H2O
2OH-+ 2OH-
2H2O1
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Balancing in base
Cr(OH)3 + ClO CrO4-2 + Cl2
Have at it
4Cr(OH)3 + 6ClO + 8OH- 4CrO4-2 + 3Cl2 + 10H2O
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Note that the activity series is simply the
oxidation half-reactions of the
metals ordered from the highest to lowest
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Using the hydrogen electrode as a referenceUsing the hydrogen electrode as a reference2H2H++ (1 M) + 2e (1 M) + 2e-- H2 (1 atm, 25 °C) E°red = 0 V
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Cell EMF or Voltage
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s) E°cell = 1.10 V
E ° ox + E ° red = E°cell
From the table:Zn+2 + 2e- Zn red= -.76VCu+2 + 2e- Cu red= .34V
Flip
Zn Zn+2 + 2e- ox = .76V
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Voltaic or Galvanic Cells
-oxidation
+reduction
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Sample Problem. Draw a voltaic cell using the following equation and label
all parts
CrCr22OO772-2-
(aq)(aq) + I + I--(aq)(aq) Cr Cr3+3+
(aq)(aq) + I + I2(s)2(s)
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Spontaneity and Extent of Redox ReactionsA positive emf indicates a spontaneous process, and a negative emf indicates a nonspontaneous one.Sample Problem: Using the standard electrode potentials, determine whether the following reaction are spontaneous A. Cu(s) + 2H+
(aq) Cu Cu2+2+(aq)(aq) + H + H2(g)2(g)
B. Cl2(g) + 2I-(aq) 2Cl 2Cl--
(aq)(aq) + I + I2(s)2(s)
Cu(s) Cu Cu2+2+(aq)(aq) + 2e + 2e- - EEoxox
= -.34V= -.34V2H+
(aq) + 2e- H H2(g) 2(g) EEredred = 0.0V= 0.0V
EEcellcell = -.34V= -.34V
Cl2(g) + 2e- 2Cl 2Cl--(aq)(aq) EEredred
= 1.36V= 1.36V 2I-
(aq) 2e 2e- - + I+ I2(s)2(s) EEoxox = -.54V= -.54V
EEcellcell = .82V= .82V
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EMF and Free Energy ChangeAny redox reaction involves free energy change
(G) which also may be used as a measure of spontaneity or work (max or min.)
n = # of moles of e-s transferred = 96500 C the charge of one mole of e-s
or 96,500 J/V-mol e-
Note that because n and are both positive values, a positive value in E leads to a negative value of G.
G° = -nE °
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Use the standard electrode potentials to calculate the standard free energy
change, G°, for the following reaction
2Br-(aq) + F2(g) Br2(l) + 2F-
(aq)
2Br-(aq) Br2(l) + 2e E°ox = -1.06
F2(g) + 2e- 2F-(aq) E°red = 2.87
2Br-(aq) + F2(g) Br2(l) + 2F-
(aq) E°cell = 1.81 V
G = -nEG = -(2 mol e- )(96,500 J/volt-mol e-)( 1.81 V)G° = -3.49 x 105 J = -349 kJ
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EMF and Equilibrium Constant“Remember in Chapter 19, we related G° to
the equilibrium constant, K?”
G° = -RT lnKG ° = -nE °
-nE° = -RT lnK
E ° = 0.0591 logK n
Simplify
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Calculate the K for the following reactionO2 (g) + 4H+
(aq) + 4Fe2+ (aq) 4Fe3+
(aq) + 2H2O (l)
O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E°red = 1.23 V
4Fe2+ (aq) 4Fe3+
(aq) + 4e- E°ox = -0.77 V
O2 (g) + 4H+ (aq) + 4Fe2+
(aq) 4Fe3+ (aq) + 2H2O (l)
E°cell = 0.46 V
log K = n E °
0.0591V 4(0.46V)
0.0591V= = 31.1
K = 1.36 x 1031
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Calculate the equilibrium constant for the reaction
IO3-
(aq) + 5Cu (s) + 12H+ (aq) I2 (s) + 5Cu2+
(aq) + 6H2O (l)
2IO3-
(aq) + 12H+ (aq) + 10e- I2 (s) + 6H2O (l) red= 1.20V
5Cu (s) 5Cu2+ (aq) + 10e- ox= -.34V
cell= .86V = .0591 log k n .86V = .0591 log k
10Log K = 145.5
K = 10145.5
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G = G° + RT lnQG° = -RT lnKStandard Conditions
Non- Standard Conditions
G = G° + 2.30 RT log Q
-nE = -nE ° + 2.30 RT log Q
E = E ° - 2.30 RT log Qn
Nernst Equation
E = E ° - 0.0591 log Qn
When T=298K
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Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) E °= 1.10 V
E = 1.10 ° - 0.0591 log
2
[Zn2+]
[Cu2+]
If [Zn2+]=0.05M, and [Cu2+] = 0.50 M
E = 1.10 ° - 0.0591 log
2
[0.05]
[.50]= 1.13V
2e- + Cu2+ (aq) Cu (s)
Zn (s) Zn2+ (aq) + 2e-
Therefore
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Calculate the emf generated by the the following reaction
CrCr22OO772-2- + 14 H + 14 H ++ + 6I+ 6I-- 2Cr 2Cr3+3+ + 3I + 3I22 + 7H + 7H22OO
[Cr[Cr22OO772-2-] = 2.0 M] = 2.0 M
[H [H ++ ] = .05 M ] = .05 M
[I[I--] = .25 M] = .25 M
[Cr[Cr3+3+] = 1.0 x 10] = 1.0 x 10-5-5MM
red= 1.33V
ox= -.54V
cell= .79V
= .79V - .0591 log (1 x 10-5)2
6 (2.0)(.05)14(.25)6
= .68V
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Concentration CellsWhat happens when both cells are identical except for
concentration. Nature will try to equalize the two cells.
E= Eº - .0591 log Qn
E = 0 - .591 log .01 = .591V.1
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anode: Pb (s) + SO42-
(aq) PbSO4 (s) + 2e + 2e-- E °= 0.356 V
cathode:PbO2(s)+ SO42-
(aq)+ H+(aq)+ 2e- PbSO4(s)+ 2H2O(l)
E °= 1.685 V Pb(s)+ PbO2(s)+ 4H+ +2SO4(aq) 2PbSO4(s)+
2H2O(l) E °= 2.041 V
Note that one advantage of the lead storage battery is that it can be recharged because the PbSO4 produced during discharge
adheres to the electrodes
Lead Storage BatteryLead Storage Battery
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Dry Cell alkalineDry Cell alkalineDry Cell alkalineDry Cell alkalineanode: Zn (s) Zn 2+ 2+
(aq)(aq) + 2e + 2e--
cathode:2NH4+
(aq)+2MnO2(s)+ 2e-
Mn2O3(s) + 2NH3(aq) + H2O (l)
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Fuel CellsFuel CellsFuel CellsFuel Cellsanode: 2H2(g)+ 4OH-
(aq) 4H2O(l) + 4e- cathode: 4e- + O2(g) + H2O(l) 4OH 4OH--
(aq)
2H2(g) + O2(g) 2H 2H22OO(l)(l)
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Electrolytic Cells: Electrolytic Cells: ElectrolysisElectrolytic Cells: Electrolytic Cells: Electrolysis
Electrolysis is driven by an outside energy Electrolysis is driven by an outside energy sourcesource
voltage source actsvoltage source acts like an electronlike an electron pumppump
voltage source actsvoltage source acts like an electronlike an electron pumppump(- red)
These electrodesare inertThese electrodesare inert
Notice
(+ ox.)
Eº = ( -)
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Electrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsElectrolysis of Aqueous SolutionsSodium cannot be prepared by electrolysis of aqueous solutions of NaCl, because wateris more easily reduced than Na+:
2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 NaNa++ + e + e- - Na Na(s)(s) E° E°redred = -2.71 = -2.71 NaNa++ + e + e- - Na Na(s)(s) E° E°redred = -2.71 = -2.71 The possible Anode reactions are:
2Cl2Cl- - Cl Cl22 E° E°oxox = -1.36 = -1.362Cl2Cl- - Cl Cl22 E° E°oxox = -1.36 = -1.362H2H22O O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23 2H2H22O O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23
2Cl2Cl--(aq)(aq) Cl Cl2(g)2(g) E° E°oxox = -1.36 = -1.362Cl2Cl--(aq)(aq) Cl Cl2(g)2(g) E° E°oxox = -1.36 = -1.36
2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 2H2H22O + 2eO + 2e- - H H22 + 2OH + 2OH-- E° E°redred = -0.83 = -0.83 E°E°cellcell >> -2.19 -2.19 E°E°cellcell >> -2.19 -2.19
Possible Cathode reactions
Therefore
(Note: Not in table on AP exam)
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Electrolysis of With Active ElectrodesElectrolysis of With Active Electrodes
Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e-- E° E°oxox = 0.28 = 0.28Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e-- E° E°oxox = 0.28 = 0.28
2H2O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23 2H2O 4H 4H++ + O + O22 + 4e + 4e-- E° E°oxox = -1.23 = -1.23
Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e--Ni(s) Ni Ni2+2+(aq)(aq) + 2e + 2e--
NiNi2+2+(aq)(aq) + 2e + 2e-- Ni Ni(s)(s)NiNi2+2+(aq)(aq) + 2e + 2e-- Ni Ni(s)(s)
anode:anode:
cathode:cathode:
Electroplating creates a silver liningElectroplating creates a silver lining
Possibilities
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Quantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysis
Calculate the mass of aluminum produced in 1.00 hr. by the electrolysis of molten AlCl3 if the current is 10.0 A. (C = amperes x seconds)
= 3.36 g of Al
(10.0A)(1.00 hr)(3600 sec)
(1 hr) (1 A-s)
(1C)
(96,500C)(1)
(3)(1molAl)
(1 mol Al)(27.0 g Al)
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Quantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysis
The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e- Mg. Calculate the mass of magnesium formed Mg. Calculate the mass of magnesium formed upon passage of 60.0 A for a period of 4000 supon passage of 60.0 A for a period of 4000 s
(60.0A)(4000s)(1C)(1) (mole Mg)(24.3g) (A•s)(96,500C) (2) (mole Mg)
30.2g Mg
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Electrical WorkElectrical WorkElectrical WorkElectrical Work
since G= wmax
and G = - nEthen wmax= - nE
The unit employed by electric The unit employed by electric utilities is kilowatt-hour utilities is kilowatt-hour (kWh =3.6 x 10(kWh =3.6 x 1066J)J)
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Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
All metals except gold and platinum are thermodynamicallycapable of undergoing oxidation in air at room temperature
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The rusting of iron is known to require oxygen; iron does not rust in water unless O2
is present.
E°red = 0.44 V
E°red = 1.23 V
Note that as the pH increases, the reduction of O2 becomes less favorable
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The corrosion of Iron:Protecting the surface with tinThe corrosion of Iron:Protecting the surface with tin The corrosion of Iron:Protecting the surface with tinThe corrosion of Iron:Protecting the surface with tin
Fe(s) Fe Fe2+2+ + 2e + 2e- - E°ox = 0.44 V
Sn(s) Sn Sn2+2+ + 2e + 2e- - E°ox = 0.14 V
Tin has lower E°ox so less likely to oxidize until the
surface is broken then it accelerates as it makes a voltaic cell with iron.
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The corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic Protection
Fe(s) Fe Fe2+2+(aq) + 2e(aq) + 2e- - E°ox = 0.44 V
Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e- - E°ox = 0.76V
GalvanizationGalvanization
sacrificialanode
Zinc is more positive andgoes first and hasan oxide coatthat seals.
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The corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic ProtectionThe corrosion of Iron:Cathodic Protection