Download Chapter 20: Electrochemistry

Chapter 20: Electrochemistry
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Balancing Redox Reactions
Galvanic/Voltaic Cells
EMF & Standard Cell Potentials
EMF & K
Nernest Eq.
pH Measurements
Electrolytic Cells
Stoichiometry of Electrolysis
Oxidation - Reduction
Reactions
(Redox Reactions)
Involve the Transfer of Electrons
Between Different Reactants
Historically:
Oxidation Meant the Formation of Oxides and
Reduction Meant the Removal of Oxygen
From the Oxide
Redox Terminology
Oxidation - Loss of Electrons
Reduction - Gain of Electrons
OIL RIG Mnemonic
OIL – Oxidation Is Loss of Electron
Rig – Reduction Is Gain of Electron
Redox Terminology
Oxidation - Loss of Electrons
Reduction - Gain of Electrons
Note Oxidation and Reduction Must
Occur Concurrently!
Oxidant - Reactant Which Oxidizes (gets Reduced)
Reductant - Reactant Which Reduces (gets
Oxidized)
Redox Example
Oxygen is a Powerful Oxidizing Agent
Consider Formation of Iron(III) Oxide
4Fe + 3O2
4Fe
4Fe+3 + 12e-
3O2 + 12e4Fe +3O2
2Fe2O3
6O-2
2Fe2O3
Identification of Redox Rxns
-Assign each element an Oxidation Number
(State). If any element changes it’s oxidation
state during a reaction, it is a redox reaction
Zero Oxidation States indicate element
is neutral
[+] Oxidation States indicate element is
“electron poor”
[-] Oxidation States indicate element is
“electron rich”
Oxidation Numbers
1. Oxd # = 0 for pure elements
2. Oxd # = charge of monatomic ion
3. Oxd # of F = -1 in compounds with other elements
4. Oxd # of Cl, Br & I = -1 in compounds
except with Oxygen & Fluorine
5. Oxd # of H= +1, except for Metal Hydrides (-1)
6. Oxd # of O = -2, except for Peroxides (-1) and
Superoxides (-1/2)
7. The Sum of the oxd # ‘s of all elements in a
compound = 0, and = the charge of a polyatomic ion
1/2 Reaction Method
Consider the reduction of ferric iron
by iodide.
Fe+3 +2I-
Fe+2 + I2
Is the above Equation Balanced?
No! - Look at the Charge
-+1 = +2
Balancing Redox Reactions
3 Techniques
1. Half Reaction Method -Good If Reaction
Can Be Separated Into 2 1/2 Reactions
2. Oxidation # Method -Good If Reaction
Can Not Be Separated Into 2 1/2 Reactions
Half Cell Method
Batteries take advantage of redox
reactions where the oxidants and the
reductants can be physically separated.
These often occur in acidic and basic
environments and so you will need to be
able to balance ½ rxns in both Acidic and
Basic Environments
Balancing Acidic Redox Rxns
1. Split Skeletal Eq. Into 1/2 Rxns
-You may need to assign oxidation #’s to all elements
being oxidized or reduced
2. Balance all elements except O & H
3. Balance O by adding water
4. Balance H by adding H+
5. Balance charge by adding e6. Multiply 1/2 rxns by appropriate integer to
cancel electrons when adding ½ reactions
7. Add 1/2 reactions and check work
Balancing Basic Redox Rxns
1-7. Balance as if Acidic
8. Add OH- to both sides to
neutralize H+ (converting it to H2O)
1/2 Reaction Method
Balance the Basic Reaction of
Thiosulfate & Iodine to Form Sulfate,
Iodide & Water
S2O3-2 + I2
SO4-2 + I-
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
1. We can separate the above reaction into
2 1/2 reactions
Reduction 1/2 Reaction
I2
I-
Oxidation 1/2 Reaction
S2O3-2
2O4-2
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
2. Balance All Elements Except Oxygen and
Hydrogen
Reduction 1/2 Reaction
I2
2I-
Oxidation 1/2 Reaction
S2O3-2
2SO4-2
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
3. Balance Oxygen by Adding Water
Reduction 1/2 Reaction
I2
2I-
Oxidation 1/2 Reaction
5H2O + S2O3-2
2SO4-2
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
4. Balance Hydrogen by Adding H+
Reduction 1/2 Reaction
I2
2I-
Oxidation 1/2 Reaction
5H2O + S2O3-2
2SO4-2 + 10H+
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
5. Balance Charge by Adding electons
Reduction 1/2 Reaction
I2 + 2e-
2I-
Oxidation 1/2 Reaction
5H2O + S2O3-2
2SO4-2 + 10H+ + 8e-
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
6. Multiply ½ rxns by appropriate integers
to cancel electrons
Reduction 1/2 Reaction
(I2 + 2e-
2I-) X 4
Oxidation 1/2 Reaction
5H2O + S2O3-2
2SO4-2 + 10H+ + 8e-
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
7. Add ½ rxns and check work
4I2 + 8e-
8I-
5H2O + S2O3-2
2SO4-2 + 10H+ + 8e-
5H2O + S2O3-2 + 4I2
2SO4-2 + 10H+ + 8I-
1/2 Reaction Method
S2O3-2 + I2
SO4-2 + I-
8. Add OH- to both sides neutralize H+
5H2O + S2O3-2 + 4I2
+10 OH10 OH- + S2O3-2 + 4I2
2SO4-2 + 10H+ + 8I+10 OH2SO4-2 + 8I- + 5H2O
Oxidation Number Method
Lets Look at the following reaction in an
acidic solution
Cr2O7-2 + HSO3Cr(H2O)5SO4+ + SO4-2
Step 1:
Write the oxidation state of all atoms which
undergo a change above that atom
6
Cr2O7-2 + HSO3-
Cr(H2O)5SO4+ + SO4-2
What is the Oxidation State of Chromium in
Dichromate?
2( ) + 7(-2) = -2
2Cr
7O2 = Charge
( ) = (-2 +14)/2 = 6
6
4
Cr2O7-2 + HSO3-
Cr(H2O)5SO4+ + SO4-2
What is the Oxidation State of Sulfur in the
Hydrogen Sulfite?
6
4
Cr2O7-2 + HSO3-
3
Cr(H2O)5SO4+ + SO4-2
What is the Oxidation State of Chromium in
the Cr(H2O)5SO4+ ?
6
4
Cr2O7-2 + HSO3-
3
6
6
Cr(H2O)5SO4+ + SO4-2
What is the Oxidation State of Sulfur
in Sulfate?
Step 2: Balance Elements in Coupled
Molecules
6
4
Cr2O7-2 + HSO3-
3
6
6
2Cr(H2O)5SO4+ + SO4-2
For every Cr2O7-2 reacted,
2 Cr(H2O)5SO4+ must be
Produced
Step 3: Write the Change in Oxidation # of
Oxidized Elements Above the Eq and
Reduced Elements Below
+2
6
4
Cr2O7-2 + HSO32(-3)= -6
3
6
6
2Cr(H2O)5SO4+ + SO4-2
Step 4. Balance Oxidizing and Reducing
Equivalents
3(+2)= +6
6
4
Cr2O7-2 + 3HSO32(-3)= -6
3
6
6
2Cr(H2O)5SO4+ +SO4-2
Step 4: Balance Oxygen by Adding Water
6H2O + Cr2O7-2 + 3HSO3-
2Cr(H2O)5SO4+ +SO4-2
Step 5: Balance Hydrogen by Adding
-add H+ in Acidic Solution
-In Basic Solution, Balance as if Acidic, then
Add Hydroxide to both sides to remove H+
6H2O + 5H+ Cr2O7-2 + 3HSO3-
2Cr(H2O)5SO4+ +SO4-2
Double Check Conservation of Charge
Redox Problems
Balance the Following Reactions
1. Cr2O7-2 + C2O4-2 --> Cr+3 + CO2 (acidic)
2. MnO4- + NO2- --> MnO2 + NO3- (basic)
3. H2S + NO3- --> NO2 + S8 (acidic)
4. H2O2 + MnO4- --> O2 + MnO2 (basic)
1. Cr2O7-2 + C2O4-2 --> Cr+3 + CO2 (acidic)
Cr2O7-2 --> 2Cr+3
Cr2O7-2 --> 2Cr+3 + 7H2O
14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O
6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O
C2O4-2 --> 2CO2
C2O4-2 --> 2CO2 + 2e6e- + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O
3( C2O4-2 --> 2CO2 + 2e-)
3C2O4-2 + 14H+ + Cr2O7-2 --> 2Cr+3 + 7H2O + 6CO2
2. MnO4- + NO2- --> MnO2 + NO3- (basic)
MnO4- --> MnO2 + 2H2O
4H+ + MnO4- --> MnO2 + 2H2O
3e- + 4H+ + MnO4- --> MnO2 + 2H2O
H2O + NO2- --> NO3H2O + NO2- --> NO3- + 2H+
H2O + NO2- --> NO3- + 2H+ + 2e2(3e- + 4H+ + MnO4- --> MnO2 + 2H2O)
3(H2O + NO2- --> NO3- + 2H+ + 2e-)
(3NO2- + 2H+ + 2MnO4- --> 2MnO2 + H2O + 3NO3-)
+2OH+2OH(3NO2- + H2O + 2MnO4- --> 2MnO2 + 2OH- + 3NO3-)
3. H2S + NO3- --> NO2 + S8 (acidic)
8H2S --> S8
8H2S --> S8 + 16H+
8H2S --> S8 + 16H+ + 16e-
NO3- --> NO2 + H2O
2H+ + NO3- --> NO2 + H2O
e- + 2H+ + NO3- --> NO2 + H2O
8H2S --> S8 + 16H+ + 16e16(e- + 2H+ + NO3- --> NO2 + H2O)
16H+ + 16NO3- + 8H2S --> S8 + 16NO2 + 16H2O
4. H2O2 + MnO4- --> O2 + MnO2 (basic)
MnO4- --> MnO2 + 2H2O
4H+ + MnO4- --> MnO2 + 2H2O
3e- + 4H+ + MnO4- --> MnO2 + 2H2O
H2O2 --> O2 + 2H+
H2O2 --> O2 + 2H+ + 2e2(3e- + 4H+ + MnO4- --> MnO2 + 2H2O)
3(H2O2 --> O2 + 2H+ + 2e-)
3 H2O2 + 2H+ + 2MnO4- --> 2MnO2 + 4H2O + 3O2
+2OH-
+2OH-
3 H2O2 + 2MnO4- --> 2MnO2 + 2H2O + 3O2 + 2OH-
Electrochemical Cells
2 Types of Electrochemical Cells
1. Galvanic Cells (batteries)- spontaneous
redox reactions, produce energy
2. Electrolytic Cells - nonspontaneous
redox reactions - require energy,
used in commercial syntheisis
(electroplating…)
Electrochemical Cells
Note: We call galvanic cells "batteries"
which is actually a misnomer.
A battery is actually a series of cells (the term
comes from a battery of cannons) and may be
constituted of galvanic or electrolytic cells
Voltaic (Galvanic) Cells
- “Classical Batteries”, Will Release
Energy
Is DG Negative or Positive?
DG < 0
-Galvanic Cells
Spontaneously Release
Energy
Activity Series
The Activity Series of Single Replacement
Reactions Describes Spontaneity
Zn(s) + CuCl2
ZnCl2 + Cu(s)
Zn(s) + CuCl2
ZnCl2 + Cu(s)
What is the net ionic equation?
Zn(s) + Cu+2
Cu(s) + Zn+2
This can be Split into 2 1/2 reactions
Zn(s)
Zn+2 + 2e-
Cu+2 + 2e-
Cu(s)
Zn(s) + CuCl2
Zn(s)
ZnCl2 + Cu(s)
Zn+2 + 2e-
Cu+2 + 2e-
Cu(s)
Electrochemical Cells physically
separate half reactions while
allowing charge transfer
Zn(s) + CuCl2
Zn(s)
ZnCl2 + Cu(s)
Zn+2 + 2e-
Zinc: Reductant or reducing agent (gets
oxidized while reducing copper)
Cu+2 + 2e-
Cu(s)
Copper: Oxidant or oxidizing agent (gets
reduced while oxidizing copper)
Zn(s) + CuCl2
Oxidation
Half Cell
ZnCl2 + Cu(s)
Reduction
Half Cell
Salt Bridge:
Allows Flow of
Ions While
Keeping Oxidant
& Reductant
Separated
Net Cell Reaction: Zn(s) + Cu+ 2(aq) --> Zn+2(aq) + Cu(s)
Shorthand Notation
2
2
Zn( s) Zn (aq) Cu (aq) Cu( s)
Anode 1/2 rxn
Cathode 1/2 rxn
Salt Bridge
-
+
Oxidation Occurs at
the Anode
Negative Electrode
Reduction Occurs at
the Cathode
Positive Electrode
Electric Potentials & Units
Work (Electrical) = (Charge)(Potential Differnce)
W = qDV
C = Coulomb - Charge of 6.2 x 1018 electrons
F = Faraday = charge of 1 mole of e1F = 96,500 coulombs
J
1V 
C
The Flow of electricity is measured in amps
1 amp = 1 C/sec
EMF (Ecell)
EMF: Electromotive Force
A galvanic cell’s maximum DV
-Open Circuit Potential (the cell
potential drops when current flows)
Wmax = -nFEcell
DGrxn = -nFEcell
Eo & DGo
Wmax =
o
-nFE
Wmax =
o
DG
o
DG
rxn
=
rxn
o
-nFE
F = Faraday’s Constant
-the charge of 1 mole of e- 96,500 J/(V x mol)
Reference Potential
What is 0 Potential?
q1q2
V k
r
V = 0 when two charged objects
are separated by infinity
We are interested in DV between
Anode and Cathode
Electric Potentials
Electrons Spontaneously Flow From Anode
(oxidation) to Cathode (reduction)
Zn Anode
V
Zn+2 + 2e-
Zn(s)
e- flows to
the lower
potential
Cu Cathode
Cu+2 + 2e-
Cu(s)
(Oxidation of Anode & Reduction of Cathode)
Standard Potentials
Reduction 1/2 Reaction
F2(g) + 2 eCl2(g) + 2 eBr2(g) + 2 eFe+3(aq) + eCu+2(aq) + 2e2H+(aq) + 2 eFe+2(aq) + 2eZn+2(aq) + 2eNa+(aq) + eK+(aq) + eLi+(aq) + e-
Eo (V)
2 F-(aq)
2 Cl-(aq)
2 Br-(aq)
Fe+2(aq)
Cu (s)
2 H2(g)
Fe (s)
Zn (s)
Na(s)
K(s)
Li(s)
+2.87
+1.36
+1.06
+0.54
+0.34
0.000
-0.44
- 0.76
-2.71
-2.93
-3.05
Standard Electrode Potentials
Eocell - Standard Electrode Potential
- the cell emf when all reactants are at
standard state conditions
Refers to Potential
Eocell = Eored(red) + Eoox(ox)
Refers to Half Reaction
At Standard State Conditions, all solutes are at 1M
concentrations and all gasses are at 1 atm pressure
Standard Electrode Potentials
How Would You Write
the Standard Cell for
the Reaction of Zinc &
Hydrogen?
2

Zn Zn (1M ) H (1M ) H 2 (1atm), Pt (s)
How Can We Set up a Scale to
Measure Electrode Potentials?
Zn Zn2 (1M ) H  (1M ) H 2 (1atm), Pt (s)
video
Standard Potentials
Reduction 1/2 Reaction
F2(g) + 2 eCl2(g) + 2 eBr2(g) + 2 eFe+3(aq) + eCu+2(aq) + 2e2H+(aq) + 2 eFe+2(aq) + 2eZn+2(aq) + 2eNa+(aq) + eK+(aq) + eLi+(aq) + e-
Eo (V)
2 F-(aq)
2 Cl-(aq)
2 Br-(aq)
Fe+2(aq)
Cu (s)
2 H2(g)
Fe (s)
Zn (s)
Na(s)
K(s)
Li(s)
+2.87
+1.36
+1.06
+0.54
+0.34
0.000
-0.44
- 0.76
-2.71
-2.93
-3.05
o
E cell
Determination of
from
Reduction Potentials
Refers to Potential
Eocell = Eored(red) + Eoox(ox)
Refers to Half Reaction
Oxidation is the opposite of reduction, so the oxidation
potential is the negative of the reduction potential
Eocell = Eored(red) - Eored(ox)
Don’t get Confused, the Concept is Simple:
The greater the “reduction potential”, the more
something wants to get reduced (gain electrons).
The greater the “oxidation potential”, the more
something wants to get oxidized (give electrons).
You simply change the sign to go from a reduction
potential to an oxidation potential because they are
opposite processes.
What is Eocell for the following cell?
Zn(s) Zn2 (aq) Cu 2 (aq) Cu(s)
(The Zinc gets Oxidized and the Cu+2 gets Reduced)
Cu+2(aq)
Zn (s)
+
2e-
Cu (s)
Zn+2(aq)
+
2e-
Eocell = 0.34 - (-.76) = 1.10V
Eored = 0.34
Eored =- .76
What is Eocell for the following cell?
Fe(s) Fe2 (aq) Zn2 (aq) Zn(s)
(The Iron gets Oxidized and the Zn gets Reduced)
But Nothing Happens
Fe+2(aq)
Fe (s)
Zn+2(aq)
+
2e-
+
2e-
Zn (s)
Eocell = -.76 – (-.44) = -.32V
Eored = -.44
Eored = -0.76
o
E
o
DG
& Spontaneity
rxn
=
o
-nFE
Positive Eocell – Spontaneous
Negative Eocell - Nonspontaneous
Use Electrode Potentials to Determine
Spontaneity for the following reaction.
. -(aq)
Cl2(g) + 2I
2Cl-(aq)+ I2(s)
Eocell = Eored + Eoox
Cl2(g) + 2I-(aq)
2Cl-(aq)+ I2(s)
Identify the 1/2 Reactions
Cl2(g) + 2e2I-(s)
2Cl-(aq)
I2(s) + 2e-
Eored = 1.36
Eored = .54
Eocell = 1.36 + (-.54) = .82V
Spontaneous
Class Problem
If Eo Is 0.82 V, Determine the Standard
Free Energy and Equilibrium Constant
Cl2(g) + 2I-
2Cl- + I2(s)
DG o  nFE o  (2mol)(96500 molJV )(.82V )  158kJ
K e
Go
 DRT
e

( 158KJ )
( 8.314J )( 298K )
K
 4.9 x1027
Nernst Eq
What Is the Effect of
Concentration on Cell EMF?
DG = DGo + RTlnQ
DG = -nFE
-nFE = -nFEo +RTlnQ
E = Eo - (RT/nF)lnQ
E = Eo - (0.0592/n)logQ
At 298 K
Class Problem
Using Standard Reduction Potentials
Calculate DG, K & E at 298 K for.
2Br-(aq) + Cl2(g)
Br2(l) + 2Cl-(aq)
Given: [Br-] = 0.1M, [Cl-] = 0.01M & PCl2 = 0.50 atm
2Br-(aq) + Cl2(g)
Br2(l) + 2Cl-(aq)
Given: [Br-] = 0.1M, [Cl-] = 0.01M & PCl2 = 0.50 atm
E = Eo - (RT/nF)lnQ
Cl2(g) + 2 eBr2(g) + 2 e-
2 Cl-(aq) 1.36V
2 Br-(aq) 1.06V
Eo = 1.36 - 1.06 = .3V
(8.314)(298) [.01]2
E  .3V 
ln
2(96500)
.5[.1]2
E = .35V
2Br-(aq) + Cl2(g)
Br2(l) + 2Cl-(aq)
DG = -nFEcell
What is n?
2Br-(aq)
Br2(l) + 2eCl2(g) +2e-(aq)
2Cl-(aq)
J
DG  (2mol)(96500 V mol
)(.35V )
DG = -67.55kJ
What is K?
K e
o
D
G
 RT
e
nFEo
RT
K = 1.4 x
e
10
10
J )(. 3V )
2 ( 96500mol
V
8.314molJK ( 298K )
Nernst Equation and Concentration
of an unknown
What is the concentration of Zn+2 in the following cell
if the potential is 1.13V?
Zn(s) Zn2 (aq) Cu 2 (1.80M ) Cu( s)
Nernst Equation and Concentration
of an unknown
What is the concentration of Zn+2 in the following cell if the cell
Potential is 1.13V?
Zn(s) Zn2 (aq) Cu 2 (1.80M ) Cu( s)
From Standard Potentials:
Zn (s)
Cu+2(aq) + 2e-
Zn+2(aq) + 2e- Eored = -.76
Cu (s)
Eored = 0.34
Eocell = -(-0.76) + 0.34 = 1.10V
Nernst Equation and Concentration
of an unknown
What is the concentration of Zn+2 in the following cell if the cell
Potential is 1.13V?
Zn(s) Zn2 (1.0) Cu 2 (1.80M ) Cu(s)
Zn( s )  Cu 2  Zn 2  Cu ( s )
RT
RT Zn 2
0
EE 
ln Q  E 
ln 2
nF
nF Cu
Zn 2 nF 0
ln 2 
E  E

Cu
RT
0
2

nF 0
E E
2 RT
Zn  Cu e

 1.8Me
2(96,500)(1.11.13)
8.314(298)
 0.17 M
Classical “Batteries”
Galvanic (Voltaic) Cells Which Can
Store Electrochemical Energy and
Release it for Work During Discharge
Lead Storage
Batteries
E0
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) +2e-
0.296V
PbO2(s) +3H+(aq)+ HSO4-(aq) + 2e- ---> PbSO4(s) + 2H2O(l)
1.628V
Pb(s) + PbO2(s) +2H+(aq)+ HSO4-(aq) ---> 2PbSO4(s) + 2H2O(l)
1.92V
How do we get a 12 Volt Car Battery?
By “stacking” six 2 volt cells in series
Edison’s Alkaline Battery
KOH Electrolyte
(5 Cells)
NiO2(s) + 2H2O(l) + 2e-  Ni(OH)2(s) + 2OH- Eored= .49V
Fe(OH)2(s) + 2e Fe(s) +2OHEored=-.877V
Edison’s Alkaline Battery
KOH Electrolyte
(5 Cells)
Nickel Oxide
Cathode
Iron
Anode
NiO2(s) + 2H2O(l) +
Fe(s) +2OH-
2e-
 Ni(OH)2(s) +
 Fe(OH)2(s) + 2e-
2OH-
Eored= .49V
Eored=+.877V
NiO2(s)+2H2O(l)+Fe(s) Ni(OH)2(s)+Fe(OH)2(s) Eocell= 1.37V
Alkaline Dry Cell
Zn(s) + 2OH- --> ZnO(s) +H2O(l) + 2e2 MnO2(s) +2H2O(aq) + 2e- ---> Mn2O3(s) +2OH-(aq)
Fuel Cells
2H2(g) + 4OH-(aq) --> 4H2O(l) + 4eO2(g) + 2H2O(l) + 4e- --> 4 OH-(aq)
Fuel Cells
PEM Electrolyzer/Fuel Cell
PEM Electrolyzer/Fuel Cell
Corrosion
Spontaneous Redox Reaction
Resulting in Formation of Oxides
From Pure Metals
Only Noble Metals do Not Corrode
Corrosion
4Fe+2(aq) + O2(g) + 4H+(aq) --> 4Fe+3(aq) + 2H2O
2Fe+3 + 4 H2O(l) --> Fe2O3.H2O + 6H+
Why does Aluminum not Rust?
It does form a layer of aluminum oxide,
but this is impervious to oxygen, and
functions as a protective coat.
Cathodic Protection
Iron can be protected by hooking up another
metal with a lower reduction potential,
which functions as a Sacrificial Anode
Cathodic Protection
Iron can be protected by hooking up another
metal with a lower reduction potential,
which functions as a Sacrificial Anode
Will Copper Work as a Sacrificial Anode? Will
Zinc Work as a Sacrificial Anode?
Will
Magnesium Work as a Sacrificial Anode?
Galvanic & Electrolytic Cells
 Galvanic (Voltaic) Cells
 Positive Potentials
 Spontaneous
 Provide Energy for Work (DG<0)
Electrolytic Cells
Negative Potentials
Non Spontaneous
Require Energy to Work (DG>0 unless
coupled to an external voltage source)
Galvanic & Electrolytic Cells
Sodium and Chlorine Spontaneously Form
Sodium Chloride in a Galvanic Cell
Ecell = 1.36V -(-2.71V) = 4.07V
Where the Sodium is being oxidized and the
Chlorine is being reduced
An Electrolytic Cell Is Formed by Applying a
Potential Greater Than 4.07 V in the Opposite
Direction, Making the Nonspontaneous Reaction
Occur
Electrolysis of Molten NaCl
Driving Non Spontaneous Reactions
+
-
2Na+(aq) + 2Cl-(aq) --> 2Na(s) + Cl2(g)
Electrolysis of aqueous NaF
What are the possible reduction 1/2 rxns?
Na+(aq)
e-
+
2H2O(l) + 2e-
Na(s)
H2(g) + 2OH-(aq)
E0red= -2.71V
E0red= -0.83V
What are the possible oxidation 1/2 rxns?
2F-(aq)
2H2O(l)
4OH-(aq)
2e-
F2(g) +
O2(g) + 4H+(aq) + 4eO2(g) + 2H2O(l) + 4e-
E0red= 2.87V
E0red= 1.23V
E0red= 0.40V
What species gets reduced?
H 2O
What species gets oxidized?
OH-
Electrolysis of aqueous NaF
The net cell reaction?
cathode: 2[2H2O(l) + 2eanode: 4OH-(aq)
cell:
2H2O(l)
H2(g) + 2OH-(aq)] E0red= -083V
O2(g) + 2H2O(l) + 4eE0red= 0.40V
2H2(g) + O2(g)
The Electrolysis of Water
E0cell= -1.23V
Electroplating & Active Electrodes
When metallic cations are being
reduced, they can bind to the cathode
Active Electrodes
Stoichiometry & Electrolysis
How do we measure the rate at which an
electric current carries charge?
1 Amp = 1C/sec
How many moles of electrons
does this represtent?
1F = 96,500 C = charge of 1 mole of e1 Amp = 1.04x10-5 mole e-/ sec
Stoichiometry & Electrolysis
The current across an electrolytic cell
provides a stoichiometric relation for the
number of electrons transferred to the
number of species oxidized and reduced
1F = 96,500 C = charge of 1 mole of e1 Amp = 1C/sec
2 Types of Problems:
Stoichiometry & Electrolysis
1. Measure Current & time, determine quantity
of species oxidized and reduced.
2. Measure quantity of any species oxidized
or reduced over a given time span, and you
can determine the current
Stoichiometry & Electrolysis
Lithium is a very active metal, and can be formed
through the electrolysis of LiCl. What mass of
Lithium and chlorine can be formed through the
electrolysis of molten LiCl if 4300.0 amps passes
through a cell operating for 24.0 hours?
m Li  (
4300C
sec
mCl2  (
4300C
sec
(24h)(
60min
h
(24h)(
60min
h
)(
60sec
min
)(
)(
60sec
min
mol e 
96500C
)( mole  )(
mol e 
96500C
Cl2
71g Cl2
)( 2mol
)(
mol )  137kg Cl 2
mol e 
)(
mol Li
6.9 g Li
mol
)  26.6kg Li
Electroplating
How many minutes will it take to plate
out 125 g Cr+3 with a constant current of
200 amps?
 molCr 3   3mol e    96500C   sec   min 
125gCr 
  molCr 3   mol e    200C   60sec   58.0 min
52
g
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