ELECTROCHEMISTRY
Electrochemistry involves the relationship between electrical energy and chemical energy.
OXIDATION-REDUCTION REACTIONS
SPONTANEOUS REACTIONSCan extract electrical energy from these.Examples: voltaic cells, batteries.
NON-SPONTANEOUS REACTIONSMust put in electrical energy to make them go.Examples: electrolysis, electrolytic cells.
QUANTITATE REACTIONS
OXIDATION-REDUCTIONOxidation = loss of electrons. Occurs at ANODE
An oxidizing agent is a substance that causes oxidation (and is itself reduced).
Reduction = gain of electrons. At CATHODE
A reducing agent is a substance that causes reduction (and is itself oxidized).
LEO goes GER
LEO: Loss of Electrons = Oxidation
GER: Gain of Electrons = Reduction
Rules for determining Oxidation States
• For pure elements, oxidation state = 0
(examples: S8, P4, O3, F2, K, Be)
• In compounds, some elements have “common” oxidation numbers:
• Assign others by difference (using charge on ion)
Alkali metals (Na+, K+..) +1
Alkaline earths (Mg2+, Ba2+…) +2
F = -1, O = -2 (but -1 in peroxides)
Cl, Br, I = -1 (except if bonded to O or halogen)
H = +1 or -1 (depends what it is bonded to)
1. Write incomplete half-reactions.
2. Balance each half-reaction separately. a. Balance atoms undergoing redox. b. Balance remaining atoms
i. Add H2O to balance oxygens. ii. Add H+ to balance hydrogens.
3. Balance charges by adding electrons.
4. Multiply half-reactions to cancel electrons.
5. Add the half-reactions.
6. In basic solutions, add OH to neutralize H+
BALANCING REDOX REACTIONS
The half-reactions for
Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
Sn2+(aq) Sn4+(aq) +2e-
2Fe3+(aq) + 2e- 2Fe2+(aq)
Oxidation: electrons are products.
Reduction: electrons are reagents.
Half Reactions
In Acid solution.:
C + HNO3 NO2 + CO2 + H2O
Basic solution.:
PbO2 + Cl + OH Pb(OH)3 + ClO
Balancing Redox Reactions
The energy released in a spontaneous redox reaction is used to perform electrical work.
Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit.
Voltaic cells are spontaneous.
If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.
Zn0(s) + Cu2+(aq) Zn2+(aq) + Cu0(s)
VOLTAIC CELL
E0cell = 1.10 V
Voltaic cells consist of
Anode: Zn(s) Zn2+(aq) + 2e-
Cathode: Cu2+(aq) + 2e- Cu(s)
The two solid metals are the electrodes (cathode and anode). Salt bridge: (used to complete the electrical circuit):
Anions and cations move to compensate excess charge.
Zn is oxidized to Zn2+ and 2e.
The electrons flow to the anode where they are used in the reduction reaction.
We expect the Zn electrode to lose mass and the Cu electrode to gain mass.
“Rules” of voltaic cells:1. At the anode electrons are products. Oxidation occurs at the anode.
2. At the cathode electrons are reagents. Reduction occurs at the cathode
3. Electrons cannot swim!
VOLTAIC CELLS
Cell voltage (EMF or Ecell) is the measure of rxn spontaneity
Ecell : Intensive property, energy per electron
Cell voltage depends on:the individual species undergoing oxidation and reduction
and their concentrations
The more spontaneous a reaction, the higher the voltage (more positive) the higher the Keq the more negative the G0
STANDARD POTENTIAL FOR AN ELECTROCHEMICAL CELL
The standard potential for an electrochemical cell is the potential (voltage) generated when reactants and products of a redox reaction are in their standard states.
Standard States:T = 25°C.Gases, P = 1 atm.[Solutes] = 1MSolids, liquids = pure
HALF-CELL POTENTIAL
The potential associated with the half-reaction.
Rules for half-cell potentials:
1. The sum of two half-cells potentials in a cell equals the overall cell potential:
E°cell = E°1/2(oxid) + E°1/2(reduc)
2. For any half-reaction: E°1/2(oxid) = E°1/2(reduc)
3. Standard half-cell is a hydrogen electrode:
H2(g,1atm) = 2H+ (aq, 1M) + 2e
E°1/2(oxid) = E°1/2(reduc) = 0 V
E0cell = 0.76 V
E0cell = 1.10 V
E01/2 (Zn Zn2+) = ?
E01/2 (Cu2+ Cu) = ?
The more positive Ered the stronger the oxidizing agent on the left.
The more negative Ered the stronger the reducing agent on the right.
A species that is higher and to the left will spontaneously oxidize one that is lower and to the right in the table.
Example: F2 will oxidize H2 or Li;
Ni2+ will oxidize Al(s).
Oxidizing and Reducing Agents
Are the following reactions spontaneous?
If so, evaluate E0cell
Cu(s) + Cl2(g) Cu2+(aq) + 2Cl(aq)
2Cu2+(aq) + 2H2O Cu(s) + O2(g) + 4H+(aq)
RELATIONSHIP BETWEEN G AND E
G = nFE
At Standard State: G° = nFE°
n = number of electrons transferred in a balanced redox reaction
F = Faraday = 96,500 coulomb/mole e-
1 coulomb = 1 Amp-sec1 J = 1 Amp-sec-V = 1 coulomb-V1 F = 96,500 J/V-mole e
Example: Is this reaction spontaneous? Cd(s) + 2H+ Cd2+ + H2 (g)
Cd2+ + 2e Cd(s) Eored
= 0.403V2H+ + 2e H2 Eo
red = 0
Reaction goes as written (reduction) for more positive redox couple.
H+ cathode: 2H+ + 2e H2
Cd anode: Cd Cd2+ + 2e
Eocell = Eored (cathode) + Eo
oxid (anode)Eocell = 0 + 0.403 = + 0.403•
Go = nFEo = 2 x 96,500 J/V-mol x 0.403V = 8.3 x 104 = 83 kJ/mol
Yes: the reaction is spontaneous!
G° = 2.303 RT log Keq
G° = nFE°
E° = 2.303 RT log Keq
nF
R = 8.314 J/K-moleF = 96,500 J/V-mole e
At 25°C = 298K:
E° = (0.0592) log Keq
n
For electrochemical cell at equilibrium: G = 0
Effect of ConcentrationStandard states:1M solution, 1 atm gas pressure
What if concentrations are different?
EnGQlnRTGG o ℑ−=+= and
QlnRTEnEn o +ℑ−=ℑ−
Qlnn
RTEE o
ℑ−=
Effect of Concentration
€
E 12=E 1
2
o −2.303 RT
nℑlog
C[ ] cD[ ]
d
A[ ] aB[ ]
b
€
E 12=E 1
2
o −0.059n
logC[ ]
cD[ ]
d
A[ ] aB[ ]
bat 298K
So for a half reaction:aA + bB + n e cC + dD
Qlnn
RTEE o
ℑ−=
€
E 12=E 1
2
0 −RTnℑ
lnC[ ]
cD[ ]
d
A[ ] aB[ ]
b
Examples
1. What is Ecell for a fuel cell running in air (PO2 = 0.2 atm), at pH = 2, with PH2 = 1 atm?
2. What is the half cell potential of the Ag/Ag+ redox couple (E0 = +0.799 V) in a 1 M NaCl solution that contains solid AgCl (Ksp = 1.1 x 10-10)?
O2(g) + 4H+(aq) + 4e 2H2O(l) Eo = +1.229V 2H+ + 2e H2 Eo = 0
Lead/Acid Battery
DURING DISCHARGE
Anode:
Pb(s) + SO42(aq) PbSO4(s) + 2e
Cathode:
PbO2(s) + SO42(aq) + 4H+ + 2e PbSO4(s) + 2H2O
Overall:
Pb(s) + PbO2(s) + 2H2SO4 2PbSO4(s) + 2H2O
Batteries
Lead-Acid Battery
DRY CELL
Anode: Zn(s)
Zn(s) Zn2+(aq) + 2e
Cathode: NH4Cl + MnO2 + graphite paste
2NH4+(aq) + 2MnO2(s) + 2e Mn2O3(s) + 2NH3(aq) + H2O
In ALKALINE CELL, NH4Cl is replaced by KOH.Provides up to 50% more energy.Zn is used as a powder mixed with the electrolyte.
Batteries
Dry Cell
Alkaline Battery
Rechargeable Nickel-Cadmium Battery
Anode: Cd metal
Cd(s) + 2OH(aq) Cd(OH)2(s) + 2e
Cathode: NiO2(s)
NiO2(s) + 2H2O + 2e Ni(OH)2(s) + 2OH(aq)
Overall:
Cd(s) + NiO2(s) + 2H2O Cd(OH)2(s) + Ni(OH)2(s)
Batteries
FUEL CELLS
H2-O2 Fuel Cell - expensive but light. Used in spacecraft.
Anode:
2H2(g) + 4OH(aq) 4H2O(l) + 4e
Cathode:
O2(g) + 2H2O(l) + 4e 4OH(aq)
Overall:
2H2(g) + O2(g) 2H2O(l)
E0cell = 1.23 V
Batteries
Corrosion
Differential aeration mechanism
CATHODIC PROTECTION OF IRON
Corrosion
CATHODIC PROTECTION OF IRON
Corrosion
ELECTROLYSIS
Electrolysis:Driving non-spontaneous reactions by applying electrical energy.
An electrolysis cell consists of two electrodes in either aqueous solution (of ions) or in a molten salt e.g. molten NaCl.
The anode is where oxidation occurs.Anions migrate to the anode and lose electrons.
The cathode is where reduction occurs.Cations migrate to the cathode and gain electrons.
ELECTROLYSIS OF MOLTEN NaCl
Cathode:
Na+ + e- → Na
Anode:2 Cl- → Cl2 + 2e-
Cathode: 2Na+ + 2e 2Na(l)Anode: 2 Cl Cl2(g) + 2e
___________________________
2 Na+ + 2 Cl 2Na(l) + Cl2(g)
ELECTROLYSIS OF NaCl, cont.
Electrolysis of aqueous NaCl:
Cathode: H2O + 2e H2(g) + 2OH
Anode: 2 Cl Cl2(g) + 2e
___________________________________
2 H2O + 2 Cl H2(g) + Cl2(g) + 2OH
so
2 Na+ + 2 OH = 2 NaOH is left behind
• It is easier to reduce H2O than Na+
• The easiest (least non-spontaneous) reaction happens
ELECTROLYSIS - examples
What products will form when an aqueous solution of ZnBr2 is electrolyzed?
What products will form when aqueous AgNO3 is electrolyzed?
ELECTROLYSIS OF AQUEOUS Na2SO4
Cathode: 4H2O + 4e 2H2(g) + 4OH
Anode: 2H2O O2(g) + 4H+ + 4e
_____________________________
6H2O 2H2(g) + O2(g) + 4H+ + 4OH
or 2H2O 2H2(g) + O2(g)
COMMERCIAL APPLICATIONS OF ELECTROLYSIS
Production of metals – Na, Al.
Purification of Metals – Cu.
Electroplating.
PURIFICATION OF COPPER
Cathode, thin sheet of pure copper:
Cu2+ + 2e- → Cu
Anode, impur ecopper:
Cu → Cu2+ + 2e-
As the reacti on proceeds, copper moves from the anode to the cathode.
Purification of Copper
Cathode: thin sheet of pure copper Cu2+ + 2e Cu(s)Anode: impure copper Cu(s) Cu2+ + 2e
As the reaction proceeds, Cu moves from anode to cathode.
ELECTROLYSIS CALCULATIONS
1 mole of e- = 1 Faraday = 96,500 Coulombs = charge on 1 mole of e-
1 Ampere = 1 coulomb/second1 coulomb = 1 Amp-sec
Electromotive Force (EMF)force that cases electrons to flow (voltage)
1 Watt = 1 Amp-Volt1 Joule = 1 coul-Volt = 1 Amp-sec-Volt = 1 Watt-sec
1 kW-hour = (1000 Watt)(3600 sec) = 3.6 x 106 Watt-sec = 3.6 x 106 Joules
ELECTROLYSIS CALCULATION
Electrolysis gives 1.00g of Cu from CuSO4
Reaction is: Cu2+ + 2e Cu
How many Faradays (F) of charge are required?
How many Coulombs is this?
ELECTROLYSIS CALCULATION
If 1.00g of Cu is obtained in 1 hour, how many amps are required?
If 2 amps were used, how long would it take to produce 1g?