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?