Slide 1 Electrochemistry Chapter 17. Slide 2 Why Study Electrochemistry? Batteries Batteries...

ElectrochemistryChapter 17

Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?

• BatteriesBatteries• CorrosionCorrosion• Industrial production of Industrial production of chemicalschemicals such as Clsuch as Cl22, ,

NaOH, FNaOH, F22 and Al and Al• Biological redox Biological redox reactions,photosynthesisreactions,photosynthesis

6CO2 + 6H2O --> C6H12O6 + 6O2

C6H12O6 + O2 --> 6CO2 + 6H2O +Energy

Redox Reactions 01Redox Reactions 01

Redox reaction are those involving the oxidation and reduction of species.

LEO – Loss of Electrons is Oxidation .

GER –Gain of Electrons Is Reduction .

Oxidation and reduction must occur together.

They cannot exist alone.


• Oxidation Half-Reaction: Zn(s) Zn2+(aq) + 2 e–. • The Zn loses two electrons to form Zn2+.



• Reduction Half-Reaction: Cu2+(aq) + 2 e– Cu(s)• The Cu2+ gains two electrons to form copper.

Electrochemical Cells

spontaneousredox reaction

anodeoxidation

cathodereduction

• Overall: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)


Electromotive Force (emf)Electromotive Force (emf)• Water only spontaneously

flows one way in a waterfall.• Likewise, electrons only

spontaneously flow one way in a redox reaction—

from higher to lower potential energy.

Electromotive Force (emf)Electromotive Force (emf)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential, and is designated Ecell.

Cell PotentialCell Potential

Cell potential is measured in volts (V).

C (Coulomb):The amount of charge transferred when a current of 1 ampere (A) Flows for one second.

Volt, a potential difference between two points which results to a current of one

ampere through a resistance of one ohm

Cell Potentials and Free-Energy Changes for Cell ReactionsCell Potentials and Free-Energy Changes for Cell Reactions

1 J = 1 C x 1 V

voltSI unit of electric potential

jouleSI unit of energy

coulombElectric charge

1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.

1 V = 1 JC

Electrochemical Cells 01Electrochemical Cells 01

• Electrodes: are usually metal strips/wires connected by an electrically conducting wire.

• Salt Bridge: is a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte.

• Anode: is the electrode where oxidation takes place, (-).

• Cathode: is the electrode where reduction takes place, (+) terminal

Cells Notation 02Cells Notation 02

Anode Half-Cell || Cathode Half-Cell

Electrode | Anode Soln || Cathode Soln | Electrode

Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)


Anode Half-Cell || Cathode Half-Cell

Electrode | Anode Soln || Cathode Soln | Electrode

Fe(s) | Fe2+(aq) || Fe3+(aq), | Pt(s)

Write the cell notation for:Write the cell notation for:

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

Electrochemical Cell PotentialsElectrochemical Cell Potentials

• The standard half-cell potentials are determined from the difference between two electrodes.

• The reference point is called the standard hydrogen electrode (S.H.E.) and consists of a platinum electrode in contact with H2 gas (1 atm) and aqueous H+ ions (1

M).

• The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00 V.

Cu2+ (aq) + 2e- Cu (s) E0 = 0.34 V∆G˚ = –nFE˚

Electrochemical Cells 06Electrochemical Cells 06---

• E0 is for the reaction as written

• The more positive E0 the greater the tendency for the substance to be reduced

• The half-cell reactions are reversible

• The sign of E0 changes when the reaction is reversed

• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

∆G˚ = –nFE˚`n = number of moles of electrons in reaction

F = 96,500 C/mole


• The standard potential of any galvanic cell is the sum of the standard half-cell potentials for the oxidation and reduction half-cells.

E°cell = E°oxidation + E°reduction

• Standard half-cell potentials are always tabulated as a reduction process. The sign must be changed for the oxidation process.


• When selecting two half-cell reactions the more negative value, or smaller E° will form the oxidation half-cell.

• Consider the reaction between zinc and silver:Ag+(aq) + e– Ag(s) E° = 0.80 V

Zn2+(aq) + 2 e– Zn(s) E° = – 0.76 V

• Therefore, zinc forms the oxidation half-cell:Zn(s) Zn2+(aq) + 2 e– E° = – (–0.76 V)

E0 = E°Ox + E°Redcell0 0 E0 = 0.76 V+ 0.80 V = 1.56 V cell

0 0

Write the cell rection and calculate the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V

Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V

Cd is the stronger oxidizer

Cd will oxidize Cr

2e- + Cd2+ (1 M) Cd (s)

Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):

Cathode (reduction):

2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

x 2

x 3


E0 = 0.74 V + (-0.4 V) cell

E0 = 0.34 V cell

Spontaneity of a Reaction 01Spontaneity of a Reaction 01

• The value of E˚cell is related to the thermodynamic quantities of ∆G˚ and K.

• The value of E˚cell is related to ∆G˚ by:

∆G˚ = –nFE˚cell

• The value of K is related to ∆G˚ by:

∆G˚ = –RT ln K

F = 96,500 C/mole

G0 = -RT ln K = -nFEcell0

Spontaneity of Redox Reactions

ΔG = -nFEcell

G0 = -nFEcell0

n = number of moles of electrons in reaction

F = 96,500J

V • mol = 96,500 C/mol

G0 = -RT ln K = -nFEcell0

Ecell0 =

RTnF

ln K(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol)ln K=

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

0.0257 Vx nE0 cellexpK =

ln K = 2.3035 Log K

º

( see slide 4)

Spontaneity of Redox Reactions

∆G˚ = –RT ln K =0.0257 V

nln KEcell

0

---

2e- + Fe2+ Fe

Oxidation:

Reduction:

What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)

=0.0257 V

nln KEcell

0

E0 = (-0.80) +( -0.45)

E0 = -1.25 V

0.0257 Vx nE0 cellexpK =

n = 2

0.0257 Vx 2-1.25 V

= exp

K = 5.67 x 10-43


E0 = -0.80 V

E0 = -0.45V

2Ag 2Ag+ + 2e-

Cell Emf Under Nonstandard Conditions

G = G0 + RT ln Q G = -nFE G0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln QRTnF

Nernst equation

At 298 K

-0.0257 V

nln QE0E = -

0.0592 Vn

log QE0E =

If we divide both sides by “-nF”

ln Q = 2.3035 Log Q

Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)

2e- + Fe2+ Fe

Cd Cd2+ + 2e-Oxidation:

Reduction:n = 2

E0 = 0.40 V + (-0.45 V)

E0 = -0.05 V -

0.0257 Vn

ln QE0E =

-0.0257 V

2ln -0.05 VE =

0.0100.60

E = 0.0026

E > 0Spontaneous

E0 = 0.40 V

E0 = -0.45 V


G = -nFEcell

ΔG < 0

Write Cell Notation and Calculate the Cell Potential for the Following CellWrite Cell Notation and Calculate the Cell Potential for the Following Cell

E = 0.088 V

-0.0592 V

nlog QE0E =

Concentration CellsConcentration Cells

• Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0.

-0.0592V

nlog QE0E =

Calculate the concentration of H+ in the following if the Cell voltage is 0.7 V .

Zn(s)|Zn2+(aq, 1 M)|| H+(aq, ?)|H2(g, 1 atm)|Pt

E °red (Zn/Zn2+)= 0.76 V.

H+ = 0.1 M

-0.0592 V

nlog QE0E = Nernst equation

Nernst Equation Could be Applied to Half Cell PotentialNernst Equation Could be Applied to Half Cell Potential

• A particularly important use of the Nernst equation is in the electrochemical determination of pH.

Pt | H2 (1 atm) | H+ (? M) || Reference Cathode

Ecell = EH2 H+ + Eref

• The Nernst equation can be applied to the

half-reaction: H2(g) 2 H+(aq) + 2 e–

The Nernst Equation 05The Nernst Equation 05

• For the half-reaction: H2(g) 2 H+(aq) + 2 e–

• E° = 0 V for this reaction (standard hydrogen electrode). According to the problem, PH2

is 1 atm.

EH2 H+= EH2 H+o – 0.0592V

n logH+2

PH2

EH2 H+=0 V – 0.0592V

n log H+2

-0.0592 V

nlog QE0E =

refcell pHV 0592.0 EE

pHV 0592.0

The Nernst Equation 07The Nernst Equation 07

• The overall potential is given by:

• Which rearranges to give an equation for the determination of pH:

pH V0.0592refcell EE

refcell pH V0592.0 EE

Replacing Standard Hydrogen Electrode With Glass Electrode (Ag/AgCl wire in dilute HCl)

pH meterpH meter

• A higher cell potential indicates a higher pH, therefore we can measure pH by measuring Ecell.

• A glass electrode (Ag/AgCl wire in dilute HCl) with a calomel reference is the most common arrangement.

Glass: Ag(s) + Cl–(aq) AgCl(s) + e– E° = –0.22 V

Calomel: Hg2Cl2(s) + 2 e– 2 Hg(l) + 2 Cl–(aq) E° = 0.28 V

pH Electrode 09pH Electrode 09

• The glass pH probe is constructed as follows:

Ag(s) | AgCl(s) | HCl(aq) | glass | H+(aq) || reference

•The difference in [H+] from one side of the glass membrane to the other causes a potential to develop, which adds to the measured Ecell.

pH V0.0592refcell EE

The following cell has a potential of 0.28 V at 25°C: Pt(s) | H2 (1 atm) | H+ (? M) || Pb2+ (1 M) | Pb(s)What is the pH of the solution at the anode?

The following cell has a potential of 0.28 V at 25°C: Pt(s) | H2 (1 atm) | H+ (? M) || Pb2+ (1 M) | Pb(s)What is the pH of the solution at the anode?

• H2(g) + Pb2+(aq) 2 H+(aq) + Pb(s)

• Eoref = - 0.13 V ( Please see page 698);

6.9 V 0.0592

)13.0(28.0

VVpH

V 0.0592refcell EE

pH

Batteries

Lead storagebattery

BatteriesBatteries

Lead Storage Battery

2PbSO4(s) + 2H2O(l)Pb(s) + PbO2(s) + 2H1+(aq) + 2HSO41-(aq)

PbSO4(s) + 2H2O(l)PbO2(s) + 3H1+(aq) + HSO41-(aq) + 2e-

PbSO4(s) + H1+(aq) + 2e-Pb(s) + HSO41-(aq)

Overall:

Anode:

Cathode:

Batteries

Leclanché cell

Dry cell

Zn (s) Zn2+ (aq) + 2e-Anode:

Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+

Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

1.5 V but deteriorates to 0.8 V with use

Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

Batteries

A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning

Anode:

Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)

2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-

2H2 (g) + O2 (g) 2H2O (l)

Figure 7.13: The essentials of a typical fuel cell.Figure 7.13: The essentials of a typical fuel cell.

© 2003 John Wiley and Sons Publishers

Replacing KOH with Proton ExchangeMembrane

Proton Exchange Membrane (PEM) Fuel CellProton Exchange Membrane (PEM) Fuel Cell

Anode side:2H2 => 4H+ + 4e-

Cathode side:O2 + 4H+ + 4e- => 2H2O

Net reaction:2H2 + O2 => 2H2O

Corrosion 01Corrosion 01

• Corrosion is the oxidative deterioration of metal.

• 25% of steel produced in USA goes to replace steel structures and products destroyed by corrosion.

• Rusting of iron requires the presence of BOTH oxygen and water.

• Rusting results from tiny galvanic cells formed by water droplets.

CorrosionCorrosion

Corrosion: The oxidative deterioration of a metal.

Corrosion Corrosion

4Fe+2(aq) + O2(g)+ 4H+(aq) 4Fe+3(aq) + 2H2O2Fe+3(aq) + 4H2O(l) Fe2O3.H2O + 6H+

…Corrosion Prevention: Galvanizing…Corrosion Prevention: Galvanizing

Corrosion Prevention 03Corrosion Prevention 03

• Galvanizing: is the coating of iron with zinc. Zinc is more easily oxidized than iron, which protects and reverses oxidation of the iron.

• Cathodic Protection: is the protection of a metal from corrosion by connecting it to a metal (a sacrificial anode) that is more easily oxidized.Attaching a magnesium stake to iron will corrode the magnesium instead of the iron

Electrolysis 01Electrolysis 01

Electrolysis of Water

• Electrolysis: is the process in which electrical energy is used to drive a nonspontaneous chemical reaction.

• An electrolytic cell is an apparatus for carrying out electrolysis.

• Processes in an electrolytic cell are the reverse of those in a galvanic cell.



• Electrolysis of Water: Requires an electrolyte species, that is less easily oxidized and reduced than water, to carry the current.

• Anode: Water is oxidized to oxygen gas.2 H2O(l) O2(g) + 4 H+(aq) + 4 e–

• Cathode: Water is reduced to hydrogen gas.4 H2O(l) + 4 e– 2 H2(g) + 4 OH–(aq)

Electrolysis Applications 01Electrolysis Applications 01

• Manufacture of Sodium (Downs Cell):

Bp (NaCl) = 801o CBp (NaCl-CaCl2) = 580oC


• Quantitative Electrolysis: The amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell.

• Reduction of 1 mol of sodium ions requires 1 mol of electrons to pass through the system.

• The charge on 1 mol of electrons is 96,500 coulombs.


• To determine the moles of electrons passed, we measure the current and time that the current flows:

Charge (C) = Current (A) x Time (s)

• Because the charge on 1 mol of e– is 96,500 C, the number of moles of e– passed through the cell is:

Molesofe– =Charge(C) 1 molee–

96,500C

Electrolysis and Mass Changes

charge (C) = current (A) x time (s)

1 mole e- = 96,500 C

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?

Anode:

Cathode: Ca2+ (l) + 2e- Ca (s)

2Cl- (l) Cl2 (g) + 2e-

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)

2 mole e- = 1 mole Ca

0.452Cs

x 1.5 hr x 3600shr 96,500 C

1 mol e-

x2 mol e-

1 mol Cax

= 0.0126 mol Ca

= 0.50 g Ca

Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery isPb(s) + HSOPb(s) + HSO44

--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?how long will the battery last?

SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Calculate moles of e-Calculate moles of e-

2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -

c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C

Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery isPb(s) + HSOPb(s) + HSO44

--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?battery last?

SolutionSolutiona)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pbb)b) Mol of e- = 4.38 molMol of e- = 4.38 molc)c) Charge = 423,000 CCharge = 423,000 C

Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)

Time (s) = 423, 000 C1.50 amp

= 282,000 sTime (s) = 423, 000 C1.50 amp

= 282,000 s About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time


• Manufacture of Sodium (Downs Cell):

Bp (NaCl) = 801o CBp (NaCl-CaCl2) = 580oC


• Manufacture of Cl2 and NaOH (Chlor–Alkali):

Chlorine bleach: Cl2 + 2 NaOH → NaCl + NaClO + H2O

• Manufacture of Aluminum (Hall–Heroult):

Anode: (positive electrode)C(s) + 2O2-

(l) ---> CO2(g) + 4e-Cathode: (negative electrode)Al3+

(l) + 3e- ---> Al(l)

Overall Reaction:2Al2O3(l) + 3C(s) ---> 4Al(l) + 3CO2(g)

Bauxite:Al2O3 + SiO2 + TiO2 + Fe2O3Hot NaOH used to dissolve alumnum Comounds and other materials separated by filration

Mixture mp. 1000 CAl2O3 mp. 2045 C

• Electrorefining and Electroplating:

Given the following reaction, which is true? Given the following reaction, which is true?

1. Plating Ag onto Cu is a spontaneous process.2. Plating Cu onto Ag is a spontaneous process.3. Plating Ag onto Cu is a nonspontaneous process.4. Plating Cu onto Ag is a nonspontaneous process.5. Energy will have to be put in for the reaction to

proceed.

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) E° = +0.46 V

Based on the standard reduction potentials, which metal would not provide cathodic protection to iron?

1. Magnesium2. Nickel3. Sodium4. Aluminum

Electrochemical Cells 06Electrochemical Cells 06---

Correct Answer:

In order to provide cathodic protection, the metal that is oxidized while protecting the cathode must have a more negative standard reduction potential. Here, only Ni has a more positive reduction potential (0.26 V) than

Fe2+ (0.44 V) and cannot be used for cathodic protection.

1. Magnesium2. Nickel3. Sodium4. Aluminum

Ni2+ is electrolyzed to Ni by a current of 2.43 amperes. If current flows for 600 s, how much Ni is plated (in grams)?

(AW Ni = 58.7 g/mol)

1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g

Correct Answer:

Fn

FWti

mass

C/mol) 96,500(2

g/mol) (58.7s) (600. A2.43mass

g 0.444mass

1. 0.00148 g2. 0.00297 g3. 0.444 g4. 0.888 g

Alkaline Dry-Cell 04Alkaline Dry-Cell 04

• Alkaline Dry-Cell: Modified Leclanché cell which replaces NH4Cl with NaOH or KOH.

Anode: Zinc metal can on outside of cell.Zn(s) + 2 OH–(aq) ZnO(s) + H2O(l) + 2 e–

Cathode: MnO2 and carbon black paste on graphite.

2 MnO2(s) + H2O(l) + 2 e– Mn2O3(s) + 2 OH–(aq)

Electrolyte: NaOH or KOH, and Zn(OH)2 paste.

Cell Potential: 1.5 V but longer lasting, higher power, and more stable current and voltage.

Batteries

Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-Anode:

Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)

Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

Mercury Battery 0.9 V

Non-rechargeable button cells for watches, hearing aids, and calculators

1. N of NO2- is reduced, Cr of Cr2O7

2- is oxidized

2. N of NO2- is oxidized, Cr of Cr2O7

2- is reduced

3. O of NO2- is oxidized, Cr of Cr2O7

2- is reduced

4. Cr3+ is reduced, N of NO2- is oxidized

5. N of NO3- is oxidized, Cr3+ is reduced

For the reaction given below, what substance is oxidized and what is reduced?

For the reaction given below, what substance is oxidized and what is reduced? 3 NO2

- + Cr2O72- + 8 H+ 2 Cr3+ + 3 NO3

- + 4 H2O

1. 1, 1, 2, 1, 1

2. 1, 1, 4, 1, 2

3. 1, 1, 2, 1, 2

4. 4, 1, 2, 1, 2

5. 4, 1, 4, 4, 2

When the following reaction is balanced, what are the coefficients for each substance?

When the following reaction is balanced, what are the coefficients for each substance?

__Ag + __O2 + __H+ __Ag+ + __H2O

1. 1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).

2. 1 M solution for Cl2(g) and for Cl–(aq).

3. 1 atm pressure for Cl2(g) and for Cl–(aq).

4. 1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).

1. 1 barr pressure for Cl2(g) and 1 M solution for Cl–(aq).

2. 1 M solution for Cl2(g) and for Cl–(aq).

3. 1 atm pressure for Cl2(g) and for Cl–(aq).

4. 1 atm pressure for Cl2(g) and 1 M solution for Cl–(aq).

1Bar = .987 atm

Which substance is the stronger oxidizing agent? Which substance is the stronger oxidizing agent?

• Br2

• O2

• NO3-

• H+

• Cl2

1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V

Calculate the emf of the following cell:

Zn(s)|Zn2+(aq, 1 M)|| H+(aq, 1 M)|H2(g, 1 atm)|Pt

E° (Zn/Zn2+)= 0.76 V.

Correct Answer:

Zn is the anode, hydrogen at the Pt wire is the cathode.

1. +0.76 V2. +1.52 V3. 0.76 V4. 1.52 V

E°cell = E°red + E°ox

E°cell = 0.00 V + (0.76 V)

E°cell = +0.76 V

The End

Batteries 06Batteries 06

• Nickel–Cadmium Battery: is rechargeable.

Anode: Cadmium metal.Cd(s) + 2 OH–(aq) Cd(OH)2(s) + 2 e–

Cathode: Nickel(III) compound on nickel metal.NiO(OH) (s) + H2O(l) + e– Ni(OH)2(s) + OH–(aq)

Electrolyte: Nickel oxyhydroxide, NiO(OH).

Cell Potential: 1.30 V

Batteries

Solid State Lithium Battery

1.5 V to about 3.7 V,

Slid Electrolyte: Inorganic ceramic and organic polymer solid-electrolyte materials are reviewed

Batteries 08Batteries 08

• Lithium Ion (Li–ion): The newest rechargeable battery is based on the migration of Li+ ions.

•Anode: Li metal, or Li atom impregnated graphite.Li(s) Li+ + e–

Cathode: Metal oxide or sulfide that can accept Li+.MnO2(s) + Li+(aq) + e– LiMnO2(s)

Electrolyte: Lithium-containing salt such as LiClO4, in

organic solvent. Solid state polymers can also be used.

Cell Potential: 3.0 V

Batteries 07Batteries 07• Nickel–Metal–Hydride (NiMH):

• Replaces toxic Cd anode witha hydrogen atom impregnated ZrNi2 metal alloy.

• Applications of NiMH type batteries includes hybrid vehicles such as the Toyota Prius and consumer electronics. 1.2 V

Anode:

Cathod:

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