electric power conversion in electrochemistry

Electric power conversion in electrochemistry

Chemical Reactions

Electric Power

Electrolysis / Power consumption

Electrochemical battery / Power generation

battery+-

inertelectrodes

powersource

vessel

e-

e-

conductivemedium

CellConstruction

Sign or polarity of electrodes

(-) (+)

Na+ Cl-

Let’s examine the electrolytic cell for molten NaCl.

+-battery

Na (l)

electrode half-cell

electrode half-cell

Molten NaCl

Na+

Cl-

Cl-

Na+

Na+

Na+ + e- Na 2Cl- Cl2 + 2e-

Cl2 (g) escapes

Observe the reactions at the electrodes

NaCl (l)

(-)

Cl-

(+)

+-battery

e-

e-

NaCl (l)

(-) (+)

cathode anode

Molten NaCl

Na+

Cl-

Cl-

Cl-

Na+

Na+

Na+ + e- Na 2Cl- Cl2 + 2e-

cationsmigrate toward

(-) electrode

anionsmigrate toward

(+) electrode

At the microscopic level

Molten NaCl Electrolytic Cellcathode half-cell (-)

REDUCTION Na+ + e- Na

anode half-cell (+)OXIDATION 2Cl- Cl2 + 2e-

overall cell reaction2Na+ + 2Cl- 2Na + Cl2

X 2

Non-spontaneous reaction!

Definitions:

CATHODE

REDUCTION occurs at this electrode

ANODE

OXIDATION occurs at this electrode

Na+ Cl-

H2O

Will the half-cell reactions be the same or different?

Water Complications in ElectrolysisIn an electrolysis, the most easily oxidized and most easily reduced reaction occurs.

When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown.

Electrode Ions ... Anode Rxn Cathode Rxn E°Pt (inert) H2O H2O(l)+ 2e- H2(g)+ 2OH-

(aq) -0.83 VH2O 2 H2O(l) 4e- + 4H+

(g) + O2(g) -1.23 V

Net Rxn Occurring:

2 H2 H22O O 2 H 2 H2(g)2(g)+ O+ O2 (g)2 (g) E° E° = - 2.06 V= - 2.06 V

battery+- power

source

e-

e-

NaCl (aq)

(-) (+)cathodedifferent half-cell

Aqueous NaCl

anode2Cl- Cl2 + 2e-

Na+

Cl-

H2O

What could be reduced at the

cathode?

Aqueous NaCl Electrolytic Cellpossible cathode half-cells (-)

REDUCTION Na+ + e- Na2H2O + 2e- H2 + 2OH-

possible anode half-cells (+)OXIDATION 2Cl- Cl2 + 2e-

2H2O O2 + 4H+ + 4e-

overall cell reaction2Cl- + 2H2O H2 + Cl2 + 2OH-

e-

Ag+

Ag

For every electron, an atom of silver is plated on the

electrode.Ag+ + e- Ag

Electrical current is expressed in terms of the

ampere, which is defined as that strength of current

which, when passed thru a solution of AgNO3 (aq) under

standard conditions, will deposit silver at the rate of

0.001118 g Ag/sec

1 amp = 0.001118 g Ag/sec

The mass deposited or eroded from an electrode depends on the quantity of

electricity.Quantity of electricity – coulomb (Q)

Q is the product of current in amps times time in

secondsQ = It

coulomb

current in amperes (amp)

time in seconds

1 coulomb = 1 amp-sec = 0.001118 g Ag

Ag+ + e- Ag

1.00 mole e- = 1.00 mole Ag = 107.87 g Ag

107.87 g Ag/mole e-

0.001118 g Ag/coul= 96,485 coul/mole e-

1 Faraday (F )mole e- = Q/F

mass = molemetal x MM

molemetal depends on the half-cell reaction

Examples using Faraday’s LawHow many grams of Cu will be deposited in

3.00 hours by a current of 4.00 amps? Cu+2 + 2e- Cu

The charge on a single electron is 1.6021 x 10-

19 coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e-.

Fig 18-26 Pg 901

The electroplating of silver.

Pg 901

Courtesy International SilverPlating, Inc.

21-8 Industrial Electrolysis Processes

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 18

of 52

A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode.

battery- +

+ + +- - -

1.0 M Au+3 1.0 M Zn+2 1.0 M Ag+

Au+3 + 3e- Au Zn+2 + 2e- Zn Ag+ + e- Ag

e-

e- e- e-

Volta’s battery (1800)

Alessandro Volta 1745 - 1827

Paper moisturized with NaCl solution

Cu

Zn

Lesson 9NEEP 42321

Galvanic cells and electrodes

To sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.

Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution.

Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential).

with Galvanic Cells

19.2

spontaneousredox reaction

anodeoxidation

cathodereduction

Fig 18-9Pg 872

Diagram of a galvanic cell containing passive electrodes. the two platinum electrodes do not take part in the redoxchemistry of this cell. Theyonly conduct electrons to andfrom the interfaces.

Electrodes are passive (not involved in the reaction)

Olmsted Williams

Cu

1.0 M CuSO4

Zn

1.0 M ZnSO4

Salt bridge – KCl in agar

Provides conduction between half-cells

CellConstruction

Observe the electrodes to see what is occurring.

Cu

1.0 M CuSO4

Zn

1.0 M ZnSO4

Cu plates out or

deposits on

electrode

Zn electrode erodes

or dissolves

cathode half-cellCu+2 + 2e- Cu

anode half-cellZn Zn+2 + 2e-

-+

What about half-cell reactions?

What about the sign of the electrodes?

What happened

at each electrode

?

Why?

Cu

1.0 M CuSO4

Zn

1.0 M ZnSO4

cathode half-cellCu+2 + 2e- Cu

anode half-cellZn Zn+2 + 2e-

-+

Now replace the light bulb with a volt meter.

1.1 volts

H2 input1.00 atm

inert metal

We need a standard electrode to make

measurements against!The Standard Hydrogen Electrode (SHE)

Pt

1.00 M H+

25oC1.00 M H+

1.00 atm H2

Half-cell2H+ + 2e- H2

EoSHE = 0.0 volts

How do we calculate Standard Redox Potentials?

We must compare the half reactions to a standard

What is that standard?

2 H3O+(aq) + 2e- H2(g) + 2 H2O(l) E°= 0.00 V

This is called the standard hydrogen electrode or SHE

Now that we have a standard, we can calculate standard redox potential by using the table of standard redox potentials

19.3

• 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

Copyright 1999, PRENTICE HALL Chapter 20 30

Cell EMFCell EMFOxidizing and Reducing AgentsOxidizing and Reducing Agents

H2 1.00 atm

Pt

1.0 M H+

Cu

1.0 M CuSO4

0.34 v

cathode half-cellCu+2 + 2e- Cu

anode half-cellH2 2H+ + 2e-

KCl in agar

+

Now let’s combine the copper half-cell with the SHE

Eo = + 0.34 v

H2 1.00 atm

Pt

1.0 M H+1.0 M ZnSO4

0.76 vcathode half-cell2H+ + 2e- H2

anode half-cellZn Zn+2 +

2e-

KCl in agar

Zn

-

Now let’s combine the zinc half-cell with the SHE

Eo = - 0.76 v

Assigning the Eo

Al+3 + 3e- Al Eo = - 1.66 v

Zn+2 + 2e- Zn Eo = - 0.76 v

2H+ + 2e- H2 Eo = 0.00 v

Cu+2 + 2e- Cu Eo = + 0.34

Ag+ + e- Ag Eo = + 0.80 v

Write a reduction half-cell, assign the voltage measured, and the sign of the electrode to the

voltage.

Incr

easi

ng a

ctiv

ity

Measuring Standard Reduction Potential

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 34 of

52

cathode cathode anodeanode

What is 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 VCd 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 = Ecathode - Eanodecell0 0

E0 = -0.40 – (-0.74) cell

E0 = 0.34 V cell

19.3

Calculating the cell potential, Eocell, at

standard conditions

Fe+2 + 2e- Fe Eo = -0.44 v

O2 (g) + 2H2O + 4e- 4 OH- Eo = +0.40 v

This is corrosion or the oxidation of a metal.

Consider a drop of oxygenated water on an iron objectFe

H2O with O2

Fe Fe+2 + 2e- -Eo = +0.44 v2x

2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 v

reverse

• Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.

G° = –nFE°cell

andG° = –RT ln Keq

therefore –RT ln Keq = –nFEocell

Equilibrium Constants in Redox Reactions

RT ln Keq RTE°cell = ––––––––– = –––– ln Keq nF nF

0.025693 VE°cell = –––––––– ln Keq n

R and F are constant, therefore at 298 K:

41

Effect of Concentration on Cell Effect of Concentration on Cell EMFEMF

• A voltaic cell is functional until E = 0 at which point equilibrium has been reached.

• The point at which E = 0 is determined by the concentrations of the species involved in the

redox reaction.

The Nernst EquationThe Nernst Equation• The Nernst equation relates emf to

concentration using

and noting that

QRTGG ln

QRTnFEnFE ln

and the previous relationship:Go = -nFEo

cell

from thermodynamics:Go = -2.303RT log K

-nFEocell = -2.303RT log K

at 25oC: Eocell = 0.0591 log K

n

where n is the number of electrons for the balanced reaction

What happens to the electrode potential if conditions are not at standard conditions?

The Nernst equation adjusts for non-standard conditions

For a reduction potential: ox + ne red

at 25oC: E = Eo - 0.0591 log (red)

n (ox)Calculate the E for the hydrogen

electrode where 0.50 M H+ and 0.95 atm H2.

in general: E = Eo – RT ln (red)

nF (ox)

1) An example:

Ni(s) | Ni2+(0.600M)|| Sn2+

(0.300M) | Sn(s)

According to the reduction potentials:

2 e- + Ni2+ Ni(s) -0.230 V

2 e- + Sn2+ Sn(s) -0.140V

One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.

Ni(s) 2 e- + Ni2+ +0.230 V

2 e- + Sn2+ Sn(s) -0.140V

Ni(s) + Sn2+ Ni2+ + Sn(s) 0.090 V = Eo

E) Calculation of the equilibrium constant

1) at equilibrium E = 0 ; Q = ____

From the Nernst Equation:

For the cell:

Ni(s) | Ni2+(0.600M)|| Sn2+

(0.300M) | Sn(s)

2) An example:

Sn(s) | Sn2+(1.0M)|| Pb2+

(0.0010M) | Pb(s)

According to the reduction potentials:

2 e- + Pb2+ Pb(s) E0=-0.126 V

2 e- + Sn2+ Sn(s) E0=-0.136V

One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!

Go = -nFEocell

Free Energy and the Cell Potential

Cu Cu+2 + 2e- -Eo = - 0.34

Ag+ + e- Ag Eo = + 0.80 v2x

Cu + 2Ag+ Cu+2 + 2AgEocell= +0.46 v

where n is the number of electrons for the balanced reaction

What is the free energy for the cell?

1F = 96,500 J/v

Electrolysis of Copper

A net reaction of zero, yet a process does take place.

A concentration cell based on the A concentration cell based on the Cu/Cu2+ half-reaction.Cu/Cu2+ half-reaction. AA, Even , Even

though the half-reactions involve the though the half-reactions involve the same components, the cell operates same components, the cell operates

because the half-cell concentrations are because the half-cell concentrations are different. different. BB, The cell operates , The cell operates

spontaneously until the half-cell spontaneously until the half-cell concentrations are equal. Note the concentrations are equal. Note the

change in electrodes (exaggerated here change in electrodes (exaggerated here for clarity) and the equal color of for clarity) and the equal color of

solutions.solutions.

Concentration Cells

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 53

of 52

Two half cells with identical electrodes but different ion concentrations.

2 H+(1 M) → 2 H+(x M)

Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)

2 H+(1 M) + 2 e- → H2(g, 1 atm)

H2(g, 1 atm) → 2 H+(x M) + 2 e-

Concentration Cells

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 54

of 52

Ecell = Ecell° - logn

0.0592 V x2

12

Ecell = 0 - log2

0.0592 V x2

1

Ecell = - 0.0592 V log x

Ecell = (0.0592 V) pH

2 H+(1 M) → 2 H+(x M)

Ecell = Ecell° - log Qn

0.0592 V

F)The pH meter is a special case of the Nernst Equation

1/2 H2 H+ + 1 e-

Q = [H+]

pH = - log [H+]

E = Eo + 0.0592 pH

The scale on the pH meter is marked off so that a change of 1 pH unit equals 0.0592 volts or 59.2

millivolts.

The pH MeterIn practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!

A stable reference electrode and a glass-membrane electrode are contained within a combination

pH electrode.

The electrode is merely dipped into a solution, and

the potential difference between the electrodes is

displayed as pH.

galvanic electrolytic

needpowersource

twoelectrodes

produces electrical current

anode (-)cathode (+)

anode (+)cathode (-)

salt bridge vessel

conductive medium

Comparison of Electrochemical Cells

G < 0G > 0

Copyright 1999, PRENTICE HALL Chapter 20 60

CorrosionCorrosionCorrosion of IronCorrosion of Iron

• Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.

• Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).

• Anode: Fe(s) Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the

oxidation of iron.• Fe2+ initially formed can be further oxidized to

Fe3+ which forms rust, Fe2O3.xH2O(s).

• Oxidation occurs at the site with the greatest concentration of O2.

21-6 Corrosion: Unwanted Voltaic Cells

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 61

of 52

O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)

2 Fe(s) → 2 Fe2+(aq) + 4 e-

2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)

Ecell = 0.841 V

EO2/OH- = 0.401 V

EFe/Fe2+ = -0.440 V

In neutral solution:

In acidic solution:

O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = 1.229 V

Copyright 1999, PRENTICE HALL Chapter 20 62

CorrosionCorrosionCorrosion of IronCorrosion of Iron

Corrosion of Iron

Corrosion of IronRust formation:

4Fe2+(aq) + O2(g) + 4H+(aq) 4Fe3+(aq) + 2H2O(l)

2Fe3+(aq) + 4H2O(l) Fe2O3H2O(s) + 6H+

(aq)

Prevention of CorrosionCover the Fe surface with a protective coatingPaint

Tin Zn

Galvanized iron

Copyright 1999, PRENTICE HALL Chapter 20 66

CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron

• Corrosion can be prevented by coating the iron with paint or another metal.

• Galvanized iron is coated with a thin layer of zinc.

• Zinc protects the iron since Zn is the anode and Fe the cathode:

Zn2+(aq) +2e- Zn(s), Ered = -0.76 V

Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V

• With the above standard reduction potentials, Zn is easier to oxidize than Fe.

Corrosion Protection

Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 67

of 52

galvanized steel (Fe)

(cathode)

(electrolyte)

(anode)

Cathodic ProtectionIn cathodic protection, an iron object to be protected is connected to a chunk of an active metal.

• The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.

• Water heaters often employ a magnesium anode for cathodic protection.