BY

MDM HAIRUL AMANI BINTI ABDUL HAMID

ELECTROCHEMICAL CELLS

CHAPTER 3

OUTLINE

Components of a cell Conventional representation of a cell Potential of cells and electrodes Thermodynamic of cells Work and free energy Standard electrode potentials Equilibrium constant Nernst equation

2

OUTLINE

Types of electrodes and general form of Nernst equation for an electrode

QUIZ 4

Types of galvanic cells and general form of Nernst equation for a galvanic cell

Applications of galvanic cell potentials Activity coefficients Equilibrium constants Solubility constants pH

TEST 2

3

ELECTROCHEMICAL CELLS

Consists of electrode in contact with an electrolyte in the electrode compartment(s)

Salt bridge concentrate electrolyte solution in agar jelly; completes the electrical circuit, and exchange of ions between 2 different electrode compartments and enable the cell to function.

4

Electrolytic cell vs Voltaic Cell

5

Redox Reaction

- Is a reaction in which there is a transfer of electron

from one species to another.

OilRig Oxidation is loss, reduction is gain Oxidation a process in which there is a loss of electron

Reducing agent/reductant is the electron donor

Reduction a process in which there is a gain of electron.

Oxidazing agent/oxidant is the electron acceptor

6

Redox Reaction

Cu2+ (aq) + 2e- Cu(s)

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

7

Half reactions

Cu2+(aq) + 2e- Cu(s) (Reduction of Cu2+)

Zn(s) Zn2+(aq) + 2e- (Oxidation of Zn)

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

The reduced and oxidised substances in half

reactions form redox couple:

Cu2+/Cu and Zn/Zn2+

8

Half reactions

In general the redox couple is written as Ox/Red with

the half reaction as:

Ox + ne- Red

Example: Redox couple

Cu2+(aq) + 2e- Cu(s) Cu2+/Cu

Zn2+(aq) + 2e- Zn(s) Zn2+/Zn

9

Reaction Quotient, Q

For half reaction:

Cu2+(aq) + 2e- Cu(s) Q= 1

(Cu2+)

Zn2+(aq) + 2e- Zn(s) Q= 1

(Zn2+)

For the standard state of pure material

(element) it has unity activity eq Cu = 1

10

Reaction Quotient, Q

Reaction quotient for the reudction of O2 to H2O in

acid solution,

O2(g) + 4H+(aq) + 4e- 2H2O(l)

Q = 2H2O since O2 behaves as perfect gas

4H+ O2 O2 = PO2

= P

P

4H+ PO2

and 2H2O = 1

11

Reactions at electrode

When spontaneous

reaction takes

place in a galvanic

cell, electron are

deposited in

anode(oxidation)

and collected from

cathode (reduction)

Zn(s) + Cu2+(aq) Zn2+ (aq) + Cu(s)

12

Galvanic cell

Electrolytic cell

Cathode has higher potential

than the anode.

Species in the electrolyte

undergoing reduction,

withdraws electrons from

cathode, leaving a relative

positive charge on the

cathode.

At anode, oxidation caused

the transfer of electrons to

the electrode, give it a

relative negative charge

Oxidation occur at the

anode, but electrons are

withdrawn from the species

since process is not

spontaneous.

Anode is relatively positive to

cathode.

At the cathode, must have

supply of electrons to drive

the reduction.

13

Galvanic vs Electrolytic cell

Galvanic cell

The difference in electrical potential between the

anode and cathode is called

cell voltage/electromotive force (emf)/cell

potential

Electrolytic cell

External source of current, so emf does not apply.

14

Varieties of electrode 1. Metal/metal ion electrode

Cu2+ + 2e- Cu

15

Varieties of electrode

2. Redox electrode

Species exist in solution in 2

oxidation state.

Electrode used is an inert

metal Pt

Eg:

Fe3+ to Fe2+

16

Varieties of electrode 3. Gas electrode

Standard hydrogen electrode

17

Varieties of electrode 3. Gas electrode (Cont)

eg. Hydrogen electrode

inert metal as electrode.

gas in equilibrium with its ion

2H+(aq) + 2e- H2 (g) (Reduction)

Redox couple: H+/H2 It may either be a cathode or an anode

The standard hydrogen electrode is attached to the electrode system under investigation.

18

Metal/Insoluble salt electrode A metal M covered by porous layer of insoluble MX

salt, immersed in a solution of X ions.

AgCl(s) + e Ag(s) + Cl-(sat)

19

Cell Notation

ZnZn2+ (1 M) Cu2+(1 M)Cu

Phase

boundary

interphase Single cell

compartment

Anode

(oxidation) Salt bridge

Cathode

(reduction)

Half Equations:

Zn(s) Zn2+(aq) + 2e Cu2+(aq) + 2e Cu(s)

20

Cell reaction

The reaction in the cell written on the assumption that right-

hand electrode is cathode and with spontaneous reaction,

reduction is taking place in the right-hand compartment

Cu2+(aq) + 2e Cu(s) (Reduction)

Zn(s) Zn2+(aq) + 2e (Oxidation)

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) (overall)

Left: Zn(s) Zn2+ + 2e

Right: Cu2+(aq) + 2e Cu(s)

Cell Notation: ZnZn2+ Cu2+Cu

21

Example 1

Write the half-equations and cell reactions for each

of the following cells:

a) AgAg+H+H2, Pt

b) PtCr2O72, Cr3+, H+BrBr2(l), Pt

22

Answer:

a) AgAg+H+H2, Pt

Ag Ag+ + e (Oxidation) (2x)

2H+ + 2e H2 (Reduction)

2Ag + 2H+ 2Ag+ + H2(g) (overall = cell reaction)

b) PtCr2O72, Cr3+, H+BrBr2(l), Pt

2Cr3+ + 7H2O Cr2O72 + 14H+ + 6e (Oxidation)

(Br2(l) + 2e 2Br) 3 (Reduction)

2Cr3+ + 7H2O + 3Br2(l) Cr2O72 + 14H+ + 6Br

23

Example2 Describe in shorthand notation a galvanic cell for

which the cell reaction is

Cu(s) + 2Fe3+(aq) Cu2+(aq) + 2Fe2+(aq)

Answer:

Oxidation: Cu(s) Cu2+(aq) + 2e

Reduction: Fe3+ + e Fe2+

Cell Notation: CuCu2+ Fe2+, Fe3+Pt

(Since both Fe2+ and Fe3+ are in solution, a Pt electrode

is used.)

24

Varieties of cells

1. Daniell cell

25

Varieties of cells 2. Electrolyte concentration cell.

electrodes are identical but these are immersed in solutions of the same electrolyte of different concentrations

Zn|Zn2+ (C1))/Anode || (Zn2+ (C2 )|Zn)/Cathode

- The source of electrical energy in the cell is the tendency of the electrolyte to diffuse from a solution of higher concentration to that of lower concentration.

- With the expiry of time, the two concentrations tend to become equal.

- At the start the emf of the cell is maximum and it gradually falls to zero.

26

Varieties of cells 3. Electrode concentration cell

The potential difference is developed between two like electrodes at different concentrations dipped in the same solution of the electrolyte.

For example, two hydrogen electrodes at different pressure in the same solution of hydrogen ions constitute a cell of this type.

Pt,H2 (Pressure p1)) |H+ ||H+, H2 (Pressure p2)Pt

In the amalgam cells, two amalgams of the same metal at two different concentrations are inserted in the same electrolyte solution.

27

Liquid Junction Potentials, Elj Electrolyte concentration cells always

have a liquid junction, electrode

concentration cells do not.

In a cell with two different electrolyte

solutions in contact, there is an

additional source of potential difference

across the interface of the two

electrolytes, and this potential is known

as the liquid junction potential.

The contribution of the liquid junction to the potential can be reduced by

joining the electrolyte compartments through a salt bridge.

The salt bridge which is an inverted tube consist of concentrated salt

solution in jelly, has two opposing liquid junction potentials at both ends

that almost cancel.

28

Cell Potential, V

- The potential difference between two electrodes.

- Measured in volts (V) (1V = 1JC-1s)

- As long as cell has not reached chemical

equilibrium, it can do work as reaction drives

electrons through external circuit

- A cell which has reached equilibrium can do no

work. The cell potential = 0.

29

Gibbs energy change in a cell Given we = non expansion work

eg. Electrical work of pushing e through a circuit

If change occurs at constant P and T

dwerev = dG (Tp = constant)

But the process is reversible, the work done must have max

value. For a measurable change

wmax = dG (Tp = constant)

For a spontaneous process, max electrical work by cell =

wmax = G

30

Relation between E and G wmax = G

For this equation to be valid, cell must be operating reversibly at specific constant composition.

Measure the cell potential when it is balanced by exactly opposing source of potential.

This results in zero current cell potential or E, electromotive force of the cell.

-nFE = G

n = no of mole

F= Faradays constant = 96500 C mol-1

E = cell potential.

31

Voltmeter A voltmeter is an instrument used for

measuring electrical potential difference

between two points in an electric circuit.

Analog voltmeters move a pointer across a

scale in proportion to the voltage of the

circuit;

digital voltmeters give a numerical display of

voltage by use of an analog to digital

converter.

The maximum possible voltage to be

measured is called the electromotive force

(emf).

The emf of a cell under standard conditions

is E0

Multimeter

Voltmeter

32

The Nernst Equation

The reaction of Gibbs energy is related to the composition of

reaction mixture by

G = Go + RT lnQ

(At equilibrium, Q = Keq and G = 0. Go = - RT lnQ )

Substituting above equation by: -nFE = G

And dividing each by nF

E = -Go - RT lnQ

nF nF

33

The Nernst Equation E = -Go - RT lnQ

nF nF

Given Go = - nFEo

E = Eo - RT lnQ or E = Eo 2.303 RT logQ

nF nF

at 25oC RT = 0.0257V

F

E = Eo 0.0257 lnQ or E = Eo 0.0592 logQ

n n

34

Standard Potentials Define a potential of one electrodes as having zero

potentials, then assign values to others on that basis.

SHE (LH) Standard Hydrogen Electrode.

Pt | H2(g)|H+(aq) E0= 0 at all temperatures

Example:

The standard potential of Cu2+/Cu couple (RH electrode)

Pt | H2(g)|H+(aq) || Cu2+|Cu

E0 (Cu2+/Cu)=0.337V

35

Standard Electrode Potentials Standard Reduction Potentials

(Eo)

is the voltage associated with

a reduction reaction at an

electrode when all solutes are

1M and all gases are at 1 atm.

Reduction reaction:

2H+(1M) + 2e- H2(1 atm)

Eo = 0

1 atm

36

Standard Electrode Potentials

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

Eocell = Eo

cathode - Eo

anode

Eocell = Eo

H+/H2 - Eo

Zn2+/Zn

0.76 = 0 - EoZn2+/Zn EoZn2+/Zn = - 0.76

Zn2+(1M) + 2e Zn Eo = - 0.76

Anode: Zn(s) Zn2+(1M) + 2e

Cathode: 2H+(1M) + 2e H2(1 atm)

Zn(s) + 2H+(1M) Zn2+(1M)+ H2(1 atm)

Eocell = 0.76 V

37

Standard Electrode Potentials

H2(1 atm) 2H+(1 M) +

Eocell = 0.34V

Pt(s)|H2(1 atm) |H+(1M)|Cu2+(1M) |Cu(s)

Eocell = Eo

cathode - Eo

anode

Eocell = Eo

Cu2+/Cu - Eo

H+/H2

0.34 = EoCu2+/Cu - 0 EoCu2+/Cu = 0.34

Cu2+(1M) + 2e Cu(s) Eo = 0.34

Anode: H2(1 atm) 2H+(1 M) + 2e

Cathode: Cu2+(1M) + 2e Cu(s)

H2(1 atm) + Cu2+(1M) Cu(s) + 2H+(1 M)

38

Standard Reduction Potential Table

Half cell reactions are

only in the form of

reduction reactions.

The more positive Eo

the greater the

tendency for substance

to be reduced

Incre

asin

g s

tre

ng

th in

oxid

isin

g a

ge

nt

Incre

asin

g s

tren

gth

in re

du

cin

g a

ge

nt

39

Standard Reduction Potential Table

Half cell reactions are

reversible

The sign Eo changes

when the reaction is

reversed

Changing stoichio-

metric coefficients of

half cell reaction does

not change value of Eo

40

Example

Consider the following reaction:

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

Determine the emf of the cell given the

concentration of Cu2+ and Zn2+ are 5.0M and

0.050M respectively at 298K.

41

Answer

Cu2+(aq) + 2e- Cu(s) EoCu2+/Cu = +0.34 V Zn2+(aq) + 2e- Zn(s) EoZn2+/Zn = - 0.76 V

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) Eocell = ?

Eocell = Eo

cathode - Eo

anode

= EoCu2+/Cu - EoZn2+/Zn

= 0.34 (-0.76)

= 1.10 V

42

Answer(cont)

Using Nernst equation:

E cell = Eo

cell 0.0592 logQ

n

E cell = Eo

cell 0.0592 log[Zn2+]

n [Cu2+]

= 1.10 - 0.0592 log [0.050]

2 [5.0]

= 1.16 V

Cell reaction becomes more spontaneous since [reactant]>[product] -- E cell > E

ocell

43

Example

Consider the following reaction:

Cr2O72- (aq) + 14H+ (aq) + 6I- (aq) 2Cr3+(aq)

3I2(s) + 7H2O(l)

Determine the emf of the cell given the

concentration:

[Cr2O72-]= 2M, [H+] =1M, [I-] = 1M and

[Cr3+] = 1.0 x 10-5M

44

Answer Cr2O7

2- (aq) + 14H+ (aq) + 6I- (aq) 2Cr3+(aq) 3I2(s) + 7H2O(l)

Eocell = Eo

cathode - Eo

anode

= EoCr2O72-

/Cr3+ - EoI2/I

-

= 1.33 0.53

= 0.8 V

45

Answer Cr2O7

2- (aq) + 14H+ (aq) + 6e 2Cr3+(aq) +7H2O(l)

6I- (aq) 3I2(s) + 6e

n = 6

Using Nernst equation:

E cell= Eocell 0.0592 logQ = E

ocell 0.0592 log[Cr

3+]2

n n [Cr2O72-][H+]14[I-]6

= 0.8 - 0.0592 log (1.0 x 10-5) 2

6 [2] [1]14[1]6

= 0.8 - 0.0592 log 5 x 10-11

6

= 0.8 (- 0.10) = 0.90 V

Since [reactant]>[product] -- E cell > Eo

cell

If value of Q increases -- value of E cell decreases. 46

Concentration Cell Cell Notation:

M(s)|M+(aq, L) M+(aq, R)| M(s)

[R] > [L], E 0

Positive potential arises because positive ions tend to

be reduced , withdrawing electrons from the electrode.

This occur in the right hand electrode compartment

which is more concentrated

M+ (m2) + e M

The reaction at the left hand electrode is:

M M+(m1) + e

Overall reaction:

M+ (m2) M+ (m1)

47

Overall reaction:

M+ (m2) M+ (m1)

If m2 > m1 the reaction is a forward direction, positive emf is produced.

If m2 < m1 electrons flow from right to left electrode, negative emf is produced.

G associated with transfer of M+ ions from M+ (m1) M+ (m1) is given by:

G = RT ln m1 m2

Since no. of mol of e- = 1,

E = RT ln m2 E is positive if m2 > m1

F m1 Go = 0 and Eo = 0 since a cell cannot drive a current

through a circuit with 2 identical electrode compartment.

48

Example:

Calculate the emf at 25oC of a concentration cell of

this type in which the molalities are 0.2m and

3.0m.

E = RT ln m2 at 25oC E = 0.0257 ln m2

F m1 m1

= 0.0257 ln 3.0

0.2

= 0.0696 V

49

Concentration Cell

Calculate the emf of the

cell in the diagram at

25oC

E = 0.0592 log 1.00

2 1.00 x 10-3

= 0.0888 V

50

Cell Equlibrium When a cell has reached equilibrium, then Q = K,

while E = 0 since chemical reaction at equilibrium do no work and generates 0 potential difference between the electrodes of the cell.

E= Eo - RT lnQ E= 0

nF

0 = Eo - RT lnK (x nF and Q = K)

nF 0 = nFEo RTlnK nFEo = RT lnK

ln K = nFEo

RT

51

Example: Calculate the equilibrium constant at 25oC for the

reaction occurring in the Daniell cell, if the standard emf is 1.10V.

Answer:

ln K = nFEo RT = 0.0257

RT F

ln K = 2 x 1.10 = 85.6

0.0257

K = e85.6 = 1.50 x 1037

For spontaneous process, Go 0 and K > 1

52

Example A reaction important in an acidic environment:

Fe(s) + 2H+(aq) + O2 (g) Fe2+ (aq) + H2O(l)

Does the equilibrium constant favour the formation of

Fe2+ (aq)?

Reduction half reaction:

Fe2+ + 2e Fe(s) Eo = - 0.44 V

(anode)

2H+(aq) + O2 (g) + 2e H2O(l) Eo = +1.23 V

(cathode)

Eo cell = Eocathode E

oanode = 1.23 (-0.44)= 1.67 V

Since Eo cell > 1, the reaction has K>1

Favour the formation of Fe2+

53

The composition dependence of

individual potentials

1. Metal electrode

-potential of an Ag+/Ag electrode is given by:

Electrode half reaction:

Ag+(aq) + e Ag(s) n = 1

E Ag+/Ag = EoAg+/Ag RT ln 1

F (Ag+)

54

2. Gas using platinum as electrode.

- eg, chlorine gas is bubbled over a platinum

electrode dipping aqueous sodium chloride at

298 K. Calculate the change in potential of

electrode when the chlorine pressure is

increased from 1.0 atm to 2.0 atm.

Electrode half reaction:

Cl2(g) + 2e 2Cl-(aq) n = 2

Q = (Cl-)2

F

55

Electrolysis

- is a chemical process that uses

electricity to cause a non-spontaneous

redox to occur.

- Take place in electrolytic cell made up of two electrodes connected to a battery

and immersed in an electrolyte.

56

Electrolytic cell

Battery function as a

source of direct current

- Forces electrons out

of anode

- Pushes electrons

towards the

cathode.

Anode: Yn- Y + ne

(oxidation)

Cathode: Xn+ + ne X

(reduction)

57

Electrolysis of Molten Salt

Cathode: 2Na+(l) + 2e 2 Na(l)

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

Overall: 2Na+(l) + 2Cl-(l) 2 Na(l) + Cl2(g)

58

Electrolysis of Aqueous solutions

- Solution contain cations and anions from the salt

and water molecules.

- Water is an electro-active substance can be oxidized or reduced depending on condition of

electrolysis.

Reduction: 2H2O(l) + 2e H2(g) + 2OH- E=-0.83V

Oxidation: 2H2O(l) O2(g) + 4H+ + 4e E=-1.23V

59

Example: Electrolysis of aqueous

solution of NaCl

At cathode:

Na+(aq) + e Na(s) Eo = - 2.71V

2H2O(l) + 2e H2(g) + 2OH-(aq) Eo = - 0.83V

At anode:

2Cl-(l) Cl2(g) + 2e Eo = -1.36V

2H2O(l) O2(g) + 4H+ + 4e Eo = -1.23V

60

61

Corrosion

62

Corrosion

Corrosion is the gradual (spontaneous)

destruction of material, usually metals, by

chemical reaction with its environment.

Electrochemical oxidation of metals in reaction

with an oxidant such as oxygen.

Example - Rusting, the formation of iron oxides,

is a well-known example of electrochemical

corrosion.

Occur on the exposed surface

63

Figure 21.20 The corrosion of iron.

)(2)(2)(4)()(2

)(24)(4)()2(

2)()()1(

22

2

22

2

lOHaqFeaqHgOsFe

lOHeaqHgO

eaqFesFe

Total

reduction) region;(cathodic

2 oxidation) region;(anodic

)(4)(.)()2()(2

1)(2 23222

2 aqHsOnHOFelOHngOaqFe (3)

)()()(2

3)(2 23222 sOnHOFelOnHgOsFe Overall

64

Figure 21.22

The effect of metal-metal contact on the corrosion of iron.

faster corrosion cathodic protection

65

Protection from corrosion

Surface coating plating with a thin layer of other metal, painting, enamel, grease, varnish

Anodizing eg aluminium oxide (oxide as the protective outer coating)

Cathode protection surface as cathode (connecting metal to a more electropositive metal)

eg steel, water, and fuel pipelines and tanks;

steel pier piles, ships, and offshore oil platforms.

Anodic protection - impresses anodic current on

the structure to be protected (opposite to the

cathodic protection)

66

Figure 21.23

The use of sacrificial anodes to prevent iron corrosion.

67

Quantitative aspect of electrolysis

First Law of electrolysis

- the quantity of a substance formed at an

electrode is directly proportional to the quantity of

electric charge that has flowed through the circuit

Second Law of electrolysis

- for a given electric charge, the amount of any

metal formed at the electrode is proportional to its

equivalent weight. (relative mass/charge on the

metal of the ion)

68

Quantity of electric charge:

Q = It

Q = electric charge in couloumbs (C)

I = current in Amperes (A)

t = time in seconds (s)

The amount of substance formed at the electrodes

may be calculated from stoichiometric of

equations, and knowing the charge on one mole

of electrons is one Faraday (F)

1 F = 96500 C 69