Syllabus

Data treatment and analysis

1. Errors in Chemical Analysis

2. Significance Tests

3. Sampling and Calibration Methods

Separation science

4. Extraction Techniques

5. Introduction to Chromatography

6. Gas Chromatography

7. High Pressure Liquid

Chromatography

8. Electrophoresis

Electrochemistry

9. Fundamentals of Electrochemistry

10. Potentiometry

11. Ion Selective Electrodes

12. Voltammetry

Nanoanalysis

13. Nanoanalysis

Equilibria

14. Chemical Equilibrium

15. Acid, Bases and Acid-Base

Equilibrium

16. Chemical Equilibrium for Complex

Systems

17. Chemical Equilibrium for Titrations

1CM2142

Fundamentals of Electrochemistry

CM2142

1. Introduction

2. Oxidation Numbers

3. Balancing Redox Equations

4. Voltaic/Galvanic Cells

5. Cell Diagrams

6. Cell Potentials

7. Free Energy and Nernst Equations

8. Electrolytic Cells

9. 2-Electrode and 3-Electrode Systems

eLecture prepared by Dr Emelyn Tan

What is Electrochemistry?

“Electrochemistry involves chemical phenomena

associated with charge separation and charge transfer,

with applications ranging from fundamental studies and

research to industrial, environmental and clinical

applications.”

Source: International Society of Electrochemistry

www.ise-online.org

CM2142 3

4

Terminology

Term Units Equation Definition

Current (i) Amperes (A) A = C / s Quantity of charge

flowing in a circuit per

second.

Potential (E) Volts (V) V = J / C Measure of work needed

to move electric charge

from 1 point to another.

Resistance (R) ohm (Ω) V = i R Ohm’s Law

Electric charge (Q) Coulombs (C) Q = n F = i t A charge of single

electron is 1.602 x 10-19

C.

1 mol of electron has

96485 C (Faraday’s

constant).

Work (w) Joules (J) w = n F E = Q E

Power (P) Watt (W) = 1 J / s P = w / t = i E Work done per unit time

CM2142

5CM2142

A = C / s

V = J / C Current

Potential

Oxidation:

- Loss of electrons

- Increase in oxidation number

- Gain of oxygen

- Loss of hydrogen

Oxidising Agent:

- Species is reduced

Reduction:

- Gain of electrons

- Decrease in oxidation number

- Loss of oxygen

- Gain of hydrogen

Reducing Agent:

- Species is oxidised

Revision: Redox Reaction

Reduction (gaining electrons) can’t happen without an

oxidation to provide the electrons.

You can’t have 2 oxidations or 2 reductions in the same

equation. Reduction has to occur at the cost of oxidation.

Hence, redox reactions!6CM2142

Oxidation Numbers (O.N.)

In order to keep track of what

loses electrons and what gains

them, assign oxidation

numbers.

Zn is oxidised as it loses two

electrons to go from neutral Zn

(O.N. = 0) metal to the Zn2+

(O.N. = 2) ion.

Each of the H+ is reduced as it

gains an electron each to go

from H+ (O.N. = 1) ions to

combine to form H2 (O.N. = 0)

gas.

7

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation

number of 0. e.g. Hg (O.N. = 0)

2. The oxidation number of a monoatomic ion is the same

as its charge. e.g. Cu2+ (O.N. = 2)

8CM2142

Assigning Oxidation Numbers

3. Non-metals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

Oxygen has an oxidation number of −2, except in the

peroxide ion (e.g. in H2O2) in which it has an oxidation number of −1.

Hydrogen has an oxidation number of +1, except when bonded to a metal as the hydride ion (e.g. in CaH2) in which it has an oxidation number of −1.

Fluorine always has an oxidation number of −1.

The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. e.g. HClO4 (Cl O.N. = 7)

9CM2142

4. The sum of the oxidation numbers in a neutral compound

is 0.

5. The sum of the oxidation numbers in a polyatomic ion is

the charge on the ion.

1 + x + 4(-2) = 0

x = +7

O.N. of Mn is +7

2x + 2(-1) = 0

x = +1

O.N. of H is +1

O.N of Mn in CsMnO4O.N of H in H2O2 O.N of l in lO6

5-

x + 6(-2) = -5

x = +7

O.N. of I is +7

10CM2142

Assigning Oxidation Numbers

Perhaps the easiest way to balance the equation of an

oxidation-reduction reaction is via the half-reaction method.

This involves treating (on paper only) the oxidation and

reduction as two separate processes, balancing these half

reactions, and then combining them to attain the balanced

equation for the overall reaction.

1 oxidation half reaction

+ 1 reduction half reaction

Overall reaction

Balancing Oxidation-Reduction

Equations

11CM2142

Half-Reaction Method

This must be done in sequential order.

1. Assign oxidation numbers to determine what is oxidised and what is reduced.

2. Write the oxidation and reduction half-reactions.

3. Balance each half-reaction.

a. Balance elements other than H and O.

b. Balance O by adding H2O.

c. Balance H by adding H+.

d. Balance charge by adding electrons.

12CM2142

4. Multiply the half-reactions by integers so that the electrons

lost and gained are the same for oxidation and reduction,

respectively.

5. Add the half-reactions, subtracting things that appear on both

sides.

6. Check that the equation is balanced on both sides according

to elements present. “Conservation of mass”

7. Check that the equation is balanced on both sides according

to charge. “Conservation of charge”

Half-Reaction Method

13CM2142

Consider the reaction between MnO4− (permanganate) and C2O4

2− (oxalate):

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)

14CM2142

+7 +3 +4+2

Manganese is reduced.

Carbon is oxidised.

C2O42− CO22 + 2e-

MnO4− Mn2+ + 4H2O+ 8H+ + 5e-

X 5

X 2

5

10 10

216 10 2 8

2MnO4− + 5C2O4

2− + 16H+ 2Mn2+ + 10CO2 + 8H2O

+ 16OH-(aq)Balancing

in Basic

Solution: 2MnO4− + 5C2O4

2− + 16H2O 2Mn2+ + 10CO2 + 8H2O + 16OH-

+ 16OH-(aq)

2MnO4− + 5C2O4

2− + 8H2O 2Mn2+ + 10CO2 + 16OH-

Energy is absorbed to drive a non-

spontaneous redox reaction

Figure 21.3

Silberberg General characteristics of voltaic and electrolytic cells.

VOLTAIC CELL ELECTROLYTIC CELL

Energy is released from

spontaneous redox reaction

Reduction half-reaction

at electrode Y++ e- Y

Oxidation half-reaction

X X+ + e- from electrode

Reduction half-reaction

in solution B++ e- B

Oxidation half-reaction

A- A + e- in solution

Overall (cell) reaction

X + Y+ X+ + Y; ∆G < 0

Overall (cell) reaction

A- + B+ A + B; ∆G > 0

Inert

electrodes

Voltaic vs. Electrolytic Cells

Cells Ecell Electrode

name

Process at

electrode

Sign of

electrode

Electrodes

Voltaic/

Galvanic

> 0

Spontaneous

Anode

Cathode

Ox.

Red.

- (e- source)

+

Redox

process

involves

electrodes

Electrolytic Tunable,

Non-

spontaneous

Anode

Cathode

Ox.

Red.

+ (anions)

- (cations)

Electrodes

are (usually)

inert

16CM2142

In spontaneous oxidation-

reduction (redox) reactions,

electrons are transferred and

energy is released.

The energy can be used to do

electrical work when the

electrons are channeled

through an external device =>

voltaic cell.

Voltaic Cells

17CM2142

Voltaic Cells • Separate the species

(i.e. Zn and Cu) into two

compartments. One for

oxidation other for

reduction.

• Connect electrodes by a

wire/voltmeter.

• Connect solutions by a

salt bridge.

Ox. Red.

‘-’ve

‘+’ve

18CM2142

Salt Bridge Once even just one electron flows from the anode to the cathode, the

charges in each beaker would not be balanced and the flow of electrons

would stop. (right half cell will be ‘-’ve, preventing further electron flow)

Therefore, a salt bridge is used, usually a U-shaped tube that contains a salt

solution, to keep the charges balanced.

Left half cell: net ‘+’ ve charge in solution so anions move toward the anode.

Right half cell: net ‘-’ ve charge in solution so cations move toward the

cathode.

Migration of ions out of the salt bridge is higher than into the salt bridge.

Diffusion from high to low ion concentration.

Besides NaNO3, other usual choices for the electrolyte in the salt bridge are

KCl or KNO3.

19CM2142

Cell Diagrams

Cell diagram: Zn(s) I Zn2+(aq) II Cu2+

(aq) I Cu(s)

Ox. first then red.

If both ox. and red. species are ions, then use Pt (or graphite) as

the electrode. e.g Zn(s) I Zn2+(aq) II Fe3+(aq), Fe2+ (aq) I Pt(s)

Cannot dip wire into solution, need an electrode.

Phase boundary Salt bridge

20CM2142

Electromotive Force (emf)

Water only spontaneously

flows one way in a

waterfall.

Likewise, electrons only

spontaneously flow one

way in a redox reaction.

Which species gives

(anode) and which takes

(cathode)?

21CM2142

Standard Reduction Potentials (Eºred)

Standard:

g = 1 atm

aq = 1 molL-1

a = 1

Higher red.

potential, more

easily reduced,

half-cell will be

the cathode.

Strong oxidising

agents.

Lower red.

potential, more

easily oxidised,

half-cell will be the

anode.

Strong reducing

agents.

Appendix 5 Skoog

CM2142 22

Standard Hydrogen Electrode:

the other half-cell Their values are referenced to a standard hydrogen electrode

(SHE).

By definition, the standard reduction potential (Eºred) for the

hydrogen ion is 0 V:

2H+ (aq, 1 M) + 2e− H2 (g, 1 atm)

aq = 1 molL-1

i.e. pH = 0 CM2142 23

V0Eo

red

by definition, Eºred of SHE = 0

Standard Hydrogen Electrode

Cu2+ is more readily

reduced cf. H+

Eºred (Cu2+) > 0

H+ is more readily

reduced cf. Zn2+

Eºred (Zn2+) < 0 24CM2142

Potential

readings can

be negative.

Standard Cell Potentials

Referenced against the SHE, all species with Eºred > 0 were

reduced and all species with Eºred < 0 were oxidised.

After a half-cell is referenced against the SHE, its Eºred can be

compared against other half-cells.

For a given voltaic cell, the cell potential at standard conditions

can be written as this equation:

Ecell = Ered (cathode) − Ered (anode)

Standard reduction potentials

Larger Eºred,

Reduction process

Smaller Eºred,

Oxidation processCM2142 25

Cell Potentials

Ecell = Ered (cathode) − Ered

(anode)

= +0.34 V − (−0.76 V)

= +1.10 V

For the oxidation (anode) in this cell,

For the reduction (cathode),

Ered = +0.34 V

Ered = −0.76 V

CM2142 26

Consider the following two electrode reactions and their standard

electrode potentials:

Al3+(aq) + 3e- → Al(s) E0 = -1.66 V

Cd2+(aq) + 2e- → Cd(s) E0 = -0.40 V

Write the cell reaction for a voltaic cell based on these two

electrodes, and calculate the standard cell potential, Eocell.

Indicate which is the positive and negative electrode.

Al is oxidised at the anode. In voltaic cells, anode is negative.

Cd2+ is reduced at the cathode. In voltaic cells, cathode is positive.

2Al(s) + 3Cd2+(aq) → 2Al3+

(aq) + 3Cd(s)

Eocell = -0.40 – (-1.66) = 1.26 V

CM2142 27

V = J / C

Free Energy

Go for a redox reaction under standard conditions, can be

written as the equation:

Go = total charge x Eocell = -RT ln K

Go = −nF x Eo

Go = −nFEo = wmax

where:

n is the number of moles of electrons transferred per mole of

reaction

F is the Faraday constant (1 F = 96,485 Cmol-1)

Eo is in V = J/C

For redox reaction to be

spontaneous, G < 0, so E > 0.

w < 0, work is done by the system.

CM2142 28

Nernst Equation

By dividing both sides of −nFE = −nFE + RT ln Q by −nF,

we obtain the Nernst equation:

Ecell = Ecell − Ecell = Ecell −RT

nF ln Q0.059

nlog Q

At 298 K,

In logx 2.303

Recall that G = G + RT ln Q

then −nFE = −nFE + RT ln Q

CM2142 29

x ln 10

A voltaic cell consists of Mn/Mn2+ and Cd/Cd2+ half-cells with

concentrations [Mn2+] = 0.75 M and [Cd2+] = 0.15 M. Use the

Nernst equation to calculate the cell potential, Ecell, at 25 oC.

(F = 96485 Cmol-1)

Data: Cd2+(aq) + 2e- → Cd(s) Eo = -0.40 V

Mn2+(aq) + 2e- → Mn(s) Eo = -1.18 V

From Ered, Cd2+ undergoes reduction and Mn oxidation.

Overall: Cd2+ + Mn → Cd + Mn2+

Ecell = -0.40 – (-1.18) = 0.78 V

Ecell = 0.78 – [(8.314 x 298)/(2 x 96485)] In (0.75/0.15) = 0.76 V

reactants

products

nF

RTFor a cell reaction, Ecell = Ecell ln

CM2142 30

Discharge of Voltaic Cell

The cell potential is initially 0.46 V, Q = 1.

As the reaction proceeds, the potential

decreases continuously until it approaches zero,

as the overall reaction approaches equilibrium.

At the end, when Ecell = 0, Q = K, K = 3.6 x 1015

Ecell = Ecell −

Cu(s) I Cu2+(aq, 1 M) II Ag+(aq, 1 M) I Ag(s)

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

E (Cu2+/Cu) = 0.34 V and E (Ag+/Ag) = 0.80 V

11

1

][Ag

][CuQ

22

2

CM2142 31

Figure 18-5

Skoog

RT

nF ln Q

Concentration 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, Ecell would be zero, but Q would not.

Therefore, as long as the concentrations are different, Ecell

will not be 0.

CM2142 32

Ecell = Ecell −RT

nF ln Q

Ni Ni

Given the following cell: Fe(s) I Fe2+(aq, 2 x 10-2 M) II Fe2+(aq, 1 M) I Fe(s)

and that the standard reduction potential of Fe2+(aq) to Fe(s) is -0.44 V,

calculate Ecell.

Oxidation: Fe(s) → Fe2+(2 x 10-2 M) + 2e-

Reduction: Fe2+(1 M) + 2e- → Fe(s)

Overall: Fe2+(1 M) → Fe2+(2 x 10-2 M)

Ecell = Eocell – RT/nF In Q

Ecell = 0 – RT/nF In (2 x 10-2 / 1)

Ecell = 0.05 V

CM2142 33

Multiplying a cell reaction by any

number does not change Ecell

CM2142

Cu(s) + 2Ag+ → Cu2+ + 2Ag(s)

Cu(s) → Cu2+(aq) + 2e-

2Ag+(aq) + 2e- → 2Ag(s)

2

2

22

2

2

1052

10

52

][Ag

][Culn

2F

RTEE

)]ln[Ag-](ln[Cu2F

RTEE

])2ln[Ag-]5(ln[Cu10F

RTEE

])10ln[Ag-](5ln[Cu10F

RTEE

)]ln[Ag-](ln[Cu10F

RTEE

][Ag

][Culn

10F

RTEE

5Cu(s) + 10Ag+ → 5Cu2+ + 10Ag(s)

5Cu(s) → 5Cu2+(aq) + 10e-

10Ag+(aq) + 10e- → 10Ag(s)

2

2

][Ag

][Culn

2F

RTEE

34

V = J / C

35CM2142

Go = -nFEo = -RT ln K

E°red values are T dependent

E°cell values are T dependent, since K is T dependent according to

the van’t Hoff equation:

In (K2/K1) = (-∆H°/R)(1/T2 – 1/T1)

E°red (Ag+/Ag) = 0.80 V at 298 K At 298 K, K = 3.39 x 1013

∆H°f (Ag+(aq)) = +105.6 kJmol-1

E°red (Ag+/Ag) = 0.795 V at 303 K At 303 K, K = 1.68 x 1013

Why is E°red (Ag+/Ag) = 0.80 V but

E°red (AgCl/Ag) = 0.22 V?

Figure 18-9

Skoog

Ag+(aq) + e- → Ag(s)

][Ag

1log

1

0.059EE o

/AgAg/AgAg

V0.22E

10 x 1.8

1log

1

0.0590.80E

K

][Cllog

1

0.059EE

o

AgCl/Ag

10-

o

AgCl/Ag

sp

o

/AgAgAgCl/Ag

36

10-

sp

sp

10 x 1.8K

AgCl(s).of K on depends ][Ag

sp

spK

][Cl

][Ag

1 ]][Cl[AgK

KCl + AgNO3 → AgCl + KNO3

AgCl(s) + e- → Ag(s) + Cl-(aq)

AgCl(s) ⇋ Ag+(aq) + Cl-(aq)-10

sp 10 x 1.8 ]][Cl[AgK

37CM2142

Calculate the electrode potential vs. SHE of a silver electrode

immersed in a 0.05 M solution of NaCl and AgCl (s) using Eored

(Ag+/Ag) = 0.80 V. Ksp (AgCl) = 1.8 x 10-10.

Ag+(aq) + e- → Ag(s)

][Ag

1log

1

0.059EE o

/AgAg/AgAg

V0.30E

10 x 1.8

0.05log

1

0.0590.80E

K

][Cllog

1

0.059EE

AgCl/Ag

10-AgCl/Ag

sp

o

/AgAgAgCl/Ag

38CM2142

2Ag+(aq) + 2e- → 2Ag(s)

][H][Ag

][Hlog

2

0.059EE

2

2

2o

cellcell

H2(g, 1 atm) → 2H+(aq, 1 M) + 2e−

H2(g, 1 atm) + 2Ag+(aq) → 2H+(aq, 1 M) + 2Ag(s)

)][H

][Hlog

][Ag

1(log

2

0.059E EE

2

2

2

o

/HH

o

/AgAgcell2

][H

][Hlog

2

0.059E

][Ag

1log

2

0.059EE

2

2o

/HH2

o

/AgAgcell2

SHE, 0 V

][Ag

1log

1

0.059EE o

/AgAgcell RE

WE

A technician is plating a faucet with 0.86 g of Cr from an electrolytic bath

containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what

current is needed?

1 A = 1 Cs-1 Current = Charge per sec

divide by M

96485 Cmol-1 e-

3 mol e-/mol Cr

divide by time

mass of Cr needed

mol of Cr needed

mol of e- transferred

current (A)

charge (C)

Cr3+(aq) + 3e- Cr(s)

0.86g x 3 mol e-

52.00 gmol-1= 0.050 mol e-

0.050 mol e- (96485 Cmol-1 e-) = 4787 C

4787 C

12.5 min x 60 s= 6.4 A

Q = nF = i t

CM2142

In an ELECTROLYTIC CELL an external source of energy is

required to force the non-spontaneous reaction to proceed. An

application of an electrolytic cell is electroplating.

39

Components involved in

Electrolytic Cells

Component

ElectrolyteChemical system capable of conducting current e.g. 0.1 M

KNO3.

Electrochemical Solution Solution in which electrolyte and analyte are dissolved.

Electrodes:

- Working electrode (WE)Electrode at which redox process of interest occurs,

potential is applied to WE.

- Reference electrode (RE) Potential is referenced against RE.

- Counter/Secondary/Auxiliary

electrode (CE)

Counters the current at the WE by allowing current to flow

through it and then the electrolyte.

CM2142 40

CM2142 41

Electrochemical Solution –

Analyte, Electrolyte, Solvent

Counter

Electrode

(CE)

Pt

wir

e

Working

Electrode

(WE)

Au

wir

e

Reference

Electrode

(RE)

Salt

bridge

Sat’d A

gC

lA

g w

ire

i

V

Potentiostat

Polarized vs. Non-polarizable Electrodes

e.g. reference electrodes such

as SCE, Ag/sat’d AgCl and

counter electrodes such as Pt

wire.

Potential unchanged upon

passage of current.

Ideal non-polarizable electrodeIdeal polarized electrode

e.g. noble metals (resistant to

corrosion) e.g. Au, carbon, Hg.

Wide potential range that has

minimal current.

CM2142

=> WE=> RE, CE

42

V = J/C A = C/s

No analyte

2-Electrode System vs. 3-Electrode System

3 Electrode system

- Electrolytic cells

- Large currents

- Large V = iRsolution (e.g. in

non-aqueous solvents)

2 electrodes system

- Voltaic cells

- Small currents, no CE

- Small V = iRsolution

CM2142

Measures V or i

Applies V or i and

measures the other.

43