Download chemistry form 6 sem 2 02

PRE UNIVERSITI
SEMESTER 2
CHAPTER 2
ELECTROCHEMISTRY
2.1 Oxidation number
 Oxidation numbers are a convenient way of determining if a
substance has been oxidised or reduced. These numbers are
assigned arbitrarily to atoms and are equal to the charge the
atom would have if its bonds were purely ionic.
1. All free atoms in element have an oxidation number of zero
Na = 0
Mg = 0
H2 =
0
Cl2 = 0
P4 = 0
2. For simple ions (and ionic compounds), the oxidation number is
the same as the charge of ion
K+ = +1
Ca2+ = +2 B3+ = +3
P3- = – 3 O2- = – 2 F- = – 1
3.
For covalent compounds, the covalent bonds are changed into
“ionic bonds” by assuming that the bonded electrons are on the
more electronegative atom. Table below shows some elements
oxidation number
Oxidation
Element
Notes
number
Group I
+1
--
Group 2
+2
--
Group 17
–1
True only to halogen without O in it
Oxygen
–2
Exception : –1 for peroxide and
+2 for F2O
Hydrogen
+1
Except : metal hydride MH (H = -1)
3. In a neutral molecule, the sum of the oxidation numbers of all
atom are equals
1C+2O=0
2 Na + O = 0
2B+3O=0
Na2O 2(Na) + (-2) = 0 B2O3 2(B) + 3(-2) = 0 CO2 1(C) + 2(-2) = 0
C = +4
Na = +1
B = +3
1S + 3O = 0
2Cl + 7O = 0
1H + 1F = 0
SO3 1(S) + 3(-2) = 0 Cl2O7 2(Cl) + 7(-2) = 0 HF 1(F) + 1(+1) = 0
S = +6
Cl = +7
F = -1
1S + 2H = 0
1N + 3H = 0
1C + 4H = 0
H2S 1(S) + 2(+1) = 0 NH3 1(N) + 3(+1) = 0 CH4 1(C) + 4(+1) = 0
S = -2
N = -3
N = -4
2H + 1S + 4O = 0
2H + 1C + 3O = 0
1H + 1Br + 3O = 0
2(+1) +1S + 4(-2) = 0
2(+1) +1C + 3(-2) = 0
1(+1) +1Br + 3(-2) = 0
H2CO3 C = +4
H2SO4 S = +6
HBrO3 Br = +5
4. In a molecular ion, the sum of the oxidation numbers of all
atoms in the formula unit equals to the charge on the ion.
CrO4-
Cr2O72-
MnO4-
1 Cr + 4 O = -1
1 Cr + 4(-2) = -1
Cr = +7
2 Cr + 7 O = -2
2 Cr + 7 (-2) = -2
Cr = +6
1 Mn + 4 O = -1
1 Mn + 4(-2) = -1
Mn = +7
C2O42-
ClO2-
HSO4-
2 C + 4 O = -2
2 C + 4(-2) = -2
C = +3
1 Cl + 2 O = -1
1 Cl + 2 (-2) = -1
Cl = +3
1 H + 1 S + 4 O = -1
1(+1) + 1S + 4(-2) = -1
S = +6
2.2
Half equation and redox reaction.
 Half equation ~ equation which shows how electrons are
accept / donate in a chemical reaction
donated ; e- is
 When a substance is oxidise, electron is ……………….
right side equation.
written at the ………..
received ; e- is
 When a substance is reduce, electron is ……………….
left side equation.
written as the ………
 Simple half equation : State the changes of oxidation number and
write the half equation.
Reaction
Oxidation
no change
Na  Na+
0  +1 oxidation Na  Na+ + e-
Mg  Mg2+
Reaction
Half equation
0  +2 oxidation Mg  Mg2+ + 2 e-
Al  Al3+
0  +3 oxidation Al  Al3+ + 3 e-
Cu2+  Cu
+2  0 reduction Cu2+ + 2 e-  Cu
Reaction
Oxi. no
change
Reaction
H2  H+
0  +1
oxidation
H2  2 H+ + 2 e-
Cl2  Cl-
0  -1
reduction
Cl2 + 2 e-  2 Cl–
I-  I2
-1  0
oxidation
2 I -  I 2 + 2 e-
O2  O2-
0  -2
reduction
O2 + 4 e-  2 O2–
Fe2+  Fe3+ +2  +3 oxidation
Fe2+  Fe3+ + e-
Pb4+  Pb2+ +4  +2 reduction
Pb4+ + 2 e-  Pb2+
Half equation
 When it comes to the reaction involving molecular ion, the overall
1.
2.
3.
4.
5.
charge has to be balanced in such order.
Write a skeleton half equation. Determine the reaction (oxidation
or reduction) using oxidation number
Balance the charge by adding electrons at the appropriate side
Balance the number of atoms other than oxygen.
Based on the changes in number of oxygen, write the number of
water molecule formed/used.
From the number of water molecule formed/used, write the
number of hydrogen ion (H+) required.
a) ClO3-  Cl- Changes in OS : +5  -1 ; reduction
Different in OS = 6, so 6 e- at the LHS of equation
ClO3- + 6 e-  ClClO3- + 6 e-  Cl- + 3 H2O
6 H+ + ClO3- + 6 e-  Cl- + 3 H2O
6 H+ + ClO3- + 6 e-  Cl- + 3 H2O
 Half equation : …………………………………………………………………
b) CrO42-  Cr3+ Changes in OS : +6  +3 ; reduction
Different in OS = 3, so 3 e- at the LHS of equation
CrO42- + 3 e-  Cr3+
CrO42- + 3 e-  Cr3+ + 4 H2O
8 H+ + CrO42- + 3 e-  Cr3+ + 4 H2O
8 H+ + CrO42- + 3 e-  Cr3+ + 4 H2O
 Half equation : …………………………………………………………………
c) Cr2O72-  Cr3+ Changes in OS : +6  +3 ; reduction
Different in OS = 3, so 3 e- at the LHS of equation
Since there are 2 Cr , so total e- = 6 ; Cr2O72- + 6 e-  2 Cr3+
Cr2O72- + 6 e-  2 Cr3+ + 7 H2O
14 H+ + Cr2O72- + 6 e-  2 Cr3+ + 7 H2O
+ + Cr O 2- + 6 e-  2 Cr3+ + 7 H O
14
H
2 7
2
 Half equation : …………………………………………………………………
d) MnO4-  Mn2+ Changes in OS : +7  +2 ; reduction
Different in OS = 5, so 5 e- at the LHS of equation
MnO4- + 5 e-  Mn2+
MnO4- + 5 e-  Mn2+ + 4 H2O
8 H+ + MnO4- + 5 e-  Mn2+ + 4 H2O
8 H+ + MnO4- + 5 e-  Mn2+ + 4 H2O
Half equation : ……………………………………………………………………..
e) NO2-  NO3-Changes in OS : +3  +5 ; oxidation
Different in OS = 2, so 2 e- at the RHS of equation
NO2-  NO3- + 2 eNO2- + H2O  NO3- + 2 eNO2- + H2O  2 H+ + NO3- + 2 e-
- + H O  2 H+ + NO - + 2 eNO
2
2
3
Half equation : ……………………………………………………………………..
f) CrO2-  CrO42- Changes in OS : +3  +6 ; oxidation
Different in OS = 3, so 3 e- at the RHS of equation
CrO2-  CrO42- + 3 e2 H2O + CrO2-  CrO42- + 3 e2 H2O + CrO2-  4 H+ + CrO42- + 3 e-
-  4 H+ + CrO 2- + 3 e2
H
O
+
CrO
2
2
4
Half equation : ……………………………………………………………………..
g) As2O3  As2O5 Changes in OS : +3  +5 ; oxidation
Different in OS = 2, so 2 e- at the RHS of equation
Since there are 2 As , so total e- = 4 ; As2O3  As2O5 + 4 eAs2O3 + 2 H2O  As2O5 + 4 eAs2O3 + 2 H2O  As2O5 + 4 e- + 4 H+
- + 4 H+
As
O
+
2
H
O

As
O
+
4
e
2 3
2
2 5
Half equation : ………………………………………………………………………
 When half equation of both oxidation and reduction reaction are
written, a redox reaction can be balanced.
Example 10 : Cu2+ (aq) + Na (s)  Cu (s) + Na+ (aq)
+ + eX2
Na

Na
Oxidation half equation : ……………………………………….………………
Cu2+ + 2 e-  Cu
Reduction half equation : ………………………………………………………
2+ + 2 Na  Cu + 2 Na+
Cu
Overall equation
: …………………………………………………………..
Example 11 : Fe2+ (aq) + MnO4- (aq)  Fe3+ (aq) + Mn2+ (aq)
Fe2+  Fe3+ + e- X 5
Oxidation half equation : ………………………………………………………
+ + MnO - + 5 e-  Mn2+ + 4 H O
8
H
4
2
Reduction half equation : ………………………………………………………
2+ 8 H+ + MnO -  Mn2+ + 4 H O + 5 Fe3+
5
Fe
4
2
Overall equation
: …………………………………………………………..
Example 12 : ClO- (aq) + SO2 (g)  Cl- (aq) + SO422 H2O + SO2  4 H+ + SO42- + 2 e-
Oxidation half equation : ……………………………………..……………..
2 H+ + ClO- + 2 e-  Cl- + H2O
Reduction half equation : ……………………………………………………..
Overall equation :
+
2H…………………………………………………
2O + ClO + SO2  2 H + SO4 + Cl
Example 13 : Cr2O72− + Cl2  ClO3−
+
Cr3+
+ + 2 ClO - + 10 e6
H
O
+
Cl

12
H
2
2
3
Oxidation half equation : ……………………………………………..
X3
H+ + Cr2O72- + 6 e-  2 Cr3+ + 7 H2O X 5
Reduction half equation :14…………………………………………….
5 Cr2O72- + 34 H+ + 3 Cl2  6 ClO3- + 17 H2O + 10 Cr3+
Overall equation : …………………………………………………
 Other than using half equation, a redox reaction can also be







balanced using the change of oxidation number.
Supposed we have a reaction : x A + y B  products
If the oxidation of reactant A increased by m while B reduced by n ;
Then
x (+ m) + y (– n ) = 0
Using a simple reaction :
ySn4+ (aq) + xFe2+ (aq)  Sn2+ (aq) + Fe3+ (aq)
–2
+4 to ……
+2 ; so the difference is ……..
For Sn ; O.N changed from ……
+1
+2 to ..….
+3 ; so the difference is ……..
For Fe ; O.N changed from ……
This will makes the equation become :
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
1 Sn4+ (aq) + 2 Fe2+ (aq)  Sn2+ (aq) +
Fe3+ (aq)
 Balanced the number of atoms on both side of the equation
Sn4+ (aq) + 2 Fe2+ (aq) 
Sn2+ (aq) + 2 Fe3+ (aq)
Example 14 : Br – (aq) + SO42- (aq)
SO2 (g) + Br2 (l)
+1
- 1 to …….
0 ; so the difference is ………
 For Br ; O.N changed from ……
-2
+6 to …….
+4 ; so the difference is ………
 For S ; O.N changed from ……
 This will make the equation become :
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
2 Br – (aq) + 1 SO42- (aq)  Br2 (l) + SO2 (aq)
2 Br – (aq) + SO42- (aq) + 4 H+  Br2 (l) + SO2 (aq) + 2 H2O
Balanced the number of atoms on both side of the equation
Example 15 : CrO42- + Cl-

Cr3+ + Cl2
For Cr ; O.N changed from +6 to +3 ; so the difference is –3
For Cl ; O.N changed from –1 to 0 ; so the difference is +1
x (+ 1) + y (– 3) = 0
So, x = 3 ; y = 1
8 H+ + CrO42- + 3 Cl-  Cr3+ + 3/2 Cl2 + 4 H2O
16 H+ + 2 CrO42- + 6 Cl-  2 Cr3+ + 3 Cl2 + 8 H2O
Example 16 : Cr2O72- + NO2-  Cr3+ + NO3For Cr ; O.N changed from +6 to +3 ; so the difference is –3
Since there are 2 Cr involved, diff. = – 6
For N ; O.N changed from +3 to +5 ; so the difference is +2
x (+ 2) + y (– 6) = 0
So, x = 3 ; y = 1
Cr2O72- + 3 NO2-  Cr3+ + NO38 H+ + Cr2O72- + 3 NO2-
 2 Cr3+ + 3 NO3- + 4 H2O
 If the redox reaction occur in a basic solution, the number of H+
shall be neutralise by the number of OH-.
Example 17 : MnO4- + SO32-  MnO2 + SO42Oxidation ½ eq :
H2O + SO32-  2 H+ + SO42- + 2 e- X 3
Reduction ½ eq :
4 H+ + MnO4- + 3 e-  MnO2 + 2 H2O X 2
Overall : 2 H+ + 2 MnO4- + 3 SO32-  2 MnO2 + 3 SO42- + H2O
H2O + 2 MnO4- + 3 SO32-  2 MnO2 + 3 SO42- + 2 OHExample 18 : Fe(OH)2 + CrO42−  Fe(OH)3 + Cr(OH)3
Oxidation ½ eq : H2O + Fe(OH)2 → Fe(OH)3 + e− + H+
X3
Reduction ½ eq : 5 H+ + 3 e− + CrO42− → Cr(OH)3 + H2O
Overall : 2 H2O + 3 Fe(OH)2 + 2 H+ + CrO42− → 3 Fe(OH)3 + Cr(OH)3
In basic : 4 H2O + 3Fe(OH)2 + CrO42− → 3Fe(OH)3 + Cr(OH)3 + 2OH−
Disproportionation reactions
~ Substances which are able to undergo self oxidation – reduction
are called disproportionation
~ Examples of disproportionation reaction.
18.
Cu+ (aq) + Cu+ (aq)  Cu2+ (aq) + Cu (s)
+1
+2
0
19.
NaOH (aq) + Cl2 (aq)  NaCl (aq) + NaOCl (aq) + H2O
0
-1
+1
20.
NaOBr (aq)
+1

NaBrO3 (aq)
+5
+
NaBr (aq)
-1
2.3 Electrode Potential
 When a strip of metal, M (s) (known as electrode) is placed in a
solution of its aqueous solution, Mn+ (aq), the following
equilibrium is established : Mn+ (aq) + n e- → M (s)
 At equilibrium, there is a separation of charge between metal
(M) and ions (Mn+) in the solution. as a result, there is a
potential difference between the metal and the solution. This
potential difference is known as electrode potential and is
written as Eo.
 Electrode potential can be measure under these circumstances
where
Metal
M
M+
M+
 Cu2+ + 2e- ↔ Cu
M+
M+
M+
M+
E Cu 2  / Cu   0.34 V
 The positive value of E0 indicates the equilibrium favours to
right position. Copper (II) ions (Cu2+), have a greater
the ………
be reduced at copper electrode.
tendency to …………...…
 Zn2+ + 2e- ↔ Zn E Zn / Zn   0.76 V
 The negative value of E0 indicates the equilibrium favours to
left position. Zinc ion (Zn2+) have a greater tendency to
the …..
be oxidised at zinc electrode.
……………..
2
Non – Metal
Cl2 (g)
Cl- (aq) [1.0 M]
E F / F   2.87 V
F2 + 2e- ↔ 2 FCl2 + 2e- ↔ 2 Cl- E Cl / Cl   1.36 V
 Positive value of ECl2/Cl- and EF2/F- indicates the equilibrium
and chlorine has a
right
favours to the………….
position.Fluorine
………………………
reduced under platinum electrode
greater tendency to be
……………
 The more positive the value, higher the tendency of nonreduced
metal to be
…………….
oxidising agent than
 In another words, fluorine is a stronger …………..
chlorine.

2

2
Mixture of aqueous ion
V
Ma+ [1.0 M] / Mb+ [1.0 M]
 A potential difference also exists between ions in an aqueous
solution. Example :
E Cr / Cr   0.41 V
3+
2+
Cr + e ↔ Cr
E Fe / Fe   0.77 V
Fe3+ + e- ↔ Fe2+
more stable than Cr2+ as
 Base on the Eo value, Cr3+ is ………
backward (Eo is negative)
equilibrium favour to …………..
less stable than Fe2+ as
 Base on the Eo value, Fe3+ is ……..
forward (Eo is positive)
equilibrium favour to …………..
3
2
3
2
2.3.1 Standard Electrode Potential
 Definition : The standard electrode potential, Eo Mn+ /
of a metal M
is the potential
………… difference between the metal M and the aqueous
…………
mol dm-3 at 298
solution of the metal ions of concentration 1.0
………………
…… K
Hydrogen Electrode
and 1.0
…. atm, measured relatively to Standard
……………………………….…
M
 Standard Hydrogen Electrode ( S.H.E.)
 It is impossible to measure the electrode potential for an
incomplete half-cell. It can only be measured for a complete
……………..
cell , i.e. only differences in electrode potentials
circuit with 2 half
……….
are measurable.
 The standard chosen for electrode potentials is the standard
hydrogen electrode (SHE). The standard electrode potentials of
other half-cells are measured relative to the SHE’s electrode
potential.
 By convention, the standard electrode potential for this
reference hydrogen half-cell is taken to be standard
…………...

2 H+ (aq) + 2 e25 oC ;
 Condition : …....
H2 (g)
1.0 atm
H2 (g) at ……
; [H+] = 1.00 M
Measuring standard electrode potential of a
metal / metal aqueous solution
 The set-up of the apparatus to measure the standard
potential electrode, Eo. is described as below :
Potentiometer
0.76 V
H2 (g)
1.0 atm
Zn (s)
H+ (aq) [1.0 M]
25oC
Salt bridge
(made of
saturated
KCl / NaCl)
Standard hydrogen electrode
Zn2+ (aq)
[1.0 M]
Zinc half cell
 The chemical cell is set-up by connecting a standard
+/H
2+/Zn
H
Zn
………….. half-cell to a standard ………….
electrode.
2
0.76 V. The potentiometer point
 The e.m.f. for the cell is ……….
H2 electrode in the external circuit,
to the direction of ……..
Zn2+ / Zn to ………….
H+ / H2 halfindicating electrons flow from …………..
cell.
 Eq. Zn half-cell : Zn (s)  Zn2+ (aq) + 2 e Eq. H half-cell
: 2 H+ (aq) + 2 e-  H2 (g)
 Overall reaction
: Zn (s) + 2 H+ (aq)  H2 (g) + Zn2+ (aq)
 The cell notation can be written as :
Zn (s) I Zn2+ (aq) II H+ (aq) , H2 (g) I Pt (s)
donated ; ….……..…
oxidation
 At zinc electrode ; electrons are ……….....
reaction occur
received ; ………….
reduction
 At platinum electrode ; electrons are …………..
reaction occur
 Since zinc is oxidised in a SHE, the standard e.m.f value is
– 0.76 V
……………
 Another example : silver / silver aqueous solution (Ag / Ag+)
 The set-up of the apparatus to measure the standard potential
electrode, Eo. is described as below :
Potentiometer
0.80 V
H2 (g)
1.0 atm
H+
(aq) [1.0 M]
25oC
Ag (s)
Salt bridge
(made of
saturated
KCl / NaCl)
Standard hydrogen electrode
Ag+ (aq)
[1.0 M]
Silver half cell
 The chemical cell set-up by connecting a standard





H+ / H2
Ag+ / Ag
………………
half-cell to a standard ……………..
electrode.
The e.m.f. for the cell is 0.80
……. V. The galvanometer point to
silver
the direction of …………..
electrode in the external circuit,
+/ H
+ / Ag
H
Ag
2
indicating electrons flow from ………….. to ………… half-cell.
 Ag half-cell
: Ag+ (aq) + e-  Ag (s)
 H2 half-cell
: H2 (g)  2 H+ (aq) + 2 e Overall :
H2 (g) + 2 Ag+ (aq)  2 Ag (s) + 2 H+ (aq)
The cell notation can be written as :
Pt (s) I H2 (g) , H+ (aq) II Ag+ (aq) I Ag (s)
received ; ……………
reduction
At silver electrode ; electrons are ………….
reaction occur
oxidation
donated ; ……………
At platinum electrode ; electrons are ……………
reaction occur
Since silver is reduced in a SHE, the standard value is
+ 0.80 V
………………
Measuring a standard electrode potential of a gaseous
substance
Cl2 / Cl–
 The chemical cell set-up by connecting a standard …………
H+ / H2
half-cell to a standard ……………
electrode. Note that the
set-up of the half-cells are the same for gaseous substances
Potentiometer
1.36 V
Cl2 (g)
1 atm
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Salt bridge
(made of
saturated
KCl / NaCl)
Standard hydrogen electrode
Cl- (aq)
[1.0 M]
Chlorine half cell
 The e.m.f. for the cell is 1.36
…….V. The galvanometer point to
Pt (Cl2) electrode in the external circuit,
the direction of ………..
H+ / H2 to ………….
Cl2 / Cl– halfindicating electrons flow from …………..
cell.
 Chlorine half-cell : Cl2 (g) + 2 e-  2 Cl– (aq)
 Hydrogen half-cell : H2 (g)  2 H+ (aq) + 2 e Overall :
H2: (g) + Cl2 (g)  2 Cl– (aq) + 2 H+ (aq)
 The cell notation can be written as :
Pt (s) I H2 (g) , H+ (aq) II Cl2 (g), Cl– (aq) I Pt (s)
 At platinum electrode in the half-cell of hydrogen ; electrons
oxidationreaction occur
are donated
………… ; …………
 At platinum electrode in the half-cell of chlorine ; electrons
received ; ……………
reduction reaction occur
are …………..
reduced by SHE, the standard value is
 Since chlorine is …………
+ 1.36 V
……….
Measuring a standard electrode potential of a mixture of metal
ions.
 The electrode potential of a mixture of ions can be measured
in the similar way, using standard hydrogen electrode (SHE)
as the other half-cell of the chemical cell
 For example, in a mixture of iron (II) and iron (III) ion
Potentiometer
0.77 V
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Salt bridge
(made of
saturated
KCl / NaCl)
Standard hydrogen electrode
Fe2+ (aq)
Fe3+ (aq)
[1.0 M]
Fe2+ / Fe3+ half cell
Fe3+ / Fe2+
 The chemical cell set-up by connecting a standard …………
H+ / H2
half-cell to a standard ……………
electrode. Note that the set-
up of the half-cells are a mixture of iron (II) and iron (III) ion
platinum as electrode.
under standard condition with ……………
0.77 V. The galvanometer point to the
 The e.m.f. for the cell is …….
Fe3+ / Fe2+ half cell in the external circuit, indicating
direction of …………...
3+ / Fe2+
H+ / H2 to Fe
electrons flow from ………….
………….
half-cell.
 Fe3+ / Fe2+ half-cell
: Fe3+ (aq) + e-  Fe2+ (aq)
 Hydrogen half-cell
: H2 (g)  2 H+ (aq) + 2 e Overall reaction :H2 (g) + 2 Fe3+(aq)  2 Fe2+ (aq) + 2 H+ (aq
The cell notation can be written as :
Pt (s) I H2 (g) , H+(aq) II Fe3+(aq), Fe2+(aq) I Pt (s)
donated ; oxidation
 At half-cell of hydrogen ; electrons are …………
……………
reaction occur
received ; …………...
reduction
 At half-cell of Fe3+/Fe2+ ; electrons are ………….
reaction occur
+ 0.77 V
reduced by SHE, the value of ……………
 Since the mixture is ………...
The calomel electrode
primary reference electrode.
 Platinum electrode is known as the ……………
However, it is relatively difficult to set up and operate under
standard condition. It is more easier and safer to use a calomel
secondary electrode. [calomel = Mercury
base alloy
electrode as a ………………
……………………..].
Diagram of a typical calomel electrode
2.4 Factors Affecting Electrode Potential
reduction as the
 By convention, the half equation is written with ………….…
forward reaction.
 The magnitude of the electrode potential depends on the position
of the above equilibrium
forward reaction is favoured
 When value is positive ; a ………………
backward
 When value is negative ; a …………………..
reaction is favoured
 Factors which affect the position of equilibrium would therefore
affect the value of electrode potential
1. Nature of metal
electropositive the metal atoms have a
 When a metal is highly ……………………..,
electron
greater tendency to become positive ions, leaving the ……………
behind on the metal electrode. The electrode potential therefore
negative and the position of equilibrium shift
become more ……………….
left (……….………
oxidation is favoured)
more to ………
Metal
Half equation
E (V)
Silver
Ag+ + e- ↔ Ag
+ 0.80
Lead
Pb2+ + 2 e- ↔ Pb
– 0.13
Zinc
Zn2+ + 2 e- ↔ Zn
– 0.76
Magnesium
Mg2+ + 2 e- ↔ Mg
– 2.38
2. Concentration of metal
• If the concentration of the hydrated metal ions is increased in the
right
equilibrium, the position of equilibrium will shift to the …………,
positive
forward ; electrode potential become more ……………
favouring ………….
Pb2+ (aq) + 2 ePb (s)
E = – 0.13 V [ Conc = 1.0 M ]
• If concentration Pb2+ changed to 0.001 M ; equilibrium shift to
backward ; E < – 0.13 V
……………….
• If concentration Pb2+ changed to 10.0 M ; equilibrium shift to
forward
……………….
; E > – 0.13 V
3 Temperature
 Most of the reduction processes are exothermic process.
Increasing the temperature will cause the equilibrium to
endothermic process ; which is to
shift to the position of ………………..
left Thus, the electrode potential becomes more
the ……
negative / less positive
…………………………….
4. Pressure for gaseous species
 From what we’ve learned from chemical equilibria, when
pressure increased, equilibrium will shift to the position
less mole of gas ; while decreasing pressure will
with ……..
more
cause equilibrium to shift to position with ……….mole
of
gas.
 Eg : Cl2 (g) + 2 e2 Cl- (aq)
E = + 1.36 V
 Increasing pressure will cause equilibrium shift to
right side ; E > + 1.36 V
……………..
 Decreasing pressure will cause equilibrium shift to
left side
……………..
; E < + 1.36 V
2.5 The electrochemical series (ECS)
 When a series of standard reduction potential of different
substances are determined and are arranged in order, a
electrochemical series is obtained.
Below are some important facts about electrochemical series.
 Half cell of the standard electrode potential is always written
as reduction
………… processes. Due to this reason, sometimes it is
Standard reduction potential (SRP)
also known as ……………………………………
 The positive / negative sign shows how substances favour to
each of the reaction.
 If the EO is positive, substances favour a forward
……..… reaction.
oxidising agent. The
In another words, it serve well as ………….
oxidising agent
more positive the value ; the stronger the …………..
 If the EO is negative, substances favour backward
………..… reaction.
reducing agent. The
In another words, it serve well as ……………
reducing agent
more negative the value ; the stronger the …………
 The number of electron involve does not affect the standard
electrode potential value. If
 Cl2 (g) + 2 e-  2 Cl- (aq) Eo = + 1.36 V;
then
1.36 V
 ½ Cl2 (g) + e-  Cl- (aq) Eo = +……….
 Some substances have more than one Eo value. For example
Fe2+ ; H2O2 ; NO2- ; Cu+. These substances can act as an
oxidising or reducing agent. Examples
 In Fe2+
Fe3+ (aq) + e-  Fe2+ (aq)
(Fe2+ act as reducing agent)
Fe2+ (aq) + 2 e-  Fe (s)
(Fe2+ act as oxidising agent)
2.6 Redox reaction and electromotive forces (e.m.f.)
 In standard hydrogen electrode, we had seen on how to
measure the standard electrode potential of 3 types of halfcell, which are metal / metal ion half-cell ; non-metal / ion halfcell ; ion / ion half-cell
 Imagine if we replace the hydrogen half-cell with other halfcell, will we still get the same value?
 The potential difference between 2 half-cells can be
measured using the same way. There are various types of the
set-up of a complete chemical cell other than the one
introduced during measuring the standard electrode potential,
such as Daniel cell (diagram below)
 A Daniel cell is built using a copper and
V
C
u
Z
n
zinc half-cell
complete the cell
 A porous pot is used to……………………
And to separate between the 2 electrolytes
………………………………………………
 Substance which has a higher position in
ECS (more negative the value of Eo) is the
anode of the cell whereas the substance
………
which has a lower position in ECS (more
cathode
positive the value of Eo) is the ……………
of the cell.
 The half equation occur at
Anode : Zn  Zn2+ + 2 e- Eo =+0.76 V
Cathode : Cu2+ + 2 e-  Cu Eo = +0.34 V
Overall : Zn + Cu2+  Zn2+ + Cu Ecell =+1.10 V
+ 1.10 V
The e.m.f. of cell is ………….
Cu2+
Zn2+
Cell notation is written as
Zn (s) I Zn2+ (aq) II Cu2+ (aq) I Cu (s)
A U tube cell is built using iron (II) ion, Fe2+
and bromine water, Br2
complete the cell
H2SO4 is used to ………………………………….…
Br2/Br Fe3+/Fe2+
And
to separate between the 2 electrolytes
…..............…………………………………………….
The substance which has more negative / less
G
anode of the
positive the value of Eo) is the …………
cell whereas the substance which has a more
positive / less negative value of Eo is the
cathode of the cell.
…………...
The half equation occur at
Anode : Fe2+  Fe3+ + e- Eo = - 0.77 V
Pt
Cathode : Br2 + 2 e-  2 Br - Eo = +1.07 V
Overall : Br2 + 2 Fe2+  2 Fe3+ + 2 Br+ 0.30 V
The e.m.f. of cell is …………..
Cell notation is written as
Pt (s) I Fe2+ (aq) , Fe3+ (aq) II Br2 (l) , Br- (aq) I Pt (s)
Note the following in a chemical cell :
POSITIVE In another words,
 The e.m.f. of a cell is always ……………
more positive standard
we must always subtract the ..………………
less positive standard electrode
electrode potential with a ……………….
potential.
anode to ……………
cathode in the external
 Electrons flow from …………
circuit
Oxidation reaction occur at anode while …………
reduction reaction
 ……………
occur at cathode of the cell.
G
Fe
/ Fe2+
half cell
Mg / Mg2+ half cell
 SRP for both cell : Fe2+ + 2e-  Fe
Mg2+ + 2e-  Mg
E0 = - 0.44 V
E0 = - 2.38 V
Since E0 for Mg2+/Mg is more negative than Fe2+/Fe, so Mg2+/Mg
will be oxidised, so SPR of Mg2+/Mg is reversed
Oxidation ½ eq : Mg  Mg2+ + 2 e-
E0 = + 2.38 V
Reduction ½ eq : Fe2+ + 2e-  Fe
E0 = – 0.44 V
Overall eq : Fe2+ + Mg  Fe + Mg2+ Ecell = + 1.94 V
Cell diagram : Mg (s) I Mg2+ (aq) II Fe2+ (aq) I Fe (s)
G
MnO4- / Mn2+ half cell
 SRP :
Ti3+ / Ti2+ half cell
MnO4- + 5 e- + 8 H+  Mn2+ + 4 H2O E0 = + 1.52 V
Ti3+ + e-  Ti2+
E0 = – 0.37 V
Since E0 for Ti3+/Ti2+ is more negative than MnO4-/Mn2+, so Ti3+/Ti2+
will be oxidised, so SPR of Ti3+/Ti2+ is reversed
Oxidation ½ eq : Ti2+  Ti3+ + e-
E0 = + 0.37 V
Red ½ eq : MnO4- + 5 e- + 8 H+  Mn2+ + 4 H2O E0 = + 1.52 V
Overall: MnO4- + 5 Ti2+ + 8 H+  Mn2+ + 4 H2O + 5 Ti3+ Ecell = + 1.89V
Cell diagram: Pt(s) I Ti2+(aq), Ti3+ (aq)II MnO4- (aq), Mn2+(aq) I Pt(s)
2.7 Feasibility of a redox reaction
 If a reaction occurs on its own record when the reactants are
mixed, the reaction is a spontaneous
...……………… reaction
 Compare the following reaction to distinguish between a
spontaneous reaction and not spontaneous reaction.
Immerse a zinc plate into HCl 1.0 M
Observation
-Bubbling is observed
-Zinc plate is corroded by HCl
Immerse a copper plate into HCl 1.0 M
Observation
- No changes occur
 We can use e.m.f. to predict the feasibility of the reaction.
Supposedly, in the reaction above, the 2 half equation for
the reactions can be written as
-Since Zn react with H+, so the 2 half
equation can be written
Zn  Zn2+ + 2eE0 = + 0.76 V
2 H+ + 2e-  H2
E0 = + 0.00 V
Zn + 2 H+  Zn2+ + H2 E = + 0.76 V
Since Ecell is positive, the reaction is
spontaneous
-Since Cu react with H+, so the 2 half
equation can be written
Cu  Cu2+ + 2eE0 = - 0.34 V
2 H+ + 2e-  H2
E0 = + 0.00 V
Cu + 2 H+  Cu2+ + H2 E = - 0.34 V
Since Ecell is negative, the reaction is
non–spontaneous
 It can also be used to deduce the strength as an oxidising




Stronger
oxidising agent

agent in halogen.
Halogens are strong oxidising agent. This is supported with
the value of standard reduction potential where
½ F2 (aq) + e-  F– (aq) Eo = + 2.87 V
½ Cl2 (aq) + e-  Cl– (aq) Eo = + 1.36 V
½ Br2 (aq) + e-  Br– (aq) Eo = + 1.07 V
½ I (aq) + e-  I– (aq)
Eo = + 0.54 V
Observation
Chlorine in
Tetrachloromethane
is added to aqueous
potassium bromide
(KBr).
Bromine in
tetrachloromethane
is added to aqueous
potassium iodide (KI)
Iodine is
tetrachloromethane
is added to aqueous
potassium chloride
(KCl)
Pale yellow solution in
CCl4 turned brown when
shaken with KBr.
Brown solution in CCl4
turned purple when
shaken with KI
Half equation & overall equation
Cl2 + 2e-  2Cl-
Eo = + 1.36 V
2Br-  Br2 + 2e-
Eo = - 1.07 V
Cl2 + 2Br-  Br2 + 2Cl–
Ecell = + 0.29 V
Br2 + 2e-  2Br-
Eo = + 1.07 V
2I-  I2 + 2e-
Eo = - 0.54 V
Br2 + 2I-  I2 + 2Br–
Ecell = + 0.53 V
I2 + 2e-  2I-
Eo = + 0.54 V
No changes occur. Purple
2Cl-  CI2 + 2e- Eo = - 1.36 V
solution remain after
I2 + 2 Cl-  Cl2 + 2 I–
shaken with KCl
Ecell = - 0.82 V
a) Iron nail are placed in zinc sulphate
b) Copper is placed in concentrated nitric
acid solution (Assume NO2 (g) is produced)
Reactant : Fe and Zn2+
Reactant : Cu and NO3Suitable half equation
Suitable half equation
Fe  Fe2+ + 2eEo = + 0.44 V
Cu  Cu2+ + 2eEo = - 0.34 V
2+
o
Zn + 2e  Zn
E = - 0.76 V NO – + 2H+ +e– NO + H O E0= +0.81 V
3
2
2
Fe + Zn2+  Fe2+ + Zn Ecell = - 0.32V Cu + 2 NO3– + 4H+  2NO2 +2H2O + Cu2+
Since Ecell is negative, reaction is not
Ecell = + 0.47 V
spontaneous (cannot react)
Since Ecell is positive, reaction is
spontaneous (can react)
c) Chlorine gas is bubbled into acidified
potassium dichromate
d) potassium iodide is added to acidified
potassium manganate (VII) solution
Reactant : Cl2 and Cr2O72–
Suitable half equation
Cl2 + 2H2O  2HOCl + 2H+ + 2e–
Eo = – 1.64 V
Cr2O72- + 6e- + 14H+ 2Cr3+ + 7H2O
Eo = + 1.33 V
Cr2O72- + 6HOCl + 2H+ 3Cl2 + H2O
Ecell = – 0.31V
Since Ecell is negative, reaction is non
spontaneous (cannot react)
Reactant : I- and MnO4Suitable half equation
2I-  I2 + 2 eEo = - 0.54 V
MnO4- + 8 H+ + 5e-  Mn2+ + 4 H2O
Eo = + 1.52 V
10 I- + 2 MnO4- + 16H+  2 Mn2+ + 8 H2O
+ 5 I2
Ecell = + 0.98 V
Since Ecell is positive, reaction is
spontaneous (can react)
e) Calcium metal is added to water
f) Acidified potassium dichromate solution
is added to a solution of iron (II) sulphate
Reactant : Ca and H2O
Suitable half equation
Reactant : Fe2+ and Cr2O722H2O + 2e-  H2 + 2OH- Eo = - 0.83 V Suitable half equation
Ca  Ca2+ + 2eEo = + 2.87 V Cr2O72- + 6e- + 14H+ 2Cr3+ + 7H2O
Ca + 2H2O  H2 + Ca2+ + 2 OHEo = + 1.33 V
Ecell = + 2.04 V Fe2+  Fe3+ + eEo = - 0.77 V
Since Ecell is positive, reaction is
Cr2O72- + 6Fe2+ + 14H+
spontaneous (can react)
2Cr3+ + 7H2O + 6Fe3+
Ecell = + 0.56 V
Since Ecell is positive, reaction is
spontaneous (can react)
 Among the oxidation states available in d-orbital, +2 and +3
oxidation states are the most common states available in the dblock elements.
 The stability of the oxidation state can be explained in terms of
electrochemistry. The standard reduction potential of a few
transition metals is given in the table below.
Half equation of
reduction
Cr3+ + e-  Cr2+
Eo (V)
Stable ion
– 0.41
Cr3+
Ti3+ + e-  Ti2+
– 0.37
Ti3+
V3+ + e-  V2+
– 0.26
V3+
Fe3+ + e-  Fe2+
+ 0.77
Fe2+
Mn3+ + e-  Mn2+
+ 1.51
Mn2+
Co3+ + e-  Co2+
+ 1.82
Co2+
 The action of dilute acids on metal are usually carried out in
the presence of oxygen. We must therefore determine
whether oxygen has any effect on such reactions. For
example, in oxidation of iron (II) ion
Action of acids on iron (II) ion in the absence Action of acids on iron (II) ion in the
of air (oxygen)
presence of air
2+
3+ + eE0 = - 0.77 V
Fe2+  Fe3+ + eE0 = - 0.77 V Fe  Fe
+
–
2H2O E0 = +1.23 V
2 H+ + 2 e-  H2
E0 = + 0.00 V O2 +2+4H + 4e 
4Fe + O2 + 4H+  4Fe3+ + 2H2O
2 Fe2+ + 2H+  2 Fe3+ + H2
Ecell = + 0.46 V
Ecell = - 0.77 V
2+
2+
2+
Ti
V
Cr
 …………….. and …………… can also react in the same way
as iron does.
 For the case of cobalt and manganese, it does not react in
the same way as iron does. Consider the reaction of cobalt
(II) ion with acid in the absence / presence of oxygen
Action of acids on cobalt in the absence of air
Action of acids on cobalt in the presence of air
E0 = - 1.82 V
Co2+  Co3+ + eE0 = - 1.82 V Co2+  Co3+ + e2 H+ + 2 e-  H2
E0 = + 0.00 V O2 + 4H+ + 4e–  2H2O E0 = +1.23 V
4Co2+ + O2 + 4H+  4Co3+ + 2H2O
2 Co2+ + 2H+  2 Co3+ + H2
Ecell = - 0.59 V
Ecell = - 1.82 V
•Graph below shows the relative stability of ions which exist in different oxidation
state
 Other than the presence of oxygen, the presence of ligands can
also affect the stability of ions.
 Consider the following electrode reactions for cobalt :
[Co(NH3)6]3+ + e- ↔ [Co(NH3)6]2+
E0 = + 0.10 V
O2 + 4 H+ + 4 e- ↔ 2 H2O
E0 = + 1.23 V
[Co(H2O)6]3+ + e- ↔
E0 = + 1.82 V
[Co(H2O)6]2+
 In water, Co2+ is stable toward oxidation, even in the presence of
oxygen, since the E for the reaction is -0.59 (based on the
calculation above). Therefore, +2 is more stable than +3 oxidation
state in aqueous solution (ligand is water)
 When aqueous NH3 is added to the solution of Co2+, the complex
ion [Co(NH3)6]2+ is formed (since NH3 is a strong ligand, water
molecule can easily displaced by NH3 ligand).
Eq : [Co(H2O)6]2+ + 6 NH3  [Co(NH3)6]2+ + 6 H2O
 When [Co(NH3)6]2+ is formed, it can react easily with acids with the
presence of air
 When [Co(NH3)6]2+ is formed, it can react easily with acids
with the presence of air
[Co(NH3)6]2+  [Co(NH3)6]3+ + eE0 = - 0.10 V
O2 + 4 H+ + 4 e-  2 H2O
E0 = + 1.23 V
4 [Co(NH3)6]2+ + O2 + 4 H+  2 H2O + 4 [Co(NH3)6]3+
Ecell = + 1.13 V
 Therefore, although Co3+ is not stable in the presence of air,
but Co3+ is stable in ammonia aqueous solution. Hence, this
is one of the ways to prepare an ion solution which is not
stable in air.
2.7 Nernst Equation and Its application
 All the electrochemical cells that discussed so far are
25 oC ; ………
1.00 atm ; …….
1.00 M]
standard Eo. [At ……..
 If concentration of ions and temperature change, it will affec
the value of electrode potential. At this moment, we can use
an equation to study the changes of concentration of ions
using Nernst Equation.
E  E0
[reac tan t ions ] x
RT

ln
nF
[product ions ] y
R = 8.31 J mol-1 K-1
T = 250C = 298 K
F (Faraday constant)
F = 96500 C mol-1
x
(
8
.
31
)(
298
)(
2
.
303
)
[
reac
tan
t
ions
]
E  E0 
lg
n (96500)
[product ions ]y
x
0
.
059
[
reac
tan
t
ions
]
E  E0 
lg
n
[product ions ]y
Ag / Ag+ (1.5 mol dm-3)
Ag+ + e-  Ag E0 = + 0.80 V
 1
0
.
059
[
Ag
]
E  E0 
lg
n
1
0.059 (1.5)1
E   0.80 
lg
1
1
E = + 0.81 V
Cl2 / Cl- (0.50 mol dm-3)
Cl2 + 2e-  2 Cl- E0 = + 1.36 V
1
0
.
059
[
Cl
]
E  E0 
lg 2 2
n
[Cl ]
0.059
(1)1
E  1.36 
lg
2
(0.50) 2
E = + 1.38 V
Cu / Cu2+ (2.0 mol dm-3)
Cu2+ + 2e-  Cu E0 = + 0.34 V
2 1
0
.
059
[
Cu
]
E  E0 
lg
n
1
0.059 (2.0)1
E   0.34 
lg
2
1
E = + 0.35 V
Fe3+ (0.800 mol dm-3) / Fe2+ (1.30 mol
dm-3)
Fe3+ + e-  Fe2+ E0 = + 0.77 V
0.059 [Fe3 ]1
E  E 
lg
n
[Fe2 ]1
0
0.059 (0.800)1
E   0.77 
lg
1
(1.30)
E = + 0.758 V
Ti3+ (1.20 mol dm-3) / Ti2+ (0.700 mol
dm-3)
Ti3+ + e-  Ti2+ E0 = - 0.37 V
3 1
0
.
059
[
Ti
]
E  E0 
lg 2 1
n
[Ti ]
E   0.37 
0.059
(1.20)
lg
1
(0.700)
E = - 0.356 V
MnO4- (1.10 mol dm-3) ; H+ (0.800 mol
dm-3) / Mn2+ (17.0 mol dm-3)
MnO4- + 8 H+ + 5e-  4 H2O + Mn2+
E0 = + 1.52 V

 8
0
.
059
[
MnO
][
H
]
4
E  E0 
lg
n
[Mn 2 ]
0.059 (1.10)(0.800)8
E  1.52 
lg
5
(17.0)
E = + 1.50 V
13.7.1 Nernst Equation and e.m.f. of a chemical cell.
 Consider the following redox reaction in an chemical cell
p A + q B ↔ r C + s D
 At 25oC, Nernst Equation :
E  E0
[reac tan t ions ] x
RT

ln
nF
[product ions ] y
E  E cell
0.059 [A]p [B]q

lg
n
[C]r [D]s
 *For pure solids and liquids, it will not appear in the
equation
a) Cr (s) Cr3+ (0.010 mol dm-3) Ni2+ (0.20 mol dm-3) Ni (s)
Oxidation
Reduction
Overall
: Cr  Cr3+ + 3 e–
E0 = + 0.74 V
: Ni2+ + 2 e–  Ni
E0 = – 0.25 V
: 2 Cr + 3 Ni2+  3 Ni + 2 Cr3+ Ecell = + 0.49 V
E  E cell
0.059
[ Ni2 ]3

lg
6
[Cr 3 ]2
0.059
(0.20)3
E   0.49 
lg
6
(0.010) 2
E = + 0.51 V
b) Mg (s) Mg2+ (0.500 mol dm-3) Fe3+ (1.80 mol dm-3) , Fe2+
(0.750 mol dm-3) Pt (s)
Oxidation
: Mg  Mg2+ + 2e–
Reduction : Fe3+ + e-  Fe2+
Overall : Mg + 2 Fe3+  Mg2+ + 2 Fe2+
E  E cell
E0 = + 2.38 V
E0 = + 0.77 V
Ecell = + 3.15 V
0.059
[Fe3 ]2

lg
2
[Mg 2 ][ Fe2 ]2
0.059
(1.80) 2
E   3.15 
lg
2
(0.500)(0.750) 2
E = + 3.18 V
c) Pt (s) Sn2+ (0.300 mol dm-3), Sn4+ (0.500 mol dm-3) 
Mn3+ (1.20 mol dm-3) , Mn2+ (0.250 mol dm-3) Pt (s)
Oxidation
: Sn2+  Sn4+ + 2e–
Reduction : Mn3+ + e-  Mn2+
Overall : Sn2+ + 2 Mn3+  Sn4+ + 2 Mn2+
E  Ecell
2
E0 = – 0.15 V
E0 = + 1.49 V
Ecell = + 1.34 V
3 2
0.059 [ Sn ][ Mn ]

lg
4
2 2
2
[ Sn ][ Mn ]
0.059
(0.300)(1.20) 2
E   1.34 
lg
2
(0.500)(0.250) 2
E = + 1.37 V
2.7.1 Nernst Equation and Equilibrium Constant, KC
 Consider the following reaction :
Cu2+ (aq) + Zn (s)  Zn2+ (aq) + Cu (s)
E0 = + 1.10 V
 The Kc of the reaction can be expressed as
[ Zn 2 ]
Kc 
[Cu 2 ]
decrease while
 As time past, the concentration of [Cu2+] ………………
increase
[Zn2+] ……………..
Using standard reduction potential of copper and zinc
 Cu2+ (aq) + 2 e-  Cu (s)
Eo = + 0.34 V ; when [Cu2+]
decrease ; equilibrium shift to ………
decrease
left Eo ……………
………………
Eo = – 0.76 V ; when [Zn2+]
increase
increase
right Eo …………...
………………
; equilibrium shift to ………
 Zn2+ (aq) + 2 e-  Zn (s)
V
+ 0.34–
Half cell of Cu
0
Time / s
– 0.76 – Half cell of Zn
 Applied to Nernst equation where
E  E
0
[reac tan t ions ] x
RT

ln
nF
[product ions ] y
 When system achieved equilibrium at room temperature,
Nernst equation is simplified to :
y
0
.
059
[
product
ion
]
E  E0 
lg
n
[reactat ion ]x
 Since
[ Zn 2 ]
Kc 
[Cu 2 ]
and at equilibrium, Ecell = 0, so
0.059
0 V E 
lg K C
n
0
Consider the following reaction :
2 Fe3+ (aq) + Cu  2 Fe2+ (aq) + Cu2+ (aq)
a) Calculate the E0 of the cell
b) Calculate the Kc for the reaction.
Fe3+ + e-  Fe2+
E0 = + 0.77 V
Cu  2 e- + Cu2+
E0 = - 0.34 V
2 Fe3+ + Cu  Cu2+ + 2 Fe2+
Ecell = + 0.43 V
KC 
2 2
2
[Fe ] [Cu ]
[Fe3 ]2
Consider the following reaction :
Fe3+ (aq) + Ag (s)  Fe2+ (aq) + Ag+ (aq)
a) Calculate the E0 of the cell
b) Calculate the Kc for the reaction.
Fe3+ + e-  Fe2+
E0 = + 0.77 V
Ag  e- + Ag+
E0 = – 0.80 V
Fe3+ + Ag  Fe2+ + Ag+
Ecell = – 0.03 V
[Ag  ][ Fe2 ]
KC 
[Fe3 ]
0.059
0 V E 
lg K C
2
0 V   0.03 
0.059
lg K C
1
0.059
0 V   0.43 
lg K C
2
0 V   0.03 
0.059
lg K C
1
lg KC = 14.57
KC = 3.77 x 1014 mol dm-3
lg KC = – 0.5085
KC = 0.310 mol dm-3
0
2.7.2
Nernst Equation and solubility product, Ksp
 Similar to calculating equilibrium constant, solubility product of a
sparingly soluble salt can also be calculated in the same way
mentioned above. Consider the following reaction of dissociation of
AgCl (s)
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
 Half-cell : AgCl (s) + e- ↔ Ag (s) + Cl- (aq)
Eo = + 0.22 V
 Half-cell : Ag (s) ↔ Ag+ (aq) + e-
Eo = - 0.80 V
 Overall Eq: AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
E0 = - 0.58 V
Using Nernst equation
and given Ksp = [Ag+][Cl–]
y
0
.
059
[
product
ion
]
E  E0 
lg
n
[reactat ion ]x
0.059 [Ag  ][Cl ]
EE 
lg
n
1
0
0.059
EE 
lg K sp
n
0
 At equilibrium ; E = 0 ; so replace in the equation
0   0.58 V 
0.059
lg K sp
1
lg Ksp = – 9.83 mol2 dm-6
Ksp = 1.48 x 10–10 mol2 dm-6
 The question may be extend to calculate the solubility
from the solubility product calculated
 Note that Eo is always negative for sparingly soluble salt
very small
as Ksp is ………………
 Example : Given HgCl2 + 2 e-  Hg + 2 Cl-
and Hg2+ + 2 e-  Hg
Eo = + 0.27 V
Eo = + 0.85 V.
Calculate the Ksp for HgCl2.
Oxidation
Hg  Hg2+ + 2 e–
E0 = – 0.85 V
Reduction
HgCl2 + 2 e-  Hg + 2 Cl-
E0 = + 0.27 V
Overall
HgCl2  Hg2+ + 2 Cl0.059
EE 
lg K sp
n
0
0   0.58 V 
0.059
lg K sp
2
lg Ksp = – 19.66
Ksp = 2.18 x 10-20 mol3 dm-9
Ecell = – 0.58 V
13.7.3
Nernst Equation and pH of a Solution
 Under SHE, if the solution of H+ is not 1.00 mol dm-3, the
Ecell can be calculated using Nernst equation.
H2 (g)
1.0 atm
H+ (aq)
[x M]
25oC
M (s)
Salt bridge
(made of
saturated
KCl / NaCl)
Standard hydrogen electrode
M2+ (aq)
[1.0 M]
Zinc half cell
 Assuming the metal is oxidise by SHE, the cell notation is
written as
M (s) I M2+ (aq) II H+ (aq) (x mol dm-3) , H2 (g) I Pt (s)
 The overall reaction can be written as
M (s) + 2 H+ (aq)  M2+ (aq) + H2 (g)
x
0
.
059
[
reac
tan
t
ions
]
E  E0 
lg
n
[product ions ]y
 2
0
.
059
[
H
]
E  E0 
lg
2
[M 2 ]1
0.059
[H  ]
E  E 
  2 lg
2
[1]
0
E  E 0  0.059 pH
E0 = + a V
Example : The e.m.f. of of the following cell at 25oC is 0.093 V
Pb (s) I Pb2+ (1.00 mol dm-3) II H+ (test solution), H2 (g) I Pt (s)
Calculate the pH of the solution
Overall equation : Pb + 2 H+  Pb2+ + H2 E0 = + 0.13 V
E = E0 – 0.059 pH
0.059 pH = + 0.13 – 0.093
pH = 0.63
Calculate the e.m.f. of the chemical cell compared relatively to
SHE at [H+] = 0.0030 mol dm-3 for a calcium half cell.
2.9
Type of cell.
 A battery is a galvanic cell, or a series of combined galvanic cells,
that can be used as a source of direct electric current at a
constant voltage.
 Although the operation of a battery is similar in principle to that of
the galvanic cells. A battery has the advantage of being completely
self-contained and requiring no auxiliary components such as salt
bridges.
 Here we will discuss several types of batteries that are in
widespread use.
 Lithium ion battery
 Fuel Cell
2.9.1
Lithium ion battery
 Figure below shows a schematic diagram of a lithium-ion battery.
During the discharge of the battery,
the half-cell reactions are
Anode : Li(s) → Li+ + eCathode : Li+ + CoO2 + e- → LiCoO2 (s)
Overall : Li (s) + CoO2 → LiCoO2 (s)
Ecell = + 3.4 V
 The anode is made of a conducting carbonaceous material, usually
graphite, which has tiny spaces in its structure that can hold both Li
atoms and Li+ ions.
 The cathode is made of a transition metal oxide such as
CoO2, which can also hold Li+ ions. Because of the high
reactivity of the metal, non - aqueous electrolyte (organic
solvent plus dissolved salt) must be used.
 The advantage of the battery is that lithium has the most
negative standard reduction potential and hence the greatest
reducing strength. Furthermore, lithium is the lightest metal
so that only 6.941 g of Li (its molar mass) are needed to
produce 1 mole of electrons.
 A lithium-ion battery can be recharged literally hundreds of
times without deterioration. These desirable characteristics
make it suitable for use in cellular telephones, digital
cameras, and laptop computers.
2.9.2 Fuel Cell
 Fossil fuels are a major source of energy, but conversion of
fossil fuel into electrical energy is a highly inefficient process.
Consider the combustion of methane:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) H = - x kJ mol-1
 To generate electricity, heat produced by the reaction is first
used to convert water to steam, which then drives a turbine
that drives a generator.
 An appreciable fraction of the energy released in the form of
heat is lost to the surroundings at each step; even the most
efficient power plant converts only about 40 percent of the
original chemical energy into electricity.
 Combustion reactions are redox reactions, it is more desirable
to carry them out directly by electrochemical means, thereby
greatly increasing the efficiency of power production
 This objective can be accomplished by a device known as a
fuel cell, a galvanic cell that requires a continuous supply of
reactants to keep functioning.
 In its simplest form, a hydrogen-oxygen fuel cell consists of an
electrolyte solution, such as potassium hydroxide solution,
and two inert electrodes. Hydrogen and oxygen gases are
bubbled through the anode and cathode compartments where
the following reactions take place.
Anode : 2 H2 (g) + 4 OH- (aq) → 4 H2O (l) + 4 e- Eo = + 0.83 V
Cathode : O2 (g) + 2 H2O (l) + 4e- → 4 OH-(aq)
Eo = + 0.40 V
Overall : 2 H2 (g) + O2 (g) → 2 H2O(l)
Eocell = + 1.23 V
 The electrodes used have a two-fold function. They serve as
electrical conductors, and they provide the necessary
surfaces for the initial decomposition of the molecules into
atomic species, prior to electron transfer. They are also known
as electrocatalysts. Metals such as platinum, nickel, and
rhodium are good electrocatalysts.
2.9.2.1 Propane-oxygen fuel cell
 In addition to the H2-O2 system, a number of other fuel cells
have been developed. Among these is the propane-oxygen
fuel cell.
 The half-cell reactions are
Anode : C3H8 (g) + 6 H2O (l) → 3 CO2 (g) + 20 H+ (aq) + 20 eCathode : 5 O2 (g) + 20 H+ (aq) + 20 e- → 10 H2O (l)
Overall : C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)
 Unlike batteries, fuel cells do not store chemical energy.
Reactants must be constantly resupplied, and products
must be constantly removed from a fuel cell. In this respect,
a fuel cell resembles an engine more than it does a battery.
However, the fuel cell does not operate like a heat engine
and therefore is not subject to the same kind of
thermodynamic limitations in energy conversion
 Properly designed fuel cells may be as much as 70 percent
efficient, about twice as efficient as an internal combustion
engine. In addition, fuel-cell generators are free of the noise,
vibration, heat transfer, thermal pollution, and other
problems normally associated with conventional power
plants.
2.10 Electrolysis ~ decomposition of a substance by
direct current electricity.
 Electrolyte – substance that can conduct electricity when in
aqueous solution or in molten state
 Electrolytic cell – cell consisting of 2 electrodes immersed
in an electrolyte for carrying out electrolysis In an
electrolytic cell, the following apply
anode whereas the negative
 Positive terminal is called as …………
cathode
terminal is called as …………
 At anode, oxidation
…………. process occur where
as at cathode, reduction
………..…process occur
cathode while
 Cations are attracted to …………..
anode
anions are attracted to ……………
anode to
 Electrons flow from the …………
cathode in the external circuit.
……..…..
A
2.11
Faraday’s Law
 Faradays Law stated that 1 Faraday is the quantity of electricity
(9.65 x 104 C) that must be supplied to an electrolytic cell in
order to produce one mole of electrons for reactions in the cell.
 The extension of Faraday Law is stated in Faraday’s First Law,
where it stated that the mass of a substance produced at an
proportional to the quantity of
electrode during electrolysis is …………….…
electricity (in Coulumb) passed. From the statement above, the
factors that influence the mass of a substance liberated during
electrolysis are
• The greater the number of electrons transferred, the greater
the mass of the product.
• The longer the time taken, the greater the electrical current
produced, the more the mass produced
 Based on the statement above, the quantity of electrical current
can be calculate according to time where
Q = It
Q = electric current
I = current
t = time in second
Example 1 : Calculate the mass of silver
formed when a current of 0.200 A is applied to
a electrolytic cell filled with aqueous silver
nitrate for 2 hour.
Q = It @ Q = (0.200)(2 x 60 x 60)
= 1440 C
Eq : Ag+ + e-  Ag
mol of e- = Q / F @ mol = 1440 / 96500
mol of e- = 0.0149 mol
Since 1 e- = 1 Ag ; mol Ag = 0.0149 mol
Mass Ag = 0.0149 x 108
= 1.61 g
Example 3 : Calculate the mass of chromium
formed when 1.20 A of current is directed into
molten chromium (III) chloride for 3 hours.
Q = It @ Q = (1.20)(3 x 60 x 60)
= 12960 C
Eq : Cr3+ + 3 e-  Cr
mol e- = Q / F @ mol = 12960 / 96500
mol of e- = 0.134 mol
Since 3 e- = 1 Cr ; mol Cr = 0.0448 mol
Mass Cr = 0.0448 x 52.0
= 2.33 g
Example 2 : An aqueous solution of copper (II)
sulphate is electrolysed using a current of 0.50
A for 4 hours. Calculate the mass of Copper,
Cu deposited at cathode.
Q = It @ Q = (0.50)(4 x 60 x 60)
= 7200 C
Eq : Cu2+ + 2 e-  Cu
mol of e- = Q / F @ mol = 7200 / 96500
mol of e- = 0.0746 mol
Since 2 e- = 1 Cu ; mol Cu = 0.0373 mol
Mass Cu= 0.0373 x 63.5
= 2.4 g
Example 4 : Calculate the time taken to
produce 18.0 g of silver from silver nitrate by a
current of 0.900A
Mol of Ag = 18.0 / 108 = 0.167 mol
Eq : Ag+ + e-  Ag
1 Ag = 1 e- ; so mol e- = 0.167 mol
Q = mol e- x F @ Q = 0.167 x 96500
Q = 16083 C
Q = It @ t =16083 / 0.900
t = 1.78 x 104 s
Example 5 : Calculate the time required to Example 6 : Calculate the time required to
form 200g of lead from molten lead (II)
form 10 g of aluminium from molten
bromide by a current of 1.50 A.
aluminium oxide by a current of 10 A
Mol of Pb = 200 / 207
= 0.966 mol
Eq : Pb2+ + 2 e-  Pb
1 Pb = 2 e- ; so mol e- =
1.93 mol
Q = mol e- x F @
Q = 1.93 x 96500
Q = 186473 C
Q = It @ 186473 / 1.50
t = 1.24 x 105 s
Mol of Al = 10 / 27
= 0.370 mol
Eq : Al3+ + 3 e-  Al
1 Al = 3 e- ; so mol e- =
1.11 mol
Q = mol e- x F @
Q = 1.11 x 96500
Q = 107222 C
Q = It @ t = 107222 / 10
t = 1.1 x 104 s
14.2.1 Faraday’s Second Law
 ~ stated that if the same quantity of electricity is passed
through different electrolytes, the mass of substance
liberated at electrode is inversely proportional to the
charge of ions.
1 Faraday
Silver (I) nitrate
Copper (II)
sulphate
Chromium (III)
chloride
Sulphuric acid

Half equation occur at cathode for
Type of
electrode
Half equation
Mol of metal
deposited
1 mol
Silver
Ag+ (aq) + e-  Ag (s)
Copper
Cu2+ (aq) + 2 e-  Cu (s)
1/2 mol
Carbon in CrCl3
Cr3+ (aq) + 3 e-  Cr (s)
1/3 mol
Carbon in
H2SO4 (occur at
anode)

Mol of non-metal
2 H2O  O2 + 4 H+ + 4 e–
1/4 mol
1 mol of Ag+
From the diagram above, 1 F will discharge ….
1/2 mol of Cu2+ ion ; ……
1/3 mol of Cr3+ ion ; ….…
1/4 mol
ions ; ……
of O2.
Example 7 : Calculate the mass of copper
deposited under the same cell if the
amount of silver formed under the same
amount of quantity charge is 1.8 g.
Mol of Ag = 1.8 / 108
= 0.01667 mol
Eq : Ag+ + e-  Ag
Since 1 Ag = 1 e- ;
mol e- = 0.01667 mol
For Cu ; Cu2+ + 2 e–  Cu
Since 2 mol e- = 1 mol Cu
So mol of Cu = 0.008333 mol
Mass of Cu = 0.008333 x 63.5
= 0.53 g
An electric current produced 0.56 g of
aluminium from molten aluminium
oxide. If the same current was used to
electrolysed molten lead (II) bromide,
calculate the mass of lead deposited.
Mol of A1 = 0.56 / 27
= 0.02074 mol
Eq : Al3+ + 3 e-  Al
Since 1 Al = 3 e- ;
mol e- = 0.06222 mol
For Pb ; Pb2+ + 2 e–  Pb
Since 2 mol of e = 1 mol of Pb
So mol of Pb = 0.03111 mol
Mass of Pb = 0.03111 x 207
= 6.44 g

Predicting the product for electrolysis
• In electrolysis, there may be more than one type of cation /
anion inside the electrolytes.
• Under such circumstance, since an electrode can only
discharge one cation / anion, the ion must be choose under
certain guidelines.
• The selectivity of ions are based on electrochemical series
No matter it is an electrolytic cell or chemical cell,
oxidation reaction occur ; electrons are ………………
donated
• At anode, ………….
• At cathode, reduction
…………. reaction occur ; electrons are received
………………
molten state or
 Electrolytes can only discharge under 2 conditions : …..…...
aqueous solution.
…………
• When in molten state, the electrolytes contain only the cation and
anion of the substance involve
 Molten lead (II) bromide : PbBr2 (l)  Pb2+ + 2 Br–
 Molten aluminium oxide : Al2O3 (l)  2 Al3+ + 3 O2–
 Molten barium chloride : BaCl2 (l)  Ba2+ + 2 Cl–
 Molten silver (I) iodide : AgI (l)  Ag+ + I–

Electrolyte
PbBr2 (l)
Al2O3 (l)
NaCl (aq)
Subst.
present
Half equation at anode
Substance at
anode
Pb2+
Br –
Br2 + 2e-  2BrE0 = + 1.07 V
Rev:
2Br-  2e- + Br2
Pb2+ + 2e-  Pb
bromine
E0 = – 0.13 V
Al3+
O2–
O2 + 4e-  2O2E0 = ? V
Rev:
2O2-  O2 + 4e-
Al3+ + 3e-  Al
E0 = – 1.67 V
Alumin
um
Na+
Cl–
Cl2 + 2e-  2ClE0 = +1.36 V
Rev:
2Cl-  Cl2 + 2e-
Na+ + e-  Na
E0 = – 2.71 V
Sodium
oxygen
chlorine
Half equation at
cathode
Substance
at cathode
lead
No matter it is an electrolytic cell or chemical cell,
oxidation reaction occur ; electrons are ………………
donated
• At anode, ………….
• At cathode, reduction
…………. reaction occur ; electrons are received
………………
• When in aqueous solution, not only it contains the cation and anion
of substance involve, but it also involves water. thus there is a
selectivity of ion occur
• In the terms of E0, a more positive
……….. value will be selected for
negative value will be selected
discharge at cathode, while a more ………….
for discharge at anode.
1.23
 At anode
: 4 H+ + O2 + 4 e–  2 H2O
E0 =+………….
V
0.83
 At cathode : 2 H2O + 2 e–  2 OH– + H2
E0 = –………….
V
 However, water is a weak electrolytes. At 250C and 1 atm, the E0
value varies with the solution used. So in deciding which E0 value we
should used, we need to consider if the solution is different or not.
 Under neutral condition, where [H+] = [OH-], using Nernst Equation,
Eo values are (At [H+] = [OH-]1.0 x 10-7 mol dm-3)
0.81 V
 At anode :
E0 = +
….…….
4 H+ + O2 + 4 e–  2 H2O
– 0.41 V
 At cathode : 2 H2O + 2 e–  2 OH– + H2
E0 = ….…….

Electrolyte
NaCl (aq)
Subst.
present
Half equation at anode
Na+
Cl–
H2O
Cl2 + 2e-  2ClE0 = +1.36 V
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23 V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
Substance
Substance
Half equation at cathode
at cathode
at anode
Na+ + e-  Na
E0 = – 2.71 V
oxygen 2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
Hydroge
n
Electrolyte
CuSO4 (aq)
PbI2 (aq)
KOH (aq)
Ions
Half equation at anode Substance
present
Half equation at
cathode
Substance
Cu2+
SO42–
H2O
S2O82- +2e-  2SO42E0 = +2.01 V
4 H+ + O2 + 4 e– 
oxygen
2 H2O
E0 = +1.23 V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
Cu2+ + 2e-  Cu
E0 = + 0.34 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
Copper
Pb2+
I–
H2O
I2 + 2 e-  2 IE0 = +0.54 V
4 H+ + O2 + 4 e– 
2 H2O
E0 = +1.23 V
Rev: 2 I–  I2 + 2 e–
Pb2+ + 2e-  Pb
E0 = – 0.13 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
Lead
K+
OH–
H2O
O2 + 2 H2O + 4 e 4 OH–
E0 = + 0.40 V
4 H+ + O2 + 4 e– 
oxygen
2 H2O
E0 = +1.23 V
Rev: 4 OH– 
2 H2O + O2 + 4 e–
iodine
K+ + e-  K
E0 = – 2.92 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
hydroge
n

Effect on concentration towards the selectivity of the ions to
discharge
• In general, an ion with a very high concentration is
preferentially discharged.
• For example if Pb2+ ion and Cu2+ ion are mixed under the
Cu2+ ion is preferred to be
same concentration, ………
discharge at cathode as it has a lower position in
electrochemical series.
• However, if the concentration of Pb2+ ion concentration is
Pb2+ ion is more
raised much higher than Cu2+ ion, ……….
readily to be discharged.
• Another example is potassium chloride in aqueous
H2O will be
solution. Under dilute solution of KCl, ………..
oxygen gas is given out, as it
selected at anode and ………...…
has a lower position in electrochemical series.
• However, if concentrated KCl is used as electrolyte, the

Cl– is selected to be
concentrated of Cl- increase, and ……
chlorine gas is given out.
discharge and ……..….…
• Still, if the position in electrochemical series differ too
much, like K+ and water in the example above, K+ ion
will not
……………
be discharge as the position is much too high.
hydrogen gas is given out at cathode.
At the end, …………….
• If chloride ion is replaced with fluoride ion, F– ion, and
concentration of F– ion is increased, water
…....... is still be
very high position in
preferred as F– ion has a …………….
electrochemical series
In the table below, predict the element that is expected to
form when electrolyse.
Electrolyte
Concentrate
NaCl
Concentrated
PbBr2
Concentrated
LiF
Ions
present
Half equation at
anode
Na+
Cl–
H2O
Cl2 + 2e-  2ClE0 = +1.36 V
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23 V
Rev: 2 Cl- 
Cl2 + 2 e–
Pb2+
Br –
H2O
Li+
F–
H2O
Substance
Half equation at
cathode
Na+ + e-  Na
E0 = – 2.71 V
– 
2
H
O
+
2
e
2
chlorine
2 OH– + H2
E0 = – 0.83 V
4 H+ + O2 + 4 e–
 2 H2O
Pb2+ + 2e-  Pb
E0 = +1.23 V
E0 = – 0.13 V
Br2 + 2e-  2Br- bromine 2 H2O + 2 e– 
E0 = + 1.07 V
2 OH– + H2
Rev:
E0 = – 0.83 V
2Br-  2e- + Br2
Li+ + e-  Li
F2 + 2e-  2FE0 = +2.87 V
E0 = – 3.04 V
4 H+ + O2 + 4 e–
2 H2O + 2 e– 
oxygen
 2 H2O
– + H
2
OH
2
E0 = +1.23 V
E0 = – 0.83 V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
Substance
Hydroge
n
lead
Hydrogen
2.12.3 Overvoltage
 Overvoltage ~ the difference between electrode potential
and discharge potential. In another words, Overvoltage is
the voltage that must be applied to an electrolytic cell in
addition to the theoretical voltage to cause an electrode
reaction to occur.
 Example of over voltage phenomenon is the electrolysis of
aqueous sodium chloride. Consider the electrolysis of
sodium chloride in aqueous solution where the substance
presence are Na+, Cl-, H2O
Half equation for substance attracted to
anode
Half equation for substance attracted to
cathode
Cl2 (g) + 2 e- → 2 Cl- (aq) E0 = + 1.36 V Na+ (aq) + e- → Na (s)
E0 = - 2.71 V
4 H+ + O2 (g) + 4 e- → 2 H2O (l)
2 H2O (l) + 2 e- → 2 OH- + H2 (g)
E0 = + 0.81 V
E0 = - 0.41 V
(conc. H+ = 1 x 10-7 mol dm-3)
(conc. OH- = 1 x 10-7 mol dm-3)




From the E0 values at anode, it is suggested that H2O should
be preferentially oxidized at the anode. However, by
experiment we find that the gas liberated at the anode is
Cl2, not O2.
In studying electrolytic processes, we sometimes find that
the voltage required for a reaction is considerably higher
than the electrode potential indicates.
The overvoltage is the difference between the electrode
potential and the actual voltage required to cause
electrolysis. In this case, overvoltage for O2 formation is quite
high. Therefore, under normal operating conditions Cl2 gas
is actually formed at the anode instead of O2.
As for the selectivity at cathode, H2O is selected since the E0
value for H2O is less negative than Na+.

Thus the half equation occur at both anode and cathode
are
Anode
Cathode
Overall


As the overall reaction shows, the concentration of the Cl2
ions decreases during electrolysis and that of the OH- ions
increases.
Therefore, in addition to H2 and Cl2, the useful by-product
NaOH can be obtained by evaporating the aqueous
solution at the end of the electrolysis.
2.12.4 Electrorefining and electroplating
 The purification of a metal by means of electrolysis is
called electrorefining. For example, impure copper
obtained from ores is converted to pure copper in an
electrolytic cell that has impure copper as the anode and
pure copper as the cathode. The electrolyte is an aqueous
solution of copper sulphate



At the impure Cu anode, copper is oxidized along with more
easily oxidized metallic impurities such as zinc and iron.
Less easily oxidized impurities such as silver, gold, and
platinum fall to the bottom of the cell as anode mud, which is
reprocessed to recover the precious metals. At the pure Cu
cathode, ions are reduced to pure copper metal, but the less
easily reduced metal ions (and so forth) remain in the
solution
Half equations occur for electrorefining process of copper
above are
• At anode
Cu (s) → Cu2+ (aq) + 2 e• At cathode
Cu2+ (aq) + 2 e- → Cu (s)
Thus, the net cell reaction simply involves transfer of copper
metal from the impure anode to the pure cathode, hence
purified the copper.



Closely related to electrorefining is electroplating, the
coating of one metal on the surface of another using
electrolysis. For example, steel automobile bumpers are
plated with chromium to protect them from corrosion, and
silver-plating is commonly used to make items of fine table
service.
The object to be plated is carefully cleaned and then set up
as the cathode of an electrolytic cell that contains a solution
of ions of the metal to be deposited (as shown in diagram
above)
Half equations occur for electroplating process of silver
above are
• At anode
Ag (s) → Ag+ (aq) + e• At cathode
Ag+ (aq) + e- → Ag (s)
2.13
Industrial Electrolysis :
 In this Chapter, we shall discussed the manufacturing of
aluminium and chlorine gas using the principle of electrolysis
Part 1 : Getting pure aluminium oxide (alumina) from bauxite.
 1st step:
Removal of impurities from the ore by dissolving
powdered bauxite in hot concentrated sodium hydroxide
solution. Al O + 2 NaOH + 3 H O  2 NaAl(OH)
2
3
2
SiO2 + 2 NaOH  Na2SiO3 + H2O

4
2nd step: Insoluble impurities are filtered off. Filtrate contain
aluminium and silicon ions. Aluminium ion is precipitated as
aluminium hydroxide which is filtered out later as white
gelatinous precipitate.
Use acid : 2 [Al(OH)4]- + 2 H+  2 Al(OH)3 + 2 H2O
Or use CO2 : 2 [Al(OH)4]- + CO2  2 Al(OH)3 + H2O + CO32-

3rd step: Aluminium hydroxide is filtered, washed, dried and
finally heated out to 12000C to produce pure aluminium oxide
(alumina), Al2O3
2 Al(OH)3  Al2O3 + 3 H2O
Part 2 : Extracting aluminium out from aluminium oxide
 Hall-Heroult process:- A process of electrolysing aluminium
oxide (alumina) to extract out aluminium.
 Aluminium metal is extracted by the cell electrolytic reduction of
alumina. Melting point of alumina is 20300C. To lower the
temperature of the electrolyte, alumina is dissolved in molten
cryolite (Na3AlF6), to maintain a temperature at about 9600C.
 When alumina dissolve in molten cryolite :

Al2O3 (s)  2 Al3+ (l) + 3 O2– (l)

Electrolyte mixture is then placed in carbon-lined iron vat
(cathode). The heating effect of the electric current melts
the electrolyte mixture, producing Na+, Al3+, O2- and F- ions.
Na+ + e-  Na E0 = – 2.71 V
Al3+ + 3e-  Al E0 = – 1.66 V
F2 + 2 e-  2 F- E0 = + 2.87 V
O2 + 4e-  2O2- E0 = + 1.++V
Half equation occur at cathode
Half equation occur at anode
Al3+ + 3e-  Al E0 = – 1.66 V
2O2-  O2 + 4e- E0 = + 1.++V


Aluminium alloy parts are anodized to greatly increase the
thickness of this layer for corrosion resistance. The
corrosion resistance of aluminium alloys is significantly
decreased by certain alloying elements or impurities :
copper, iron, and silicon, tend to be most susceptible.
By making an aluminium the anode of cell in which dilute
sulphuric acid is the electrolytes, it is possible to produce a
thicker and harder film of aluminium oxide on the surface of
metal.
Al
C
2.13.1.2
Recycling aluminium
 Environmental pollution arises as cans are littered
everywhere.
 The best solution of environmental pollution is recycling.
 The benefits of recycling can be seen by comparing the
energy consumed in the extraction of aluminium from the
bauxite or using Hall process with that consumed when
aluminium is recycle.




Pure aluminium has a rather low melting print of 6600C, thus
requiring only 26.1 kJ mol-1 of energy.
On comparison between the Hall process and recycling,
Energy used in recycling
= 26.1  297  100%
= 8.8%
 This means that about 91% of the energy is saved for
every 1 mole of the aluminium produced through
recycling

In industrial process, chlorine gas, together with sodium
metal, is prepared using molten sodium chloride (brine)
using mercury-cathode cell.
Half equation occur at cathode
Na+ + e-  Na


E0 = – 2.71 V
Half equation occur at anode
2Cl-  Cl2 + 2e- E0 = – 1.36 V
Mercury is specially used to attract the sodium formed in
cathode and form an alloy named amalgam
This method is not environment friendly as the mercury
used is poisonous.



Similar to the mercury-cathode cell, the electrolytes used in
brine (sodium chloride)
diaphragm cell is also …………………………..
The process inside the diaphragm cell is known as the
chlor-alkali process.
When sodium chloride dissociates under the effect of an
electric current, the chloride ions are discharged. Half
equation : 2 Cl-  Cl2 + 2 e-






Titanium is chosen as the anode because it resists
corrosion by the very reactive chlorine
At cathode, since the sodium ion (Na+) is attracted to
cathode through the diaphragm, the selectivity to discharge
is between sodium ion and water molecule.
Standard reduction potential of sodium :
Na+ + e-  Na
E0 = – 2.71 V
Standard reduction potential of water :
2 H2O + 2 e–  2 OH– + H2
E0 = – 0.83 V
water is discharge
water
Since …………
has a higher E0 value, …………
hydrogen gas is produced.
and …………………
The level of brine (left or anode position) is always placed
higher than the water (right or cathode position) to
prevent water from crossing to brine portion. This will
………………………..…………………………………………......
dilute the solution and chlorine will not be discharged.
13.8 Corrosion of metal
 Corrosion is the oxidative deterioration of a metal, such as
metal
metal oxide
the conversion of ………...…
to ……………..
 2 main important components for rusting are
oxygen
water
…………………..
and ……………………
 A possible mechanism for rusting, consistent with the known
facts, is illustrated in Figure below
In alkaline / neutral condition
At anode :
At cathode :
Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)
Eo = + 0.40 V
2 Fe (s) + O2 (aq) + 2 H2O  2 Fe2+ + 4 OH– [or 2 Fe(OH)2]
Ecell = + 0.84 V
Fe(OH)2 (aq) + OH–  Fe(OH)3 + e – O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)
Eo = + 0.40 V
Eo = +0.56 V
4 Fe(OH)2 (aq) + O2(aq) + 2H2O(l)  4 Fe(OH)3 (aq) Ecell = + 0.96 V
Forming rust : 2 Fe(OH)3 (s)  Fe2O3.x H2O + (3 – x) H2O
 In acidic condition
At anode :
At cathode :
Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)
Eo = + 1.23 V
2 Fe (s) + O2 (aq) + 4 H+  2 Fe2+ + 2 H2O
At anode :
Fe2+ (aq)  Fe3+ (aq) + eEo = -0.77 V
Ecell = + 1.67 V
At cathode :
O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)
Eo = + 1.23 V
4 Fe2+ (aq) + O2 (aq) + 4 H+(aq) → 4 Fe3+ (aq) + 2H2O(l)
Ecell = + 0.46 V
Forming rust : 2 Fe3+ (aq) + 4 H2O (l)  Fe2O3.H2O (s) + 6 H+ (aq)
13.8.2
Prevention of rusting
 Various methods are used to prevent / slowing down rusting.
Methods
Explanation
Stainless steel
• Iron is alloyed with nickel and chromium to form …………………….
The chromium forms an impervious oxide layer on the surface of
Alloying
iron increasing its resistance to corrosion. Chromium at the same
decorated / coated
time ………………..
………the steel
 Painting the iron object
Barrier
 Use grease / oil to coat the moving parts of machine
 Coating ironwith chromium (plating) or zinc (galvanising)
anodic protection
 Also known as ……………
higher
 Metal with a ……………….
position in electrochemical series is
Sacrificial
metal Will be
‘connected’ to iron. Under such way, reactive
………………...
oxidised first before iron.
Additional Slide
(from old syllabus)
 13.7.4
Effect of pH on Electrode Potential of a Half cell
 Some reaction involve H+ ions. Examples of the are
14 H+ + Cr2O72- + 6 e-  2 Cr3+ + 7 H2O
8 H+ + MnO4- + 5 e-  Mn2+ + 4 H2O
6 H+ + ClO3- + 6 e-  Cl- + 3 H2O
 Under standard conditions, the [H+] is 1.00 mol dm-3. varying the
concentration of [H+] and hence its pH, would change the
electrode potential of the half cell.
 Consider the following half cell reaction :
MnO4- (aq) + 8 H+ (aq) + 5 e-  Mn2+ (aq) + 4 H2O (l) E = + 1.52 V
 Using Nernst Equation, Ecell can be expressed as
0.059 [MnO 4 ][ H  ]8
E   1.52 
lg
5
[Mn 2 ]
 When [MnO4-] = [Mn2+] = 1.00 mol dm-3- ; Ecell =
0.059
E   1.52 
 8 lg [H  ]
5
E   1.52  0.0944 pH
[H+] / mol dm-3
1.0
0.1
0.01
0.00001
pH
0
1
2
5
Ecell (V)
+ 1.52
+ 1.43
+ 1.33
+1.05
It is then compare to the standard electrode potential of chlorine, bromine
and iodine
Cl2 + 2 e-  2 Cl- Eo = + 1.36 V @ 2 Cl-  Cl2 + 2 e- Eo = - 1.36 V
Br2 + 2 e-  2 Br- Eo = + 1.07 V @ 2 Br-  Br2 + 2 e- Eo = - 1.07 V
I2 + 2 e-  2 I- Eo = + 0.54 V @ 2 I-  I2 + 2 eEo = - 0.54 V
Cl- , Br- , I- are
+ 1.43
• Under pH = 1 , Ecell of manganate (VII) ion is ................
V ……………
oxidising
oxidise by manganate (VII) ion as it is still a strong ……………
able to …………
agent.
1.33 V. Only ……………
Br- , I• Under pH = 2, Ecell of manganate (VII) ion is +
..............
Cl- cannot oxidise as the
oxidise by manganate (VII) ion. ……
are able to …………
spontaneous (Ecell =-…………)
0.03 V
reaction is not ………………
+ 1.05 V. Only ……
I- are
• Under pH = 5, Ecell of manganate (VII) ion is ..............
- , Broxidise by manganate (VII) ion. Cl
able to …………
……….
cannot oxidise as the
-0.03 V for Brreaction is not spontaneous
……………… (Ecell = ……………………)
-0.31 V for Cl-
Primary cell
Dry cell
Alkaline cell
Diagram
Anode
Equation occur at
anode
Cathode
Equation occur at
cathode
Electrolytes
zinc
Zn 
Zn2+
zinc
+ 2e
–
Manganese (IV) oxide
2 MnO2 + 2 NH4+ 2 e- 
Mn2O3 + 2 NH3 + H2O
Ammonium chloride and
zinc chloride paste
Zn + 2OH– 
H2O + ZnO + 2 e –
Manganese (IV) oxide
2 MnO2 + 2 H2O + 2 e- 
Mn2O3 + 2 OH–
Potassium hydroxide
Primary cell
Mercury cell
Lead – acid accumulator
Diagram
Anode
Equation
occur at
anode
Zinc
Zn + 2OH– 
H2O + ZnO + 2 e –
Cathode
Mercury (II) oxide
Equation
occur at
cathode
HgO + H2O + 2 e- 
Hg + 2 OH–
Electrolytes
Potassium hydroxide
Lead
Pb + HSO4– 
PbSO4 + H+ + 2 e–
Lead (IV) oxide
PbO2 + 3 H+ + HSO4– + 2 e– 
PbSO4 + 2 H2O
Sulphuric acid
Fuel Cell
Lithium ion cell
Anode
Hydrogen
Lithium metal
Equation occur
at anode
2 H2 + 4 OH–  2 H2O + 4 e–
Li (s)  Li+ (aq) + e-
Cathode
Oxygen
Manganese (IV) oxide
Diagram
Equation occur
at cathode
Electrolytes
O2+ 2 H2O + 4 e–  4 OH–
Hot potassium hydroxide (aq)
MnO2 (s) + Li+ + e-  LiMnO2
Lithium chlorite (VII), LiClO4
13.9 Dental Filling
amalgam
 The material commonly used to fill decaying teeth is an ………………….
mercury
(a ……………………
base alloy). The component in dental filling of
tin
mercury
silver
amalgam are ………………….
, ………………
and …………..
 The standard electrode potential of these electrode system are :
 Hg22+ (aq) / Ag2Hg3 (s)
Eo = + 0.85 V
 Sn2+ (aq) / Ag3Sn (s)
Eo = – 0.55 V
 Sn2+ (aq) / Sn8Hg (s)
Eo = – 0.13 V
 The diagram shows the reaction take place when gold is contact with
dental amalgam
dental amalgam, which result a electrochemical cell. ……………………
gold
act as the anode of the cell, while …………
act as the cathode and
saliva act as the electrolyte.
………..
positive
 Since tin is more electro…………………
than gold, hence tin will corrode
to form Sn2+ and mixed with saliva. This will result an unpleasant taste in
the mouth.
 If the dental amalgam is in contact with an aluminium foil, an
electrochemical cell will also produced. Unlike gold, aluminium is more
electropositive than any of the electrode above, which makes
…………………..
anode
amalgam
aluminium serves as an ………………
of the cell, while …………………….
as the cathode of cell. This will result a weak current flow between the
electrode and cause an unpleasant sensation in the tooth.