Inorganic Chemistry : Electrochemistry

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
2. For simple ions (and ionic compounds), the oxidation number is
the same as the charge of ion
Na = Mg = H2 = Cl2 = P4 =
K+ = Ca2+ = B3+ = P3- = O2- = F- =
0 0 0 0 0
+1 +2 +3 – 3 – 2 – 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
Element
Oxidation
number
Notes
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)

Na2O B2O3 CO2
SO3 Cl2O7 HF
H2S NH3 CH4
H2CO3 H2SO4 HBrO3
3. In a neutral molecule, the sum of the oxidation numbers of all
atom are equals
2 Na + O = 0
2(Na) + (-2) = 0
Na = +1
2 B + 3 O = 0
2(B) + 3(-2) = 0
B = +3
1 C + 2 O = 0
1(C) + 2(-2) = 0
C = +4
1S + 3O = 0
1(S) + 3(-2) = 0
S = +6
2Cl + 7O = 0
2(Cl) + 7(-2) = 0
Cl = +7
1H + 1F = 0
1(F) + 1(+1) = 0
F = -1
1S + 2H = 0
1(S) + 2(+1) = 0
S = -2
1N + 3H = 0
1(N) + 3(+1) = 0
N = -3
1C + 4H = 0
1(C) + 4(+1) = 0
N = -4
2H + 1C + 3O = 0
2(+1) +1C + 3(-2) = 0
C = +4
2H + 1S + 4O = 0
2(+1) +1S + 4(-2) = 0
S = +6
1H + 1Br + 3O = 0
1(+1) +1Br + 3(-2) = 0
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
- Cr2O7
2- MnO4
-
C2O4
2- ClO2
- HSO4
-
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
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
 When a substance is oxidise, electron is ………………. ; e- is
written at the ……….. side equation.
 When a substance is reduce, electron is ………………. ; e- is
written as the ……… side equation.
 Simple half equation : State the changes of oxidation number and
write the half equation.
Reaction
Oxidation
no change
Reaction Half equation
Na  Na+
Mg  Mg2+
Al  Al3+
Cu2+  Cu
donated
right
received
left
0  +1 oxidation Na  Na+ + e-
0  +2 oxidation Mg  Mg2+ + 2 e-
0  +3 oxidation Al  Al3+ + 3 e-
+2  0 reduction Cu2+ + 2 e-  Cu

Reaction
Oxi. no
change
Reaction Half equation
H2  H+
Cl2  Cl-
I-  I2
O2  O2-
Fe2+  Fe3+
Pb4+  Pb2+
0  +1 oxidation H2  2 H+ + 2 e-
0  -1 reduction Cl2 + 2 e-  2 Cl–
-1  0 oxidation 2 I-  I2 + 2 e-
0  -2 reduction O2 + 4 e-  2 O2–
+2  +3 oxidation Fe2+  Fe3+ + e-
+4  +2 reduction Pb4+ + 2 e-  Pb2+

 When it comes to the reaction involving molecular ion, the overall
charge has to be balanced in such order.
1. Write a skeleton half equation. Determine the reaction (oxidation
or reduction) using oxidation number
2. Balance the charge by adding electrons at the appropriate side
3. Balance the number of atoms other than oxygen.
4. Based on the changes in number of oxygen, write the number of
water molecule formed/used.
5. From the number of water molecule formed/used, write the
number of hydrogen ion (H+) required.

a) ClO3
-  Cl-
 Half equation : …………………………………………………………………
b) CrO4
2-  Cr3+
 Half equation : …………………………………………………………………
c) Cr2O7
2-  Cr3+
 Half equation : …………………………………………………………………
Changes in OS : +5  -1 ; reduction
Different in OS = 6, so 6 e- at the LHS of equation
ClO3
- + 6 e-  Cl-
ClO3
- + 6 e-  Cl- + 3 H2O
6 H+ + ClO3
- + 6 e-  Cl- + 3 H2O
6 H+ + ClO3
- + 6 e-  Cl- + 3 H2O
Changes in OS : +6  +3 ; reduction
CrO4
2- + 3 e-  Cr3+
CrO4
2- + 3 e-  Cr3+ + 4 H2O
8 H+ + CrO4
2- + 3 e-  Cr3+ + 4 H2O
8 H+ + CrO4
2- + 3 e-  Cr3+ + 4 H2O
Since there are 2 Cr , so total e- = 6 ; Cr2O7
2- + 6 e-  2 Cr3+
Cr2O7
2- + 6 e-  2 Cr3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 e-  2 Cr3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 e-  2 Cr3+ + 7 H2O

d) MnO4
-  Mn2+
Half equation : ……………………………………………………………………..
e) NO2
-  NO3
-
Half equation : ……………………………………………………………………..
f) CrO2
-  CrO4
2-
Half 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
Changes in OS : +3  +5 ; oxidation
Different in OS = 2, so 2 e- at the RHS of equation
NO2
-  NO3
- + 2 e-
NO2
- + H2O  NO3
- + 2 e-
NO2
- + H2O  2 H+ + NO3
- + 2 e-
NO2
- + H2O  2 H+ + NO3
- + 2 e-
CrO2
-  CrO4
2- + 3 e-
2 H2O + CrO2
-  CrO4
2- + 3 e-
2 H2O + CrO2
-  4 H+ + CrO4
2- + 3 e-
2 H2O + CrO2
-  4 H+ + CrO4
2- + 3 e-

g) As2O3  As2O5
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)
Oxidation half equation : ……………………………………….………………
Reduction half equation : ………………………………………………………
Overall equation : …………………………………………………………..
Example 11 : Fe2+ (aq) + MnO4
- (aq)  Fe3+ (aq) + Mn2+ (aq)
Oxidation half equation : ………………………………………………………
Reduction half equation : ………………………………………………………
Overall equation : …………………………………………………………..
Since there are 2 As , so total e- = 4 ; As2O3  As2O5 + 4 e-
As2O3 + 2 H2O  As2O5 + 4 e-
As2O3 + 2 H2O  As2O5 + 4 e- + 4 H+
As2O3 + 2 H2O  As2O5 + 4 e- + 4 H+
Na  Na+ + e-
Cu2+ + 2 e-  Cu
X 2
Cu2+ + 2 Na  Cu + 2 Na+
Fe2+  Fe3+ + e-
8 H+ + MnO4
- + 5 e-  Mn2+ + 4 H2O
X 5
5 Fe2+ 8 H+ + MnO4
-  Mn2+ + 4 H2O + 5 Fe3+

Example 12 : ClO- (aq) + SO2 (g)  Cl- (aq) + SO4
2-
Oxidation half equation : ……………………………………..……………..
Reduction half equation : ……………………………………………………..
Overall equation : …………………………………………………
Example 13 : Cr2O7
2− + Cl2  ClO3
− + Cr3+
Oxidation half equation : ……………………………………………..
Reduction half equation : …………………………………………….
Overall equation : …………………………………………………
2 H2O + SO2  4 H+ + SO4
2- + 2 e-
2 H+ + ClO- + 2 e-  Cl- + H2O
H2O + ClO- + SO2  2 H+ + SO4
2- + Cl-
6 H2O + Cl2  12 H+ + 2 ClO3
- + 10 e-
14 H+ + Cr2O7
2- + 6 e-  2 Cr3+ + 7 H2O
X 3
X 5
5 Cr2O7
2- + 34 H+ + 3 Cl2  6 ClO3
- + 17 H2O + 10 Cr3+

 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)
 For Sn ; O.N changed from …… to …… ; so the difference is ……..
 For Fe ; O.N changed from …… to ..…. ; so the difference is ……..
 This will makes the equation become :
 Balanced the number of atoms on both side of the equation
+4 +2 – 2
+2 +3 + 1
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
1 Sn4+ (aq) + 2 Fe2+ (aq)  Sn2+ (aq) + Fe3+ (aq)
Sn4+ (aq) + 2 Fe2+ (aq)  Sn2+ (aq) + 2 Fe3+ (aq)

Example 14 : Br – (aq) + SO4
2- (aq) SO2 (g) + Br2 (l)
 For Br ; O.N changed from …… to ……. ; so the difference is ………
 For S ; O.N changed from …… to ……. ; so the difference is ………
 This will make the equation become :
- 1 0 + 1
+6 +4 - 2
x (+ 1) + y (– 2) = 0
So, x = 2 ; y = 1
2 Br – (aq) + 1 SO4
2- (aq)  Br2 (l) + SO2 (aq)
2 Br – (aq) + SO4
2- (aq) + 4 H+  Br2 (l) + SO2 (aq) + 2 H2O

Balanced the number of atoms on both side of the equation
Example 15 : CrO4
2- + Cl-  Cr3+ + Cl2
Example 16 : Cr2O7
2- + NO2
-  Cr3+ + NO3
-
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+ + CrO4
2- + 3 Cl-  Cr3+ + 3/2 Cl2 + 4 H2O
16 H+ + 2 CrO4
2- + 6 Cl-  2 Cr3+ + 3 Cl2 + 8 H2O
For 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
Cr2O7
2- + 3 NO2
-  Cr3+ + NO3
-
8 H+ + Cr2O7
2- + 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
- + SO3
2-  MnO2 + SO4
2-
Example 18 : Fe(OH)2 + CrO4
2−  Fe(OH)3 + Cr(OH)3
Oxidation ½ eq : H2O + SO3
2-  2 H+ + SO4
2- + 2 e-
Reduction ½ eq : 4 H+ + MnO4
- + 3 e-  MnO2 + 2 H2O
Overall : 2 H+ + 2 MnO4
- + 3 SO3
2-  2 MnO2 + 3 SO4
2- + H2O
H2O + 2 MnO4
- + 3 SO3
2-  2 MnO2 + 3 SO4
2- + 2 OH-
X 3
X 2
Oxidation ½ eq : H2O + Fe(OH)2 → Fe(OH)3 + e− + H+
Reduction ½ eq : 5 H+ + 3 e− + CrO4
2− → Cr(OH)3 + H2O
Overall : 2 H2O + 3 Fe(OH)2 + 2 H+ + CrO4
2− → 3 Fe(OH)3 + Cr(OH)3
In basic : 4 H2O + 3Fe(OH)2 + CrO4
2− → 3Fe(OH)3 + Cr(OH)3 + 2OH−
X 3

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)
19. NaOH (aq) + Cl2 (aq)  NaCl (aq) + NaOCl (aq) + H2O
20. NaOBr (aq)  NaBrO3 (aq) + NaBr (aq)
+1 +2 0
0 +1-1
+1 +5 -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
 Cu2+ + 2e- ↔ Cu
 The positive value of E0 indicates the equilibrium favours to
the ……… position. Copper (II) ions (Cu2+), have a greater
tendency to …………...… at copper electrode.
 Zn2+ + 2e- ↔ Zn
 The negative value of E0 indicates the equilibrium favours to
the ….. position. Zinc ion (Zn2+) have a greater tendency to
…………….. at zinc electrode.
V34.0E Cu/Cu2 
V76.0E Zn/Zn2 
M
M+
M+
M+ M+
M+
M+
right
be reduced
left
be oxidised

Non – Metal
F2 + 2e- ↔ 2 F-
Cl2 + 2e- ↔ 2 Cl-
 Positive value of ECl2/Cl- and EF2/F- indicates the equilibrium
favours to the…………. position. ……………………… has a
greater tendency to …………… under platinum electrode
 The more positive the value, higher the tendency of non-
metal to …………….
 In another words, fluorine is a stronger ………….. agent than
chlorine.
V87.2E F/F2

V36.1E Cl/Cl2

Cl2 (g)
Cl- (aq) [1.0 M]
be reduced
Fluorine and chlorineright
be reduced
oxidising

Mixture of aqueous ion
 A potential difference also exists between ions in an aqueous
solution. Example :
Cr3+ + e- ↔ Cr2+
Fe3+ + e- ↔ Fe2+
 Base on the Eo value, Cr3+ is ……… stable than Cr2+ as
equilibrium favour to ………….. (Eo is negative)
 Base on the Eo value, Fe3+ is …….. stable than Fe2+ as
equilibrium favour to ………….. (Eo is positive)
V41.0E 23
Cr/Cr

V77.0E 23
Fe/Fe

Ma+ [1.0 M] / Mb+ [1.0 M]
more
backward
less
forward
V

2.3.1 Standard Electrode Potential
 Definition : The standard electrode potential, Eo
M
n+
/ M of a metal M
is the ………… difference between the metal M and the …………
solution of the metal ions of concentration ……………… at …… K
and …. atm, measured relatively to ……………………………….…
 Standard Hydrogen Electrode ( S.H.E.)
 It is impossible to measure the electrode potential for an
…………….. half-cell. It can only be measured for a complete
circuit with 2 ………. , i.e. only differences in electrode potentials
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 …………...
 2 H+ (aq) + 2 e- H2 (g)
 Condition : ….... oC ; H2 (g) at …… atm ; [H+] = 1.00 M
potential aqueous
1.0 mol dm-3 298
1.0
Standard Hydrogen Electrode
incomplete
half cell
standard
25
1.0

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 :
Standard hydrogen electrode Zinc half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Zn (s)
Zn2+ (aq)
[1.0 M]
0.76 V

 The chemical cell is set-up by connecting a standard
………….. half-cell to a standard …………. electrode.
 The e.m.f. for the cell is ………. V. The potentiometer point
to the direction of …….. electrode in the external circuit,
indicating electrons flow from ………….. to …………. half-
cell.
 Eq. Zn half-cell :
 Eq. H half-cell :
 Overall reaction :
 The cell notation can be written as :
 At zinc electrode ; electrons are ………..... ; ….……..…
reaction occur
 At platinum electrode ; electrons are ………….. ; ………….
reaction occur
 Since zinc is oxidised in a SHE, the standard e.m.f value is
……………
H+/H2Zn2+/Zn
0.76
H2
H+ / H2Zn2+ / Zn
Zn (s)  Zn2+ (aq) + 2 e-
2 H+ (aq) + 2 e-  H2 (g)
Zn (s) + 2 H+ (aq)  H2 (g) + Zn2+ (aq)
Zn (s) I Zn2+ (aq) II H+ (aq) , H2 (g) I Pt (s)
donated oxidation
received reduction
– 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 :
Standard hydrogen electrode Silver half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Ag (s)
Ag+ (aq)
[1.0 M]
0.80 V

 The chemical cell set-up by connecting a standard
……………… half-cell to a standard …………….. electrode.
 The e.m.f. for the cell is ……. V. The galvanometer point to
the direction of ………….. electrode in the external circuit,
indicating electrons flow from ………….. to ………… half-cell.
 Ag half-cell :
 H2 half-cell :
 Overall :
 At silver electrode ; electrons are …………. ; ……………
reaction occur
 At platinum electrode ; electrons are …………… ; ……………
reaction occur
 Since silver is reduced in a SHE, the standard value is
………………
donated oxidation
received reduction
+ 0.80 V
H+ / H2Ag+ / Ag
0.80
H+ / H2
silver
Ag+ / Ag
Ag+ (aq) + e-  Ag (s)
H2 (g)  2 H+ (aq) + 2 e-
H2 (g) + 2 Ag+ (aq)  2 Ag (s) + 2 H+ (aq)
Pt (s) I H2 (g) , H+ (aq) II Ag+ (aq) I Ag (s)

Measuring a standard electrode potential of a gaseous
substance
 The chemical cell set-up by connecting a standard …………
half-cell to a standard …………… electrode. Note that the
set-up of the half-cells are the same for gaseous substances
Cl2 / Cl–
H+ / H2
Standard hydrogen electrode Chlorine half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Cl2 (g)
1 atm
Cl- (aq)
[1.0 M]
1.36 V

 The e.m.f. for the cell is …….V. The galvanometer point to
the direction of ……….. electrode in the external circuit,
indicating electrons flow from ………….. to …………. half-
cell.
 Chlorine half-cell :
 Hydrogen half-cell :
 Overall : :
 At platinum electrode in the half-cell of hydrogen ; electrons
are ………… ; ………… reaction occur
 At platinum electrode in the half-cell of chlorine ; electrons
are ………….. ; …………… reaction occur
 Since chlorine is ………… by SHE, the standard value is
……….
1.36
Pt (Cl2)
H+ / H2 Cl2 / Cl–
Cl2 (g) + 2 e-  2 Cl– (aq)
H2 (g)  2 H+ (aq) + 2 e-
H2 (g) + Cl2 (g)  2 Cl– (aq) + 2 H+ (aq)
Pt (s) I H2 (g) , H+ (aq) II Cl2 (g), Cl– (aq) I Pt (s)
donated oxidation
received reduction
reduced
+ 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
Standard hydrogen electrode Fe2+ / Fe3+ half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq) [1.0 M]
25oC
Potentiometer
Fe2+ (aq)
Fe3+ (aq)
[1.0 M]
0.77 V

 The chemical cell set-up by connecting a standard …………
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
under standard condition with …………… as electrode.
 The e.m.f. for the cell is ……. V. The galvanometer point to the
direction of …………... half cell in the external circuit, indicating
electrons flow from …………. to …………. half-cell.
 Fe3+ / Fe2+ half-cell :
 Hydrogen half-cell :
 Overall reaction :
The cell notation can be written as :
 At half-cell of hydrogen ; electrons are ………… ; ……………
reaction occur
 At half-cell of Fe3+/Fe2+ ; electrons are …………. ; …………...
reaction occur
 Since the mixture is ………... by SHE, the value of ……………
Fe3+ / Fe2+
H+ / H2
platinum
0.77
Fe3+ / Fe2+
Fe3+ / Fe2+
H+ / H2
Fe3+ (aq) + e-  Fe2+ (aq)
H2 (g)  2 H+ (aq) + 2 e-
H2 (g) + 2 Fe3+(aq)  2 Fe2+ (aq) + 2 H+ (aq
Pt (s) I H2 (g) , H+(aq) II Fe3+(aq), Fe2+(aq) I Pt (s)
donated oxidation
received reduction
reduced + 0.77 V

The calomel electrode
 Platinum electrode is known as the …………… reference electrode.
However, it is relatively difficult to set up and operate under
standard condition. It is more easier and safer to use a calomel
electrode as a ……………… electrode. [calomel = ……………………..].
Diagram of a typical calomel electrode
primary
secondary Mercury base alloy

2.4 Factors Affecting Electrode Potential
 By convention, the half equation is written with ………….… as the
forward reaction.
 The magnitude of the electrode potential depends on the position
of the above equilibrium
 When value is positive ; a ……………… reaction is favoured
 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
 When a metal is highly …………………….., the metal atoms have a
greater tendency to become positive ions, leaving the ……………
behind on the metal electrode.The electrode potential therefore
become more ………………. and the position of equilibrium shift
more to ……… (……….……… is favoured)
reduction
forward
backward
electropositive
electron
negative
left oxidation

Metal Half equation E (V)
Silver + 0.80
Lead – 0.13
Zinc – 0.76
Magnesium – 2.38
2. Concentration of metal
• If the concentration of the hydrated metal ions is increased in the
equilibrium, the position of equilibrium will shift to the …………,
favouring …………. ; electrode potential become more ……………
Pb2+ (aq) + 2 e- Pb (s) E = – 0.13 V [ Conc = 1.0 M ]
• If concentration Pb2+ changed to 0.001 M ; equilibrium shift to
………………. ; E – 0.13 V
• If concentration Pb2+ changed to 10.0 M ; equilibrium shift to
………………. ; E – 0.13 V
Ag+ + e- ↔ Ag
Pb2+ + 2 e- ↔ Pb
Zn2+ + 2 e- ↔ Zn
Mg2+ + 2 e- ↔ Mg
right
forward positive
backward <
forward >

3 Temperature
 Most of the reduction processes are exothermic process.
Increasing the temperature will cause the equilibrium to
shift to the position of ……………….. process ; which is to
the ……Thus, the electrode potential becomes more
…………………………….
4. Pressure for gaseous species
 From what we’ve learned from chemical equilibria, when
pressure increased, equilibrium will shift to the position
with …….. mole of gas ; while decreasing pressure will
cause equilibrium to shift to position with ……….mole of
gas.
 Eg : Cl2 (g) + 2 e- 2 Cl- (aq) E = + 1.36V
 Increasing pressure will cause equilibrium shift to
…………….. ; E + 1.36V
 Decreasing pressure will cause equilibrium shift to
…………….. ; E + 1.36V
endothermic
left
negative / less positive
less
more
right side
left side
>
<

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 ………… processes. Due to this reason, sometimes it is
also known as ……………………………………
 The positive / negative sign shows how substances favour to
each of the reaction.
 If the EO is positive, substances favour a ……..… reaction.
In another words, it serve well as …………. agent. The
more positive the value ; the stronger the ………….. agent
 If the EO is negative, substances favour ………..… reaction.
In another words, it serve well as …………… agent. The
more negative the value ; the stronger the ………… agent
 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
 ½ Cl2 (g) + e-  Cl- (aq) Eo = ………. V
reduction
Standard reduction potential (SRP)
forward
oxidising
oxidising
backward
reducing
reducing
+ 1.36

 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 half-
cell, which are metal / metal ion half-cell ; non-metal / ion half-
cell ; ion / ion half-cell
 Imagine if we replace the hydrogen half-cell with other half-
cell, 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
zinc half-cell
 A porous pot is used to……………………
………………………………………………
 Substance which has a higher position in
ECS (more negative the value of Eo) is the
……… of the cell whereas the substance
which has a lower position in ECS (more
positive the value of Eo) is the ……………
of the cell.
 The half equation occur at
Anode :
Cathode :
Overall :
The e.m.f. of cell is …………. V
Cell notation is written as
C
u
Z
n
V
complete the cell
And to separate between the 2 electrolytes
anode
cathode
Zn  Zn2+ + 2 e- Eo =+0.76 V
Cu2+ + 2 e-  Cu Eo = +0.34 V
Zn + Cu2+  Zn2+ + Cu Ecell =+1.10 V
+ 1.10
Zn (s) I Zn2+ (aq) II Cu2+ (aq) I Cu (s)
Cu2+ Zn2+

A U tube cell is built using iron (II) ion, Fe2+
and bromine water, Br2
H2SO4 is used to ………………………………….…
…..............…………………………………………….
The substance which has more negative / less
positive the value of Eo) is the ………… of the
cell whereas the substance which has a more
positive / less negative value of Eo is the
…………... of the cell.
The half equation occur at
Anode :
Cathode :
Overall :
The e.m.f. of cell is …………..V
Cell notation is written as
G
complete the cell
And to separate between the 2 electrolytes
anode
cathode
Fe2+  Fe3+ + e- Eo = - 0.77 V
Br2 + 2 e-  2 Br - Eo = +1.07 V
Br2 + 2 Fe2+  2 Fe3+ + 2 Br-
+ 0.30
Pt (s) I Fe2+ (aq) , Fe3+ (aq) II Br2 (l) , Br- (aq) I Pt (s)
Pt
Br2/Br-
Fe3+/Fe2+

Note the following in a chemical cell :
 The e.m.f. of a cell is always …………… In another words,
we must always subtract the ..……………… standard
electrode potential with a ………………. standard electrode
potential.
 Electrons flow from ………… to …………… in the external
circuit
 …………… reaction occur at anode while ………… reaction
occur at cathode of the cell.
POSITIVE
more positive
less positive
anode cathode
Oxidation reduction

 SRP for both cell : Fe2+ + 2e-  Fe E0 = - 0.44 V
Mg2+ + 2e-  Mg 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
Fe / Fe2+ half cell Mg / Mg2+ half cell

G
MnO4
- / Mn2+ half cell Ti3+ / Ti2+ half cell
 SRP : 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 ...……………… reaction
 Compare the following reaction to distinguish between a
spontaneous reaction and not spontaneous reaction.
Immerse a zinc plate into HCl 1.0 M Immerse a copper plate into HCl 1.0 M
Observation Observation
spontaneous
-Bubbling is observed
-Zinc plate is corroded by HCl
- 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
 It can also be used to deduce the strength as an oxidising
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
-Since Zn react with H+, so the 2 half
equation can be written
Zn  Zn2+ + 2e- E0 = + 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+ + 2e- E0 = - 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
Stronger
oxidisingagent

Observation Half equation & overall equation
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.
2Br-  Br2 + 2e- Eo = - 1.07 V
Cl2 + 2e-  2Cl- Eo = + 1.36 V
Cl2 + 2Br-  Br2 + 2Cl–
Ecell = + 0.29 V
Brown solution in CCl4
turned purple when
shaken with KI
2I-  I2 + 2e- Eo = - 0.54 V
Br2 + 2e-  2Br- Eo = + 1.07 V
Br2 + 2I-  I2 + 2Br–
Ecell = + 0.53 V
No changes occur. Purple
solution remain after
shaken with KCl
2Cl-  CI2 + 2e- Eo = - 1.36 V
I2 + 2e-  2I- Eo = + 0.54 V
I2 + 2 Cl-  Cl2 + 2 I–
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)
c) Chlorine gas is bubbled into acidified
potassium dichromate
d) potassium iodide is added to acidified
potassium manganate (VII) solution
Reactant : Fe and Zn2+
Suitable half equation
Fe  Fe2+ + 2e- Eo = + 0.44 V
Zn2+ + 2e-  Zn Eo = - 0.76 V
Fe + Zn2+  Fe2+ + Zn Ecell = - 0.32V
Since Ecell is negative, reaction is not
spontaneous (cannot react)
Reactant : Cu and NO3
-
Cu  Cu2+ + 2e- Eo = - 0.34 V
NO3
– + 2H+ +e– NO2 + H2O E0= +0.81 V
Cu + 2 NO3
– + 4H+  2NO2 +2H2O + Cu2+
Ecell = + 0.47 V
Since Ecell is positive, reaction is
spontaneous (can react)
Reactant : Cl2 and Cr2O7
2–
Cl2 + 2H2O  2HOCl + 2H+ + 2e–
Eo = – 1.64 V
Cr2O7
2- + 6e- + 14H+ 2Cr3+ + 7H2O
Eo = + 1.33 V
Cr2O7
2- + 6HOCl + 2H+ 3Cl2 + H2O
Ecell = – 0.31V
Since Ecell is negative, reaction is non
spontaneous (cannot react)
Reactant : I- and MnO4
-
2I-  I2 + 2 e- Eo = - 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

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
2H2O + 2e-  H2 + 2OH- Eo = - 0.83 V
Ca  Ca2+ + 2e- Eo = + 2.87 V
Ca + 2H2O  H2 + Ca2+ + 2 OH-
Ecell = + 2.04 V
Reactant : Fe2+ and Cr2O7
2-
Cr2O7
2- + 6e- + 14H+ 2Cr3+ + 7H2O
Eo = + 1.33 V
Fe2+  Fe3+ + e- Eo = - 0.77 V
Cr2O7
2- + 6Fe2+ + 14H+
2Cr3+ + 7H2O + 6Fe3+
Ecell = + 0.56 V

 Among the oxidation states available in d-orbital, +2 and +3
oxidation states are the most common states available in the d-
block 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
Eo (V) Stable ion
Cr3+ + e-  Cr2+ – 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
 …………….. 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 iron (II) ion in the absence
of air (oxygen)
Action of acids on iron (II) ion in the
presence of air
Fe2+  Fe3+ + e- E0 = - 0.77 V
2 H+ + 2 e-  H2 E0 = + 0.00 V
2 Fe2+ + 2H+  2 Fe3+ + H2
Ecell = - 0.77 V
Fe2+  Fe3+ + e- E0 = - 0.77 V
O2 + 4H+ + 4e–  2H2O E0 = +1.23 V
4Fe2+ + O2 + 4H+  4Fe3+ + 2H2O
Ecell = + 0.46 V
Ti2+ V2+ Cr2+

Action of acids on cobalt in the absence of air Action of acids on cobalt in the presence of air
Co2+  Co3+ + e- E0 = - 1.82 V
2 H+ + 2 e-  H2 E0 = + 0.00 V
2 Co2+ + 2H+  2 Co3+ + H2
Ecell = - 1.82 V
Co2+  Co3+ + e- E0 = - 1.82 V
O2 + 4H+ + 4e–  2H2O E0 = +1.23 V
4Co2+ + O2 + 4H+  4Co3+ + 2H2O
Ecell = - 0.59 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- ↔ [Co(H2O)6]2+ E0 = + 1.82 V
 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+ + e- E0 = - 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
standard Eo. [At ……..oC ; ……… atm ; ……. M]
 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.
25 1.00 1.00
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE 
R = 8.31 J mol-1 K-1
T = 250C = 298 K
F (Faraday constant)
F = 96500 C mol-1
y
x
0
]ionsproduct[
]ionsttanreac[
lg
)96500(n
)303.2)(298)(31.8(
EE 
y
x
0
]ionsproduct[
]ionsttanreac[
lg
n
059.0
EE 

Ag / Ag+ (1.5 mol dm-3) Cu / Cu2+ (2.0 mol dm-3)
Cl2 / Cl- (0.50 mol dm-3) Fe3+ (0.800 mol dm-3) / Fe2+ (1.30 mol
dm-3)
1
]Ag[
lg
n
059.0
EE
1
0


Ag+ + e-  Ag E0 = + 0.80 V
1
)5.1(
lg
1
059.0
80.0E
1

E = + 0.81 V
1
]Cu[
lg
n
059.0
EE
12
0


Cu2+ + 2e-  Cu E0 = + 0.34 V
1
)0.2(
lg
2
059.0
34.0E
1

E = + 0.35 V
2
1
20
]Cl[
]Cl[
lg
n
059.0
EE 

Cl2 + 2e-  2 Cl- E0 = + 1.36 V
2
1
)50.0(
)1(
lg
2
059.0
36.1E 
E = + 1.38 V
12
13
0
]Fe[
]Fe[
lg
n
059.0
EE 


Fe3+ + e-  Fe2+ E0 = + 0.77 V
)30.1(
)800.0(
lg
1
059.0
77.0E
1

E = + 0.758 V

Ti3+ (1.20 mol dm-3) / Ti2+ (0.700 mol
dm-3)
MnO4
- (1.10 mol dm-3) ; H+ (0.800 mol
dm-3) / Mn2+ (17.0 mol dm-3)
12
13
0
]Ti[
]Ti[
lg
n
059.0
EE 


Ti3+ + e-  Ti2+ E0 = - 0.37 V
)700.0(
)20.1(
lg
1
059.0
37.0E 
E = - 0.356 V
]Mn[
]H][MnO[
lg
n
059.0
EE 2
8
40



MnO4
- + 8 H+ + 5e-  4 H2O + Mn2+
E0 = + 1.52 V
)0.17(
)800.0)(10.1(
lg
5
059.0
52.1E
8

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 :
 *For pure solids and liquids, it will not appear in the
equation
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE 
sr
qp
cell
]D[]C[
]B[]A[
lg
n
059.0
EE 

a) Cr (s) Cr3+ (0.010 mol dm-3) Ni2+ (0.20 mol dm-3) Ni (s)
Oxidation : Cr  Cr3+ + 3 e– E0 = + 0.74 V
Reduction : Ni2+ + 2 e–  Ni E0 = – 0.25 V
Overall : 2 Cr + 3 Ni2+  3 Ni + 2 Cr3+ Ecell = + 0.49 V
E = + 0.51 V
23
32
cell
]Cr[
]Ni[
lg
6
059.0
EE 


2
3
)010.0(
)20.0(
lg
6
059.0
49.0E 

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– E0 = + 2.38 V
Reduction : Fe3+ + e-  Fe2+ E0 = + 0.77 V
Overall : Mg + 2 Fe3+  Mg2+ + 2 Fe2+ Ecell = + 3.15 V
E = + 3.18 V
222
23
cell
]Fe][Mg[
]Fe[
lg
2
059.0
EE 


2
2
)750.0)(500.0(
)80.1(
lg
2
059.0
15.3E 

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– E0 = – 0.15 V
Reduction : Mn3+ + e-  Mn2+ E0 = + 1.49 V
Overall : Sn2+ + 2 Mn3+  Sn4+ + 2 Mn2+ Ecell = + 1.34 V
E = + 1.37 V
224
232
]][[
]][[
lg
2
059.0



MnSn
MnSn
EE cell
2
2
)250.0)(500.0(
)20.1)(300.0(
lg
2
059.0
34.1 E

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
 As time past, the concentration of [Cu2+] ……………… while
[Zn2+] ……………..
Using standard reduction potential of copper and zinc
 Cu2+ (aq) + 2 e-  Cu (s) Eo = + 0.34 V ; when [Cu2+]
……………… ; equilibrium shift to ……… Eo ……………
 Zn2+ (aq) + 2 e-  Zn (s) Eo = – 0.76 V ; when [Zn2+]
……………… ; equilibrium shift to ……… Eo …………...
]Cu[
]Zn[
Kc 2
2



decrease
increase
decrease left decrease
increase right increase

V
+ 0.34–
0
Time / s
– 0.76–
Half cell of Cu
Half cell of Zn

 Applied to Nernst equation where
 When system achieved equilibrium at room temperature,
Nernst equation is simplified to :
 Since and at equilibrium, Ecell = 0, so
y
x
0
]ionsproduct[
]ionsttanreac[
ln
nF
RT
EE 
x
y
0
]ionreactat[
]ionproduct[
lg
n
059.0
EE 
]Cu[
]Zn[
Kc 2
2



C
0
Klg
n
059.0
EV0 

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.
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
Cu  2 e- + Cu2+ E0 = - 0.34 V
2 Fe3+ + Cu  Cu2+ + 2 Fe2+
Ecell = + 0.43 V
23
222
C
]Fe[
]Cu[]Fe[
K 


C
0
Klg
2
059.0
EV0 
CKlg
2
059.0
43.0V0 
lg KC = 14.57
KC = 3.77 x 1014 mol dm-3
Fe3+ + e-  Fe2+ E0 = + 0.77 V
Ag  e- + Ag+ E0 = – 0.80 V
Fe3+ + Ag  Fe2+ + Ag+
Ecell = – 0.03 V
]Fe[
]Fe][Ag[
K 3
2
C 


CKlg
1
059.0
03.0V0 
CKlg
1
059.0
03.0V0 
lg KC = – 0.5085
KC = 0.310 mol dm-3

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–]
x
y
0
]ionreactat[
]ionproduct[
lg
n
059.0
EE 
1
]Cl][Ag[
lg
n
059.0
EE 0


sp
0
Klg
n
059.0
EE 

 At equilibrium ; E = 0 ; so replace in the equation
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
as Ksp is ………………
spKlg
1
059.0
V58.00 
very small

 Example : Given HgCl2 + 2 e-  Hg + 2 Cl- Eo = + 0.27 V
and Hg2+ + 2 e-  Hg 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 Cl- Ecell = – 0.58 V
sp
0
Klg
n
059.0
EE 
spKlg
2
059.0
V58.00 
lg Ksp = – 19.66
Ksp = 2.18 x 10-20 mol3 dm-9

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.
 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)
Standard hydrogen electrode Zinc half cell
Salt bridge
(made of
saturated
KCl / NaCl)
H2 (g)
1.0 atm
H+ (aq)
[x M]
25oC
M (s)
M2+ (aq)
[1.0 M]

 The overall reaction can be written as
M (s) + 2 H+ (aq)  M2+ (aq) + H2 (g) E0 = + a V
y
x
0
]ionsproduct[
]ionsttanreac[
lg
n
059.0
EE 
12
2
0
]M[
]H[
lg
2
059.0
EE 


]1[
]H[
lg2
2
059.0
EE 0


pH059.0EE 0


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.
 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.
During the discharge of the battery,
the half-cell reactions are
Anode : Li(s) → Li+ + e-
Cathode : Li+ + CoO2 + e- → LiCoO2 (s)
Overall : Li (s) + CoO2 → LiCoO2 (s)
Ecell = + 3.4 V

 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) Eo
cell = + 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 e-
Cathode : 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
 Positive terminal is called as ………… whereas the negative
terminal is called as …………
 At anode, …………. process occur where
as at cathode, ………..…process occur
 Cations are attracted to ………….. while
anions are attracted to ……………
 Electrons flow from the ………… to
……..….. in the external circuit.
A
anode
cathode
oxidation
reduction
cathode
anode
anode
cathode

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
electrode during electrolysis is …………….… to the quantity of
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 = electric current I = current t = time in second
proportional
Q = I t

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.
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.
Example 3 : Calculate the mass of chromium
formed when 1.20 A of current is directed into
molten chromium (III) chloride for 3 hours.
Example 4 : Calculate the time taken to
produce 18.0 g of silver from silver nitrate by a
current of 0.900A
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
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
Since 2 e- = 1 Cu ; mol Cu = 0.0373 mol
Mass Cu= 0.0373 x 63.5
= 2.4 g
Q = It @ Q = (1.20)(3 x 60 x 60)
= 12960 C
Eq : Cr3+ + 3 e-  Cr
mol e- = Q / F @ mol = 12960 / 96500
Since 3 e- = 1 Cr ; mol Cr = 0.0448 mol
Mass Cr = 0.0448 x 52.0
= 2.33 g
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
form 200g of lead from molten lead (II)
bromide by a current of 1.50 A.
Example 6 : Calculate the time required to
form 10 g of aluminium from molten
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
 From the diagram above, 1 F will discharge …. mol of Ag+
ions ; …… mol of Cu2+ ion ; …… mol of Cr3+ ion ; ….… mol of
O2.
Type of
electrode
Half equation
Mol of metal
deposited
Silver
Copper
Carbon in CrCl3
Carbon in
H2SO4 (occur at
anode)
Mol of non-metal
Ag+ (aq) + e-  Ag (s) 1 mol
Cu2+ (aq) + 2 e-  Cu (s) 1/2 mol
Cr3+ (aq) + 3 e-  Cr (s) 1/3 mol
2 H2O  O2 + 4 H+ + 4 e–
1/4 mol
1
1/2 1/3 1/4

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.
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 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
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,
• At anode, …………. reaction occur ; electrons are ………………
• At cathode, …………. reaction occur ; electrons are ………………
 Electrolytes can only discharge under 2 conditions : …..…... state or
………… 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–
oxidation donated
reduction received
molten
aqueous

Electrolyte
Subst.
present
Half equation at anode
Substance at
anode
Half equation at
cathode
Substance
at cathode
PbBr2 (l)
Al2O3 (l)
NaCl (aq)
Pb2+
Br –
Br2 + 2e-  2Br-
E0 = + 1.07
V
Rev:
2Br-  2e- + Br2
bromine
Pb2+ + 2e-  Pb
E0 = – 0.13 V
lead
Al3+
O2–
O2 + 4e-  2O2-
E0 = ? V
Rev:
2O2-  O2 + 4e-
oxygen
Al3+ + 3e-  Al
E0 = – 1.67 V
Aluminu
m
Na+
Cl–
Cl2 + 2e-  2Cl-
E0 = +1.36
V
Rev:
2Cl-  Cl2 + 2e-
chlorine
Na+ + e-  Na
E0 = – 2.71 V Sodium

 No matter it is an electrolytic cell or chemical cell,
• At anode, …………. reaction occur ; electrons are ………………
• At cathode, …………. reaction occur ; electrons are ………………
• 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 ……….. value will be selected for
discharge at cathode, while a more …………. value will be selected
for discharge at anode.
 At anode : E0 = …………. V
 At cathode : 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)
 At anode : E0 = ….……. V
 At cathode : E0 = ….……. V
oxidation donated
reduction received
positive
negative
4 H+ + O2 + 4 e–  2 H2O + 1.23
2 H2O + 2 e–  2 OH– + H2
– 0.83
4 H+ + O2 + 4 e–  2 H2O + 0.81
2 H2O + 2 e–  2 OH– + H2
– 0.41

Electrolyte
Subst.
present
Half equation at anode
Substance
at anode
Half equation at cathode
Substance
at cathode
NaCl (aq)
Na+
Cl–
H2O
Cl2 + 2e-  2Cl-
E0 = +1.36
V
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23
V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
oxygen
Na+ + e-  Na
E0 = – 2.71 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
Hydroge
n

Electrolyte
Ions
present
Half equation at anode Substance
Half equation at
cathode
Substance
CuSO4 (aq)
PbI2 (aq)
KOH (aq)
Cu2+
SO4
2–
H2O
S2O8
2- +2e- 
2SO4
2-
E0 = +2.01 V
4 H+ + O2 + 4 e– 
2 H2O
E0 = +1.23 V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
oxygen
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 I-
E0 = +0.54 V
4 H+ + O2 + 4 e– 
2 H2O
E0 = +1.23 V
Rev: 2 I–  I2 + 2
e–
iodine
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– 
2 H2O
E0 = +1.23 V
Rev: 4 OH– 
2 H2O + O2 + 4
oxygen
K+ + e-  K
E0 = – 2.92 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
hydrogen

 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
same concentration, ……… ion is preferred to be
discharge at cathode as it has a lower position in
electrochemical series.
• However, if the concentration of Pb2+ ion concentration is
raised much higher than Cu2+ ion, ………. ion is more
readily to be discharged.
• Another example is potassium chloride in aqueous
solution. Under dilute solution of KCl, ……….. will be
selected at anode and ………...… gas is given out, as it
has a lower position in electrochemical series.
Cu2+
Pb2+
H2O
oxygen

• However, if concentrated KCl is used as electrolyte, the
concentrated of Cl- increase, and …… is selected to be
discharge and ……..….… gas is given out.
• Still, if the position in electrochemical series differ too
much, like K+ and water in the example above, K+ ion
…………… be discharge as the position is much too high.
At the end, ……………. gas is given out at cathode.
• If chloride ion is replaced with fluoride ion, F– ion, and
concentration of F– ion is increased, …....... is still be
preferred as F– ion has a ……………. position in
electrochemical series
 In the table below, predict the element that is expected to
form when electrolyse.
Cl–
chlorine
will not
hydrogen
water
very high

Electrolyte
Ions
present
Half equation at
anode
Substance
Half equation at
cathode
Substance
Concentrate
NaCl
Concentrated
PbBr2
Concentrated
LiF
Na+
Cl–
H2O
Cl2 + 2e-  2Cl-
E0 = +1.36 V
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23 V
Rev: 2 Cl- 
Cl2 + 2 e–
chlorine
Na+ + e-  Na
E0 = – 2.71 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
Hydroge
n
Pb2+
Br –
H2O
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23
V
Br2 + 2e-  2Br-
E0 = + 1.07
V
Rev:
2Br-  2e- + Br2
bromine
Pb2+ + 2e-  Pb
E0 = – 0.13 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
lead
Li+
F–
H2O
F2 + 2e-  2F-
E0 = +2.87 V
4 H+ + O2 + 4 e–
 2 H2O
E0 = +1.23 V
Rev: 2 H2O 
4 H+ + O2 + 4 e–
oxygen
Li+ + e-  Li
E0 = – 3.04 V
2 H2O + 2 e– 
2 OH– + H2
E0 = – 0.83 V
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
4 H+ + O2 (g) + 4 e- → 2 H2O (l)
E0 = + 0.81 V
(conc. H+ = 1 x 10-7 mol dm-3)
Na+ (aq) + e- → Na (s) E0 = - 2.71 V
2 H2O (l) + 2 e- → 2 OH- + H2 (g)
E0 = - 0.41 V
(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.
 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.
 3rd step: Aluminium hydroxide is filtered, washed, dried and
finally heated out to 12000C to produce pure aluminium oxide
(alumina), Al2O3
Al2O3 + 2 NaOH + 3 H2O  2 NaAl(OH)4
SiO2 + 2 NaOH  Na2SiO3 + H2O
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 + CO3
2-
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.
Half equation occur at cathode Half equation occur at anode
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
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.
 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.
Half equation occur at cathode Half equation occur at anode
Na+ + e-  Na E0 = – 2.71
V
2Cl-  Cl2 + 2e- E0 = – 1.36 V

 Similar to the mercury-cathode cell, the electrolytes used in
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 :
brine (sodium chloride)
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 :
 Standard reduction potential of water :
 Since ………… has a higher E0 value, ………… is discharge
and ………………… is produced.
 The level of brine (left or anode position) is always placed
higher than the water (right or cathode position) to
………………………..…………………………………………......
Na+ + e-  Na E0 = – 2.71 V
2 H2O + 2 e–  2 OH– + H2 E0 = – 0.83
V water water
hydrogen gas
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
the conversion of ………...… to ……………..
 2 main important components for rusting are
………………….. and ……………………
 A possible mechanism for rusting, consistent with the known
facts, is illustrated in Figure below
metal metal oxide
oxygen water

At anode :
Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V
At cathode :
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 –
Eo = +0.56 V
O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)
Eo = + 0.40 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 alkaline / neutral condition

 In acidic condition
At anode :
Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V
At cathode :
O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)
Eo = + 1.23 V
2 Fe (s) + O2 (aq) + 4 H+  2 Fe2+ + 2 H2O Ecell = + 1.67 V
At anode :
Fe2+ (aq)  Fe3+ (aq) + e-
Eo = -0.77 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
Alloying
• Iron is alloyed with nickel and chromium to form …………………….
The chromium forms an impervious oxide layer on the surface of
iron increasing its resistance to corrosion. Chromium at the same
time ……………….. ………the steel
Barrier
 Painting the iron object
 Use grease / oil to coat the moving parts of machine
 Coating ironwith chromium (plating) or zinc (galvanising)
Sacrificial
 Also known as …………… protection
 Metal with a ………………. position in electrochemical series is
‘connected’ to iron. Under such way, ………………... Will be
oxidised first before iron.
Stainless steel
decorated / coated
anodic
higher
reactive metal

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+ + Cr2O7
2- + 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
 When [MnO4
-] = [Mn2+] = 1.00 mol dm-3- ; Ecell =
]Mn[
]H][MnO[
lg
5
059.0
52.1E 2
8
4



]H[lg8
5
059.0
52.1E 
 pH0944.052.1E 

[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 e- Eo = - 0.54 V
• Under pH = 1 , Ecell of manganate (VII) ion is ................ V …………… are
able to ………… by manganate (VII) ion as it is still a strong ……………
agent.
• Under pH = 2, Ecell of manganate (VII) ion is .............. V. Only ……………
are able to ………… by manganate (VII) ion. …… cannot oxidise as the
reaction is not ……………… (Ecell = …………)
• Under pH = 5, Ecell of manganate (VII) ion is .............. V. Only …… are
able to ………… by manganate (VII) ion. ………. cannot oxidise as the
reaction is not ……………… (Ecell = ……………………)
+ 1.43 Cl- , Br- , I-
oxidise oxidising
+ 1.33 Br- , I-
oxidise Cl-
spontaneous - 0.03 V
+ 1.05 I-
oxidise Cl- , Br-
spontaneous -0.03 V for Br-
-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+ + 2 e –
Manganese (IV) oxide
2 MnO2 + 2 NH4
+ 2 e- 
Mn2O3 + 2 NH3 + H2O
Ammonium chloride and
zinc chloride paste
zinc
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
Cathode
Equation
occur at
cathode
Electrolytes
Zinc
Zn + 2OH– 
H2O + ZnO + 2 e –
Mercury (II) oxide
HgO + H2O + 2 e- 
Hg + 2 OH–
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
Diagram
Anode Lithium metal
Equation occur
at anode
Li (s)  Li+ (aq) + e-
Cathode Manganese (IV) oxide
Equation occur
at cathode MnO2 (s) + Li+ + e-  LiMnO2
Electrolytes Lithium chlorite (VII), LiClO4
Hydrogen
2 H2 + 4 OH–  2 H2O + 4 e–
Oxygen
O2+ 2 H2O + 4 e–  4 OH–
Hot potassium hydroxide (aq)

13.9 Dental Filling
 The material commonly used to fill decaying teeth is an ………………….
(a …………………… base alloy). The component in dental filling of
amalgam are …………………. , ……………… and …………..
 The standard electrode potential of these electrode system are :
 Hg2
2+ (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, which result a electrochemical cell. ……………………
act as the anode of the cell, while ………… act as the cathode and
……….. act as the electrolyte.
 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
………………….. than any of the electrode above, which makes
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.
amalgam
mercury
mercury silver tin
dental amalgam
gold
saliva
positive
electropositive
anode amalgam

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