C10 ELECTROCHEMISTRY

Spontaneous & Non-Spontaneous reactions

 Standard cell potential can be used to predict
the spontaneity of a chemical reaction.

Redox reaction E cell or Eocell
spontaneous positive

non spontaneous negative
at equilibrium 0

51

Example 1:

Predict the spontaneity of following:
Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

Answer:

Reduction

Pb2+(aq) + 2Cl-(aq) → Pb(l) + Cl2(g)

Oxidation

Cathode: Pb2+(aq) + 2e → Pb(s) Eored = -0.13 V
Anode :
Overall : 2Cl-(aq) → Cl2(g) + 2e Eoox = -1.36 V

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

Eocell = Eored + Eoox Non-spontaneous reaction

= (-1.36) + (-0.13) No Reaction 52
= -1.49 V

Example 2:
Predict whether the following reaction occurs spontaneously :

2Ag(s) + Br2(l) 2Ag+(aq) + 2Br-(aq)

EAog+ /Ag = +0.8 V standard reduction
potential
EBor2 - = +1.07 V
/Br

Answer :

2Ag(s) 2Ag+(aq) + 2e Eoox = - 0.80 V

Br2(l) + 2e 2Br - Eored = +1.07 V
(aq)

2Ag(s) + Br2(l) 2Ag+(aq) + 2Br-(aq) Eocell = + 0.27 V

The reaction is spontaneous 53

Exercise

A cell consists of silver and tin in a solution of 1 M silver ions
and tin(II) ions. Determine the spontaneity of the reaction and
calculate the cell voltage of this reaction.

Answer:

Ag+(aq) + e- → Ag(s) E0 = +0.80 V (cathode)
Sn2+(aq) + 2e- → Sn(s) E0 = -0.14 V (anode)

Eocell = Eocathode - Eoanode

= +0.80 – (-0.14)
= +0.94 V

E0cell is positive value, therefore the reaction is spontaneous.

54

10.2
NERNST EQUATION

55

Learning outcomes:

At the end of this topic, students should be able to:
(a) Write Nernst equation for a given galvanic cell.
(b) Explain the factors that affect cell potential.

(c) Use Nernst equation, Ecell = E°cell – RT ln Q to determine:

(i) cell potential (emf) nF

(ii) concentration or partial pressure of a species

(iii) equilibrium constant, K when Ecell = 0

(d) Predict the spontaneity of a cell reaction.

56

Nernst equation

• Nernst equation below is used to calculate the Ecell
for any chosen concentration and temperature:

Ecell = Eocell – RT ln [ product ]x
nF [ reactant]y

57

• If T=298 K , R= 8.314 J K-1 mol-1 and 1 F = 96500C

Ecell = Eocell – 0.0257 [ product ]x
2.303 log

n [ reactant]y

Ecell = Eocell – 0.0592 [ product ]x
n log

[ reactant]y

Ecell = Eocell – 0.0592 log Q
n

Where 58
n = no of e- involved

[product]x
Q = reaction quotient = [reac tan t]y

Factor That Affect Cell Potential

= ° - lnQ


 The Nernst equation says that a cell
potential under any condition depends on
the potential at standard-state
concentrations and a term for the potential
at nonstandardard-state concentration.

59

From the equation,
 When Q < 1 and thus [reactant] > [product],

In Q < 0, so > °

 When Q = 1 and thus [reactant] = [product],
In Q = 0, so = °

 When Q > 1 and thus [reactant] < [product],
In Q > 0, so < °

60

Example 1

(a) Calculate the Ecell for the following cell at
25oC:

Zn(s) / Zn2+ (aq, 0.02M) // Cu2+(aq, 0.40 M) / Cu(s)

[Ans:+1.14 V]

(b) What could be done to obtain a higher Ecell
value than that of (a)

61

Answer: Zn2+(aq) + Cu(s)

Zn(s) + Cu2+(aq)

Eocell = Eored + Eoox @ Eocell = Eocathode – Eoanode

= +0.34 V + 0.76 V = +0.34 V - (- 0.76 V)

= +1.10 V = +1.10 V

Ecell = Eo cell – 0.0592 log [ Zn2+] 62
n [ Cu2+]

Ecell = +1.10 V – 0.0592 log (0.02)
2 (0.40)

= +1.10 V – (-0.0385)

= +1.14 V

b) = ° - 0.0592 log [ Zn2+]
[ Cu2+]

 Increase the concentration of Cu2+
 Decrease the concnetration of Zn2+

63

Example 2: Eo = 0.25 V
Eo = 0.74 V
Given two half-equations :

Ni2+(aq) + 2e  Ni(s)
Cr3+(aq) + 3e  Cr(s)

Calculate the cell potential at 25oC if :
[Ni2+] = 1.0 x 104 M
[Cr3+] = 2.0 x 103 M

[Ans : +0.43 V ]

64

Answer:

2Cr(aq) + 3Ni+(aq) → 2Cr3+(aq) + 3Ni(s)

E°cell = E°cathode - E°anode
= -0.25V – (-0.74V)
= +0.49V

Ecell = Eo cell – 0.0592 log [Cr3+]2
n [Ni+]3

= 0.49 – 0.0592 log (2.0x10-3)2
6 (1.0 x 10-4)3

= +0.43V 65

Example 3

An electrochemical cell is represented as,
Pt(s) | H2 (g,1 atm) | H+ (aq, x M) || Ag+ (aq, 1.0 M) | Ag(s)

if the value of the cell potential is 1.198V, determine
the concentration of the H+ solution at 25oC.

[Ans:1.884 x 10-7 M]

66

Answer:

H2(g) + 2Ag+(aq) → 2H+(aq) + 2Ag(s)

E°cell = E°red + E°ox
= +0.80 + 0.00
= +0.80V

Ecell = E°cell – 0.0592 log [H+]2
n [Ag+]2(PH2 )

1.198V = 0.80 V – 0.0592 log [H+]2 67
2 1.0)2(1)

[H+] = 1.884 x 10-7 M

Example 4 :

Predict whether the following reaction would proceed
spontaneously at 25oC. The cell notation is:
Cu(s)|Cu2+(aq,0.5M) ||Fe3+(aq,0.2M) ,Fe2+(aq,0.1M) |Pt(s)

68

Answer: 2(Fe3+(aq) + e  Fe2+(aq))
Cathode: Cu(s)  Cu2+(aq) + 2e
Anode: Cu(s) + 2Fe3+(aq)  2Fe2+(aq) + Cu2+(aq)
Overall:

Eocell = E°red + E°ox
= +0.77+(0.34)
= +0.43 V

Ecell = Eocell  0.0592 log [Cu2+][Fe2+]2
n [Fe3+]2
= +0.43  0.0592 log (0.5)(0.1)2
2 (0.2)2
= +0.46 V (the reaction proceed spontaneously)

69

At equilibrium:

~ No net reaction occur (Q=K)

~ Ecell = 0

Ecell = Eocell – 0.0592 log K
n

0 = Eocell – 0.0592 log K
n

Eocell = 0.0592 log K

n 70

Example 5

Calculate the equilibrium constant (K) for the following

reaction at 25oC.

Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

Answer

At equilibrium, E cell = 0

Eo cell = Eo cathode - Eo anode
= +0.80 – ( +0.34)
= +0.46 V

71

E cell = Eo cell – 0.0592 log K
2

0 = 0.46 – 0.0592 log K
2

0.0592 log K = 0.46
2
log K = 15.54
K = 3.5 x 1015

72

10.3
ELECTROLYTIC CELL

73

Learning outcomes :

At the end of this topic, students should be able to:
(a) Draw an electrolytic cell.
(b) Describe the operation of an electrolytic cell.
(c) Explain the influence of the following factors on the

selective discharge of a species at the electrode :
(i) standard reduction/electrode potential of the species
(ii) concentration of the species
(iii) nature of electrodes

74

(d) Explain the electrolysis of the following electrolytes using
inert electrodes:
(i) molten salt
(ii) water
(iii)concentrated and dilute aqueous NaCl
(iv)aqueous Na2SO4

(e) Predict the products of electrolysis using appropriate
examples.

(f) Define Faraday’s first law of electrolysis.
(g) Apply Faraday’s first law in calculation.

75

Predicting the products of electrolysis

Factors influencing the products

Reduction/oxidation potential of the
species in electrolyte
Concentrations of ions

Types of electrodes used – active or
inert

76

Electrolysis of molten salt

 Electrolysis of molten salt requires high temperature.
 Electrolysis of molten NaCl

Cation : Na+ Anion : Cl-

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

2Na+ (l) + 2e- → 2Na (l)

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

77

• Electrolysis of molten NaCl gives sodium metal
deposited at cathode and chlorine gas evolved at
Anode.

• Electrolysis of molten NaCl is industrially important.
The industrial cell is called ‘Downs Cell’

78

Exercise
Determine the products formed at cathode
and anode of electrolysis of molten PbI2.

Answer:

Cation : Pb2+ Anion : I-

Anode : 2I-(l) → I2(g) + 2e-
Cathode : Pb2+(l) + 2e- → Pb(l)

Overall reaction : 2I-(l) + Pb2+(l) → I2(g) + Pb(l)

79

ELECTROLYSIS OF WATER

A glass electrolytic
cell with separated
gas compartments
is used to keep the
H2 and O2 gases
from mixing.

80

Anode : 2H2O(l) → O2 (g) + 4H+(aq) + 4e

• Oxidation of water produce oxygen gas at anode

Cathode : 2H2O(l) + 2e → H2(g) + 2OH-(aq)

• Reduction of water produce hydrogen gas at cathode

Overall Reaction : 2H2O(l) → 2H2(g) + O2 (g)

81

Electrolysis of Aqueous Salt

• Electrolysis of aqueous salt is more complex

because the presence of water.
• Aqueous salt solutions contains anion, cation and water.
• Water is an electro-active substance that may be

oxidised or reduced in the process depending on

the condition of electrolysis.

Reduction :

2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V

Oxidation : 4H+ (aq) + O2 (g) + 4e- E0 = -1.23 V
2H2O (l)
82

Electrolysis of
Aqueous Solution

Electrolysis of Electrolysis of Electrolysis of
diluted NaCl concentrated diluted/concentrated
NaCl solution
solution Na2SO4 solution

83

Electrolysis of Aqueous NaCl

• NaCl aqueous solution contains Na+ (cation), Cl- (anion)
and water molecules

• On electrolysis,
 the cathode attracts Na+ ion and H2O molecules
 the anode attracts Cl- ion and H2O molecules

• The electrolysis of aqueous NaCl depends on the
concentration of electrolyte.

84

Electrolysis of diluted NaCl solution

Cathode E0 = -2.71 V
E0 = -0.83 V
Na+ (aq) + e- → Na (s)
2H2O (l) + 2e- → H2 (g) + 2OH- (aq)

 E0 for water molecules is more positive.
 H2O easier to reduce.

Anode E0 = +1.36 V
E0 = +1.23 V
Cl2 (g) + 2e- → 2Cl- (aq)
O2 (g) + 4H+ (aq) + 4e- → 2H2O (l)

 In dilute solution, water will be selected for oxidation

because of its lower Eo. 85

Reactions involved

Cathode: 2H2O (l) + 2e- → H2 (g) + 2OH- (aq)
4H2O (l) + 4e- → 2H2 (g) + 4OH- (aq)

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

Cell 6H2O(l) → O2(g) + 2H2(g) + 4OH-(aq) + 4H+(aq)
reaction:

4 H2O

2H2O(l) → O2(g) + 2H2(g)

86

Electrolysis of Concentrated NaCl solution

Cathode E0 = -2.71 V
E0 = -0.83 V
Na+ (aq) + e- → Na (s)
2H2O (l) + 2e- → H2 (g) + 2OH- (aq)

 E0 for water molecules is more positive
 H2O easier to be reduce

Anode

Cl2 (g) + 2e- → 2Cl- (aq) E0 = +1.36 V

O2 (g) + 4H+ (aq) + 4e- → 2H2O (l) E0 = +1.23 V

 In concentrated solution, chloride ions will be
oxidised because of its high concentration.

87

Reactions involved

Cathode: 2H2O (l) + 2e- → H2 (g) + 2OH- (aq)
Anode : 2Cl- (aq) → Cl2 (g) + 2e-

Cell 2H2O(l) + 2Cl- → Cl2(g) + H2(g) + 2OH-(aq)
reaction:

88

Electrolysis of aqueous Na2SO4 solution

• Na2SO4 aqueous solution contains Na+ ion, SO42- ion
and water molecules

• On electrolysis,
 the cathode attracts Na+ ion and H2O molecules
 the anode attracts SO42- ion and H2O molecules

89

Cathode E0 = -2.71 V
E0 = -0.83 V
Na+ (aq) + e- → Na (s)

2H2O (l) + 2e- → H2 (g) + 2OH- (aq)

 E0 for water molecules is more positive
H2O easier to reduce

Anode

O2 (g) + 4H+ (aq) + 4e- →2H2O (l) E0 = +1.23 V

 E0 for water molecules is less positive
 Sulphur anions cannot be oxidised because the
sulphur atom in the anion is already at the maximum
oxidation state +6.
H2O easier to oxidise
90

Equation

Cathode: 2H2O (l) + 2e- → H2 (g) + 2OH- (aq)

Anode: 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
2H2O(l) → O2(g) + 2H2(g)
Cell
Reaction:

 Cathode = H2 gas is produced and solution become

basic at cathode because OH- ions are formed

 Anode = O2 gas is produced and solution become

acidic at anode because H+ ions are formed

91

The nature of electrodes matters.

 An active metal electrodes is one that can itself
participate in the oxidation or reduction half
reaction.

92

Example :

Cu

? Cu2+

CuSO4

From the figure above, predict the electrode reactions
and the overall reaction when the anode is made of
Copper.

93

ANSWER:

Cathode :Cu2+(aq) + 2e → Cu(s)

Anode :Cu(s) → Cu2+(aq) + 2e

Overall Reaction : Cu2+(aq) + Cu(s) → Cu(s) + Cu2+(aq)

 Copper is transferred from anode to the cathode
through the solution as Cu2+ .

 The concentration of CuSO4 remains unchanged.

94

Faraday’s Law of Electrolysis

Describes the relationship between the amount of
electricity passed through an electrolytic cell and
the amount of substances produced at electrode.

Faraday’s First Law

States that the quantity of substance formed at an
electrode is directly proportional to the quantity of
electric charge supplied.

95

Faraday’s 1st Law

mαQ

Q = electric charge in coulombs (C)
m = mass of substance discharged

96

Q = It

Q = electric charge in coulombs (C)
I = current in amperes (A)
t = time in second (s)

Faraday constant (F) is the charge on 1 mole of electron

1 F = 96 500 C

97

Example

An aqueous solution of CuSO4 is electrolysed using a
current of 0.150 A for 5 hours. Calculate the mass

of copper deposited at the cathode.

Answer

Electric charge, Q = Current (I) x time (t)
Q = (0.150 A) x ( 5 x 60 x 60 )s
Q = 2700 C

1 mole of electron Ξ 1 F Ξ 96 500 C

No. of e- passed through = 2700 = 0.028 mol
96 500
98

Cu2+ (aq) + 2e-  Cu (s)
From equation:

2 mol electrons  1 mol Cu
0.028 mol electrons  0.014 mol Cu
Mr for Cu = 63.5
Mass of Copper deposited = 0.014 x 63.5

= 0.889 g

99