9.

Oxidation and Reduction

 DEFINITIONS OF OXIDATION AND
REDUCTION
OXIDATION—loss of electron(s) by a species; increase in
oxidation number; gain of oxygen; decrease in hydrogen.

REDUCTION—gain of electron(s); decrease in oxidation

number; loss of oxygen; increase in hydrogen.
How to remember the definitions

LEO the lion says GER!

o l x a l e
s e i i e d
e c d n c u
t a t c
r t r t
o i o i
n o n o
s n s n

GER!
Another way to remember

OIL RIG
x
i
s o
s
e
d
s a
i
d s u n
a c
t t
i i
o o
n n
 Examples
Fe → Fe2+ + 2e
Iron is oxidized through loss of electrons

Cu2+ + 2e- → Cu

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

In this reaction the iron is reduced because the
oxidation number is reduced from +3 to 0.

The CO is oxidized because the oxidation number

of carbon has increased from +2 in CO to +4 in
CO2.
 REDOX REACTIONs
Reduction (gaining electrons) always happens
with an oxidation (loss of electrons) to provide
the electrons. These reactions involving transfer
of electrons are termed REDOX reactions. This
term is coined from REDuctionOXidation.

Try: What substance is oxidized in the following:

1. C + O2 → CO2
2. 2Mg + O2 → 2MgO
3. N2 + 3H2 → 2NH3
4. Zn + HCl → ZnCl2 + H2
 Non-Redox Reactions

These are reactions in which there are no changes in

the oxidation on numbers of species.
Neutralization: NaOH + HCl → NaCl + H2O
Precipitation. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
 Review of Oxidation numbers
Review of Rules for assigning oxidation numbers.

1. All elements in the uncombined state have an

oxidation number of zero.

Na, Be, K, Pb, H2, O2 = 0

2. In monatomic ions, the oxidation number is equal to
the charge on the ion.

Li+, O.N. Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In
peroxides, e.g. H2O2 it is –1. It is +2 in OF2.
4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.

5. Group IA metals have O.N. +1, IIA metals have O.N. +2

Aluminium has an oxidation no. of +3

6. The sum of the oxidation numbers of all the atoms in a

molecule or ion is equal to the charge on the
molecule or ion.

7. The oxidation numbers of all the halogens when they

are bonded to metals is equal to -1. The oxidation
number of fluorine is always -1
What are the oxidation numbers of all the
atoms in HCO3- ?
Let x = oxidation number of C
HCO3-
O = -2 from rules

H = +1 from rules
3x(-2) + 1 + x = -1
Where x = O.N. of C
C = +4
 Naming common substances
using oxidation numbers
Name the metal followed by the oxidation state
and then the word ion.

Cu2+ = Copper(II) ions

Cu+ = Copper (I) ions
Fe2+ = Iron (II) ions
Iron (III) ions
MnO4- =Tetraoxomanganate(VII) ion
Cr2O72- = Heptaoxodichhromate(VI) ion
 Name the following compounds
a) HNO3
b) Na2SO3
c) Cu(NO3)2
d) FeSO4
e) Cu2O
f) SO3
g) PbCO3
 OXIDIZING AND REDUCING
AGENTS
OXIDIZING AGENT—is electron acceptor, or
oxygen donor; an oxidizing agent is reduced
in a redox reaction.
Common Oxidizing agents: Acidified KMnO4
acidified K2Cr2O7, conc. HNO3

REDUCING AGENT—is electron donor, or

oxygen acceptor; a reducing agent is oxidized
in a redox.
E.g. NH3, H2S, KI, FeCl2.
 TRY

Identify the oxidizing and reducing agents in the

following reactions
CH4 + 2O2 → CO2 + 2H2O
2Fe + 3Cl2 → 2FeCl3
2HNO3 + 3H2S → 2NO + 3S + 4H2O
2HI + 2HNO2 → I2 + 2NO + 2H2O
CuO(s) + H2(g) → Cu(s) + H2O(l)
4CuO(s) + CH4(g) → 4Cu(s) + 2H2O(l) + CO2(g)
2NH3 + 3CuO→ N2 + 3Cu + 3H2O
Tests for Oxidizing and Reducing
Agents
Test for Oxidizing agent: Add sample to a solution
of KI, a colour change to brown shows the
sample is an oxidizing agent.

Test for Reducing agent: Add sample to acidified

KMnO4 a discharge of the purple colour of KMnO4
shows sample is an oxidizing agent.

Or add acidified potassium dichromate to sample,

the change of colour from orange to green shows
sample is a reducing agent.
 Rules for Balancing Redox
Equations
The Ion-Electron Method

Procedure

•Write the net ionic equation (skeletal equation) for

the reaction.

•Divide the equation into the oxidation and reduction

half-equations

•Balance the atoms on each half equation.

•Balance the charge using electrons

 Balancing Equations in Acidic Medium

1. Balance oxygen by adding one water molecule

for each oxygen atom required to the side
needing oxygen and two hydrogen ions to the
opposite side.
2. Balance hydrogen by adding one H+ ion to the
side that needs hydrogen.
3. Balance charge by adding electrons to the side
with excess positive charge to give the same net
charge on both sides of the equation.
4. Multiply each half equation by integers to
equalize number of electrons.
5. Add the two half-equations and cancel out
common terms appearing on both sides of the
equation.
 Example in acidic Medium
Balance the oxidation of Fe2+ to Fe3+ by Cr2O72- in acid
solution?
1. Write the unbalanced equation for the reaction ion ionic form.
Fe2+ + Cr2O72- Fe3+ + Cr3+

2. Separate the equation into oxidation and reduction half-

reactions. +2 +3
Oxidation: Fe2+ Fe3+
+6 +3
Reduction: Cr2O7 2- Cr3+

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+
 Continuation
4. Add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72- 2Cr3+ + 7H2O

14H+ + Cr2O72- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the
charges.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-
reactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
 Balancing Redox Equations
7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both
sides must cancel. You should also cancel like species.

Oxidation: 6Fe2+ 6Fe3+ + 6e-

Reduction: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

Charge:14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

Balance the following equations in acidic
medium:

a) MnO4- + SO32-→ Mn2+ + SO42-

b) MnO4- + I- → Mn2+ + I2

c) H2S + Cr2O72- → S + Cr3+

d) Cl-(aq) + Cr2O72-(aq) → Cl2(g) + Cr3+ (aq)

e) NO3-(a) + S → NO2(g) + SO42-(aq)

TRY
I- + IO3- →I2
HNO2 → NO + NO2
NO2 + H2O → HNO3 + NO
CH3CH2OH + Cr2O72- → CH3COOH + Cr3+
MnO4- + C2H6O → MnO2 + C2H4O
Mn2+(aq) + BiO3¯(aq) → MnO4¯(aq) + Bi3+
 REDOX TITRATION
These are titrations in which reactions are carried
out between oxidizing and reducing agents.
Common oxidizing agents used in redox titration:
Acidified KMnO4
Acidified K2Cr2O7
Iodine solution, I2.

Common reducing agents in redox titration:

Sodium oxalate, Na2C2O4
Sodium thiosulphate
Ferrous ammonium sulphate
 Common Equations in Redox
Titration

1. 2MnO4- + 10C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O

2. MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

3. Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O

4. I2 + 2S2O32- → 2I- + S4O62-

 Potassium Permanganate Titrations
The balanced redox equations for the reactions of KMnO4 with
some reagents are shown below:

2MnO4- + 5C2O42- +16H+  2Mn2+ + 10CO2 + 8H2O

2MnO4- + 5H2O2 +6H+  2Mn2+ + 5O2 + 8H2O
MnO4- + 5Fe2+ +8H+  Mn2+ + 5Fe3+ + 4H2O

The MnO4- ion acts as the oxidising agent and accepts electrons
from the other reactant which acts the reducing agent. The MnO4-
ion is purple while the Mn2+ ion is colourless. When the
permanganate is in the conical flask, the end point is indicated by
a change of the purple colour to colourless. However, when the
permanganate is in the burette, the end point is marked by the
appearance of the first trace of permanent pink colour.
 More about permanganate
titrations
In all MnO4- titrations, the MnO4- ion acts as its a self
indicator.
The Mn2+ acts as a catalyst for the reaction.
Titration is carried out in acidic medium to prevent the
formation of MnO2 which will form a cloudy precipitate
making the detection of the end point difficult.
H2SO4 is the most suitable acid used as it does not
interfere with the reactions of permanaganate.
When titrating MnO4- with C2O42-, the titration should be
carried out at between 60-70C to speed up the
reaction as the oxalate ion is organic.
 Example
The mass percent of iron in a soluble iron(II) compound is
measured using a titration based on the balanced equation below.

5Fe2+(aq) + MnO4–(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

12g of an impure sample of the iron(II)sulphate was dissolved in

dilute sulphuric acid to form 1dm3 of solution. 16.20cm3 of 0.02M
KMnO4 were required to completely react with 25cm3 of the iron(II)
solution.
a) Calculate moles of KMnO4 that reacted.
b) Calculate the moles of FeSO4 that reacted
c) Calculate the mass of FeSO4 in 1dm3 of the solution.
d) Calculate the percentage purity of the iron(II)sulphate
 Solution
Amount of MnO4- = Cx V(dm3)
= 0.02moldm-3 x 0.0162dm3 = 0.000324mols
From the reaction equation MnO4- : Fe2+ = 1:5
There moles of iron(II) = 5 x moles MnO4-.
Moles Fe2+ = 5x0.000324 = 0.00162mols
n 0.00162
CFe2 =   0.065moldm 3
V(dm3 ) 0.025

Mass of pure iron(II) sulphate in 1dm3 = nxM

= 0.065 x 151.8 = 9.87g
9.87
Percentage Purity= x100=82%
12
 TRY
1. When 13.20g of impure KMnO4 was used to prepare 1dm3 of
solution. 15.00cm3 of this solution was completely neutralised
by 25cm3 of 0.050M Na2C2O4 solution. Calculate the
percentage purity of the KMnO4. [ Molar mass KMnO4 =
160g/mol ]

2. 2. When 7.80g of the salt FeSO4.xH2O was dissolved in distilled

water to form 1dm3 of solution.17.80cm3 of this solution was
completely neutralised by 10cm3 of 0.01M KMnO4 solution.
(a) Calculate the concentration of Fe2+ in moldm-
3.

(b) What is the molar mass of the hydrated salt?

(c) Calculate the percentage of Fe in the salt.
(d) Deduce the value of x
[ Fe=55.8; S=32; O=16.0; H=1.00 ]
 Titration with iodine
In the titration of iodine with thiosulphate, the thiosulphate is
added until the solution turns pale yellow. Starch indicator is then
added. Starch forms a deep blue complex with the iodine.

Starch + iodine  Starch-Iodine Complex (deep blue)

The titration is continued until the deep blue complex breaks

down to give a colourless solution. This marks the end of the
reaction. The equation of the reaction of iodine with Na2S2O3 is

2S2O32-(aq) + I2(aq)  S4O62-(aq) + 2I-(aq)

Iodine is not very soluble in water. However, it

dissolves readily in a solution of potassium iodide.
 TRY
The amount of iodine in an impure sample may be determined by
dissolving the sample in KI solution, filtering off any impurities,
and titrating the aqueous solution against aqueous Na2S2O3.

25cm3 of a solution prepared by dissolving 12g of impure iodine in

KI solution to form 1dm3 was completely reacted with 28.20cm3 of
0.073M Na2S2O3.

Calculate
a) the concentration of iodine in moldm-3.
b) the mass of pure iodine in 1dm3.
c) the percentage purity of the sample of iodine.
2S2O32-(aq) + I2(aq)  S4O62-(aq) + 2I-(aq)
 Try

The composition of a mixture of FeSO4 and

Fe2(SO4)3 may be determined by reacting an
aqueous solution of the mixture against aqueous
KMnO4. The Fe2+ is oxidised by the KMnO4 but the
Fe3+ does not take part in the reaction.
25cm3 of a solution prepared by dissolving 10.0g
of a mixture of FeSO4 and Fe2(SO4)3 was
neutralized by 18.4cm3 of 0.02M KMnO4 solution.
Calculate the percentage of FeSO4 in the mixture.

5Fe2+ + 8H+ + MnO4-  5Fe3+ + Mn2+ + 4H2O

 Half-Cells
When a strip of a metal is placed into a solution
of its ions for some time equilibrium is set up
between the metal atoms and its ions in solution.
M (S) M n+ (aq) +ne-

The metal strip, dipping in a solution of its ions is

called a half-cell.
 ELECTROCHEMICAL CELL

When two half-cells are

connected so that electron
transfer occurs through an
external wire the
arrangement is called a
galvanic cell, voltaic cell, or
electrochemical cell.
 Terminology of Electrochemical cells
Anode: This is the negatively charged electrode at which
oxidation takes place.

Cathode: This is the positively charge electrode at which

reduction takes place.

Electrolyte: this is the aqueous solution containing the

ions of the electrode (usually 1M).

Salt bridge: This is an electrolyte embedded in a gel that

facilitates the bidirectional movement of ions.

Conducting wire: an external wire through which electrons

flow.
Processes in an electrochemical
cell
 Cell diagram

The arrangement of two half-cells forming an

electrochemical cell can be represented by a cell
diagram.
Zn(s) |Zn 2+ (aq) (1M)||Cu 2+ (aq) (1M)|Cu(s)

The anode is written on the left hand side while

the cathode is written on the right hand side.
The solid metal is separated from its aqueous
solution by a vertical line.
The salt bridge separating the half-cells is written
as two vertical lines
 EMF of cells
When two half cells are connected, electric current is generated,
known as the emf of the cell. For example, when a zinc half cell is
coupled with a copper half cell a current of 1.1 volts is generated.
 Processes involved in an ECC
The anode gets oxidized and goes into solution as ions.
The electrons released flow through a conducting wire to
the cathode where they are used to reduce the cations of
the cathode.
To maintain electrical neutrality in the half-cells ions in
solution undergo bidirectional migration through the salt
bridge.

It is cumbersome to compare values of emf of cells.

Therefore by convention the standard hydrogen electrode
is chosen for the measurements of all emf values. Values
obtained with the SHE at 100kPa pressure and 298K are
known standard electrode potentials. They are written with
superscript .
 The Standard Hydrogen
Electrode (SHE)
This consists of a piece of platinum plate dipping in a 1 molar
solution of hydrochloric acid, through which hydrogen gas at
100kPa is passed through at 298K.

Equation for half-cell: 2H+(aq) + 2e-  H2(g)

 Why Platinum?
Platinum is chosen because it is a fairly inert metal that will
not ionize.
The reaction on the electrode happens rapidly as the large
surface area helps with the adsorption of hydrogen gas.
Adsorption only occurs on the surface.
As the electrode is immersed in the acid, an equilibrium is
set up between the adsorbed layer of H2 gas and the H+
ions.
2H+(aq) + 2e- H2(g) Eo = 0.00V
The hydrogen half-cell is arbitrarily assigned an electrode
potential of 0.00V.
This gives us a means to measure and compare the
electrode potential of any other half-cell to which it is
connected.
Let’s refer to the activity series to make sense of all of this.
Mg -2.37 V strongest reducing
agent
Al -1.66 V (most readily oxidized)
Zn -0.76 V
Fe -0.44 V
Pb -0.13 V
Hydrogen falls here 0.00 V
Cu +0.34
Ag +0.80 weakest reducing agent
(least readily oxidized)
Notice that H2 would be the cathode for the metals above it
and the anode for the metals below it on the series.
 Measurement of E
The given half-cell is connected to the SHE.
The electric potential is measured using a voltmeter.

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

Elements with a greater tendency to oxidize relative to the
hydrogen have negative electrode potentials while those with
lesser tendency to oxidize have positive electrode potentials
 Calculation of Electrode potentials
The emf of a given cell is calculated from the equation:
E(cell) = E(Cathode) - E(anode)

Or E(cell) = E(Right) - E(Left)

Since the cathode is the right hand side electrode, and the anode
the left hand side electrode.

For example, when the zinc half-cell is connected to the SHE the
voltmeter reads 0.76V

E(cell) = E(Cathode) - E(anode)

E(cell) = 0- 0.76
Thus the electrode potential of the zinc half-cell is -0.76V
 Factors affecting the magnitude
of electrode potentials
1. Concentration of ions – the higher the concentration,
the higher (more positive) the E0.

2. Pressure – for gaseous species the higher the

pressure, the higher the E0.

3. Temperature – the higher the temperature the lower

(more negative) the E0.
 Electrochemical series
This is an arrangement of the reduction potential of the elements
in table from negative to positive.
Those with a higher tendency to be oxidized relative to hydrogen
have negative values.
Those in which hydrogen has a greater tendency to oxidize
relative to them are given positive electrode potentials
Elements with highly negative reduction potentials are strong
reducing agents.
The more negative the reduction potential the more reactive the
metal.
Elements higher in the table (more negative potential) can
displace the ion of any element lower (more positive potential)
from solution.
Elements with very large difference in potentials react more
vigorously.
Table of electrochemical series
 Calculation of emf of cells
Once the electrode potentials are known, the emf of
A given cell formed by coupling the half-cees can be
calculated using the equation:

Eocell = EoR - EoL

The Eo reduction values provided in the table of

electrode potentials ot activity series are than substituted
to get the Eo Cell.
 Example: What is the standard emf of an electrochemical
cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution
and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V

Solution
The anode is the electrode with the more negative potential
0 = E0
Ecell - E 0
cathode anode

0 = -0.40 - (-0.74)
Ecell

0 = +0.34 V
Ecell
 Alternative method of Calculation

Calculate the cell potential for a cell made from silver and zinc
electrodes.

From the standard reduction table

Zn2+ + 2 e-  Zn - 0.76 v
Ag+ + e-  Ag + 0.80v

Identify the anode reaction and reverse it. The anode reaction is the
more negative. Remember to reverse the sign of Eo.

Zn  Zn+2 + 2 e- + 0.763 v (anode)

Write the two reactions and balance the charge then add them up.

Zn  Zn+2 + 2 e- Eo = + 0.76 v (anode)

2Ag+ + 2e-  2Ag Eo = + 0.80 v (cathode)
Zn + 2Ag+  Zn2+ + 2Ag Cell potential = 1.56 volts
 EMF and spontaneity of redox
reactions
When the calculated emf is positive, the reaction is spontaneous
as written.
If the calculated emf is negative, the reaction is non-spontaneous
in the forward direction. It rather takes place in the reverse
direction.
1. Deduce whether a reaction will occur when zinc sulphate
solution is poured into a silver can.
Ag+ + e-  Ag E°= +0.80 V
Zn2+ + 2e-  Zn E°= - 0.76 V
2. Deduce whether a reaction will occur when 1M calcium chloride
solution is poured into a copper can.
Cu2+ + 2e-  Cu E°= +0.34 V
Ca2+ + 2e-  Ca E°= - 2.87 V
3. Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
 Try
1. What is the emf of a cell formed by coupling a magnesium half-
cell with a zinc half-cell?

2. Consider the redox reaction below:

2Al(s) + 3Zn2+(aq)  2Al3+(aq) + 3Zn(s)

Write the cell diagram for the reaction and calculate the emf of the
cell.
3. For the reaction
M(s) + Cu2+(aq)  M2+(aq) + Cu(s) E° = 0.740 volt at 25°C
Use the E of the copper half cell form the data booklet to
calculate the standard electrode potential for the reduction half
reaction:
M2+(aq) + 2e-  M(s)
 Try
4. a) Draw an electrochemical cell for the spontaneous reaction.
Zn + 2 Ag+  Zn2+ + 2Ag
b) Calculate the emf of the cell

5. Consider the cell shown

a) Write the balanced net ionic equation for the reaction

b) Calculate the cell potential
6. Consider the diagram below

When the switch is closed, the mass of the Sn electrode increases.

The half-reactions are shown below.
Sn2+ (aq) + 2e– ® Sn(s) E˚ = –0.14 V
X3+(aq) + 3 e– ® X(s) E˚ = ?
a) Label the electrode that is the cathode. Justify your answer.
b) If the standard cell potential E˚cell is +0.60 V, what is the standard
potential, in volts for the X3+/X electrode?
 Electrode Potentials and Free
energy
The free energy of a reaction is related to the standard
electrode potential by the equation

∆G =-nFE
F is Faraday’s constant, 96500C.

Thus knowing the electrode potential the free energy of

a redox process can be calculated.
The process is spontaneous when ∆G is negative (E
is positive) and non-spontaneous when ∆G is positive
(E is negative) . When E is zero, ∆G is zero showing
that the process is at equilibrium.
This equation allows us to calculate the free-energy
change of a reaction from standard electrode
potential data.
The more positive the value for E, the more favorable
the reaction or more negative the delta G.

Example: Consider the reaction

Pb2+ + Zn = Pb + Zn+
a) Calculate the standard emf of the cell.
b) Calculate the free energy change for the reaction
and hence explain whether the reaction is
spontaneous.
 Corrosion

This is a redox process in which atoms at the

surface of metals get oxidized. This could be
through dry corrosion or wet corrosion.
In dry corrosion metals at the surface react with
oxygen in air to form an oxide which flakes off
exposing a fresh surface for further oxidation.
In wet corrosion the metal is oxidized by aqueous
solutions of chemicals like acids. The metal gets
oxidized and dissolves in the solution.
 Rusting of iron

Rusting is a redox reaction in which iron is oxidized in

the presence of moisture and oxygen. At the anodic
regions on the surface of the metal iron is oxidized
Fe → Fe2+ + 2e-
The electrons released are carried to the cathodic
regions where they are used to reduce water
2H2O + 4e-  4OH-
The Fe2+ reacts with the OH- ions
Fe2+ + 2OH-  Fe(OH)2
The Fe2+ is further oxidized to Fe3+ ions with the help
of oxygen leading to the formation of hydrated
iron(III)oxide called rust.
Fe(OH)2 + O2 + H2O  Fe2O3.xH2O
 Processes of rusting

The processes can be summarized in the diagram

below. However, it is important to note that
rusting is a complex process and different
mechanisms may be encountered in textbooks.
 Prevention of rusting
Rust can also be prevented by placing a sacrificial
electrode in electrical contact with the iron. The sacrificial
anode has a greater tendency to oxidize, and hence will
quickly oxidize to prevent the iron from rusting.
 ELECTROLYSIS

The electrolysis of an aqueous solution involves

ions from the solvent water as well as the ions
from the ionic compound.
More than one type of ion then migrates to the
anode and the cathode.
Selective discharge is a process by which certain
ions are discharged at each electrode.
The products of electrolysis are determined by
the following factors that influence the discharge
of ions:
 The relative E˚ of the ions
 The relative concentrations of the ions in
the electrolyte
 The nature of the electrode.
 ELECTROLYSIS OF WATER

The ionization of pure water is very low.

However, the addition of ions increases its
conductivity so some ionic compound like NaOH
is added when electrolysis is performed.
1. Ions present: H2O(l) ↔ H+ + OH-
NaOH(aq) → Na+ + OH-
cathode anode
2. At the cathode:
 Na+ and H+ accumulate.
 E˚ for Na+ is -2.71V and E˚ for H+ is 0.00V
 H+ is preferentially reduced and H2 is
discharged. 2H+(aq) + 2e- → H2(g)

At the Anode:
 OH- is discharged as the only ion
 4OH-(aq) → 2H2O(l) + O2(g) + 4e-
3. The overall equation:
2H2O(l) → 2H2(g) + O2(g)
4. The observed changes at electrodes:
 A colorless gas evolved at both
electrodes –
○ O2 at anode and H2 at cathode
 The ratio of volume of gases is 2H2 to
1O2.
 The pH at the anode decreases as OH- is
discharged (taken away from the
solution).
 The pH at the cathode increases as H+ is
discharged (taken away from the
 ELECTROLYSIS OF NaCl(aq)

NaCl(aq) is known as brine.

The electrolysis of brine leads to the production
of H2, Cl2 and NaOH.
The electrolysis of NaCl(aq) and NaCl(l) are different
processes leading to different products.
1. Ions present: NaCl(aq) → Na+ + Cl-
H2O(l) ↔ H+ + OH-
cathode anode
2. At the cathode:
 Na+ and H+ accumulate.
 E˚ for Na+ is -2.71V and E˚ for H+ is 0.00V
 H+ is preferentially reduced and H2 is discharged. 2H+(aq) +
2e- → H2(g)
At the Anode:
 Cl- and OH- accumulate. The E˚ for OH- is lower than Cl- so
we would expect OH- to be preferentially oxidized, but it is
more complicated.
○ When the concentration of Cl- is low, OH- is discharged
leading to the release of O2.
○ But when the concentration of NaCl is greater than 25% by
mass of the solution, Cl- is discharged leading to the
release of Cl2. 2Cl- → Cl2 + 2e-
○ The industrial electrolysis of brine uses saturated
solutions of aqueous NaCl to produce Cl2.
3. The overall equation when Cl- is discharged:
2NaCl(aq) + 2H2O(l) → 2H2(g) + Cl2(g) + 2Na+ + 2OH-
4. The observed changes at electrodes:
 A gas evolves at both electrodes –
○ Cl2 at anode and H2 at cathode
 Cl2(g) is identified at the anode through its
strong smell and bleaching effect on blue
litmus paper.
 An increase in pH of the electrolyte occurs
due to the loss of H+.
 ELECTROLYSIS OF CuSO4(aq)

CuSO4(aq) is bright blue due to the hydrated Cu2+ ion.

The electrolysis leads to different products depending
upon the nature of the electrodes: either carbon or
copper.
1. Ions present: CuSO4 (aq)→ Cu2+ + SO42-
H2O(l) ↔ H+ + OH-
cathode anode
(A) With Carbon (graphite) as the electrodes
2. At the cathode:
 Cu2+ and H+ accumulate.
 E˚ for Cu2+ is +0.34V and E˚ for H+ is 0.00V
 Cu2+ is preferentially reduced and Cu(s) is
discharged. Cu2+(aq) + 2e- → Cu(s)
At the Anode:
 SO42- and OH- accumulate. The E˚ for OH- is
lower than SO42- so OH- is discharged.
4OH-(aq) → 2H2O(l) + O2(g) + 4e-
3. The overall equation is:
2Cu2+(aq) + 2H2O(l) → 2Cu(s) + O2(g) + 4H+
4. The observed changes at electrodes:
 Pinky-brown color develops as copper is
deposited on the cathode.
 A colorless gas O2 is evolved at the anode.
 A decrease in pH of the electrolyte occurs due
to the loss of OH-.
 Loss of intensity of blue color as Cu2+ is
discharged.
(B) With copper as the electrodes
2. At the cathode:
 Cu2+ and H+ accumulate.
 E˚ for Cu2+ is +0.34V and E˚ for H+ is 0.00V
 Cu2+ is preferentially reduced and Cu(s) is
discharged. Cu2+(aq) + 2e- → Cu(s)
At the Anode:
 The Cu electrode itself is oxidized, supplying
electrons for the reaction and dissolving as
Cu2+.
Cu(s) → Cu2+(aq) + 2e-
3. So the net reaction is the movement of Cu2+ from
where it is produced at the anode to the cathode where
it is discharged as Cu(s).
4. The observed changes at electrodes:
 Pinky-brown color develops as copper is
deposited on the cathode.
 Disintegration of the Cu anode.
 No change in pH.
 No change in intensity of blue color as the
concentration of Cu2+ remains constant.
The three factors influencing the amount of
products are:
 The charge on the ion
 The current
 The duration of the electrolysis
There are two types of problems involving
electrolysis.
 Applications of electrolysis

Extraction of highly reactive metals. E.g Ca, K, Na,

Mg, Al
Production of gases like hydrogen and chlorine
Refining of metals . e.g. copper.
Electroplating
Electroanalysis: The purity of a metal may be
determined by making the metal the anode.
Electrolysis is then carried out for a period of time.
The gain in weight at the cathode is compared to the
loss in weight at the anode to determine the purity.
 FARADAY’S LAWS OF
ELECTROLYSIS
Faraday’s first Law: this stated that the amount of
a substance, n, discharged electrode is directly
proportional to the quantity of charge, Q, that has
passed through the electrolyte.
Mathematically, n  Q
The quantity of charge is obtained by multiplying
the current in amperes (A) by the time in
seconds(s).
Q = Ixt
Thus quantity discharged, n Ixt
 Faraday’s Second Law

This states that 1mole of every substance is discharged by a

whole number of faradays. The faraday is the charge carried by
1mole of electrons. The value is approximately 96500C/mol.
1 mole of electrons carry a charge of 96500C

Note: From the two laws the quantity of a substance deposited

does not depend on the concentration of the electrolyte, the
surface area of the electrodes or the temperature of the
electrolyte. It only depends on the duration of electrolysis and the
strength of the current applied.
 Deductions
1mole of a singly charged ion is discharged by 1faraday
=96500C
1mole of a doubly charged ion requires 2 faradays to
discharged =2x96500C
1mole of a triply charged ion requires 3 faradays to
discharged=3x96500C
Ion Number of faradays Charge required to discharged
1mole
Li+, Na+, K+, Cl-, Br-, I- 1F 96500C

Be2+, Mg2+, Ca2+, Ba2+, 2F 2x96500C=193000C

O2-
Al3+ 3F 3x96500C=289500C
 Examples
1. What quantity of charge is required to discharge
the following:
a) 0.08mols of potassium
b) 1.20mols of calcium
c) 16g of aluminium
d) 48g of copper
e) 22.4dm3 of hydrogen gas
f) 1dm3 of oxygen
 Sample calculations
1. Calculate the mass of calcium deposited when
molten calcium chloride is electrolysed for
30minutes using a current of 1.84A. [Ca=40g/mol,
F=96500C]
Solution
Q=It =1.84Ax (30x60)s =3312C
Ca2+ + 2e- Ca
2 Faradays liberate 1mole Ca
2x96500C liberate 1 mole Ca
Therefore, 3312C will liberate?
Mass = molesxMolar mass calcium =
 Examples
What length of time in minutes is required to plate out 2g of gold from
a solution of gold(III)trioxonitrate(V) using a current of 0.4A?
[Au=197g/mol, F=96500C]
Solution:
Au3+ +3e- Au
3F liberate 1mole Au =197g
3x96500C liberate 197g Au
Charge required to liberate 2g Au will be …
Q = It
T = Q/I= (Complete the work)
 Sample questions
1. How many grams of Cu will be deposited from a
solution of CuSO4 by a current of 1.8 A flowing for 2h
30 minutes?
2. Find the mass of zinc produced at the cathode by
passing a current of 2.00 A through aqueous zinc
sulphate for 3 hours.
3. A current is passed through an electrolytic cell
containing molten AlCl3 for 5 hours. If 5litres of
chlorine are collected,
a) what is the current in amperes?
b) What mass of aluminium is produced at the cathode?
4. Consider the electrolysis of CaCl2 (l). How many grams
of Ca (s) can be produced by passing 2.5 A through the
solution for 1h 20minutes?
 Sample questions

5. How much copper will be plated out when a current

of 10.0 amps is passed through a solution of Cu2+ ions
for 30.0 minutes? Cu2+ + 2e- → Cu(s)
6. How long must a current of 5.00 A be applied to a
solution of Ag+ to produce 10.5 g of silver?
7. How many faradays of electricity ( or moles of
electrons) are required to produce 1.2dm3 of oxygen
gas from electrolysis of aqueous NaOH
8. How long would it take in minutes to produce 60.0 g
of Cr from a solution of CrCl3 by a current of 1.8 A?
[F=96500C, Vm = 22.4dm3/mol
 Procedure
1. The material to be electroplated is made the
cathode, while the metal that will be electroplated
on the material is made the cathode.
2. The electrolyte is an aqueous solution of the metal
used for electroplating.
3. The electrodes are connected to an external DC
source.
4. During electroplating, the metals ions in solution
gain electrons at the cathode and get deposited on
it as metal, thus increasing the size of the cathode.
5. The anode dissolves to replace the ions that have
been deposited at the cathode.
 Electroplating a spoon with
silver
In electroplating silver on an iron spoon, the spoon is made the
cathode while the silver metal forms the anode. The solution is
that of silver compound.

At anode: Ag(s)  Ag+(aq) + e-

At cathode: Ag+(aq) + e-  Ag(s)
 Purification of Metals
The impure metal is made the anode
A small piece of pure metal is made the cathode
When the circuit is completed, the impure metal
oxidizes at the anode.
M(s) → Mn+(aq) + ne-
The ions go to the cathode where the pick up
electrons and are reduced to the pure metal and
deposited on the cathode.
Mnn+(aq) + ne- → M(s)
The impurities fall under the anode, and is
collected as “anode mud”
 Purification of copper
Impure copper made the anode
Small piece of pure copper made the cathode
Solution is that of copper sulphate

Anode reaction: Cu(s) → Cu2+(aq) + 2e-

Cathode reaction: Cu2+(aq) + 2e- → Cu(s)
Size of anode decreases, size of cathode increases.
 Why electroplate?

Purpose of electrolysis
• Decorative purposes
• Corrosion control
• Improvement of function

Process can be controlled by altering the

current and time according to how thick a
layer of metal is desired.
 Citations

International Baccalaureate Organization. Chemistry

Guide, First assessment 2016. Updated 2015.

Brown, Catrin, and Mike Ford. Higher Level Chemistry.

2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.

Most of the information found in this power point comes

directly from this textbook.

The power point has been made to directly complement

the Higher Level Chemistry textbook by Catrin and Brown
and is used for direct instructional purposes only.