Redox Equilibrium Handbook Part 2

REDOX EQUILIBRIUM |1

STANDARD ELECTRODE POTENTIAL

What is Electrode Potential ?
Electrode potential is the potential difference produced when an equilibrium is
established between metal M and the aqueous solution containing metal Mn+ ions in a
half-cell.

➢ If a metal is immersed in a solution
containing the metal ions, the metal
atoms tend to release electrons to form
metal ions.

➢ The electrons released accumulate on
the metal, making the metal negatively charged.

➢ The metal is surrounded by positive ions. A small number of metal ions will
receive electrons and become metal atoms again.

➢ Thus, an equilibrium is achieved between metal atoms and its metal ions in a
half cell.

➢ This condition results in a potential difference between the metal (electrode)
and its solution (electrolyte).

➢ The potential difference is called electrode potential.
➢ The strength of an oxidising agent and a reducing agent depends on the

standard electrode potential, E⁰.

REDOX EQUILIBRIUM |2
1. The electrode potential value of an electrode system is determined based on

the difference of electrode potential between two half-cells.
2. The electrode potential measured under standard conditions is called standard

electrode potential, E⁰ and has the unit of volt.
3. The standard hydrogen electrode (S.H.E) was chosen as the reference

electrode.
4. Standard conditions to measure the standard electrode potential, E0 :

i. concentration of ions in aqueous solutions is 1.0 mol dm−3
ii. gas pressure of 1 atm or 101 kPa.
iii. temperature at 25 ⁰C or 298K.
iv. platinum is used as an inert electrode when a half-cell is not a metal

electrode.

STANDARD HYDROGEN ELECTRODE

Label the Standard Hydrogen Electrode below:

REDOX EQUILIBRIUM |3
1. What is the function of platinum(IV)oxide?
The function of platinum(IV) oxide, PtO2 is to increase the surface area of
platinum, Pt to adsorb hydrogen gas, H2; so that hydrogen molecules are in closer
contact with hydrogen ions, H+ in the solution.
2. Half equation of the hydrogen half-cell:

2H+(aq) + 2e− ⇌ H2(g)

3. The standard hydrogen electrode potential, E0 is given the value of 0.00 V.
4. The reference half reaction is as follows:

H+(aq) + e− ⇌ ½ H2(g) E0 = 0.00 V

STANDARD ELECTRODE POTENTIAL VALUE, E0

An unknown standard electrode potential can be obtained by constructing a voltaic
cell consisting of a reference half cell and another unknown half -cell.

Standard electrode potential value, E0 of zinc :

REDOX EQUILIBRIUM |4
1. Observation:

❖ Gas bubbles are produced around the platinum electrode.
❖ Zinc electrode becomes thinner.
2. Standard hydrogen potential is : 0.00 V.
3. Voltmeter reading of the cell: 0.76 V .
4. Electrode potential of zinc : 0.76 V.
5. Zinc, Zn has a greater tendency to release electrons compared to hydrogen.
Hence, zinc, Zn becomes the negative terminal.
6. Electrons move from zinc electrode, Zn (negative terminal) to platinum
electrode, Pt (positive terminal) through the connecting wires.
7. Half equation at the negative terminal (oxidation reaction)

Zn(s) → Zn2+(aq) + 2e−
8. Half equation at the positive terminal (reduction reaction)

2H+(aq) + 2e− → H2(g)
9. A standard electrode potential for zinc half-cell is written as reduction.

Zn2+(aq) + 2e− ⇌ Zn(s) E0 = -0.76 V
10.The negative symbol shows that zinc electrode is the negative terminal when

connected to the standard hydrogen electrode.

REDOX EQUILIBRIUM |5

OXIDISING AGENTS AND REDUCING AGENTS BASED ON THE VALUE OF
STANDARD ELECTRODE POTENTIAL

The standard electrode potential, E0 is also known as the standard reduction
potential.
All half-cell equations are written as reduction.
E0 value is a measure of the tendency of a substance to accept or donate electrons.

Oxidising agent + electron ⇌ Reducing agent.

The more positive the E⁰ value is,(the lower position in the series) the easier for the
substance on the left side (atom or ion) of the half equation to undergo reduction. It
will also become the stronger oxidising agent.
The more negative the E⁰ value is, (the higher position in the series) the easier for the
chemical substances on the right side of the equation to undergo oxidation. It will also
become the stronger reducing agent.

The relationship between E0 value to the strength of a substance as an
oxidizing agent and a reducing agent

Molecules or ions with a more positive or less
negative standard electrode potential value E0.

Has a greater tendency Easier to undergo A stronger
to accept electrons. reduction reaction. oxidising agent.

REDOX EQUILIBRIUM |6

Atoms or ions with a more negative or less
positive standard electrode potential value, E0.

Has a greater tendency Easier to undergo A stronger
to release electrons. oxidation reaction. reducing agent.

Example 1:

Half Cell Equations E0 / V (298 K)
Zn2+ (aq) + 2e− ⇌ Zn(s) -0.76 V
2H+ (aq) + 2e− ⇌ H2(g) 0.00 V
Cu2+ (aq) + 2e− ⇌ Cu(s) +0.34 V

Referring to the E⁰ value:
Copper(II) ions are most easily reduced compared to hydrogen ions.
Zinc ions is the most difficult to be reduced.
The strength of oxidising agent is in the order of:
Cu2+ > H+ > Zn2+
Negative E⁰ value indicates that zinc atom is more easily oxidised followed by
hydrogen gas and copper atom being most difficult to be oxidised.
The strength of reducing agent is in the order of:
Zn > H2 > Cu

REDOX EQUILIBRIUM |7

Example 2: Determine whether silver, Ag or magnesium, Mg is an oxidising agent or
reducing agent.

Half Cell Equations E0 / V (298 K)
Mg2+ (aq) + 2e− ⇌ Mg(s) -2.38 V
+0.80 V
Ag+ (aq) + e− ⇌ Ag(s)

E0 value of Ag+ ion is more positive. E0 value of Mg is more negative.
Silver ion, Ag+ on the left side is a stronger Magnesium atom, Mg on the right side is a
oxidising agent. stronger reducing agent.
It is easier for Ag+ ion to receive electrons and It is easier for magnesium atom, Mg to release
undergo reduction. electrons and undergo oxidation.
Conversely, silver atom, Ag on the right side is Conversely, magnesium ion, Mg2+ on the left
difficult to release electrons. side is difficult to accept electrons.

Example 3:

Diagram above shows copper wire, Cu in dipped in silver nitrate, AgNO3 solution.

REDOX EQUILIBRIUM |8

a) By referring to the standard electrode potential series, identify the oxidising
agent and reducing agent in the reaction.

Reducing agent:

➢ E0 value of Cu is more negative or less positive than E0 of Ag.

➢ Copper, Cu is a stronger reducing agent compared to silver, Ag.

➢ Therefore, copper atom, Cu has a greater tendency to release electrons to

form copper(II) ions, Cu2+. The more negative the E⁰ value is, the easier
➢ Copper, Cu undergoes oxidation reaction. for the substance on the right of the half

equation to undergoes oxidation.

Cu2+ (aq) + 2e− ⇌ Cu(s) E⁰ = +0.34 V

Oxidising agent:

➢ The E0 value of Ag+ ion is more positive than E0 of Cu2+ ion.

➢ The silver ion, Ag+ is a stronger oxidising agent than copper(II) ions, Cu2+.

➢ Therefore, silver ion, Ag+ has a greater tendency to receive electrons to form

silver atom, Ag.

➢ Silver ion, Ag+ undergoes reduction. The more positive the E⁰ value is, the easier
for the substance on the left of the half

equation to undergoes reduction.

Ag+ (aq) + e− ⇌ Ag (s) E⁰ = +0.80 V

b) Does a reaction occur between copper, Cu and silver nitrate, AgNO3?
➢ Yes. Copper, Cu can displace silver, Ag from silver nitrate, AgNO3 solution

because copper, Cu is a stronger reducing agent compared to silver, Ag.
➢ Displacement reaction equation:

Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

REDOX EQUILIBRIUM |9
Example 4:
A student immersed copper wire into a beaker containing magnesium nitrate solution.
Does a reaction between copper, Cu and magnesium nitrate, Mg(NO3)2 take place?
The E0 value of Cu is more positive than E0 of Mg.

➢ Copper, Cu is a weaker reducing agent than magnesium, Mg.
➢ Therefore, copper atom, Cu has a lower tendency to release electrons to form

copper(II) ion, Cu2+.
➢ Copper, Cu does not undergo oxidation.
The E0 value of Mg2+ ion is more negative than E0 of Cu2+ ion:
➢ Magnesium ion, Mg2+ is a weaker oxidising agent than copper(II) ions, Cu2+.
➢ Therefore, magnesium ions, Mg2+ are difficult to receive electrons to form

magnesium atom, Mg.
➢ Magnesium ion, Mg2+ does not undergo reduction.

Copper, Cu cannot displace magnesium, Mg from magnesium nitrate, Mg(NO3)2
solution because copper, Cu is a weaker reducing agent compared to magnesium, Mg.
The displacement reaction does not occur.

R E D O X E Q U I L I B R I U M | 10

VOLTAIC CELL

What is a simple chemical cell?

Also known as voltaic cell or Can be constructed by immersing two different
galvanic cell metals in an electrolyte and connected by
connecting wires.

A redox reaction occurs in the The potential difference detected by the
cell causes the flow of voltmeter indicates the presence of electrical
electron.
current.

Redox Reaction in a Voltaic Cell

Simple chemical cell for a pair of Mg/Cu metals

Negative Terminal (Anode) Positive Terminal (Cathode)

Magnesium ribbon becomes the negative Copper becomes the positive terminal
terminal because its E⁰ value is more because its E⁰ value is more positive.
negative (stronger reducing agent than
copper) Reaction that occurs at the negative
terminal is reduction.
Reaction that occurs at the negative
terminal is oxidation. Copper(II) ions receives electron and
becomes copper atom.
Magnesium atoms release electrons and form
magnesium ions Half equation:

Half equation: Cu2+(aq) + 2e− → Cu(s)

Mg(s) → Mg2+(aq) + 2e−

R E D O X E Q U I L I B R I U M | 11

Observation

Magnesium ribbons becomes thinner Copper plate becomes thicker

Electrons flow from the negative terminal to the positive terminal; while the current flows
from the positive terminal to the negative terminal.
The electric current can be detected by a bulb / ammeter/voltmeter/galvanometer.
Energy change in a voltaic cell is from chemical energy to electrical energy

Simple chemical cell for a pair of Zn/Ag metals

Negative Terminal (Anode) Positive Terminal (Cathode)

Zinc becomes the negative terminal because Silver becomes the positive terminal
its E⁰ value is more negative (stronger because its E⁰ value is more positive.
reducing agent than silver)
Reaction that occurs at the negative
Reaction that occurs at the negative terminal is reduction.
terminal is oxidation.
Silver ions receives electron and becomes
Zinc atoms release electrons and form zinc silver atom.
ions
Half equation:
Half equation:
Ag+(aq) + e− → Ag(s)
Zn(s) → Zn2+(aq) + 2e−

Observation

Zinc electrode becomes thinner Silver electrode becomes thicker

R E D O X E Q U I L I B R I U M | 12

Daniell Cell

What is a Daniell Cell?
It is a chemical cell which is constructed by combining two half-cells with different E⁰ values.
Daniell cell is an example of voltaic cell which consist of :

➢ Zinc electrode dipped in zinc sulphate solution.
➢ Copper electrode dipped in copper(II) sulphate solution.
➢ The two solutions are connected by a porous pot or a salt bridge.
The function of porous pot or salt bridge is to allow the movement of ions between the two
solutions and complete the circuit.
Draw a Daniell Cell using a salt bridge and a porous pot.

R E D O X E Q U I L I B R I U M | 13

E⁰ for both half cell Zn2+(aq) + 2e− ⇌ Zn(s) E0 = - 0.76 V
Cu2+(aq) + 2e− ⇌ Cu(s) E0 = + 0.34 V

Negative terminal Electrode and reason Zn because it has a more negative E0 value.
(anode) Reaction takes place Oxidation (release electrons)

Half Equation Zn(s) → Zn2+(aq) + 2e−

Observation Zinc electrode becomes
thinner/smaller/dissolved in the solution

Positive terminal Electrode and reason Cu because it has more positive E⁰ value.

Reaction takes place Reduction (receive electrons)

Half Equation Cu2+(aq) + 2e− → Cu(s)

Observation Copper electrode becomes thicker / brown solid
deposited.
Intensity of the blue solution decreases
Blue solution becomes paler.

Overall ionic equation Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Flow of electron Electrons flow from zinc electrode, Zn to copper electrode, Cu through a
connecting wire.

Cell notation Zn(s) Ι Zn2+(aq, 1.0 mol dm–3) ΙΙ Cu2+(aq, 1.0 mol dm–3) Ι Cu(s)

Voltage reading E0 cell = (+0.34) – (- 0.76) = + 1.10 V

R E D O X E Q U I L I B R I U M | 14

The voltaic cell can be written in the form of

cell notation. Anode is written on the left
side of the cell notation and cathode on the

right side.

Complete the following on the reaction in a voltaic cell. (Activity 1G-Textbook)

E⁰ for both half cell Fe2+(aq) + 2e− ⇌ Fe(s) E⁰ = +0.44
Ag+(aq) + e− ⇌ Ag(s) E⁰ = +0.80

Negative terminal Electrode and reason Iron because it has a more negative E0 value.
(anode) Reaction takes place Oxidation (release electrons)

Half Equation Fe(s) → Fe2+(aq) + 2e−

Observation Iron electrode becomes thinner/smaller/dissolved
in the solution.

Green solution becomes darker/ intensity of green
solution increases.

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Positive terminal Electrode and reason Silver because it has more positive E⁰ value.

Overall ionic Reaction takes place Reduction (receive electrons)
equation
Flow of electron Half Equation Ag+(aq) + e− → Ag(s)
Cell notation
Voltage reading Observation Silver electrode becomes thicker/grey solid
deposit.

Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)

Electrons flow from iron electrode, Fe to silver electrode, Ag through a
connecting wire.
Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s)

E0 cell = (+0.80) – (- 0.44) = + 1.24 V

TO DETERMINE THE VOLTAGE OF A VOLTAIC CELL BY USING DIFFERENT PAIRS OF METALS

Aim To determine the voltage of a voltaic cell by using different pairs of metals
(Refer textbook for complete report)

Variables

Operational
Definition of
Chemical Cell

Materials Copper plate, iron nail, zinc plate, magnesium ribbon, sandpaper,
Apparatus 1.0 moldm-3 iron(II) sulphate solution ,1.0 moldm-3 copper(II) sulphate solution, 1.0
moldm-3 zinc sulphate solution, 1.0 moldm-3 magnesium sulphate solution.

Voltmeter, 250 cm3 beaker, connecting wires with crocodile clips, porous pot.

Procedures R E D O X E Q U I L I B R I U M | 16

1. All metals plates are cleaned with sandpaper.
2. A porous pot is filled with zinc sulphate solution until it is two-third full.
3. A zinc plate is dipped into zinc sulphate solution.
4. A beaker is filled with copper(II) sulphate solution.
5. A copper plate is dipped into the copper(II) sulphate solution.
6. The porous pot is placed into the beaker.
7. The circuit is completed by connecting the metals to a voltmeter using

connecting wires.
8. The reading of the voltmeter is recorded.
9. Repeat steps 2-8 by replacing zinc plate and zinc sulphate solution in the

porous pot with:
i. Iron nail and iron(II) sulphate solution
ii. Magnesium ribbon and magnesium sulphate solution.

Diagram

Set Pair of metals Voltage(V) Observations at the electrodes
I Zinc + copper
Result 1.1 Zinc plate becomes thinner.
II Iron + copper
Brown solid is deposited on the copper
III Magnesium + plate
copper
0.7 Iron nail becomes thinner.

Brown solid is deposited on the copper
plate

2.7 Magnesium ribbon becomes thinner.

Brown solid is deposited on the copper
plate

Discussion R E D O X E Q U I L I B R I U M | 17

1. For each set of the experiment:

i. Write the half equation for oxidation reaction, half equation for
reduction reaction and the overall ionic equation.

ii. Write the cell notations for the voltaic cells.

iii. calculate the theoretical voltage of the cells using the standard
electrode potential value of the half-cells.

2. Deduce the relationship between the pair of metals and the voltage of the
cells.

(Answer the discussion in a foolscap paper and paste it together with the lab report.)

Write the cell notation for the following chemical cells.
Refer to standard electrode potential, E⁰ Series.

Chemical Cell Ionic Equation Cell Notation

Zn → Zn2+ + 2e- Zn + Cu2+ → Zn2+ + Cu Zn(s) Ι Zn2+(aq) ΙΙ Cu2+(aq) Ι Cu(s)
(a)
Pb + 2Ag+ → Pb2+ + 2Ag Pb(s) Ι Pb2+(aq) ΙΙ Ag+(aq) Ι Ag(s)
Cu2+ + 2e- → Cu
Pb → Pb2+ + 2e- Cu + 2Ag+ → Cu2+ + 2Ag Cu(s) Ι Cu2+(aq) ΙΙ Ag+(aq) Ι Ag(s)
(b)
Ag+ + e- → Ag Al + 3Ag+ → Al3+ + 3Ag Al(s) Ι Al3+(aq) ΙΙ Ag+(aq) Ι Ag(s)
Cu → Cu2+ + 2e-
(c) Mg + Sn2+ → Mg2+ + Sn Mg(s) Ι Mg2+(aq) ΙΙ Sn2+(aq) Ι Sn(s)
Ag+ + e- → Ag
Al → Al3+ + 3e- Ni + 2Ag+ → Ni2+ + 2Ag Ni(s) Ι Ni2+(aq) ΙΙ Ag+(aq) Ι Ag(s)
(d)
Ag+ + e- → Ag 10Cl- + 2MnO4- + 16H+ → 5Cl2 Pt(s) Ι Cl-(aq), Cl2(aq) ΙΙ MnO4-(aq), Mn2+(aq), H+(aq) Ι
(e) Sn2+ / Sn and Mg2+/Mg.
(f) Ni2+ /Ni and Ag+ / Ag + 2Mn2+ + 8H2O Pt(s)
Cl2 / Cl− and MnO4−
(g) /Mn2+.

R E D O X E Q U I L I B R I U M | 18
Calculate the voltage for the following cell. (Self-Assess 1.3 -Textbook)

(a) Ni(s) | Ni2+(aq) || Pb2+(aq) | Pb(s). E0 cell = (-0.13) – (- 0.25) = + 0.12 V

(b) Pt(s) | I−(aq), I2(aq) || Ag+(aq) | Ag(s). E0 cell = (+0.80) – (+0.54) = + 0.26 V
(c) Pt(s) | Fe2+(aq), Fe3+(aq) || Ag+(aq) | Ag(s). E0 cell = (+0.80) – (+0.77) = + 0.03 V

(d) Pt(s) | Br−(aq), Br2(aq) || Cl2(aq), Cl−(aq) | Pt(s). E0 cell = (+1.36) – (+ 1.07) = + 0.29 V

R E D O X E Q U I L I B R I U M | 19

ELECTROLYTIC CELL

ELECTROLYTE & NON-ELECTROLYTE

Substances can be classified into three types based on electrical conductivity.

Type of Definitions Notes

substance

Solid ionic compound does not

Electrolytes are substances that can conduct electricity.

conduct electricity in either the The ions in ionic lattice structure

molten state or aqueous solution and Ionic cannot move freely
undergo chemical changes (undergo Compound Ionic compounds in molten or
decomposition into their constituent aqueous solution can conduct

elements.) electricity.

Can conduct electricity due to the There is presence of free moving

Electrolyte presence of ions that move freely. ions.

Electrical conductivity increases as Acids and Glacial acids/alkalis (absence of

temperature increases. Alkalis water) cannot conduct electricity

Examples are ionic compound, acids, because there are no free moving

and alkalis. ions

In the presence of water, acid and

alkali ionise in water to produce

free moving ions, so they can

conduct electricity.

Substances that cannot conduct Covalent compound cannot conduct electricity

Non- electricity in all states. because the particles in covalent compounds are

electrolyte Examples are covalent compound. neutral molecules. There are no free moving ions

in the compound.

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Substances that conduct electricity in Substances that conduct electricity without

solid or molten state, but do not undergoing decomposition. Can conduct

undergo chemical changes. electricity due to the presence of electrons that
Conductor Electrical conductivity decreases as move freely.

temperature increases.
Examples of conductors are metals

and graphite.

Classify the substances in the text box below into electrolyte and non-electrolyte.

Solid lead(II) chloride, molten aluminium oxide, lead(II) nitrate solution, solid sodium
chloride, sodium chloride solution, molten lead(II) chloride, glucose solution, glacial
ethanoic acid, dilute ethanoic acid, molten naphthalene, ethanol, tetrachloromethane,
sodium hydroxide solution, aqueous ammonia

Electrolyte Non-electrolyte

molten aluminium oxide, Solid lead(II) chloride,

lead(II) nitrate solution, solid sodium chloride,
sodium chloride solution, glucose solution,
molten lead(II) chloride,
glacial ethanoic acid,
dilute ethanoic acid, molten naphthalene,
sodium hydroxide solution,
ethanol,
aqueous ammonia tetrachloromethane,

R E D O X E Q U I L I B R I U M | 21

Explain why solid ionic substances In solid ionic substances, the ions are held together in fixed

do not conduct electricity? positions by strong attraction forces between particles.

There are no free moving ions presences in solid ionic
substances.

Explain why molten sodium Sodium chloride is an ionic compound. Molten sodium chloride
chloride can conduct electricity can conduct electricity because of the presence of the free
moving ions in the molten state.
while molten naphthalene
cannot? Naphthalene is a covalent compound and exists as a neutral
molecule. It does not contain any ions. Therefore, molten
naphthalene cannot conduct electricity.

ELECTROLYSIS OF MOLTEN COMPOUND

The diagram below shows the set-up of apparatus of electrolysis of molten lead(II) bromide.
Name the main apparatus and materials in the diagram.

Crucible Carbon electrodes

Electrolyte : Molten lead(II)
bromide, PbBr2

Electrolytic cell:
A type of cell which uses electrical energy to produce chemical reactions.
Uses source of electrical energy (dry cell) to change electrical energy to chemical energy.
Reaction in electrolytic cell will not occur without an external source of electrical energy.

R E D O X E Q U I L I B R I U M | 22

Example of electrolytic cell

Electrolysis of a molten Electrolysis of an aqueous Electrolysis of an aqueous
compound solution solution (collecting gases)

An electrolytic cell consists of battery (dry cell), electrodes and an electrolyte.

Electrode The metal/carbon (conductor) which are dipped into electrolyte and will passes
electricity through them.
Has 2 types :

➢ Inert electrodes – does not take part in the chemical reaction during the
electrolysis [Example : carbon and platinum]

➢ Active electrodes – take part in the chemical reaction during the
electrolysis [Example : copper, iron, zinc]

Anode In an electrolytic cell, anode is the electrode that is connected to the positive terminal
of the battery.
Oxidation reaction occurs here.

Cathode In an electrolytic cell, cathode is the electrode that is connected to the positive
terminal of the battery.
Reduction reaction occurs here.

Substances that can conduct electricity in either the molten state or aqueous solution
Electrolyte

and undergo chemical changes.

R E D O X E Q U I L I B R I U M | 23

ELECTROLYSIS PROCESS

Electrolysis is a process whereby compounds (electrolytes) in the molten state or an aqueous

solution decompose into their constituent elements by passing electricity through them.

Solid sodium chloride (ionic compounds) contains sodium
ions and chloride ions which are in fixed position and not
freely moving.

In solid state, sodium ions and chloride ions are strongly
attracted by electrostatic forces in a lattice.

The ions are free to move when the compound is heated
(molten) or dissolved in water (aqueous solution)

During electrolysis cations are attracted to the cathode
and anions are attracted to the anode.

Write the ionization equation of the molten compound below.

Molten Compound Ionic equation

Zinc chloride ZnCl2 Zn2+ + 2Cl-
Lead (II) bromide PbBr2 Pb2+ + 2Br-
Silver iodide AgI Ag+ + I-

Aluminum oxide Al2O3 2Al3+ + 3O2-
Magnesium chloride MgCl2 Mg2+ + 2Cl-
Potassium iodide KI K+ + I-

Copper(II) chloride CuBr2 Cu2+ + 2Cl-

R E D O X E Q U I L I B R I U M | 24
Ionisation / decomposition of the electrolytes

Anions (negative ions) move to Cations(positive ions) move to
the ANODE. the CATHODE.

Anions will undergo oxidation at Cations will undergo reduction at
the anode by the cathode by

releasing/losing/donating receiving/accepting electrons.
electrons.

Given below is a list of ionic compounds in molten state. Identify the cation and anion in each
electrolyte.

Molten Electrolyte Cation Anion

Name Formula Name Formula

Sodium chloride Sodium ion Na+ Chloride ion Cl-

Lead(II) oxide Lead(II) ion Pb2+ Oxide ion O2-

Potassium bromide Potassium ion K+ Bromide ion Br-

R E D O X E Q U I L I B R I U M | 25

Given below is a list of electrolytes and products discharged at both electrodes. Based on the
given substance discharged at the electrode, write a half equation to represent the reaction
occurring at the electrode.

Molten Electrolyte Ions attracted at the electrodes and the half equation for the reaction
Aluminium oxide occurs
Potassium iodide
Lead(II) iodide Anode Cathode
Zinc bromide
Ion attracted: oxide ion. Ion attracted: aluminium ion.
Half equation: Half equation:
2O2- → O2 + 4e- Al3+ + 3e- → Al

Ion attracted: iodide ion. Ion attracted: potassium ion.
Half equation: Half equation:
2I- → I2 + 2e- K+ + e- → K

Ion attracted: iodide ion. Ion attracted: Lead(II) ion.
Half equation: Half equation:
2I- → I2 + 2e- Pb2+ + 2e- → Pb

Ion attracted: bromide ion. Ion attracted: zinc ion.
Half equation: Half equation:
2Br- → Br2 + 2e- Zn2+ + 2e- → Zn

ELECTROLYSIS OF MOLTEN LEAD(II) BROMIDE, PbBr2

Aim To study the electrolysis of molten lead(II) bromide, PbBr2.
Materials Lead(II) bromide powder, PbBr2.

Apparatus Beaker, crucible, light bulb, battery, carbon electrodes, connecting wire with
crocodile clip, switch, tripod stand, pipe clay triangle and Bunsen burner.

Procedures

R E D O X E Q U I L I B R I U M | 26

Draw the apparatus setup diagram.
Diagram

Electrode Observation Inferences
Anode
Result Cathode Brown gas is Bromine gas is produced
Discussion released at the
anode.

Grey solid is formed Lead metal, Pb is formed
at the cathode

1. State the name of the ions that move to the cathode and anode during
electrolysis.
Lead(II) ion moves to cathode and bromide ion moves to anode.

2. Write the half equations for the reactions at the:
(a) Cathode: Pb2+ + 2e- → Pb

(b) Anode: 2Br- → Br2 + 2e-

3. Identify the products formed at the cathode and anode.
Cathode : Lead
Anode : Bromine gas

R E D O X E Q U I L I B R I U M | 27

4. Explain how the products at the cathode and anode are formed.

Cathode : Lead(II) ion gains two electrons to form lead atom.

Anode: Bromide ion loses one electron to form bromine atom. Two bromine
atoms combine to form bromine molecule. (OR two bromide ions lose two
electrons to form bromine molecule)

5. Write the overall ionic equation that represents the electrolysis of molten
lead(II) bromide, PbBr2.

Pb2+ + 2Br- → Pb + Br2

Electrolysis of lead(II) bromide

The flow chart below is used to predict the products formed at the electrodes during the
electrolysis of molten lead(II) bromide. Complete the chart.

Molten lead(II) bromide

(Ions that are present) Consists of Cathode

Anode Lead(II) ions and Bromide ions Lead(II) ions

Bromide ions ( Movement of ions)

Pb2+ + 2e- → Pb (Half equation) 2Br- → Br2 + 2e-

(Products formed) Bromine gas

Lead metal

Full ionic equation : Pb2+ + 2Br- → Pb + Br2

R E D O X E Q U I L I B R I U M | 28
Using a suitable thinking map, predict the observation and explain how the products of
electrolysis are formed at the anode and cathode for:

(a) molten zinc oxide, ZnO

(b) molten magnesium chloride, MgCl2.

R E D O X E Q U I L I B R I U M | 29

Important notes on some observation of electrolysis (molten and aqueous)

Product of Observation Confirmatory test
electrolysis

Chlorine Greenish yellow gas is Collect the gas released in a test tube.
gas released.
Place a damp blue litmus paper near the mouth of the
test tube.
The gas will turn moist blue litmus paper red and then

bleaches it

Brown gas is released. Collect the gas released in a test tube.
Place a damp blue litmus paper near the mouth of the
Bromine
gas test tube.
The gas will turn moist blue litmus paper red.

Iodine Purple gas is released (if in Collect the gas released in a test tube.
vapour
aqueous, it will dissolve in Place a damp blue litmus paper near the mouth of the
water and produce a brown test tube.

solution) The gas will turn moist blue litmus paper red.

Iodine solution :

Add a few drops of starch solution.
Starch solution turns dark blue

Colourless gas is released (in Collect the gas released in a test tube.

Oxygen gas aqueous: bubbles of Put a glowing wooden splinter into the test tube. The
colourless gas are released) glowing wooden splinter will light up / rekindles.

Hydrogen Colourless gas is released (in Collect the gas released in a test tube.
gas aqueous: bubbles of Put a burning/lighting wooden splinter at the mouth of

colourless gas are released) the test tube.
A ‘pop’ sound will be heard.

If metal is Grey (most of the metals)
formed
solid is formed // grey solid
are deposited.

Copper Brown solid formed //
metal deposited

formed

R E D O X E Q U I L I B R I U M | 30
Complete the table below on electrolysis of molten compound.
(a) Lithium chloride, LiCl

List all the ions present in Lithium ion and chloride ion
the electrolyte.

Electrode Anode Cathode

Ion attracted to the Chloride ion Lithium ion
electrode.
Li+ + e- → Li
Half equation 2Cl- → Cl2 + 2e- Grey solid is formed

Observation Greenish yellow gas is produced Lithium metal

Name of the product Chlorine gas One lithium ion receives one
electron to form a lithium atom.
Explain how the product is Two chloride ions release two
formed. electrons to form a chlorine Reduction

molecule.

Name of the reaction Oxidation

Ful ionic equation 2Cl- +2Li+ → Cl2 + 2Li

(b) Magnesium bromide

List all the ions present in Magnesium ion and bromide ion
the electrolyte.

Electrode Anode Cathode

Ion attracted to the Bromide ion Magnesium ion
electrode.
Mg2+ + 2e- → Mg
Half equation 2Br- → Br2 + 2e- Grey solid is formed

Observation Brown gas is released Lithium metal
One magnesium ion receives one
Name of the product Bromine gas
electron to form a magnesium
Explain how the product is Two bromide ions release two atom.
formed. electrons to form a bromine
Reduction
molecule.

Name of the reaction Oxidation

Ful ionic equation 2Br- + Mg2+ → Br2 + Mg

R E D O X E Q U I L I B R I U M | 31

(c) Lead(II) iodide

List all the ions present in Lead(II) ion and iodide ion
the electrolyte.

Electrode Anode Cathode

Ion attracted to the Iodide ion Lead(II) ion
electrode.
Pb2+ + 2e- → Pb
Half equation 2I- → I2 + 2e- Grey solid is formed

Observation Purple gas is released Lithium metal

Name of the product Iodine gas One lead(II) ion receives two
electrons to form a lead atom.
Explain how the product is Two iodide ions release two
formed. Reduction
electrons to form an iodine
molecule.

Name of the reaction Oxidation

Ful ionic equation 2I- + Pb2+ → I2 + Pb

R E D O X E Q U I L I B R I U M | 32

ELECTROLYSIS OF AQUEOUS SOLUTION

Molten salt consists of one type of cation and one
type of anion only.
So, both of the ions will be discharged at the cathode
and anode.

➢ An aqueous solution is produced when a solute is dissolved in water.
➢ An aqueous solution contains two types of cations (cation of the salt and hydrogen

ion).
➢ It also contains two types of anions (anion of the salt and hydroxide ion)
➢ H+ ions and OH- ions are always present together with the ions produced from the

dissociation of salts in aqueous solution.
➢ This is because water dissociates partially to form hydrogen ion and hydroxide ion.

Aqueous solution Salt Dissociation Water dissociation Cation Anion
NaCl → Na+ + Cl- H2O → H+ + OH- present present
Sodium chloride H2SO4 → 2H+ + SO42- H2O → H+ + OH- H+ , Na+ OH- , Cl-
solution HCl → H+ + Cl- H2O → H+ + OH-
CuSO4 → Cu2+ + SO42- H2O → H+ + OH- H+ OH- , SO42-
Sulphuric acid
solution H+ OH- , Cl-

Hydrochloric acid H+ , Cu2+ OH- , SO42-
solution

Copper (II)
sulphate solution

➢ During electrolysis of an aqueous salt solution, two types of cations will move to
the cathode while two types of anions will move to the anode.

➢ However, ONLY one type of cation and anion will be discharged at each electrode
at a time.

R E D O X E Q U I L I B R I U M | 33

There are three factors that affect the formation of products at the electrodes for the
electrolysis of an aqueous solution, which are:

(a) E0 Value

(b) Concentration of solution

(c) Type of electrode used.

FACTOR 1: E⁰ VALUE

 This factor will be considered if the electrolyte solution used is dilute and the
electrodes used are inert.

 Example of inert electrodes : carbon and platinum.

 The tendency of ions to be selectively discharged at electrodes depends on their E⁰
value in the Standard Electrode Potential Series.

 At anode: Anions with a MORE NEGATIVE OR LESS POSITIVE E0 VALUE in the
Standard Electrode Potential series will be easier to be discharged and oxidised.

 At cathode: Cations with a MORE POSITIVE OR LESS NEGATIVE E0 VALUE in the
Standard Electrode Potential Series will be easier to be discharged and reduced.

EXPERIMENT: ELECTROLYSIS OF AN AQUEOUS SOLUTION USING CARBON ELECTRODES

Aim To study the electrolysis of the aqueous solution of copper(II) sulphate, CuSO4 and
Materials dilute sulphuric acid, H2SO4 with carbon electrodes.

Apparatus

Procedures

R E D O X E Q U I L I B R I U M | 34

Draw the apparatus setup diagram.
Diagram

Electrode Observation Inferences
Anode
Result

Cathode

Discussion 1. State the ions present in the aqueous solution of:
(a) Copper(II) sulphate, CuSO4 solution.
Cu2+ ion, SO42- ion, H+ ion, OH- ion
(b) Sulphuric acid, H2SO4.
H+ ion, SO42- ion, OH- ion

2. For each of the copper(II) sulphate, CuSO4 solution and sulphuric acid, H2SO4:

R E D O X E Q U I L I B R I U M | 35

(a) name the ions that move to the cathode and anode during electrolysis.

(b) identify the ions discharged at the cathode and anode. Give a reason for
your answer.

(c) name the products formed at the cathode and anode.

(d) write the half equations of the reactions that take place at the cathode
and anode.

(e) explain how the products at the cathode and anode are formed.

(f) write the overall ionic equation that represents electrolysis.

CuSO4 Solution:

No Cathode Anode

(a) Cu2+ ion and H+ ion SO42- ion dan ion OH-

Cu2+ ion OH- ion.

(b) E0 value of Cu2+ ion is more positive E0 value of OH- ion is less positive

compared to E0 of H+ ion compared to E0 of SO42- ion

(c) Copper Oxygen

(d) Cu2+ + 2e- → Cu 4OH- → O2 + 2H2O + 4e-

(e) Cu2+ ion gains two electrons to form Four OH− ions lose four electrons to

copper atom form oxygen molecule and water

(f) 2Cu2+ + 4OH- → 2Cu + O2 + 2H2O

H2SO4 Solution:

No Cathode Anode

(a) H+ ion SO42- ion dan ion OH-

H+ ion OH- ion.

(b) Only H+ ion presents at cathode E0 value of OH- ion is less positive
compared to E0 of SO42- ion

(c) Hydrogen Oxygen

(d) 2H+ + 2e- → H2 4OH- → O2 + 2H2O + 4e-

(e) Two H+ ions gain two electrons to Four OH− ions lose four electrons to
form hydrogen molecule form oxygen molecule and water

(f) 4H+ + 4OH- → 2H2 + O2 + 2H2O

R E D O X E Q U I L I B R I U M | 36

Complete the following table on electrolysis of aqueous solution.

Electrolyte Copper(II) sulphate solution

Draw the apparatus set up

Ions exist in the solution Copper(II) ion, Cu2+, hydrogen ion, H+, sulphate ion, SO42− and hydroxide ion,
OH−

Electrode Anode Cathode

Ions move to the electrode. Sulphate ion, SO42− and hydroxide ion, Copper(II) ion, Cu2+ and hydrogen ion,
H+
Ion chosen to be discharged OH−
at the electrode.
Hydroxide ion, OH− Copper(II) ion, Cu2+

Reason for discharged of ion E0 value of hydroxide ion, OH− is less E0 value of copper(II), Cu2+ is more
at the electrode positive than the E0 value of sulphate positive than E0 value of hydrogen

Half equation at the ion, SO42− ion, H+
electrode
4OH− → O2 + 2H2O + 4e− Cu2+ + 2e− → Cu

Observation at the Colourless gas Brown solid deposited
electrode bubbles are released

Name of the product formed Oxygen gas Copper

Confirmatory test for the Collect the gas released in a test
product formed if any tube. Put a glowing wooden splinter

into the test tube. The glowing
wooden splinter will rekindle.

Explain how the product Four hydroxide ions, OH− release four Copper(II) ion, Cu2+ receives two
formed at the electrode electrons to form copper atom
electrons to form oxygen and water
molecules

Reaction occurs at the Oxidation Reduction
electrode

Full ionic equation 2Cu2+ + 4OH- → 2Cu + O2 + 2H2O

R E D O X E Q U I L I B R I U M | 37

Electrolyte Sodium nitrate solution
Draw the apparatus set up
*Notes:

NO3-(aq) + 4H+(aq) + 3e- -> NO(g) + 2H2O(l) E⁰ value = +0.96V

Ions exist in the solution Sodium ion, Na+, hydrogen ion, H+, nitrate ion NO3− and hydroxide ion, OH−
Electrode
Anode Cathode

Ions move to the electrode. Nitrate ion NO3− and hydroxide ion, Sodium ion, Na+ and hydrogen ion, H+
OH−

Ion chosen to be discharged Hydroxide ion, OH− hydrogen ion, H+
at the electrode.

Reason for discharged of ion E0 value of hydroxide ion, OH− is less E0 value of hydrogen ion, H+ is more
at the electrode positive than the E0 value of Nitrate positive than E0 value of Sodium ion,

Half equation at the ion NO3− Na+
electrode
4OH− → O2 + 2H2O + 4e− 2H+ + 2e- → H2

Observation at the Colourless gas Colourless gas
electrode bubbles are released bubbles are released

Name of the product formed Oxygen gas Hydrogen gas

Confirmatory test for the Collect the gas released in a test Collect the gas released in a test
product formed if any tube. Put a glowing wooden splinter tube. Put a burning wooden splinter
into the test tube. Pop sound will be
into the test tube. The glowing
wooden splinter will rekindle. heard.

Explain how the product Four hydroxide ions, OH− release four Two hydrogen ions receive 2 electrons
formed at the electrode to form hydrogen molecule
electrons to form oxygen and water
molecules

Reaction occurs at the Oxidation Reduction
electrode

Full ionic equation 4OH− + 4H+ → O2 + 2H2O + 2H2

Electrolyte R E D O X E Q U I L I B R I U M | 38
Draw the apparatus set up
Potassium chloride solution

Ions exist in the solution Potassium ion, K+, hydrogen ion, H+, chloride ion Cl− and hydroxide ion, OH−

Electrode Anode Cathode

Ions move to the electrode. Chloride ion Cl− and hydroxide ion, Potassium ion, K+, hydrogen ion, H+
OH−

Ion chosen to be discharged Hydroxide ion, OH− hydrogen ion, H+
at the electrode.

Reason for discharged of ion E0 value of hydroxide ion, OH− is less E0 value of hydrogen ion, H+ is more
at the electrode positive than E0 value of Potassium
positive than the E0 value of Chloride
ion Cl− ion, K

Half equation at the 4OH− → O2 + 2H2O + 4e− 2H+ + 2e- → H2
electrode

Observation at the Colourless gas Colourless gas
electrode bubbles are released bubbles are released

Name of the product formed Oxygen gas Hydrogen gas

Confirmatory test for the Collect the gas released in a test Collect the gas released in a test
product formed if any tube. Put a glowing wooden splinter tube. Put a burning wooden splinter
into the test tube. Pop sound will be
into the test tube. The glowing
wooden splinter will rekindle. heard.

Explain how the product Four hydroxide ions, OH− release four Two hydrogen ions receive 2 electrons
formed at the electrode to form hydrogen molecule
electrons to form oxygen and water
molecules

Reaction occurs at the Oxidation Reduction
electrode

Full ionic equation 4OH− + 4H+ → O2 + 2H2O + 2H2

R E D O X E Q U I L I B R I U M | 39

Electrolyte Silver nitrate solution
Draw the apparatus set up
*Notes:

NO3-(aq) + 4H+(aq) + 3e- -> NO(g) + 2H2O(l) E⁰ value = +0.96V

Ions exist in the solution Silver ion, Ag+, hydrogen ion, H+, nitrate ion NO3− and hydroxide ion, OH−
Electrode
Anode Cathode

Ions move to the electrode. Nitrate ion NO3− and hydroxide ion, Silver ion, Ag+, and hydrogen ion, H+
OH−

Ion chosen to be discharged Hydroxide ion, OH− Silver ion, Ag+
at the electrode.

Reason for discharged of ion E0 value of hydroxide ion, OH− is less E0 value of silver ion, Ag+, is more
at the electrode positive than the E0 value of Nitrate positive than E0 value of hydrogen

Half equation at the ion NO3− ion, H+
electrode
4OH− → O2 + 2H2O + 4e− Ag+ + e- → Ag

Observation at the Colourless gas Silver solid is deposited
electrode bubbles are released

Name of the product formed Oxygen gas Silver

Confirmatory test for the Collect the gas released in a test
product formed if any tube. Put a glowing wooden splinter

into the test tube. The glowing
wooden splinter will rekindle.

Explain how the product Four hydroxide ions, OH− release four Silver ion receive one electron to
formed at the electrode form silver atom
electrons to form oxygen and water
molecules

Reaction occurs at the Oxidation Reduction
electrode

Full ionic equation 4OH− + 4Ag+ → O2 + 2H2O + 4Ag

R E D O X E Q U I L I B R I U M | 40

FACTOR 2: CONCENTRATION OF SOLUTION

➢ This factor will be considered if the solution used is concentrated and the electrodes
used are inert.

➢ At anode:

• This factor is only considered for the selection of ions at the anode if the
aqueous solution contains halide ions.

• Halide ions with a higher concentration in the electrolytes will be discharged
and oxidised at the anode, even though the E0 value of the halide ions are more
positive.

➢ At cathode:

• Cations with a more positive or less negative E0 value in the standard
electrode potential series will be easier to be discharged and reduced.

State the selected ions to be discharged at the anode and cathode for the following
concentrated solutions.

Electrolyte Anode Cathode

Concentrated hydrochloric acid , HCl Chloride ion, Cl- Hydrogen ion, H+
solution using carbon electrodes.

Concentrated potassium iodide, KI Iodide ion, I- Hydrogen ion, H+
solution using carbon electrodes.

Concentrated sodium chloride, NaCl Chloride ion, Cl- Hydrogen ion, H+
solution using carbon electrodes.

Concentrated copper(II) nitrate, Cu(NO3)2 Hydroxide ion, OH- Copper(II) ion, Cu2+
solution using carbon electrodes.

R E D O X E Q U I L I B R I U M | 41

EXPERIMENT: THE EFFECTS OF CONCENTRATIONS OF IONS IN SOLUTIONS ON THE SELECTIVE
DISCHARGE OF IONS

Aim To study the effects of the concentrations of ions in hydrochloric
acid, HCl on the selective discharge of ions at the electrodes.
Problem
Statement
Hypothesis

Variables

Set-up apparatus

Procedures

R E D O X E Q U I L I B R I U M | 42

Observation & Discussion

Electrolyte 0.001 mol dm-3 HCl 1.0 mol dm-3 HCl

Ions attracted to Hydrogen ion, H+ Hydrogen ion, H+
the cathode.

Reaction occurs at Reduction Reduction
cathode

Half equation at 2H+ + 2e- → H2 2H+ + 2e- → H2
cathode

Observation at Bubbles of colourless gas is released Bubbles of colourless gas is released
cathode

Confirmatory test Collect the gas released in a test Collect the gas released in a test
at cathode (if any)
tube. Put a burning wooden splinter tube. Put a burning wooden splinter
into the test tube. Pop sound will be into the test tube. Pop sound will be

heard. heard.

Name the product Hydrogen gas Hydrogen gas
at cathode.

Explain how the Two hydrogen ions receive two Two hydrogen ions receive two

product is formed electrons and become a hydrogen electrons and become a hydrogen
at anode molecule. molecule.

Ions attracted to Hydroxide ion, OH- and chloride ion, Hydroxide ion, OH- and chloride ion,
the anode. Cl- Cl-

Ion chosen to be Hydroxide ion, OH- Chloride ion, Cl-
discharged at
E0 value of hydroxide ion, OH− is less Concentration of chloride ion, Cl- is
anode & reason positive than the E0 value of Chloride higher than concentration of
hydroxide ion, OH-
ion Cl−

Reaction occurs at Oxidation Oxidation
anode

Half equation at 4OH− → O2 + 2H2O + 4e− 2Cl- → Cl2 + 2e-
anode

Observation at Colourless gas Greenish yellow gas is released
anode bubbles are released

Confirmatory test Collect the gas released in a test Collect the gas released in a test
at anode (if any) tube.
tube. Put a glowing wooden splinter
into the test tube. The glowing Place a damp blue litmus paper near
the mouth of the test tube.
wooden splinter will rekindle.
The gas will turn moist blue litmus
paper red and then bleaches it

R E D O X E Q U I L I B R I U M | 43

Name the product Oxygen gas Chlorine gas
at anode.
Four hydroxide ions, OH− release four Two chloride ions, Cl- release two
Explain how the electrons to form oxygen and water electrons to form chlorine molecules
product is formed
molecules
at anode

Complete the following table for the electrolysis of 2.0 mol dm-3 sodium iodide solution using
carbon electrodes.

Set-up apparatus Carbon electrodes

2.0 mol dm-3
Sodium iodide solution

Electrode Anode Cathode

Ions attracted to the Hydroxide ion, OH- & iodide ion, I- Sodium ion, Na+ & Hydrogen ion, H+
electrode.

Ion chosen to be Iodide ion, I- Hydrogen ion, H+
Concentration of iodide ion, I- is E0 value of hydrogen ion, H+ is more
discharged at the higher than hydroxide ion, OH-
electrode & reason positive than E0 value of Sodium
ion, Na+

Reaction occurs at the Oxidation Reduction
electrode

Half equation at 2I- → I2 + 2e- 2H+ + 2e- → H2
electrode

Observation at Brown solution is formed.(iodine Bubbles of colourless gas is
electrode gas dissolves in potassium iodide released

solution)

Confirmatory test at the Collect some of the brown solution Collect the gas released in a test
electrode into a test tube. Put a few drops of
tube. Put a burning wooden
starch solution. The solution will splinter into the test tube. Pop
turn blue black.
sound will be heard.

R E D O X E Q U I L I B R I U M | 44

Name the product at Iodine Hydrogen gas
the electrode.
Two hydrogen ions receive 2
Explain how the product Two iodide ions, I- release two electrons to form hydrogen
is formed at the electrons to form iodine molecules
molecule
electrode.

Complete the following table for the electrolysis of 2.0 mol dm-3 copper(II) chloride solution
using carbon electrodes.

Carbon electrodes

Set-up apparatus

2.0 mol dm-3
Copper(II) chloride

solution

Electrode Anode Cathode

Ions attracted to the Chloride ion, Cl- & Hydroxide ion, Copper(II) ion, Cu2+ and hydrogen
electrode. OH- ion, H+

Ion chosen to be Chloride ion, Cl- Copper(II) ion, Cu2+
E0 value of copper(II), Cu2+ is more
discharged at the Concentration of chloride ion, Cl- is positive than E0 value of hydrogen
electrode & reason higher than concentration of
hydroxide ion, OH- ion, H+

Reaction occurs at the Oxidation Reduction
electrode

Half equation at 2Cl- → Cl2 + 2e- Cu2+ + 2e− → Cu
electrode

Observation at electrode Greenish yellow gas is released Brown solid deposited

Confirmatory test at the Collect the gas released in a test
electrode tube.

Place a damp blue litmus paper
near the mouth of the test tube.

The gas will turn moist blue litmus
paper red and then bleaches it

R E D O X E Q U I L I B R I U M | 45

Name the product at the Chlorine gas Copper
electrode.
Two chloride ions, Cl- release two Cu2+ ion gains two electrons to
Explain how the product electrons to form chlorine form copper atom
is formed at the molecules
electrode.

FACTOR 3: TYPE OF ELECTRODE

 The common materials used as electrodes are carbon and platinum because they are
inert. Both materials do not react with the electrolytes or the products of electrolysis.
They only act as conductor and does not undergo any chemical changes.

 The types of electrodes used can determine the type of ions discharged at the anode.

 If active electrode (such as nickel, copper, silver) is used at the anode, the electrode
will ionize. (The electrode at anode will release electrons and become a positive ion,
the anode then will dissolve and become thinner)

 At anode:

➢ This factor is only considered for active electrodes (example: copper or silver).

➢ No anions are discharged.

➢ Metal atoms at the anode release electrons to form metal ions.

 At cathode:

➢ Cations with a more positive or less negative E0 value in the standard electrode
potential series will be easier to be discharged and reduced.

EXPERIMENT: THE EFFECTS OF TYPE OF ELECTRODE ON THE SELECTION OF IONS TO BE
DISCHARGED

Aim To study the effects of the type of electrode used on the selective discharge of
ions at the electrodes.
Problem
Statement Does the type of electrode used affect the selective discharge of ions at the
Hypothesis electrodes?

Variables When carbon electrode is used, colourless gas bubbles released at anode while
copper electrode is used, anode copper becomes thinner.

a. Manipulated variable: carbon electrode and copper electrode.
b. Responding variable: Observation at anode.
c. Fixed variable: Copper(II) sulphate solution

R E D O X E Q U I L I B R I U M | 46

Materials 0.5 mol dm−3 of copper(II) sulphate, CuSO4 solution, carbon electrode, copper
Apparatus electrode.
Procedures
Beaker, battery, , connecting wires with crocodile clips, switch, ammeter, test
Observation tube, wooden splinter, electronic scale and sandpaper.

1. Clean two carbon electrodes with sandpaper.

2. Pour 0.5 mol dm−3 of copper(II) sulphate solution, CuSO4 into a beaker until
half full.

3. Connect carbon electrodes to ammeter, switch and batteries using
connecting wire.

4. Dipped the carbon electrodes into copper(II) sulphate solution to complete
the circuit.

5. Turn on the switch to allow electricity to pass through the electrolyte for
15 minutes.

6. Observe and record the changes that occur at the electrode, anode and
cathode.

7. Repeat steps 1 to 6 using copper electrodes to replace carbon electrodes.

Type of Cathode Observation Electrolyte
electrode Anode
Blue colour of
Carbon Brown solid is copper(II) sulphate
deposited solution becomes
Bubbles of colourless paler / Intensity of
gas is released Copper cathode blue colour of
becomes thicker copper(II) sulphate
Copper solution decreases.

Copper anode Blue colour of
becomes thinner copper(II) sulphate
solution remains the

same

Discussion Explanation

Electrolysis Ions present Copper(II) ions, Cu2+, sulphate ion, SO42-, Hydrogen
of CuSO4 in the ion, H+ and Hydroxide ion, OH-
solution solution
Anode Cathode
using carbon Electrode
electrodes
Name of the
product Oxygen gas Copper

R E D O X E Q U I L I B R I U M | 47

Reaction Oxidation. Reduction
occurs at
OH- ion is chosen to be Cu2+ is chosen to be
the discharged because E0 discharged because E0
electrode & value of OH- ion is less value of Cu2+ ion is more
explanation positive than E0 value of positive compared to E0

SO42- ion. of H+ ion.

Four OH− ions lose four Cu2+ ion gains two
electrons to form oxygen electrons to form copper

molecule and water atom

Half 4OH- → O2 + 2H2O + 4e- Cu2+ + 2e- → Cu
equation

Explanation

Electrode Anode Cathode

Name of the Copper(II) ions Copper
product

Electrolysis Reaction Oxidation Reduction
of CuSO4 occurs at
solution Copper electrode is an Cu2+ is chosen to be
the active electrode, so it discharged because E0
using copper electrode & value of Cu2+ ion is more
electrodes explanation will oxidize. positive compared to E0

Copper atom releases 2 of H+ ion.
electrons to form
copper(II) ions. Cu2+ ion gains two
electrons to form copper

atom

Half Cu → Cu2+ + 2e- Cu2+ + 2e- → Cu
equation

The blue colour of copper(II) sulphate solution remains the same
because the concentration of copper(II) ions, Cu2+ remain the
same.

The rate of copper atom at being ionised / oxidised at anode is the
same with the rate of copper(II) ions, Cu2+ being
discharged/reduced at cathode

R E D O X E Q U I L I B R I U M | 48
Complete the table below on electrolysis using different metals as the electrodes.
Electrolysis of silver nitrate solution, AgNO3 using silver electrode.

Draw the apparatus set up.

State all the ions present in the electrolyte:
Silver ion, Ag+, hydrogen ion, H+, nitrate ion, NO3- and hydroxide ion, OH-

Anode Cathode

Observation: Observation:
The size of silver electrode become The size of silver electrode becomes thicker.
thinner/smaller. / Grey solid is deposited.
/The silver electrode dissolves.

Name of the product: Name of the product:
Silver ion Silver

Explain the reaction that occurs: Explain the reaction that occurs:
Silver electrode is an active electrode.
So, silver electrode will be oxidized. Silver ion is chosen to be discharged at
One silver atom release one electron to form cathode because E0 value of silver ion, Ag+ is
one silver ion, Ag+. more positive than E0 value of hydrogen ion,
H+.

Silver ion will be reduced.

One silver ion will receive one electron to
form one silver atom.

Write the half equation: Write the half equation:
Ag → Ag+ + e- Ag+ + e- → Ag

R E D O X E Q U I L I B R I U M | 49

Full ionic equation :
Ag + Ag+ → Ag+ + Ag

State the electron flow:
Electron flow from anode to cathode through connecting wires.

Electrolysis of magnesium sulphate solution using magnesium (anode) and carbon (cathode)
electrode.

Draw the apparatus set up:

Magnesium Carbon

Magnesium
sulphate solution

State all the ions present in the electrolyte:
Magnesium ion, Mg2+, hydrogen ion, H+, sulphate ion, SO42- and hydroxide ion, OH-

Anode Cathode

Observation: Observation:

The size of magnesium electrode become Bubbles of colourless gas is released at
thinner/smaller. cathode.

/The magnesium electrode dissolves.

Name of the product: Name of the product:
Magnesium ion Hydrogen gas

Explain the reaction that occurs: Explain the reaction that occurs:
Magnesium electrode is an active electrode.
So, magnesium electrode will be oxidized. Magnesium ion, Mg2+ and hydrogen ion, H+ are
One magnesium atom release two electrons attracted to the cathode.
to form one magnesium ion, Mg2+.
Hydrogen ion is chosen to be discharged at
cathode because E0 value of hydrogen ion, Ag+
is more positive than E0 value of magnesium
ion, H+.

Hydrogen ion will be reduced.

Two hydrogen ions will receive two electrons
to form one hydrogen molecule

Write the half equation: Write the half equation:
Mg → Mg2+ + 2e- 2H+ + 2e- → H2

R E D O X E Q U I L I B R I U M | 50

Full ionic equation :
Mg+ 2H+ → Mg2+ + H2
State the electron flow:
Electron flow from anode to cathode through connecting wires.

ELECTROPLATING OF METALS

Electroplating is a process to coat an object with a thin protective layer of a metal.
The purpose of electroplating is to make the object:

➢ More attractive and improve the appearance of the object.
➢ Prevent corrosion / rusting.
In electroplating,
➢ The anode is the electroplating metal.
➢ The cathode is the object to be electroplated.
➢ The electrolyte must contain the ions of the plating metal.
For a good quality of electroplating,
➢ The current must be small.
➢ Use a low concentration of electrolyte.
➢ Rotate the object to be plated during electroplating.