PRA U STPM CHEMISTRY PENGGAL 2

CONTENTS

Chapter Chemical Energetics 1 Chapter Group 14 163

7 • ••••••••••• ••••••••••••••• • • • • • • • • • • • • • • • • • •• • • • • • • •1• •1• • ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• ••••••••••••••• • • • • • • •

7.1 Enthalpy Changes of Reaction, ∆H 2 11.1 Physical Properties of Group 14

7.2 Hess’ Law 19 Elements 164

7.3 Born-Haber Cycle 21 11.2 Tetrachlorides and Oxides of

7.4 The Solubility of Solids in Liquids 27 Group 14 Elements 168

Summary 28 11.3 Relative Stability of +2 and +4 174
STPM Practice 7 29 Oxidation States of Group 14

Elements

Chapter 11.4 Silicon, Silicone and Silicates 177

8• Electrochemistry 38 • • • • • • • 11.5 Tin Alloys • • • • • • • • • • • • • • • • • • • • • • • • • • 184 • • • • • • •
• • •S•u• •m• •m• •a•r•y• • • • • • • 1• 8• •4•
••••••••••••••••••••••••••••••••••••••••••••• ••

8.1 Half-cell and Redox Equations 39 STPM Practice 11 185

8.2 Standard Electrode Potential 50 Chapter
8.3 Non-standard Cell Potentials 69 Group 17 189
8.4 Fuel Cells 79 12 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

8.5 Electrolysis 82 12.1 Physical Properties of Selected

8.6 Applications of Electrochemistry 91 Group 17 Elements 190

Summary 97 12.2 Reactions of Selected Group 17

STPM Practice 8 98 Elements 193

12.3 Reactions of Selected Halide Ions 199

Chapter 12.4 Industrial Applications of

9 Periodic Table: Periodicity 108 Halogens and Their Compounds 201

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •S•u• •m• •m• •a•r•y• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2• 0• •2• • • • • • • •

9.1 Physical Properties of Elements STPM Practice 12 203

of Period 2 and Period 3 109 Chapter

9.2 Reactions of Period 3 Elements 131 13 Transition Elements 207
with Oxygen and Water • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

9.3 Acidic and Basic Properties of 13.1 Physical Properties of First Row

Oxides and Hydrolysis of Oxides 136 Transition Elements 208

Summary 138 13.2 Chemical Properties of First Row

STPM Practice 9 139 Transition Elements 216

13.3 Nomenclature and Bonding of

Chapter Group 2 144 Complexes 229

10 • •••••••• ••• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •1•3• .•4• • •EU• •lse•em•s•eo•n•f t•Fs•i•ar•sn•td•R•T•o•hw•e•iT•rr•Ca•n•o•smi•t•ip•oo•n•u•n•d• s• • • • • • • 2• 3• •6• • • • • • • •

10.1 Selected Group 2 Elements and Summary 237

Their Compounds 145 STPM Practice 13 237

10.2 Anomalous Behaviour of Beryllium 156

10.3 Uses of Group 2 Compounds 159 • STPM Model Paper (962/2) 241

Summary 159 • Appendix 247

STPM Practice 10 159 • Glossary 256
• Answers 259

• Index 279

viii

8CHAPTER ELECTROCHEMISTRY

Concept Map Electrochemistry

Daniell Disproportionation Oxidation Corrosion Standard Electrode
Cell Number • Rusting of iron Potential Series
• Redox reactions • Preventing rusting

Electrochemical Cells Electrode Potential Constructing Redox Electrolysis
• Cells
• Construction • Standard hydrogen Equations • Molten lead(II) bromide

• Fuel cells electrode • Methods electrolysis
• Factors affecting
• Batteries for electric cars • Measuring standard • Oxidation number
electrolysis products
electrode potential • Ion-electron • Quantitative electrolysis

• Balancing redox equations Applications of
Electrolysis in Industries
Factors Affecting Standard in basic solutions • Extraction of aluminium
• Anodisation
Electrode Potential • Manufacturing chlorine
• Effluent treatment
• Concentration • Electroplating

• Nernst equation Uses of The Standard Electrode Potential
• Pressure Series
• Nernst equation and • Compare strength of oxidising agents and

– Electrochemical cell reducing agents
– Equilibrium constant • Predicting feasibility of redox reactions
– Solubility product • Predict stability of aqueous ions
• pH • Calculating e.m.f. of electrochemical cell
• Complex formation

Learning  Outcomes

Students should be able to: and higher voltage, as exemplified by hydrogen-oxygen fuel
cell.
Half-cell and redox equations
• explain the redox processes and cell diagram (cell notation) Electrolysis
• compare the principles of electrolytic cell to electrochemical
of the Daniell cell; construct redox equations.
cell;
Standard electrode potential
• describe the standard hydrogen electrode; • predict the products formed during electrolysis;

• use the standard hydrogen electrode to determine standard • state the Faraday’s first and second laws of electrolysis;
electrode potential (standard reduction potential), E °;
• state the relationship between the Faraday constant, the
• calculate the standard cell potential using the E ° values, and Avogadro constant and the electronic charge;
write the redox equations;
• calculate the quantity of electricity used, the mass of material
• predict the stability of aqueous ions from E ° values; and/or gas volume liberated during electrolysis.

• predict the power of oxidising and reducing agents from E ° Applications of electrochemistry
values; • explain the principles of electrochemistry in the process and

• predict the feasibility of a reaction from E °cell value and from prevention of corrosion (rusting of iron);
the combination of various electrode potentials: spontaneous
and non-spontaneous electrode reactions. • describe the extraction of aluminium by electrolysis, and state
the advantages of recycling aluminium;
Non-standard cell potentials
• calculate the non-standard cell potential, Ecell, of a cell using • describe the process of anodisation of aluminium to resist
corrosion;
the Nernst equation.
• describe the diaphragm cell in the manufacture of chlorine
Fuel cells from brine;
• describe the importance of the development of more efficient
• describe the treatment of industrial effluent by electrolysis to
batteries for electric cars in terms of smaller size, lower mass remove Ni2+, Cr3+ and Cd2+;

• describe the electroplating of coated plastics.

Chemistry Term 2  STPM  Chapter 8 Electrochemistry 

8.1 Half-cells and Redox
Equations

1 Electrochemistry deals with reactions involving the transfer of 8

electrons from one chemical species to another, and the electrical + Current –
Electron
energy that is produced or used during these reactions.
2 Such reactions are called redox reactions, which is a short-form for

reduction and oxidation.
3 Oxidation is defined as a process of electron loss. For example,
  Fe ⎯→ Fe2+ + 2e–
  2I– ⎯→ I2 + 2e–
4 Reduction is defined as a process of electron gain. For example,
  Fe3+ + e– ⎯→ Fe2+
  Cl2 + 2e– ⎯→ 2Cl–
5 When electrons are made to move through a solid conductor, an

electrical current is generated and flows in the opposite direction.

Daniell Cell

1 The conversion of chemical energy into electrical energy is The Daniell cell converts
demonstrated by the Daniell cell. A simplified diagram of the Daniell chemical energy into electrical
cell is shown below. energy.

V+

Zn Salt bridge Cu

ZnSO4 (aq) CuSO4(aq)



2 The cell consists of two half-cells. The zinc half-cell consists of a zinc

rod partially immersed in aqueous zinc sulphate, and the copper

half-cell consists of a copper electrode partially immersed in aqueous

copper sulphate.

3 The two half-cells are joined using a salt bridge which is a glass tubing Sometimes, saturated potassium
nitrate is used as the salt bridge.
filled with saturated potassium chloride. Electricity flows from copper to
zinc.
4 When the circuit is closed, an electric current flow from the copper Electrons flow from zinc to
copper.
electrode to the zinc electrode as indicated by the voltmeter. As Blue colour of CuSO4 fades.

a result, there is a corresponding flow of electrons from the zinc 39

electrode to the copper electrode.

5 As time passes, the zinc electrode decreases in size, while the copper

electrode increases in size, and the blue colour of the copper sulphate

solution slowly fades.

6 As current is drawn from the cell, the zinc electrode in the zinc half-

cell dissolves to produce aqueous Zn2+ ions.

Zn(s) → Zn2+(aq) + 2e–

This is called a half-equation.

  Chemistry Term 2  STPM  Chapter 8 Electrochemistry

7 The electrons travel through the external circuit to the copper

half-cell, where the Cu2+ aqueous ions accept the electrons and get

Info Chem converted to copper which adhere to the copper electrode.

The blue colour of aqueous Cu2+(aq) + 2e– → Cu(s)
copper(II) sulphate is due to
the presence of Cu2+ aqueous 8 As more and more Cu2+ aqueous ions are removed from the

TipsCioun(sHo2Or Em)6x2o+raeaqmcuoerroeucstlyiotnhse. electrolyte, the blue colour fades.

9 Thus, there is a flow of electrons from the zinc electrode to the

copper electrode, and correspondingly a current flow from the copper

electrode to the zinc electrode through the connecting wires.

8 Electricity flows from the positive 10 The copper electrode forms the positive terminal and the zinc
terminal to the negative terminal.
electrode forms the negative terminal of the Daniell cell.

[By convention, electric current flows from the positive terminal to

the negative terminal of a cell].

Electrons

Negative terminal Zinc Copper Positive terminal

Exam Tips Current

The cell notation for the Daniell 11 The overall cells reaction that produces the electricity is:
cell is:
Zn(s) Zn2+(aq) Cu2+(aq) Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

In the reaction:

(a) Zinc undergoes oxidation by losing electrons.

(b) Copper(II) ions get reduced by gaining electrons.

(c) Zinc supplies electrons to copper(II) ions. Zinc is called a

reducing agent.

(d) Copper(II) ions remove (or accept) electrons from zinc.

Copper(II) ion is an oxidising agent.

12 Reducing agents are electron donors.

Oxidising agents are electron acceptors.

13 Note that zinc, as a reducing agent, undergoes oxidation, while

copper(II), as an oxidising agent, undergoes reduction.

14 By definition, anode is the electrode where oxidation occurs, and

cathode is the electrode where reduction occurs.

15 Thus, zinc is the anode of the cell, and copper is the cathode of the

Daniell cell.

Electrons

Anode/negative Zinc Copper Cathode/positive

Current

16 Note that oxidation and reduction must occur simultaneously. We

cannot have one without the other. The electrons supplied by the

reducing agent must be received by the oxidising agent.

Reducing agent gets oxidised. 17 The function of the salt bridge is to maintain electrical neutrality
Oxidising agent gets reduced.
of the system. As zinc dissolves to form Zn2+(aq) ions and Cu2+(aq)

ions are reduced to copper, there is surplus of positive charge in the

zinc half-cell while there is a surplus of negative charge in the copper

half-cell.

40

Chemistry Term 2  STPM  Chapter 8 Electrochemistry 

18 The salt bridge allows the migration of Zn2+(aq) ions into the copper

half-cell and migration of SO42–(aq) ions from the copper half-cell
into the zinc half-cell at the same time.

ZnSO4 (aq) CuSO4 (aq) The salt bridge completes
the circuit and helps to maintain
Zn2+ SO42– SO42– Zn2+ electrical neutrality and also
prevents the electrolytes from the
19 The Daniell cell can be represented by the following cell diagram: two half-cells from mixing. 8

Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) Another name for the cell
diagram is cell notation.

20 Summary:

Oxidation is the process of electron loss.

Reduction is the process of electron gain.

Oxidising agents are electron acceptors.

Reducing agents are electron donors.

Example 8.1

Identify the oxidising agents and reducing agents in the following
reactions by writing the appropriate half-equations.
(a) Cl2 + 2Fe2+ → 2Cl– + 2Fe3+
(b) Cu + 2Ag+ → Cu2+ + 2Ag

Solution
(a) Cl2 + 2e– → 2Cl–
Fe2+ → Fe3+ + e–
Cl2 is the oxidising agent (electron acceptor).
Fe2+ is the reducing agent (electron donor).

(b) Cu → Cu2+ + 2e–
Ag+ + e– → Ag
Cu is the reducing agent (electron donor).
Ag+ is the oxidising agent (electron acceptor).

Quick Check 8.1

Identify the oxidising agents and reducing agents in the following reactions.

(a) Mg + Fe2+ → Mg2+ + Fe (c) Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
(b) Cl2 + Sn2+ → 2Cl– + Sn4+ (d) S2O82– + 2I– → 2SO42– + I2

Oxidation Number (O.N.) or Oxidation State 2015/P2/Q19(b)

1 All atoms, either in elements, compounds or ions, can be assigned All elements have ‘zero’ oxidation
an oxidation number or oxidation state. number.

2 All atoms in their respective elemental state are assigned oxidation
numbers of ‘zero’.

41

  Chemistry Term 2  STPM  Chapter 8 Electrochemistry

For example:

Element H2 Na Al Fe C Cl2 S8 N2

Oxidation number/state 0 0 0 0 0 0 0 0

3 In simple ions, the oxidation number of the atom/element concerned
is the same as the charge on the ion.

For example:

Oxidation number of simple ions. Element Cu+ Mg2+ Al3+ Sn4+ Cl– O2– N3–

Oxidation number +1 +2 +3 +4 –1 –2 –3

8 Sum of all the oxidation numbers 4 For a polyatomic ion (or molecular ion), the sum of all the oxidation
must give the charge of the ion. numbers of the elements in the ion is equal to the net charge on the
ion.

For example:
SO42– : (oxidation number of S) + 4(oxidation number of O) = –2

Cr2O72– : 2(oxidation number of chromium) + 7(oxidation number of O) = –2

MnO4– : (oxidation number of Mn) + 4(oxidation number of O) = –1

5 For a neutral covalent molecule, the more electronegative element

would be assigned a negative oxidation number, while the less

electronegative element will have a positive oxidation number. The

sum of the oxidation numbers of all the atoms in the molecule is

zero. For example:

CO2 : (oxidation number of C) + 2(oxidation number of oxygen) = 0

NaAl(OH)4 : (O.N. of Na) + (O.N. of Al) + 4(O.N. of O) + 4(O.N. of H) = 0

6 Certain elements have fixed oxidation number in all their compounds.
For example, all elements in Group 1, Group 2 and Group 13 have
fixed oxidation number of +1, +2 and +3 respectively.

Compound Na2O MgCl2 Al2(SO4)3
O. N. of the metallic element +1 +2 +3

Fluorine exhibits a constant 7 Fluorine (F), the most electronegative element, has a fixed oxidation
oxidation number of –1 in all its state of –1 in all its compounds.
compounds.
Compound AlF3 Na3AlF6 SbF5
O. N. of F –1 –1 –1

8 Hydrogen (H) is assigned an oxidation number of +1 for all its
compounds, except when it combines with elements from Group 1,
2 and 13 (to form metal hydrides), where it has an oxidation number
of –1.

For example:

Some elements have variable Compound CH4 SiH4 NH3 H2O HCl HNO3
oxidation states
Oxidation number of hydrogen +1 +1 +1 +1 +1 +1

Compound Lithium hydride, Magnesium Aluminium
Oxidation number of LiH hydride, MgH2 hydride, AlH3
hydrogen –1
–1 –1

42

Chemistry Term 2  STPM  Chapter 8 Electrochemistry 

9 Oxygen has an oxidation number of –2, except in peroxides (such as Lewis diagram of the peroxide
barium peroxide, BaO2), where it has an oxidation number of –1. ion, O22–:

Compound Na2O MgO SO3 P4O10 xx ••

Oxidation number of oxygen –2 –2 –2 –2 •O– x

Compound Hydrogen peroxide, Barium peroxide, xx O– ••
Oxidation number of oxygen H2O2 BaO2
–1 –1 xx ••

However, oxygen exhibits oxidation number of +2 in its combination 8
with fluorine to form fluorine oxide, F2O.

Example 8.2

Determine the oxidation number of the underlined element in the

following species.

(a) NiO2 (c) H2CO3 (e) CH3Cl

(b) MnO4– (d) S2O32– (f) NH3

Solution

(a) Let the oxidation number of Ni = x

x + 2(–2) = 0

x = +4

(b) Let the oxidation number of Mn = x

x + 4(–2) = –1

x = +7

(c) Let the oxidation number of C = x

2(+1) + x + 3(–2) = 0

2 + x – 6 = 0

x = +4 Info Chem

(d) Let the oxidation number of S = x Actually this is the average
oxidation state of S. The two
2x + 3(–2) = –2 sulphur atoms in S2O32– have
oxidation numbers of –2 and +6
2x = +4 respectively. The average
= —–2—2+—6 = +2.
x = +2

(e) Let the oxidation state of carbon = x

x + 3(+1) + (–1) = 0

∴ x = –2

(f) Let the oxidation state of nitrogen be x.

x + 3(+1) = 0

∴ x = –3

Quick Check 8.2

Determine the oxidation numbers of the following underlined elements.

1 SO3 6 S4O62– (Determine the average oxidation number)

2 FeO42– 7 Fe(CN)64–
3 ClO4– 8 Cu2SO4

4 VO2+ 9 MnO42–
5 VO2+ 10 K4Fe(CN)6

43

  Chemistry Term 2  STPM  Chapter 8 Electrochemistry

Oxidation Number and Redox Reactions

1 Redox reactions can also be defined in terms of oxidation number.

Definition of oxidation and 2 Oxidation is defined as a reaction where there is an increase in the
reduction in terms of oxidation
number oxidation number of an atom/element.

8 Reduction is defined as a reaction where there is a decrease in the

oxidation number of an atom/element.

3 Examples of oxidation are:

Al ⎯→ Al3+ + 3e–

Oxidation number 0 +3

Fe2+ ⎯→ Fe3+ + e–

Oxidation number +2 +3

2SO42– ⎯→ S2O82– + 2e–

Oxidation number +6 +7

4 Examples of reduction are:
Cl2 + 2e– ⎯→ 2Cl–


Oxidation number 0 –1

H2 O 2 + 2H+ + 2e– ⎯→ 2H2O

Oxidation number –1 –2

Quick Check 8.3

Determine whether the underlined elements in the following equation have undergone oxidation or
reduction, and name the oxidising and reducing agent.
(a) Cr2O72– + 8H+ + 3NO2– + 3H2O → 2Cr3+ + 7H2O + 3NO3–
(b) 2MnO4– + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
(c) 10HNO3 + I2 → 10NO2 + 2HIO3 + 4H2O
(d) H2SO4 + 6HI → 3I2 + S + 4H2O
(e) 4Co3+ + 2H2O → 4Co2+ + O2 + 4H+
(f) 5H2O2 + 2IO3– + 2H+ → I2 + 5O2 + 6H2O

Definition of disproportionation Disproportionation

1 Disproportionation is a redox reaction where the same substance
gets oxidised and reduced simultaneously.

2 Take the following redox reaction:
2Fe3+ + Sn2+ → 2Fe2+ + Sn4+

The two half-equations are:
Fe3+ + e– → Fe2+
Sn2+ → Sn4+ + 2e–

In the above half-equations:
(a) Fe3+ undergoes reduction, i.e. Fe3+ is the oxidising agent (electron
acceptor).
(b) Sn2+ undergoes oxidation, i.e. Sn2+ is the reducing agent (electron
donor).

44

Chemistry Term 2  STPM  Chapter 8 Electrochemistry 

3 Now, let’s look at another redox reaction:

2Cu2+ ⎯→ Cu2+ + Cu

Oxidation number +1 +2 0

The two half-equations are:

Cu+ ⎯→ Cu2+ + e–

Cu+ + e– ⎯→ Cu

In the above half-equations:

(a) Cu+ is oxidised to Cu2+, i.e. Cu+ is the reducing agent (electron

donor). 8

(b) Cu+ is reduced to Cu, i.e. Cu+ is the oxidising agent (electron

acceptor).

(c) the oxidation number of copper is increased and decreased at

the same time. The reverse reaction:
Cu2+ + Cu → 2Cu+
(d) Cu+ is oxidised and reduced at the same time.
is not disproportion because Cu2+
(e) Cu+ is the oxidising agent as well as the reducing agent. is the oxidising agent, while Cu is
the reducing agent.
(f) Such reaction is called disproportionation.

Example 8.3

Determine if the following reaction is disproportionation.

2H2O2 → 2H2O + O2
Solution

2H2O2 → 2H2O + O2

Oxidation number –1 –2 0

The oxidation number of oxygen in H2O2 is decreased to –2 (in H2O)
and increased to 0 (in O2). Hence, it is disproportionation.
H2O2 acts as an oxidising agent as well as a reducing agent in the
reaction.

Quick Check 8.4

Determine which of the following reactions are disproportionation.
1 Cl2 + 2OH– → Cl– + ClO– + H2O
2 3BrO– → 2Br– + BrO3–
3 2CrO42– + 2H+ → Cr2O72– + H2O
4 C12H22O11 → 12C + 11H2O
5 I2 + H2O → HI + HIO
6 Na2S2O3 + 2H+ → 2Na+ + SO2 + S + H2O
7 3BrO– → 2Br– + BrO3–
8 3Fe2+ → Fe + 2Fe3+
9 5Cl– + ClO3– + 3H2O → 3Cl2 + 6OH–
10 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O

45

  Chemistry Term 2  STPM  Chapter 8 Electrochemistry

Constructing Redox Equations

1 Redox equations, like any other chemical equations, must be balanced
both in terms of mass and charge.

2 There are two main methods for balancing redox equations: The
oxidation number method and the ion-electron method.

The Oxidation Number Method

1 Redox reactions involve transfer of electrons from one chemical
species to the other.

8 2 The same number of electrons that is released by a reducing agent
must be accepted by an oxidising agent.
Exam Tips
3 The total increase in the oxidation number of one species
Exam Tips = the total decrease in the oxidation numbers of the other species.

Always equate the unknown 4 The net change in the oxidation numbers of all the species in a
with the largest coefficient balanced redox equation is zero.
equal to 1.
5 In places where there is not enough oxygen atoms, it is balanced by
adding ‘H2O’.

6 To balance hydrogen atoms, add ‘H+’ to the appropriate side of the
equation.

7 Consider a general redox reaction:
  aA + bB ⎯→ Products
Let the change in the oxidation number of A in the reaction be +x

unit, and the change in the oxidation number of B be –y unit, then:
  a(+x) + b(–y) = 0

8 The following example serves to illustrate this method.
  H2O2 + Fe2+ ⎯→ Fe3+ + H2O

(a) Let the equation be:

aH2O2 + bFe2+ ⎯→ cFe3+ + dH2O

Oxidation number –1 +2 +3 –2

(b) The change in the oxidation state of oxygen is –1.

The change in the oxidation state of iron is +1.

(c) The total change in the oxidation state of oxygen is 2a(–1), since

there are 2a atoms of oxygen.

The total change in the oxidation number of iron is b(+1).

(d) Hence, 2a(–1) + b(+1) = 0

b – 2a = 0

If, a = 1

Then, b = 2

(e) Substitute into the equation, we get:

  H2O2 + 2Fe2+ ⎯→ c Fe3+ + dH2O
(f) Balancing both sides:

  c = 2 and d = 2

  H2O2 + 2Fe2+ ⎯→ 2Fe3+ + 2H2O
(g) The left-hand side of the equation is short of 2 hydrogen atoms.

46

Chemistry Term 2  STPM  Chapter 8 Electrochemistry 

This is balanced by adding 2 H+ to the left-hand side of the
equation.
(h) The balanced equation is:
H2O2 + 2H+ + 2Fe2+ ⎯→ 2Fe3++ 2H2O

Example 8.4

Balance the following equations using the oxidation number method. 8
(a) Cr2O72– + Fe2+ → Cr3+ + Fe3+
(b) ClO3– + I– → Cl– + I2

Solution
(a) Let the equation be:

aCr2O72– + bFe2+ → cCr3+ + dFe3+

Oxidation number +6 +2 +3 +3

The total change in the oxidation number of Cr

= 2a(3 – 6) = –3(2a)

The total change in the oxidation number of iron = b(+1)
 ∴ –6a + b = 0
If, a = 1 Then, b = 6

Substitute into the equation:
 Cr2O72– + 6Fe2+ → cCr3+ + dFe3+
Solving: c = 2 and d = 6

Hence, the equation becomes:
 Cr2O72– + 6Fe2+ → 2Cr3+ + 6Fe3+

Add 7H2O to the right-hand side to balance the oxygen atoms:
 Cr2O72– + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O

Add 14H+ on the left-hand side to balance the hydrogen atoms:

   Cr2O72– + 6Fe2+ + 14H+ →2Cr3+ + 6Fe3+ + 7H2O
(b) Let the equation be:

aClO3– + bI– → cCl– + dI2

   Oxidation number +5 –1 –1 0

The change in the oxidation number of Cl = a[(–1) – (+5)]

= –6a

Total change in the oxidation number if iodine = b[(0) – (–1)]
= +b

 ∴ b + (–6a) = 0
 If, a = 1
Then, b = 6

Substitute into the equation:
 ClO3– + 6I– → cCl– + dI2

Balancing both sides:
c = 1 and d = 3

47

ANSWERS

Chapter 7 Chemical Energetics 3 (a) C(Diamond) C(Graphite)

Quick Check 7.1 –396 kJ –394 kJ

1 (a) Ag+(aq) + Cl–(aq) → AgCl(s) ∆H° = –66 kJ mol–1 CO2
(b) +1.32 kJ

2 (a) +354 kJ (b) –885 kJ (b) –2 kJ mol–1 (c) Graphite

3 (a) –8.55 kJ (b) 29.12 °C 4 –848 kJ
5 +45.0 kJ mol–1
4 (a) –100 kJ mol–1 (b) 45.02 kJ 6 (a) 2C(s) + —21 O2(g) + 3H2(g) → C2H5OH(l)

5 (a) –150.3 kJ (b) –58.94 kJ (c) 970.1 g (b) –279 kJ mol–1

6 (a) 831.82 kJ (b) –2868.34 kJ 7 –1260 kJ mol–1

Quick Check 7.2 8 –334.6 kJ
1 (a) Ca(s) + C(s) + —32 O2(g) → CaCO3(s) 9 ((ba)) C–22H005.O2HkJ(lm) +ol–—112 O2(g) → CH3CHO(l) + H2O(l)

(b) Cu(s) + S(s) + 5H2(g) + —92 O2(g) → CuSO4•5H2O(s) 10 –466 kJ mol–1
(c) 2C(s) + 2H2(g) + O2(g) → CH3COOH(l)
(d) —12 N2(g) + O2(g) → NO2(g) Quick Check 7.7
(e) —12 H2(g) + —12 I2(s) → HI(g)
(f) C(s) + 2Cl2(g) → CCl4(l) 1 (a) Na+(g) + Br–(g) → NaBr(s)
(g) Cd(s) + 2C(s) + N2(g) → Cd(CN)2(s) (b) Ag+(g) + Cl–(g) → AgCl(s)
(c) 2Na+(g) + O2–(g) → Na2O(s)
Quick Check 7.3 (d) Ca2+(g) + 2F–(g) → CaF2(s)
(e) 2Al3+(g) + 3O2–(g) → Al2O3(s)
1 –876.0 kJ
2 +127 kJ mol–1 Quick Check 7.8 (b) MgSO4 (c) Al2O3
1 (a) MgO (e) Na2O
(d) LiF

Quick Check 7.4 Quick Check 7.9 2 –494 kJ mol–1
1 (a) Na(s) + —14 O2(g) → —21 Na2O(s) 1 –2050.6 kJ mol–1
3 –325 kJ mol–1
(b) H2C2O4(s) + —21 O2(g) → 2CO2(g) + H2O(l)
(c) SO2(g) + —21 O2(g) → SO3(g) Quick Check 7.10 (b) Soluble
(d) NH3(g) + —43 O2(g) → —12 N2(g) + —23 H2O(l) (b) Insoluble
2 –720 kJ mol–1 1 (a) –18 kJ mol–1
2 –86.7 kJ mol–1
3 91.65 °C 3 12.8 °C
4 (a) +81 kJ mol–1
4 (a) 23.76 kJ
(b) 570.24 kJ mol–1 STPM Practice 7

5 (a) C2H5OH + 3O2 → 2CO2 + 3H2O Objective Questions
(b) –1337.6 kJ mol–1
(c) 2.91 × 104 kJ 1 B 2 B 3 B 4 A 5 A
6 D 7 B 8 C 9 B 10 C
Quick Check 7.5 11 D 12 A 13 B 14 C 15 C
16 C 17 C 18 A 19 B 20 D
1 29.1 °C 21 B 22 B 23 B 24 A 25 C
2 (a) CH3COOH + NaOH → CH3COONa + H2O 26 C 27 D 28 A 29 B 30 C
31 C 32 A 33 B 34 A 35 D
(b) 1421.2 kJ 36 B 37 C 38 A 39 B
(c) –56.8 kJ mol–1
Structured and Essay Questions
Quick Check 7.6 1 (a) Ag+(g) + Br–(g) → Ag+Br–(s)
1 –110 kJ mol–1
2 –110 kJ mol–1 (b) (i) (+285) + (+112) + (+731) + (–325) + ∆H = –100
∆H = –903 kJ mol–1

(ii) Silver bromide has significant covalent character.
(iii) Less exothermic. The size of I– is larger than Br–.
(c) Less. The size of I– is larger than that of Br –. The

electron is less strongly held by the nucleus.

259

  Chemistry Term 2  STPM  Answers

2 (a) Born-Haber cycle (b) There is close agreement between the theoretical

(b) ∆H1 = Enthalpy of atomisation of Mg (assuming that the bond is 100% ionic) and
∆H2 = (First + second) ionisation energy of Mg
∆H3 = Enthalpy of atomisation of oxygen experimental values for NaCl. On the other hand,
∆H4 = (First + second) electron affinity of oxygen
∆H5 = Lattice energy of MgO there is a large difference between the theoretical
∆H6 = Enthalpy of formation of MgO
(c) (+150) + (736 + 1450) + (+248) + (+702) + ∆H5 = and experimental values for AgCl. This is because the

–603 difference in electronegativity between Na and Cl is

large. Hence, the bond in NaCl is 100% ionic. However,

the difference in electronegativity between Ag and Cl

is small leading to a significant amount of covalent

∆H5 = –3889 kJ mol–1 character in AgCl. The presence of covalent character
(d) Less. The size of Ca2+ is larger than Mg2+.
in AgCl causes the discrepancy in the theoretical and

3 (a) The standard enthalpy change of formation of a experimental values.

compound is the heat change when one mole of the 7 (a) Heat change when 1 mole of a substance is burned

compound is formed from its constituent elements completely in oxygen under standard conditions.
(b) (i) C8H18 + —225–O2 → 8CO2 + 9H2O
under standard conditions. C8H18 + —225–O2 → 8CO2 + 9H2O
(b) (i) –12–N2(g) + –21–O2(g) → NO(g) ∆H = + 90 kJ mol–1 (ii)
(ii) A lot of energy is required to break the strong
∆Hf°: –250 0 8(–394) 9(–286)
triple bond in the N  N molecules. ∆H = [8(–394) + 9(–286)] – [–250]

(c) 2Al(s) + 6HCl(aq) → Al2Cl6(aq) + 3H2(g) –1007 = –5476 kJ mol–1
3H2(g) + 3Cl2(g) → 6HCl(g) 3(–185)
8 (a) (i) Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2
6HCl(g) → 6HCl(aq) 6(–73) (ii) Q = 30 × 4.18 × 5.2 = 652.08 J (released)
∆H = – 652.08 × —0.01—2–0– = –32 604 J mol–1
—2AAl2—lC(sl—6)(+a—q3)—C→l2—(Ag—)l2→C—l6A(—sl)2 —Cl6—(s)— — —∆H—=—–—13—54—+k6J—4m6—ol––1 (b) (i) NaHCO3 + —21 H2SO4 → —21 Na2SO4 + H2O + CO2
4 (a) The heat energy released when one mole of a substance (ii) Q = 30 × 4.18 × 3.9 = 489.06 J (absorbed)
∆H = +489.06 × —0.01—2–0– = + 24 453 J mol–1
is completely burned in excess oxygen under standard (c) (i) 2NaHCO3 → Na2CO3 + CO2 + H2O

conditions.
(b) (i) C2H6(g) + —72 O2(g) → 2CO2(g) + 3H2O(l)
(ii) q = mc∆T

= 40.5 × 4.2 × (85 – 25) (ii) 2NaHCO3 ⎯x→ Na2CO3 + CO2 + H2O

= 10 206 kJ 2(+24 453) –32 604
No. of moles of C2H6 required = —180—00 × —1105—240—06

= 8.28 mol Na2SO4 + 2CO2 + 2H2O

Volume required = 8.28 × 24 dm3 By Hess’ Law:

= 198.72 dm3 x – 32 604 = 2(24 453)

(c) (i) CH4(g) +—122OO22((gg)) → CO2(g) + 2H2O(l) x = +81 510 J
H2(g) + → H2O(l)
= +81.51 kJ
(ii) For =CH– 4— 1:16–
∆H 9 (a) Heat changes when one mole of ethane is formed from

× 890 kJ = –55.63 kJ g–1 carbon and hydrogen under standard conditions.

For H2: 2C(s) + 3H2(g) → C2H6(g)
∆H = —21 × (–286) = –143 kJ g–1
(iii) Weight by weight, hydrogen is a better fuel than (b) 2C(s) + 3H2(g) ⎯x → C2H6(g)

2(–394) + 3(–286) –1560

methane. 2CO2 + 3H2O

5 (a) (i) –2 By Hess’ Law:
(ii) N2H4(l) + O2(g) → N2(g) + 2H2O(g)
(iii) QTem= p3—3e.2–5ra×tu(r–e5c4h0a)n=ge–5=9—.5619.2–k.15–J = 9.46 °C x – 1560 = –1646
(b) (i) N2(g) + 2H2(g) → N2H4(l)
x = –86 kJ mol–1

(ii) N2H4(l) + O2(g) → N2(g) + 2H2O(g) 10 (a) (i) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆H = –540 kJ (ii) Heat absorbed by water = mc∆T

= 200 × 4.18 × (75 – 25)

∆Hf : a 0 0 2(–286 + 44) = 41 800 J

= 41.8 kJ
Heat released by ethanol = —160—80 × 41.8 kJ
2 (–286 + 44) – a = –540

a = +56 kJ mol–1 = 61.48 kJ

6 (a) Lattice energy is the energy released when one mole of ∆H° = – —426– × 61.48 kJ mol–1

an ionic solid is formed from its constituent gaseous = –1414 kJ mol–1

ions under standard conditions.

260