CIE Chemistry Revision Guide for A2 Level

7/26/2019 CIE Chemistry Revision Guide for A2 Level

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CHEMISTRYRevision guide

for A2 level

For CIE examination


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IntroductionThis book is for A2 level only and the practical work is not mentioned in this

book.

Each section starts with a list of learning objectives, which are directly related

to the CIE syllabus. Questions throughout the text reinforce students

understanding and offer excellent opportunities for independent study.


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For CIE examinations UNIT 6 IONIC COMPOUNDS AND ELECTROCHEMISTRY

1

UNIT 6 IONIC COMPOUNDS AND ELECTROCHEMISTRY

Section 1 Ionic compounds

Learning objectives:

1 explain and use the terms: (iii) lattice energy (H negative, i.e. gaseous ions to solid

lattice) explain, in qualitative terms, the effect of ionic charge and of ionic radius on the

numerical magnitude of a lattice energy apply Hess' Law to construct simple energy

cycles, and carry out calculations involving such cycles and relevant energy terms, with

particular reference to:

(iii) the formation of a simple ionic solid and of its aqueous solution

vv(iv) Born-Haber cycles (including ionisation energy and electron affinity) interpret and

explain qualitatively the trend in the thermal stability of the nitrates and carbonates in

terms of the charge density of the cation and the polarisability of the large anion

interpret and explain qualitatively the variation in solubility of the sulfates in terms ofrelative magnitudes of the enthalpy change of hydration and the corresponding lattice

energy

2 explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical

magnitude of a lattice energy

3 apply Hess' Law to construct simple energy cycles, and carry out calculations involving

such cycles and relevant energy terms, with particular reference to: (iii) the formation of a

simple ionic solid and of its aqueous solution (iv) Born-Haber cycles (including ionisation

energy and electron affinity)

4 interpret and explain qualitatively the trend in the thermal stability of the nitrates and

carbonates in terms of the charge density of the cation and the Polaris ability of the large

anion

5 interpret and explain qualitatively the variation in solubility of the sulfates in terms of

relative magnitudes of the enthalpy change of hydration and the corresponding lattice

energy

Key ideas

Lattice enthalpy (energy)

1 What is the lattice energyLattice enthalpy (energy) is a measure of the strength of the forces between the ions in an

ionic solid.

Mg2+(g) + O2-(g)

MgO(s)

3923 kJ mol-1

Na+(g) + Cl-(g)

NaCl(s)

787 kJ mol-1


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The greater the lattice enthalpy, the stronger the forces and the higher the BP./MP.

2 Definitions

So lattice enthalpy could be described in either of two ways.

You could describe it as the enthalpy change when 1 mole of sodium chloride (or whatever)

was formed from its scattered gaseous ions. In other words, you are looking at a downward

arrow on the diagram.

In the sodium chloride case, that would be -787 kJ mol-1.

Or, you could describe it as the enthalpy change when 1 mole of sodium chloride (or

whatever) is broken up to form its scattered gaseous ions. In other words, you are looking

at an upward arrow on the diagram.

In the sodium chloride case, that would be +787 kJ mol-1.

The CIE definition

The enthalpy change when 1 mole of an ionic compound (solid lattice) is formed from its

gaseous ions under standard conditions (at 298K and 1 atm). (The downward arrow on thediagram)

i.e.:

NaCl

Na+(g) + Cl-(g) NaCl(s) = -787 kJ mol-1

Na2O,

2Na+(g) + O2-(g) Na2O(s) = -2590 kJ mol-1

MgO,

Mg2+(g) + O2-(g) MgO(s) = -3923 kJ mol-1

NaCl MgO

787

3923

1413

3600

801

2852

Lattice enthalpy Boiling point Melting point


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3 Factors affecting the Lattice enthalpy

(1)The charges on the ions

The greater the charges, the higher the lattice enthalpy

(2)The radius of the ions

The larger the ionic radius, the smaller the lattice enthalpy change

4 Calculate the lattice enthalpy

(1) Theoretical values for lattice energy (Not required by CIE)

Z+the relative charge of the cationZ-the relative charge of the anion

r0r++ r-

(2) Experimental valuesBorn Haber cycles

a) Some enthalpy changesdefinitions

Standard enthalpy change of atomisation ( )

The enthalpy change when 1mol of gaseous atoms are formed from the element in its

standard conditions

i.e. 1/2 Br2(l) Br(g)

The first electron affinity: ( )

The enthalpy change when one electron is added to each gaseous atom in 1 mol, to form

1 mol of 1- gaseous ions.

i.e. Cl(g)+ e- Cl-(g)

The second electron affinity: ( )

The enthalpy change when one electron is added to each 1- gaseous ion in 1 mol, to

form 1 mol of 2- gaseous ions.

i.e. Cl-(g)+ e- Cl2-(g)

(b) Calculate the lattice enthalpy by drawing a BornHaber cycle

i.e. Calculate the lattice energy of the KClkJ/mol

This method is based on the Hesslaw.

Hesslaw - The total enthalpy change for a chemical reaction is independent of the route by

which the reaction takes place, provided the initial and final conditions are the same.


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Standard enthalpy change of formation of KCl -435.8 kJ mol-1

Standard enthalpy change of atomisation of K 90.0 kJ mol-1

The first ionisation energy of K 425.0 kJ mol-1

Standard enthalpy change of atomisation of Cl 120.9 kJ mol-1

The first electron affinity of Cl -349.0 kJ mol-1

How to draw a Born Haber cycle

According to the Hesslaw, the relationship between these enthalpy changes is shown

below.

(KCl) = (K) + (K) + (Cl) + (Cl) + (KCl)

(KCl) = (KCl)( (K) + (K) + (Cl) + (Cl))

= - 435.8(90.0 + 425.0 + 120.9 + (-349.0))

= - 725.4 kJ mol-1

Thermal stabilities of the carbonates and nitrates of Group II

1 The trends (Learned in AS)

The carbonates and nitrates of Group II become more stable to heat as you go down thegroup.

enthalpy

High

Low

K+(g) + Cl-(g)

K(s) + 1/2Cl2(g) NaCl(s)

Step 1

Construct the equation which

is responsible for the

(KCl) and set this

equation at the bottom of the

diagram.

Construct another equation

which is responsible for the

(KCl) and set this

equation at the top of the

diagram.

diagram.

K(g)

(KCl)

(KCl)

Cl(g)

Step 2

List the enthalpy changes of

the metal element in the order

of atomisation and ionisation.

Step 3

List the enthalpy changes of

the non - metal element in the

order of atomisation and

electron affinity.

(K)

(K) (Cl)

(Cl)


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2 Explanations

(1) The smaller the positive ion, the higher the charge density, and the greater polarisingability it will have on the carbonate/nitrate ion.

(2) As the positive ions get bigger as you go down the Group, they have less effect on the

carbonate/nitrate ions near them.

(3) The distorted carbonate/nitrate breaks up into CO2and O2-more readily

Solubility of the sulphate

1 The trend in solubility of the sulphate

The sulphates of Group II are less soluble in water as you go down the group.

2 Explanations

(1) Enthalpy changes during the dissolving

Lattice enthalpy Standard enthalpy change of hydration

Enthalpy

changeof solution

-91 MgSO4(2959) Mg2+(-1890)

-18 CaSO4(2704) Ca2+(-1562) - 1160

-2 SrSO4(2572) Sr2+(-1414)

+26 BaSO4(2459) Ba2+(-1273)

The enthalpy change of solution of the sulphate decreases down the Group, which

makes these sulphates less soluble.

Possible answers to explain the solubilities of the sulphates:

lattice energy decreases

solvation/hydration energy (of cation) decreases

but hydration energy decreases more than lattice energy/is not able to overcome LE

Hsolubecomes more endothermic/positive/less exothermic

X2+

XSO4 X2+

(aq) + SO42-

(aq)dissolving

+ +=


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Section 2 Electrochemistry

Learning objectives:

1 Define the terms:

(i) Standard electrode (redox) potential(ii) Standard cell potential

2 Describe the standard hydrogen electrode

3 Describe methods used to measure the standard electrode potentials of:

(i) metals or non-metals in contact with their ions in aqueous solution

(ii) ions of the same element in different oxidation states

4 Calculate a standard cell potential by combining two standard electrode potentials

5 Use standard cell potentials to:

(i) Explain/deduce the direction of electron flow from a simple cell

(ii) Predict the feasibility of a reaction

6 Construct redox equations using the relevant halfequations7 Predict qualitatively how the value of an electrode potential varies with the concentration

of the aqueous ion

8 State the possible advantages of developing other types of cell, e.g. the H2/O2 fuel cell

and improved batteries (as in electric vehicles) in terms of smaller size, lower mass and

higher voltage

9 Predict the identity of the substance liberated during electrolysis from the state of

electrolyte (molten or aqueous), position in the redox series (electrode potential) and

concentration

10 State the relationship, F = Le, between the Faraday constant, the Avogadro constant

and the charge on the electron

11 Calculate:

(i) The quantity of charge passed during electrolysis

(ii) The mass and/or volume of substance liberated during electrolysis, including those in

the electrolysis of H2SO4(aq); Na2SO4(aq)

12 Describe the determination of a value of the Avogadro constant by an electrolytic

method

Key ideas

Standard electrode (redox) potential(1) What is the electrode potential

There is equilibriumpresent in the system shown above.

Cu2+(aq) + 2e- Cu(s) The equation is called half cell.

Cu2+

(aq)

Copper bar


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Electrode potential is measured in volts (V) and is numerical indication of how favourable

(or easy) the reductionis. So that the halfcell must be written in the reduction direction.

(2) Standard electrode potentialHowever, the value of the electrode potential is not measurable. We can only get the

differences in the electrode potential between two halfcells.

If the other halfcell consists of a 1.00 mol L-1solution of H+ions in contact with hydrogen

gas at 1 atm, all at temperature of 298K, then the voltage measured is called standard

electrode potential for the reaction.

Definition of the standard electrode potential:

The voltage measured under standard conditions when the half cell is incorporated into

an electrochemical cell with the other half cell being a standard hydrogen electrode.

(3) Standard hydrogen electrode potential (S.H.E.)

= 0 V

(4) Measuring a standard electrode potential

i.e. Measuring a metal ion/metal standard electrode potential, such as

H+

1 mol dm-

Platinum wire

Supply of hydrogen gas at a

pressure of 1 atmosphereThe half cell

2H+

+ e-

H2

Platinum electrode

Standard conditions: all concentrations are 1.00 mol dm-3

the temperature is 298K

the pressures of any gases used are 1 atm

The direction, Cu(s) Cu2+

(aq) + 2e-, is not acceptable because its an oxidation

direction.


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i.e. Measuring a ion/ion standard electrode potential, such as

Graphite electrode

1 mol dm-3

Fe2+

(aq)

1 mol dm-3

Fe3+

(aq)

H+

1 mol dm-

Platinum wire

Supply of hydrogen gas at a

pressure of 1 atmosphere

Platinum electrode

Salt bridge

+ 0.77V

The internal resistance of the standard hydrogen electrode is quite high due to the adsorbed

H2(g) so that we have to choose a high-resistance voltmeter to measure voltage accurately.

If the voltmeter has a high resistance, little current will flow through the cell, and so little

contamination from the ions in the salt bridge will occur.

The set can also be simply expressed as following:

(+) CuCu2+

(1 mol dm-3

)H+(1 mol dm

-3)H2(g, 1atm),Pt ()

means the boundary between two different phases and is the salt bridge.


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i.e. Measuring a nonmetal /anion standard electrode potential, such as

(5) The meaning of the standard electrode potential

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

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

Cu2+(aq) + 2e- Cu(s) = + 0.34V

H+(aq) + e- 1/2H2(s) = + 0.00V

Zn2+(aq) + 2e- Zn(s) = - 0.76V

Mg2+(aq) + 2e- Mg(s) = - 2.38V

This set can also be simply expressed as following:

(+) Pt, Cl2(g, 1atm)Cl-(1 mol dm

-3)H

+(1 mol dm

-3)H2(g, 1atm),Pt ()

H+

1 mol dm-

Platinum wire

Supply of hydrogen gas at

a pressure of 1

Platinum electrode

Salt bridge

+ 1.36V

Cl-1 mol dm

-

Platinum wire

Supply of chlorine gas

at a pressure of 1

atmosphere

Platinum electrode

This set can also be simply expressed as following:

(+) CFe2+

(1 mol dm-3

), Fe3+

(1 mol dm-3

)H+(1 mol dm

-3)H2(g, 1atm),Pt ()

The oxidising ability

of the ion/element on

the left increases as

the increases

Stronger oxidising

agent

Weaker oxidising

agent

The reducing ability

of the ion/element on

the right increases

as the gets more

negative.

Weaker

reducing agent

Stronger

reducing agent


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Using the standard electrode potential

(1) Predict the cell voltage and electron flowi.e. Label the electron flow on the graph and predict the cell voltage.

How to answer this question?

Step 1 list the half cells in the electrochemical cell

Cu2+(aq) + 2e- Cu(s) = + 0.34V

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

Step 2 The electrons transfer from the electrode with a higher to the electrode

with a lower . So that in this cell the electrons move from the copper

electrode to the Cl2/ Cl-electrode.

Step 3 The cell voltage is the difference of the between the two half cells.

Ecell= 1.36 0.34 = 1.02 V

(2) Predict the direction/possibility of a chemical reaction.

i.e. Predict feasibility of the chemical reaction below:

Cu + 2H+ Cu2++ H2

Cu electrode

1 mol dm-3

Cu2+

(aq)

Cl(aq) 1 mol dm

-

Platinum wire

Supply of chlorine gasat

a pressure of 1 atm

Platinum electrode

Salt bridge

A


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Step 1 give out the half cells (reduction potential) included in the reaction.

Cu2+(aq) + 2e- Cu(s) = + 0.34V

H+(aq) + e- 1/2H2(s) = + 0.00V

Step 2 reverse the sign of one of the values

In the chemical reaction Cu is converted in Cu2+which is an oxidation. The electrode

potential we used is the reduction potential so that reverse the sign of the (Cu2+/Cu)

value, (Cu2+/Cu).

Step 3 work out the standard cell potential

= (H+/H2) (Cu2+/Cu)

= 0 0.34 = 0.34V

< 0

The chemical reaction (forward) is impossible.

(3) Concentrations affecting the direction of the chemical reactions

The is measured under standard conditions only, but lots of reactions dont take

place under standard conditions.

i.e. Concentrated HCl is used to make chlorine gas by reacting with manganese(IV)

oxide.

4HCl(conc.)+ MnO2(s) MnCl2(aq)+ Cl2(g)+ H2O(l)

However, the chlorine gas is not released by using a dilute HCl to instead the conc. HCl.

Give out an explanation in the terms of cell potential.

> 0 forward reaction possible but backward impossible

< 0 forward reaction impossible but backward possible

Reverse the sign of the value which has an opposite direction to the halfcell.


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CIE Chemistry Revision Guide for A2 Level