Unit 2 Equilibria, energetics and elements
(F325) Energy
Topics covered in this module:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
1|Page
Lattice enthalpy
Constructing Born-Haber cycles
Born-Haber cycle calculations
Further examples of Born-Haber cycles
Enthalpy change of solution
Understanding hydration and lattice enthalpies
Entropy
Free energy
Redox
Cells and half cells
Cell potentials
The feasibility of reactions
Storage and fuel cells
Hydrogen for the future
1.
Lattice enthalpy
Unit 2 – Equilibria, energetics and elements
Done
Energy

Explain and use the term lattice enthalpy.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 lattice enthalpy.
When writing an atomisation equation remember that only one mole of atoms is
formed. Students frequently get this equation wrong in exams!
Learn all key definitions for lattice enthalpy, Hess’ Law, standard enthalpy change of
formation, enthalpy change of atomisation, first and second ionisation energy, and
first and second electron affinity.
State symbols are critical and it is important to write all state symbols in your
equations.
2|Page
2.
Constructing Born-Haber cycles
Unit 2 – Equilibria, energetics and elements
Done
Energy

Use the lattice enthalpy of a simple ionic solid and relevant energy
term to construct Born-Haber cycles.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 lattice enthalpy.
When drawing a Born-Haber cycle it is important to make sure that you have
sufficient space to complete your diagram. Start with the elements in their standard
states a quarter of the way up from the bottom of the space.
It is important to show all the equations with the correct state symbols in the exam.
Marks will be lost if state symbols are missed out.
3|Page
3.
Born-Haber cycle calculations
Unit 2 – Equilibria, energetics and elements
Done
Energy

Construct Born-Haber cycles.
Carry out related calculations.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 lattice enthalpy.
Remember that exothermic energy changes have a negative value and the arrows
go downwards. Endothermic changes have a positive value and the arrows go
upwards.
4|Page
4.
Further examples of Born-Haber cycles
Unit 2 – Equilibria, energetics and elements
Done
Energy

Construct Born-Haber cycles.
Carry out related calculations.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 lattice enthalpy.
Three common mistakes made in Born-Haber cycle calculations are:
 Not doubling the enthalpy change of atomisation when you require two moles
of gaseous atoms e.g. the two moles of chlorine atoms required to make
CaCl2.
 Not doubling the electron affinity when needing two moles of negative ions
e.g. the two moles of electrons required to make the two moles of chlorine
ions in CaCl2.
 Not representing electron affinities correctly e.g. in the formation of copper (II)
oxide the first electron affinity is exothermic, but the second electron affinity is
endothermic.
5|Page
Specimen paper questions
The table below shows the enthalpy changes needed to calculate the lattice
enthalpy of calcium oxide, CaO.
(a)
process
enthalpy change/ kJ mol–1
first ionisation energy of calcium
+590
second ionisation energy of calcium
+1150
first electron affinity of oxygen
–141
second electron affinity of oxygen
+ 791
enthalpy change of formation of calcium oxide
–635
enthalpy change of atomisation of calcium
+178
enthalpy change of atomisation of oxygen
+248
(i)
Explain why the second ionisation energy of calcium is more
endothermic than the first ionisation energy of calcium.
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[2]
(ii)
Suggest why the second electron affinity of oxygen is positive.
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[2]
[Turn over]
6|Page
(b)
Complete the Born–Haber cycle for calcium oxide below.
Use the data in the table to calculate the lattice enthalpy of calcium
oxide.
energy/
kJ mol–1
Ca(s) + O(g)
Ca(g) + 0.5 O2 (g)
Ca(s) + 0.5 O 2 (g)
CaO(s)
lattice enthalpy = ............................ kJ mol–1
[5]
(c)
You will know
how to do this
part after
covering
section 6,
‘Understanding
hydration and
lattice
enthalpies’
The lattice enthalpies of calcium oxide and magnesium oxide are
different.
Comment on this difference.
In your answer you should make clear how the sizes of the lattice
enthalpies are related to any supporting evidence.
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[3]
[Total 12 marks]
7|Page
Past paper questions
The table below shows the enthalpy changes needed to calculate the enthalpy
change of formation of calcium oxide.
process
enthalpy change/kJ mol–1
lattice enthalpy for calcium oxide
–3459
first ionisation energy for calcium
+590
second ionisation energy for calcium
+1150
first electron affinity for oxygen
–141
second electron affinity for oxygen
+798
enthalpy change of atomisation for oxygen
+249
enthalpy change of atomisation for calcium
+178
(a)
(i)
Explain why the first ionisation energy of calcium is endothermic.
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[1]
(ii)
Explain why the first electron affinity for oxygen is exothermic.
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[1]
[Turn over]
8|Page
(b)
(i)
Draw a Born-Haber cycle for calcium oxide.
Include
•
•
correct formulae and state symbols
energy changes in kJ.
[3]
(ii)
You will know
how to do this
part after
covering
section 6,
‘Understanding
hydration and
lattice
enthalpies’
Use your Born-Haber cycle in (i) to calculate the enthalpy change
of formation for calcium oxide.
enthalpy change of formation = .........................................
[2]
(iii)
The lattice enthalpy for iron(II) oxide is –3920 kJ mol–1.
Suggest a reason for the difference in lattice enthalpy between
calcium oxide and iron(II) oxide.
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[1]
[Total 8 marks]
9|Page
Examiner’s comments
Candidates of all abilities were able to draw a recognisable Born–Haber cycle but
frequently candidates found parts (a) and (c) very demanding.
(a)
The majority of candidates were unable to explain the direction of energy transfer
in first ionisation energy and the first electron affinity. It was not sufficient to state that
first ionisation energy required energy and first electron affinity released energy. To gain
credit candidates had to refer to the attraction between electrons and the nucleus as well
as electron loss in part (i) and electron gain (ii). There was evidence that a significant
proportion of candidates did not read the question sufficiently well and answered about
first and second ionisation energy.
(b)
The majority of candidates drew energy level diagrams but other correct cycles
were given full credit. The most common errors were the
•
use of the wrong formulae e.g. O+(g)
•
wrong stoichiometry e.g. Ca(s) + O2(g) rather than Ca(s) + ½O2(g),
•
wrong labelling of energy changes
The correct value for the enthalpy of formation was –635 kJ mol–1. The unit was
required for full credit. An error carried forward mark was allowed from incorrect cycles.
In part (iii) only a small proportion of candidates realised that Fe 2+ must have a
smaller ionic radius than Ca2+. Credit was not given to references to the wrong type of
particle for example Fe has a smaller atomic radius.
10 | P a g e
Past paper questions
The table below shows the enthalpy changes needed to construct a BornHaber cycle for sodium oxide, Na2O.
process
enthalpy change / kJ mol
first ionisation energy of sodium
+495
first electron affinity of oxygen
–141
second electron affinity of oxygen
+791
enthalpy change of formation for sodium oxide
–416
enthalpy change of atomisation for sodium
+109
enthalpy change of atomisation for oxygen
+247
(a)
–1
Use the table of enthalpy changes to complete the Born-Haber cycle by
putting in the correct numerical values on the appropriate dotted line.
2Na+(g) + O2–(g)
2Na+(g) + O(g) + 2e–
ΔH = ..........kJ
ΔH = ..........kJ
H = ..........kJ
2Na+(g) + O–(g) + e–
2Na+(g) + 12 O2(g) + 2e–
H = ..........kJ
2Na(g) +
lattice
enthalpy of
sodium oxide
1
O (g)
2 2
H = ..........kJ
2Na(s) +
1
O (g)
2 2
H = ..........kJ
Na2O(s)
[4]
11 | P a g e
(b)
Use the Born-Haber cycle to calculate the lattice enthalpy of sodium
oxide.
lattice enthalpy = ....................................kJ mol–1
[2]
(c)
Which one of the following compounds has the most exothermic lattice
enthalpy?
•
•
•
•
calcium bromide
calcium chloride
potassium bromide
potassium chloride
Explain your answer in terms of the ions present.
You will know
how to do this
part after
covering
section 6,
‘Understanding
hydration and
lattice
enthalpies’
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[4]
[Total 10 marks]
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Examiner’s comments
This question focussed on lattice enthalpy and the Born-Haber cycle. Candidates from the
whole of the ability range were able to access the questions and even the weakest
candidates often scored at least four marks.
(a)
Many candidates were able to complete the Born-Haber cycle using the
appropriate enthalpy changes. A significant proportion only scored two marks since they
did not double the first ionisation energy of sodium and the enthalpy change of
atomisation of sodium.
(b)
Candidates gained full marks for the correct lattice enthalpy of –2521 kJ mol–1 or
for a value that was an error carried forward from their enthalpy changes in the BornHaber cycle in (a). The most frequent error that was carried forward was –1917 kJ mol-1. A
common misconception was to use the negative value for the enthalpy change of
formation i.e. +416 rather than –416. Good answers were exemplified by clear working
out that allowed the Examiner to see any error carried forward.
(c)
Most candidates selected calcium chloride in Q.1(c) and then gave an explanation
based on the difference between anionic radius and cationic charge. Many candidates
referred to the wrong particles or did not specify a type of particle at all in Q.1(c). Typical
incorrect statements referred to the charge of calcium rather than the calcium ion or
referred to the difference in the atomic radii rather than the ionic radii. The use of the
wrong
particle
was
only
penalised
once
in
this
question.
Common misconceptions included using the difference in electronegativity of the
halogens and the reactivity of the metals to explain differences in lattice enthalpies. Other
candidates referred to polarisation which was ignored.
13 | P a g e
Past paper questions
The table below shows the enthalpy changes needed to calculate the lattice
enthalpy of calcium chloride, CaCl2.
enthalpy change / kJ mol–1
process
first ionisation energy of calcium
+590
second ionisation energy of calcium
+1150
electron affinity of chlorine
–348
enthalpy change of formation for calcium chloride
–796
enthalpy change of atomisation for calcium
+178
enthalpy change of atomisation for chlorine
+122
(a)
The Born-Haber cycle below can be used to calculate the lattice enthalpy
for calcium chloride.
Ca2+(g) + 2Cl(g) + 2e –
H = .................... kJ mol –1
Ca2+ (g) + Cl2(g) + 2e–
H = .................... kJ mol –1
H = .................... kJ mol –1
Ca2+(g) + 2Cl –(g)
Ca +(g) + Cl2(g) + e–
H = .................... kJ mol –1
Ca(g) + C l2(g)
lattice enthalpy of
calcium chloride
H = .................... kJ mol –1
Ca(s) + Cl2(g)
H = .................... kJ mol –1
CaCl2(s)
(i)
Use the table of enthalpy changes to complete the Born-Haber
cycle by putting in the correct numerical values on the appropriate
dotted line.
[3]
[Turn over]
14 | P a g e
(ii)
You will know
how to do this
part after
covering
section 6,
‘Understanding
hydration and
lattice
enthalpies’
Use the Born-Haber cycle to calculate the lattice enthalpy of
calcium chloride.
answer ........................... kJ mol–1
[2]
(iii)
Describe how, and explain why, the lattice enthalpy of magnesium
fluoride differs from that of calcium chloride.
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[3]
(b)
Explain why the first ionisation energy of calcium is less positive than the
second ionisation energy.
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[2]
[Total 10 marks]
15 | P a g e
Examiner’s comments
This question focused on the Born-Haber cycle and lattice enthalpy.
(a)
Many candidates could score three marks but it was rare for candidates to score
all five. The most common error was the failure to use 2 × the enthalpy change of
atomisation of chlorine and 2 x the electron affinity for chlorine. The correct answer was
–2262 kJ mol–1. A small but significant proportion of candidates gave a positive lattice
enthalpy. In order to score full marks for part (iii), candidates had to refer to the
difference in ionic radii or the charge density of both the cations and the anions. Many
candidates referred to the difference in atomic radii and this was not given credit.
Another misconception was that compounds had ionic radii. The final mark was awarded
for the connection between the change in lattice enthalpy of the two compounds and the
strength of the electrostatic attraction between the ions. A significant proportion of
candidates did not appreciate that the lattice enthalpy of magnesium fluoride was more
exothermic than that of calcium chloride and referred to a bigger or greater lattice
enthalpy with no regard for sign.
(b)
In this part, a small but significant proportion of the candidates referred to the
gain of electrons rather than the loss of electrons from a calcium atom. A common
misconception was that the nuclear charge increased as electrons were lost, rather than
the nuclear charge remained constant but attracted a smaller number of electrons.
16 | P a g e
5.
Enthalpy change of solution
Unit 2 – Equilibria, energetics and elements
Done
Energy

Use the enthalpy change of solution of a simple ionic solid and
relevant enthalpy change terms to construct Born-Haber cycles.
Carry out relevant calculations.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 enthalpy change of solution
 lattice enthalpy.
Lattice energy is an exothermic process and the sign is –ve. The process involved in
breaking down the ionic lattice is endothermic and the sign is +ve.
The enthalpy change of solution can be either +ve or –ve.
17 | P a g e
Past paper questions
Examiner’s comments
18 | P a g e
6.
Understanding hydration and lattice enthalpies
Unit 2 – Equilibria, energetics and elements
Done
Energy

Explain, in qualitative terms, the effect of ionic charge and ionic
radius on lattice enthalpy and enthalpy change of hydration.
Key areas to concentrate on…

You will need to be able to explain what lattice enthalpy means and to construct
Born-Haber cycles based upon information given – state symbols will be relevant
here!
 You may be asked about some of the AS material covered in AS Unit F322:
Module 2 - Alcohols, halogenoalkanes and analysis including:
o Q = mcΔT
o standard enthalpy change of reaction
o standard enthalpy change of combustion
o standard enthalpy change of formation
o average bond enthalpy.

Energy cycles involving Hess’ law may also feature.

You need to understand new terms such as:
 enthalpy change of solution
 enthalpy change of hydration
 lattice enthalpy.
Remember that lattice enthalpy is the enthalpy change when one mole of an ionic
compound is formed from its gaseous ions under standard conditions.
e.g. K+(g) + Cl-(g)  KCl(s)
ΔH = -711 kJ mol-1
It is important to state that ions affect the magnitude of the lattice enthalpy and
enthalpy change of hydration. Sloppy wording such as ‘chlorine’ and ‘bromine’
instead of chloride and bromide must be avoided at all costs.
19 | P a g e
Past paper questions
In this question, one mark is available for the quality of spelling, punctuation
and grammar.
The lattice enthalpy of magnesium chloride, MgCl2, can be determined using a
Born-Haber cycle and the following enthalpy changes.
name of process
enthalpy change / kJ mol–1
enthalpy change of formation of MgCl2(s)
–641
enthalpy change of atomisation of magnesium
+148
first ionisation energy of magnesium
+738
second ionisation energy of magnesium
+1451
enthalpy change of atomisation of chlorine
+123
electron affinity of chlorine
–349
•
Define, using an equation with MgCl2 as an example, what is meant by
the term lattice enthalpy.
•
Construct a Born-Haber cycle for MgCl2, including state symbols, and
calculate the lattice enthalpy of MgCl2.
•
Explain why the lattice enthalpy of NaBr is much less exothermic than
that of MgCl2.
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20 | P a g e
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[11]
Quality of Written Communication [1]
[Total 12 marks]
21 | P a g e
Examiner’s comments
This question focussed on the Born-Haber cycle and lattice enthalpy.
The question also included one mark for the quality of spelling, punctuation and
grammar. To be awarded this mark, candidates had to write at least two sentences
that addressed the question set and had no significant errors of spelling,
punctuation and grammar. A significant proportion of candidates could not be
awarded this mark because they did not write any sentences and just carried out
the calculation and the Born-Haber cycle.
Although many candidates answered the questions in the order of the bullet
points others concentrated on the qualitative parts first and left the Born-Haber
cycle and the calculation until the end.
Many candidates did not write the correct equation associated with the lattice
enthalpy of magnesium chloride. The most frequent error was the omission of
state symbols. A majority of candidates gave the correct definition for lattice
enthalpy although two common misconceptions were that it was the energy
required and that lattice enthalpy involves the gaseous elements rather than the
gaseous ions.
Most candidates drew Born-Haber cycles as energy level diagrams and labelled the
species and energy changes. One mark was awarded for all the correct formulae in
the cycle and one for the correct state symbols. Many candidates were awarded
one mark out of the two available.
The correct value for the lattice enthalpy was –2526 kJ mol–1 but many candidates
made errors because they did not double the enthalpy of atomisation of chlorine
or the electron affinity of chlorine. An error carried forward mark was allowed for
these errors.
Candidates tended to use the terms atom, ion and molecule as though they meant
the same thing when trying to explain why the lattice enthalpy of MgCl 2 was more
exothermic than that of NaBr. Another similar misconception was to refer to the
charge density of NaBr or MgCl2. Candidates were only penalised once within the
question for such errors. Good answers compared the ionic charges and the ionic
radii of the ions involved and then made a comment about the electrostatic
attraction between the ions.
22 | P a g e
7.
Entropy
Unit 2 – Equilibria, energetics and elements
Done
Energy

Explain that entropy is a measure of the disorder of a system, and
that a system becomes energetically more stable when it becomes
more disordered.
Explain: the difference in entropy of a solid and a gas; the change
when a solid lattice dissolves; the change in a reaction in which
there is a change in the number of gaseous molecules.
Calculate the entropy change for a reaction given the entropies of
reactants and products.
Key areas to concentrate on…

You must be able to:
 carry out simple entropy calculations
23 | P a g e
Past paper questions
Examiner’s comments
24 | P a g e
8.
Free energy
Unit 2 – Equilibria, energetics and elements
Done
Energy

Explain that the tendency of a process to take place depends on
the temperature, T, the entropy change in the system, ΔS, and the
enthalpy change, ΔH, with the surroundings.
Explain that the balance between entropy and enthalpy change is
the free energy change, ΔG.
State and use the relationship ΔG = ΔH – TΔS.
Explain how endothermic reactions are able to take place
spontaneously.
Key areas to concentrate on…

You must be able to:
 carry out simple entropy calculations
 understand the meaning of the term free energy change, ΔG, and the
equation ΔG = ΔH – TΔS
 understand the relationship between entropy and enthalpy.
The free energy change, ΔG, is the net driving force that determines whether a
chemical reaction will happen.
In calculations, you need to get the units the same.
ΔH is usually given in kJ mol-1
ΔS is usually given in J K-1 mol-1
First get ΔS into kJ K-1 mol-1:
 To convert J to kJ, divide by 1000.
Also remember that entropy is worked out using temperature in K:
 To convert oC to K, add 273.
 To convert K to oC, subtract 273.
25 | P a g e
Specimen paper questions
Most metals can be extracted by reduction from compounds obtained from
their naturally-occurring ores.
Metals such as calcium and magnesium are normally extracted by electrolysis
but it is feasible that calcium oxide could be reduced by carbon as shown in
the equation below.
CaO(s) + C(s)  Ca(s) + CO(g)
Use the data in the table below to help you answer parts (i)–(iii) below.
(i)
CaO(s)
C(s)
Ca(s)
CO(g)
Hfο /kJ mol–1
–635
0
0
–110
Sο/J K–1 mol–1
39.7
5.7
41.4
197.6
Calculate the standard enthalpy change for the CaO reduction in the
equation.
Hο = ............................................ kJ mol–1
[1]
(ii)
Calculate the standard entropy change for the CaO reduction in the
equation.
Sο = ......................................... J K–1 mol–1
[1]
[Turn over]
26 | P a g e
(iii)
Calculate the minimum temperature at which the carbon reduction in the
equation is feasible.
minimum temperature = ...............................
[5]
[Total 7 marks]
27 | P a g e
9.
Redox
Unit 2 – Equilibria, energetics and elements
Done
Energy

Explain the terms redox, oxidation number, half-reaction, oxidising
agent and reducing agent for simple redox reactions.
Construct redox equations using relevant half-equations or
oxidation numbers.
Interpret and make predictions for reactions involving electron
transfer.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.
Never get in a muddle: use OILRIG
Oxidation Is Loss
Reduction Is Gain
If one species gains electrons (reduction), another species loses the same number
of electrons (oxidation).
If one species decreases its oxidation number (reduction), then another species
must increases its oxidation number (oxidation).
The total increase in oxidation number equals the total decrease in oxidation
number.
28 | P a g e
Past paper questions
The carbonates and nitrates of Group 2 elements decompose when heated.
(a)
Barium nitrate decomposes when heated to make barium oxide, nitrogen
dioxide and oxygen.
2Ba(NO3)2(s) → 2BaO(s) + 4NO2(g) + O2(g)
(i)
Use oxidation states to explain why this decomposition reaction
involves both oxidation and reduction.
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[3]
(ii)
Calculate the enthalpy change of reaction, ∆Hr, in kJ mol–1, for the
thermal decomposition of barium nitrate using the enthalpy
changes of formation, ∆Hf, given in the table.
compound
∆Hf /kJ mol–1
Ba(NO3)2(s)
–992
BaO(s)
–558
NO2(g)
+33
answer ........................... kJ mol–1
[3]
[Turn over]
29 | P a g e
(b)
A student investigates the volume of gas formed when barium nitrate is
heated.
The diagram shows the apparatus the student uses.
100cm3 gas syringe
barium nitrate
heat
(i)
A 1.31 g sample of barium nitrate is completely decomposed.
Use the equation above to calculate the volume, in cm3, of gas
formed at room temperature and pressure.
1 mol of gas molecules occupies 24 000 cm3 at room temperature
and pressure.
answer ......................... cm3
[3]
[Turn over]
30 | P a g e
(ii)
Suggest one problem that the student may encounter when
carrying out the investigation.
...........................................................................................................
...........................................................................................................
[1]
[Total 10 marks]
Examiner’s comments
This question focussed on the thermal decomposition of carbonates and nitrates of Group
2 elements.
(a)
In part (i) many candidates were unable to deduce the oxidation numbers of
barium, nitrogen and oxygen in the substances given in the equation. In particular a
‘combined’ oxidation number, e.g. –12 for oxygen in Ba(NO3)2, was given. A common
misconception was to give barium different oxidation numbers in Ba(NO3)2 and BaO. The
majority of candidates could interpret oxidation and reduction in terms of changes in
oxidation numbers and so obtained an error carried forward mark.
Many candidates correctly calculated the enthalpy change of reaction in part (ii) as +1000
kJ; however a significant proportion gave the answer –1000 kJ, while others did not use
the correct number of moles as shown in the equation. Many candidates did not show
sufficient working out. A small proportion of the candidates just added the numbers up as
shown in the table.
(b)
A large proportion of candidates obtained the correct answer of 300 cm3 in part (i).
Other candidates did not appreciate that they had to calculate the volume of both oxygen
and of nitrogen dioxide and gave an answer of 60 cm3 or 240 cm3.
A common misconception was to use the Mr of Ba(NO3)2 as 522.
In part (ii) the majority of candidates realised that the syringe was too small.
31 | P a g e
Past paper questions
In order to obtain full marks in this question, you must show all your working
clearly.
In its reactions, sulphuric acid, H2SO4, can behave as an acid, an oxidising
agent and as a dehydrating agent.
The displayed formula of pure sulphuric acid is shown below.
H
O
O
S
H
O
O
Concentrated sulphuric acid will readily oxidise halide ions to the halogen.
The equation below represents the unbalanced equation for the oxidation of
iodide ions by sulphuric acid.
H+
(i)
+ SO42– +
I–
→
I2
+
H2S
+
H2O
Write the oxidation numbers of sulphur and iodine in the boxes above the
equation.
[2]
(ii)
Balance the equation above.
[1]
[Total 3 marks]
Examiner’s comments
Most candidates were able to show correct oxidation numbers for iodine but those of
sulphur proved to be more elusive. Only the very best candidates correctly constructed
the equation, the difficulty being the need for 4I2 and 8I– to account for all electrons.
Correct equations had usually been balanced using oxidation numbers.
32 | P a g e
Past paper questions
NO2 reacts with oxygen and water to form nitric acid, HNO3. In the
atmosphere, this contributes to acid rain. Construct a balanced equation for
this formation of nitric acid and use oxidation numbers to show that this is a
redox reaction.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
[Total 2 marks]
Examiner’s comments
Many candidates were able to construct a fully balanced equation for the formation of
nitric acid although some candidates did not seem to realise that the species were all
given in the opening sentence of the question. When it came to using oxidation numbers
to show that it was a redox reaction, the answers were rather poor, certainly more so
than in previous years. A large number of candidates were satisfied with showing only
that nitrogen was oxidised from +4 to +5 and did not discuss the changes to oxygen. Of
those, who did, the most common error was to assign an oxidation number of –6 to the
oxygen in HNO3.
33 | P a g e
10.
Cells and half cells
Unit 2 – Equilibria, energetics and elements
Done
Energy

Describe simple half cells made from
i.
metals or non-metals in contact with their ions in aqueous
solution;
ii.
ions of the same element in different oxidation states.
Describe how half cells can be combined to make an
electrochemical cell.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.

A way to earn some marks is by knowing the definition of standard electrode
potential.

You will need to describe how standard electrode potentials can be calculated
against the standard hydrogen electrode – both for metals and ions or pairs of
ions.
 You will need to know how to use values of electrode potentials in order to
predict if reactions will occur.
34 | P a g e
Past paper questions
Examiner’s comments
35 | P a g e
11.
Cell potentials
Unit 2 – Equilibria, energetics and elements
Done
Energy

Define the term standard electrode (redox) potential, Eο.
Describe how to measure standard electrode potentials using a
standard hydrogen half cell.
Calculate a standard cell potential by combining two standard
electrode potentials.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.

A way to earn some marks is by knowing the definition of standard electrode
potential.

You will need to describe how standard electrode potentials can be calculated
against the standard hydrogen electrode – both for metals and ions or pairs of
ions.
 You will need to know how to use values of electrode potentials in order to
predict if reactions will occur.
Remember that the particles that carry charge in the electric current are:
 electrons in the wire
 ions in the salt bridge.
The sign of the standard electrode potential of a half cell gives the sign (‘polarity’)
of the electrode compared with the hydrogen half cell.
36 | P a g e
Specimen paper questions
Use the standard electrode potentials in the table below to answer the
questions that follow.
I
Fe2+(aq) + 2e–
Fe(s)
Eο = –0.44 V
II
V3+(aq) + e–
V2+(aq)
Eο = –0.26 V
III
2H+(aq) + 2e–
H2(g)
Eο = 0.00 V
IV
O2(g) + 4H+(aq) + 4e–
2H2O(l)
Eο = +0.40 V
An electrochemical cell was set up based on systems I and II.
(i)
Write half-equations to show what has been oxidised and what has been
reduced in this cell.
oxidation:
reduction:
[2]
(ii)
Determine the cell potential of this cell.
Ecell = ......................................................... V
[1]
[Total 3 marks]
37 | P a g e
Past paper questions
The standard electrode potential of Cu2+(aq) + 2e–
(a)
Cu(s) is +0.34 V.
Define the term standard electrode potential.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
(b)
Complete the diagram to show how the standard electrode potential of
Cu2+(aq) + 2e–
Cu(s) could be measured.
copper rod
Cu2+(aq)
[3]
[Total 6 marks]
38 | P a g e
Examiner’s comments
(a)
Many candidates know this definition. Quite a large number did not include the
actual standard conditions in the definition but these were then used in the diagram in
(b). This was credited.
(b)
Diagrams of the cell were usually of a good standard. Occasionally candidates
forgot to draw in the liquid level and a few electrodes were not in contact with liquid,
particularly for the hydrogen half-cell.
39 | P a g e
Past paper questions
The standard electrode potentials for some redox systems involving vanadium
are shown below.
These are labelled A, B, C and D.
Eο/ V
(a)
A
VO2+ + 2H+ + e–
B
V3+ + e–
V
–0.26
C
V2+ + 2e–
V
–1.20
D
VO2+ + 2H+ + e–
V
VO2+ + H2O
2+
3+
+ H2O
+1.00
+0.34
Which of the vanadium species shown in A, B, C and D is the most
powerful oxidising agent?
....................................................................................................................
[1]
(b)
A student wishes to set up a cell with a standard cell potential of 0.60V.
(i)
Which two of the redox systems, A, B, C or D, should he choose?
...........................................................................................................
[1]
(ii)
Complete the labelling of the following diagram which shows the
cell with a standard cell potential of 0.60V.
V
[4]
40 | P a g e
(iii)
The emf of this cell is only 0.60 V under standard conditions. What
do you understand by the expression standard conditions?
...........................................................................................................
...........................................................................................................
...........................................................................................................
[1]
[Total 7 marks]
Examiner’s comments
(a)
A majority of candidates misinterpreted the requirement of this part of the
question. The common answer given was ‘A’ when the actual species which was the most
powerful oxidising agent, VO2+, was required.
(b)
(i)
A majority of candidates correctly chose B and D.
(ii)
A common error here was to miss off the H+ ion for D and to use vanadium
electrodes instead of platinum.
(iii)
Standard conditions are well known.
41 | P a g e
Past paper questions
The standard electrode potential of the
1
Cl2/ Cl– half-cell may be measured
2
using the following apparatus.
C
salt bridge
D
H2(g)
Cl –(aq)
B
A
(a)
Suggest suitable labels for A, B, C and D.
A ................................................................................................................
B ................................................................................................................
C ................................................................................................................
D ................................................................................................................
[2]
(b)
The half cell reactions involved are shown below.
1
Cl2 + e–
2
H+ + e–
(i)
Cl–
1
H2
2
Eο = +1.36 V
Eο = 0.00V
Use an arrow to show the direction of flow of electrons in the
diagram of the apparatus. Explain your answer.
...........................................................................................................
...........................................................................................................
[2]
[Turn over]
42 | P a g e
(ii)
The values of Eο are measured under standard conditions. What
are the standard conditions?
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
(c)
The half cell reaction for ClO3–/
ClO3– + 6H+ + 5e–
1
Cl2 is shown below.
2
1
Cl2 + 3H2O
2
Eο = +1.47 V
What does this tell you about the oxidising ability of ClO3– compared with Cl2?
Explain your answer.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 8 marks]
Examiner’s comments
(a)
Some candidates were not awarded marks for giving appropriate state symbols for
+
H and Cl2
(b)
(i)
Most candidates gave the correct direction for the flow of electrons,
although it was necessary to show them flowing through the wire and not the salt bridge.
(ii)
A surprising number of candidates omitted to state that the concentration of the
solution should be 1 mol dm–3.
(c)
Many candidates answered this correctly by comparing two electrode potentials
43 | P a g e
Past paper questions
Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid
with potassium manganate(VII). The following standard electrode potentials
relate to this reaction.
1
Cl2 + e–
2
MnO4– + 8H+ + 5e–
(a)
Cl–
Eο = +1.36 V
Mn2+ + 4H2O
Eο = +1.52 V
Define the term standard electrode potential.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
(b)
Determine the standard cell potential for a cell constructed from these
two redox systems.
[1]
[Total 4 marks]
Examiner’s comments
(a)
This was generally well answered, but many candidates failed to state that the
standard conditions are 1 atmosphere pressure, 298 K and 1 mol dm–3.
(b)
A majority of candidates scored this mark. A few omitted the unit and a few gave
the obvious wrong answer of +2.88 V. There were also a few responses of –0.16 V.
44 | P a g e
12.
The feasibility of reactions
Unit 2 – Equilibria, energetics and elements
Done
Energy

Predict, using standard cell potentials, the feasibility of reactions.
Consider the limitations of predictions made using standard cell
potentials, in terms of kinetics and concentration.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.

A way to earn some marks is by knowing the definition of standard electrode
potential.

You will need to describe how standard electrode potentials can be calculated
against the standard hydrogen electrode – both for metals and ions or pairs of
ions.
 You will need to know how to use values of electrode potentials in order to
predict if reactions will occur.
Notice that a reaction takes place between reactants taken from different sides of the
two relevant half-equations.
The equilibrium with the more negative Eο value goes to the left.
45 | P a g e
Past paper questions
Some standard electrode potentials are shown below.
+
–
Ag + e
1
Cl2 + e–
2
Cu2+ + 2e–
Fe3+ + e–
1
I2 + e–
2
(a)
Ag
Eο/V
+ 0.80
Cl–
+ 1.36
Cu
Fe2+
+ 0.34
+ 0.77
I–
+ 0.54
Define the term standard electrode potential.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
(b)
The diagram below shows an incomplete cell consisting of Cu/Cu2+ and
Ag/Ag+ half-cells.
Cu(s)
Ag+(aq)
(i)
Complete and label the diagram to show how the cell potential of
this cell could be measured.
[2]
(ii)
On the diagram, show the direction of electron flow in the circuit if
a current was allowed.
[1]
[Turn over]
46 | P a g e
(iii)
Calculate the standard cell potential.
standard cell potential = ……………………V
[1]
(iv)
Write the overall cell reaction.
...........................................................................................................
[1]
(c)
Chlorine will oxidise Fe2+ to Fe3+ but iodine will not. Explain why, using
the electrode potential data.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 10 marks]
Examiner’s comments
(a)
This should have been an easy start to the paper but many candidates simply used
the words ‘electrode potential’ in the definition, rather than the expected ‘voltage’ or
‘emf’. Standard conditions were often not given.
(b)
This was generally well answered but a few candidates missed out the salt bridge
and a number had electrons flowing from silver to copper. On the point of electron flow,
candidates should indicate this on the wire between the electrodes. Too many candidates
put an arrow somewhere between the wire and the salt bridge.
(c)
This is a difficult concept. Good candidates had no problem with this but weaker
candidates tended to discuss larger or smaller electrode potentials without referring to
oxidation or reduction. There were too many statements such as ‘iodine is oxidised by
iron’. Clearly weaker students do not fully understand the concept of electrode potential.
47 | P a g e
Past paper questions
Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid
with potassium manganate(VII). The following standard electrode potentials
relate to this reaction.
1
Cl2 + e–
2
MnO4– + 8H+ + 5e–
(a)
Cl–
Eο = +1.36 V
Mn2+ + 4H2O
Eο = +1.52 V
Use the half-equations above to:
(i)
construct an ionic equation for the reaction between hydrochloric
acid and potassium manganate(VII);
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
(ii)
determine the oxidation numbers of chlorine and manganese
before and after the reaction has taken place;
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
(iii)
state what is oxidised and what is reduced in this reaction.
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
(b)
If potassium manganate(VII) and very dilute hydrochloric acid are mixed,
there is no visible reaction. Suggest why there is no visible reaction in
this case.
....................................................................................................................
....................................................................................................................
[1]
[Total 7 marks]
48 | P a g e
Examiner’s comments
(a)
(i)
More able candidates scored well here. Many weaker candidates showed
chlorine on the left hand side and chloride on the right hand side. A few candidates left
electrons in the final equation and were unable to correctly balance the equation.
(ii)
This was generally well done. Candidates who placed chlorine on the left-hand side
were awarded a mark here, carrying forward their error from part (i).
(iii)
Many candidates failed to score here because they did not specify the actual
species that was oxidised or reduced, often merely stating ‘chlorine is oxidised’ and
‘manganese is reduced’
(b)
Many candidates commented that the concentration was too low without linking
this to a slow reaction. A few suggested that the reaction would be slow owing to a small
cell potential. Very few suggested that the reaction was no longer possible because
conditions were not standard.
49 | P a g e
13.
Storage and fuel cells
Unit 2 – Equilibria, energetics and elements
Done
Energy

Apply the principles of electrode potentials to modern storage cells.
Explain that a fuel cell uses the energy from the reaction of a fuel
with oxygen to create a voltage.
Explain the changes that take place at each electrode in a
hydrogen-oxygen fuel cell.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.

A way to earn some marks is by knowing the definition of standard electrode
potential.

You will need to describe how standard electrode potentials can be calculated
against the standard hydrogen electrode – both for metals and ions or pairs of
ions.
 You will need to know how to use values of electrode potentials in order to
predict if reactions will occur.

It is important to know the changes that can happen within fuel cells – e.g. in
hydrogen-oxygen fuel cells.
 You must also be aware of any economic, political and environmental
issues.
In exams, you will not be expected to recall a specific storage cell. However, you
might be expected to make predictions about a given cell. All relevant electrode
potentials and other data will be supplied.
50 | P a g e
14.
Hydrogen for the future
Unit 2 – Equilibria, energetics and elements
Done
Energy

Outline the development of fuel cell vehicles (FCVs) that use
hydrogen gas and hydrogen-rich fuels.
State the advantages of FCVs over conventional petrol- or dieselpowered vehicles.
Understand how hydrogen might be stored in FCVs.
Consider limitations of hydrogen fuels cells.
Comment about the contribution of the ‘hydrogen economy’ to
future energy and discuss its limitations.
Key areas to concentrate on…

You will revisit the idea of redox – first learnt in AS Unit F321: Module 1 - Atoms
and reactions. At A2 you should be able to use the terms:
 redox
 oxidation number
 half-reaction
 reducing agent
 oxidising agent appropriately.

A way to earn some marks is by knowing the definition of standard electrode
potential.

You will need to describe how standard electrode potentials can be calculated
against the standard hydrogen electrode – both for metals and ions or pairs of
ions.
 You will need to know how to use values of electrode potentials in order to
predict if reactions will occur.

It is important to know the changes that can happen within fuel cells – e.g. in
hydrogen-oxygen fuel cells.
 You must also be aware of any economic, political and environmental
issues.
51 | P a g e
Specimen paper questions
Use the standard electrode potentials in the table below to answer the questions that
follow.
I
Fe2+(aq) + 2e–
Fe(s)
Eο = –0.44 V
II
V3+(aq) + e–
V2+(aq)
Eο = –0.26 V
III
2H+(aq) + 2e–
H2(g)
Eο = 0.00 V
IV
O2(g) + 4H+(aq) + 4e–
2H2O(l)
Eο = +0.40 V
An electrochemical fuel cell was set up based on systems III and IV.
(i)
Construct an equation for the spontaneous cell reaction. Show your
working.
[2]
(ii)
Fuels cells based on systems such as III and IV are increasingly being
used to generate energy.
Discuss two advantages and two disadvantages of using fuels cells for
energy rather than using fossil fuels.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[4]
[Total 6 marks]
52 | P a g e
Sample student answers
A2 Unit F325: Equilibria, energetics and elements
Module 5: Energy
Question 1
(a)
Total marks: 15
What is meant by the term lattice enthalpy?
Marks available: 2
Student answer:
(a) The enthalpy change when 1 mole of solid ionic compound is formed from its constituent
ions in the gaseous state
Examiner comments:
(a) The change is exothermic, so don’t start off by writing ‘The energy required...’.
53 | P a g e
(b)
Construct a Born–Haber cycle for magnesium chloride. Insert all the values from the
information given below.
First ionisation energy of magnesium = +736 kJ mol–1.
Second ionisation energy of magnesium = +1452 kJ mol–1.
First electron affinity of chlorine = –349 kJ mol–1.
Enthalpy of atomisation of magnesium = +148 kJ mol–1.
Enthalpy of atomisation of chlorine = +122 kJ mol–1.
Enthalpy of formation of magnesium chloride = –641 kJ mol–1.
Marks available: 6
Student answer:
(b)
Examiner comments:
(b) Here are a few tips:



54 | P a g e
All the arrows must be in the correct direction.
As well as using minus (–) signs in front of exothermic values, it is a good idea to
insert plus (+) signs in front of endothermic changes.
Check if multiples are needed such as with the Cl changes here.
(c)
Calculate the lattice enthalpy of magnesium chloride.
Marks available: 2
Student answer:
(c)
All values used are in kJ mol–1:
–641 = +148 + 736 + 1452 + (2 × +122) + (2 × –349) + LE
LE = –641 – [+148 + 736 + 1452 + (2 × +122) + (2 × –349)]
= –2523 kJ mol–1
Examiner comments:
(c)
Always show your working. Check the final answer by re-doing the calculation step by step,
for example:


(d)
replace 2 × 122 by 244
calculate the value within the square brackets before doing a final subtraction.
Define the term enthalpy of hydration.
Marks available: 2
Student answer:
(d) Enthalpy change when 1 mole of gaseous ions is dissolved in excess water
Examiner comments:
(d) As with most definitions, the term 1 mole will usually score a mark.
(e)
Explain why 2.0 g of magnesium sulfate will dissolve in a test tube of water but 2.0 g of
barium sulfate will not.
Marks available: 3
Student answer:
(e) Mg2+ ions are smaller than Ba2+ ions. This means they have a higher charge density.
Although the lattice enthalpy of MgSO4 is more exothermic than BaSO4, meaning the solid
lattice is held together more tightly, the enthalpy of hydration of Mg2+ ions is even more
exothermic than Ba2+ ions. The increase in the magnitude of hydration is greater than the
increase in magnitude of lattice enthalpy.
Examiner comments:
(e) Make sure you describe the correct particle, e.g. not Mg atoms or simply magnesium but
Mg2+ ions.
Try to use the phrase more exothermic lattice enthalpy rather than higher lattice enthalpy –
this is because lattice enthalpy is always a negative value.
55 | P a g e
Module 5: Energy
Question 2
(a)
Total marks: 15
Sulfuroxychloride and water react as shown by the equation given below:
SOCl2(l) + H2O(l)  SO2(g) + 2HCl(g)
(i)
Calculate the entropy change for the reaction:
ΔS SOCl2(l) = 218 J K–1 mol–1
ΔS H2O(l) = 70 J K–1 mol–1
ΔS SO2(g) = 248 J K–1 mol–1
ΔS HCl(g) = 187 J K–1 mol–1
(ii)
The enthalpy change for the reaction is +47.1 kJ mol–1 which means it is endothermic.
Predict if this reaction is spontaneous at room temperature, 298 K.
Marks available:
(i) 3 (ii) 3
Student answer:
(a)
(i) [248 + (2 × 187)] – [218 + 70] = +334 J K–1 mol–1
(ii) ΔG = ΔH – TΔS
= +47100 – (298 × +334) = –52432 J mol–1
= –52.4 kJ mol–1
ΔG < 0 Therefore reaction will take place.
Examiner comments:
(a) (i) Showing your working is always useful since marks can still be allocated for the final
answer, as allowances can be made for any mistakes as errors carried forward (ecf).
(ii) It is important to convert ΔH from kJ mol–1 to J mol–1 to match the units of ΔS. However,
the answer has to be written in the correct units: kJ mol–1.
56 | P a g e
(b)
Draw a diagram to show the standard hydrogen electrode and explain how you could use
the standard hydrogen electrode to measure the standard electrode potential of
Fe3+(aq) + e–  Fe2+(aq).
Marks available: 6
Student answer:
(b)
Examiner comments:
(b) In drawing this diagram marks awarded for:
 Hydrogen electrode labelled.
 Fe2+/Fe3+ half-cell labelled.
 Salt bridge and voltmeter labelled.
 Inert platinum electrodes labelled.
 Hydrogen at 1 atm.
 Fe2+, Fe3+ and H+ all at 1.00 mol dm–3.
57 | P a g e
(c)
A student believed aluminium would react with aqueous nickel ions after consulting
standard electrode potentials. Demonstrate why he came to this conclusion and suggest
one reason why he may be proved wrong.
Al3+ + 3e–
Al E
Ni2+ + 2e–
Ni E
= -1.67V
= -0.25V
Marks available: 3
Student answer:
(c)
The Al3+/Al equilibrium has the more negative (less positive) E
value and so the
equilibrium will move to the left and Al will react to form Al3+ ions, supplying electrons as it
does so.
The Ni2+/Ni equilibrium has the more positive (less negative) E
value and so the
equilibrium will move to the right and Ni2+ ions will react to form Ni atoms, accepting
electrons as it does so.
Thus the overall reaction should be 2Al + 3Ni2+ → 2Al3+ + 3Ni so the reaction is feasible.
Examiner comments:
(c)
There are other approaches to working this out, but it is always better to try to use what we
refer to as ‘first principles’.
58 | P a g e
2
9.0
Be
beryllium
4
24.3
Mg
magnesium
12
40.1
Ca
calcium
20
87.6
Sr
strontium
38
137.3
Ba
barium
56
[226]
Ra
radium
88
1
59 | P a g e
6.9
Li
lithium
3
23.0
Na
sodium
11
39.1
K
potassium
19
85.5
Rb
rubidium
37
132.9
Cs
caesium
55
[223]
Fr
francium
87
89
actinium
[227]
Ac*
57
lanthanum
39
138.9
La*
yttrium
21
88.9
Y
scandium
45.0
Sc
•
•
•
•
•
•
73
72
neodymium
60
uranium
92
140.9
Pr
praseodymium
59
[231]
Pa
protactinium
91
140.1
Ce
cerium
58
232.0
Th
thorium
90
238.1
U
144.2
Nd
106
105
seaborgium
dubnium
104
[266]
Sg
74
93
neptunium
[237]
Np
61
promethium
144.9
Pm
107
bohrium
[264]
Bh
75
rhenium
43
186.2
Re
tungsten
technetium
42
183.8
W
25
[98]
Tc
manganese
54.9
Mn
molybdenum
24
95.9
Mo
chromium
52.0
Cr
rutherfordium
[262]
Db
tantalum
hafnium
[261]
Rf
41
180.9
Ta
niobium
23
92.9
Nb
vanadium
50.9
V
40
178.5
Hf
zirconium
22
91.2
Zr
titanium
47.9
Ti
atomic (proton) number
name
Key
relative atomic mass
atomic symbol
94
plutonium
[242]
Pu
62
samarium
150.4
Sm
108
hassium
[277]
Hs
76
osmium
44
190.2
Os
ruthenium
26
101.1
Ru
iron
55.8
Fe
58.9
Co
58.7
Ni
63.5
Cu
65.4
Zn
95
americium
[243]
Am
63
europium
152.0
Eu
109
meitnerium
[268]
Mt
77
iridium
45
192.2
Ir
rhodium
27
102.9
Rh
cobalt
96
curium
[247]
Cm
64
gadolinium
157.2
Gd
110
darmstadtium
[271]
Ds
78
platinum
46
195.1
Pt
palladium
28
106.4
Pd
nickel
97
berkelium
[245]
Bk
65
terbium
158.9
Tb
111
roentgenium
[272]
Rg
79
gold
47
197.0
Au
silver
29
107.9
Ag
copper
zinc
81
thallium
49
204.4
Tl
indium
31
114.8
In
gallium
82
lead
50
207.2
Pb
tin
32
118.7
Sn
germanium
14
72.6
Ge
silicon
6
28.1
Si
carbon
12.0
C
83
bismuth
51
209.0
Bi
antimony
33
121.8
Sb
arsenic
15
74.9
As
phosphorus
7
31.0
P
nitrogen
14.0
N
84
polonium
52
[209]
Po
tellurium
34
127.6
Te
selenium
16
79.0
Se
sulfur
8
32.1
S
oxygen
16.0
O
85
astatine
53
[210]
At
iodine
35
126.9
I
bromine
17
79.9
Br
chlorine
9
35.5
Cl
fluorine
19.0
F
86
radon
54
[222]
Rn
xenon
36
131.3
Xe
krypton
18
83.8
Kr
argon
10
39.9
Ar
neon
20.2
Ne
98
californium
[251]
Cf
66
dysprosium
162.5
Dy
99
einsteinium
[254]
Es
67
holmium
164.9
Ho
100
fermium
[253]
Fm
68
erbium
167.3
Er
101
mendelevium
[256]
Md
69
thulium
168.9
Tm
102
nobelium
[254]
No
70
ytterbium
173.0
Yb
103
lawrencium
[257]
Lr
71
lutetium
175.0
Lu
Elements with atomic numbers 112–116 have been reported but not fully
authenticated
80
mercury
48
200.6
Hg
cadmium
30
112.4
Cd
13
69.7
Ga
aluminium
5
27.0
Al
boron
10.8
B
2
0
helium
7
1
6
hydrogen
5
4.0
He
4
1.0
H
3
The Periodic Table of the Elements
The Periodic Table of the Elements