Download Part I Power generation in fuel cells

S 7C
Name:_________________ ( )
Carmel Secondary School
Final Examination (04-05)
Chemistry (Paper 1)
Date: 23 Feb., 2005
Time allowed: 3 hrs.
Total no. of pages:13
Total marks: 100
Instructions:
1. There are THREE sections in this paper, Section A, Section B and Section C.
2. Section A carries 60 marks; Section B carries 20 marks; Section C carries 20
marks.
3. All questions in Section A and B are COMPULSORY. Answers are to be written
in this Question-Answer Book.
4. Answer ONE question in Section C. Answers are to be written in the single-lined
papers.
5. Some useful constants and a Periodic Table are respectively printed on page 13 of
this Question-Answer Book.
Section A
Answer ALL questions in this Section. Write your answers in the spaces provided.
1. (a)
Chlorine consists of two isotopes having mass number 35 and 37 of relative
abundance 75% and 25%, respectively. The questions below relate to the
mass spectrum of a sample of gaseous chlorine.
(i) How many peaks would you expect in the m/e range 34-38?
What would be the m/e value of the most intense peak in this range?
(Consider singly charged ions only)
(ii) How many peaks corresponding to ions of type Cl 2+ would you expect to
observe?
(Consider singly charged ions only.)
Give the m/e value of each and identify the most intense peak.
CHMS704E1P1
P.1
(5 marks)
(b)
The thermal decomposition of ozone is represented by the equation below
and is second order with respect to ozone:
2O3(g)  3O2(g)
(i) Write down the rate expression for this reaction, and use it to define the
terms rate constant and order of reaction.
(ii) The value of the rate constant at 298 K is 3.38 x 10-5 dm3 mol-1 s-1.
(1) State the units in which the rate of the reaction is expressed.
(2) Calculate the rate of reaction when the concentration of O3 is 2.5 x
10-3 mol dm-3.
(6 marks)
2
(a)
Define the first ionization energy of the element M.
(b)
Using the axes below, sketch the way in which the ionization energy changes.
(i) as the second period (Li-Ne) is crossed; and
(ii) as Group I is descended.
CHMS704E1P1
P.2
Li
3 (a)
Na
K
Rb
Cs
(5 marks)
Salbutamol, with structure shown below, is used in Ventolin inhalers to
help asthma sufferers breathe more easily.
HOCH2
OH
HO
CH-CH2-NH-R
R is an alkyl group, which for the purpose of this question does not
affect the rest of the molecule.
(i) Name four reactive organic groups in Salbutamol.
(ii) State two of these groups which are affected by mild oxidation, and what
groups are produced.
(iii) Would you expect the salbutamol molecule to be acidic, neutral or
alkaline? Explain your answer.
(6 marks)
CHMS704E1P1
P.3
3(b)
This question is about lactic acid, CH3CH(OH)CO2H, which occurs
naturally, especially in sour milk.
(i)
What is the systematic name for lactic acid?
(ii)
Draw this molecule in a three-dimensional representation.
(iii) What type of isomerism may be shown by this compound?
(3 marks)
3(c)
Dilute acidified sodium dichromate(VI) is used to distinguish between
primary, secondary and tertiary alcohols. Draw full structural formulae of
the final organic products (if any) when the following alcohols are treated
with this reagent.
(i)
CH3CH2CH2CH2OH
(ii)
CH3CH2CHCH3
OH
CH3
(iii)
CH3-C-OH
CH3
(3 marks)
CHMS704E1P1
P.4
4.
Chloroethene, A, is the monomer from which the polymer PVC is produced.
CH2=CH-Cl
A
A can be obtained from ethene in the laboratory by the following route:
stage I
CH2=CH2
stage II
ClCH2CH2Cl
CH2=CHCl
(a)
Suggest reagents and conditions for stages I and II.
(b)
Name the type of mechanism and write the steps in the mechanism of stage I.
(c)
Draw the displayed formula of the PVC chain, showing two repeating units.
What type of polymerisation is that in this example?
(d)
Suggest a reason why PVC might break down when exposed to concentrated
aqueous sodium hydroxide whereas poly(ethene) does not. Would you expect
poly(phenylethene) to be stable or unstable in the presence of aqueous sodium
hydroxide?
CHMS704E1P1
P.5
(8 marks)
5.
The passage below deals with two types of important electrochemical cell.
Read the passage and then answer the questions followed.
Part I
Power generation in fuel cells
Recent advances in electrochemistry and technology have led to the introduction
of a compact source of power in which fuels are used to produce electricity without
the intervention of thermal devices such as boilers, turbines and generators: these fuel
cells can be much more efficient than convention thermal sources. They have already
been employed in spacecraft, and may become the basis of pollution-free power for
transport.
A fuel cell operates just like a conventional electrochemical cell with the
exception that the material to be oxidized and reduced at the electrodes is stored
outside the cell rather than forming an integral part of its construction. A fundamental
but important example of a fuel cell is the hydrogen/oxygen cell. If the electrolyte is
an acid, the reduction reaction at the cathode is
O2 + 4H+ + 4e-  2H2O
At the anode the reaction is
2H2  4H+ + 4eOnce problem with this cell is that the current density of the oxygen reaction is
very small, and this limits the current available from the cell. One solution to this
problem is to use a large surface area for the electrode. Although in such cells the
normal oxidizing fuel is oxygen, because of its easy availability, the fuel for the anode
reaction may be any of a wide variety of materials. The hydrocarbons form a
particularly important group of anode fuels. Hydrocarbon/air fuel cells normally use
platinum electrodes, concentrated phosphoric acid electrolyte, and operate at
temperatures in the region of 100oC.
Part II
Corrosion cells
An indication of the likelihood of corrosion may be made on the basis of
equilibrium electrochemistry. The half reaction Fe2+(aq) + 2e-  Fe(s) (Eo = -0.44 V)
can be driven to the left by coupling it to another half reaction of suitable potential.
In acidic solution, the reactions most often responsible for driving the iron  iron(II)
oxidation are
(1) the reduction of hydrogen ions to hydrogen gas,
(2) the reduction of oxygen to water.
CHMS704E1P1
P.6
It must also be remembered that electrode potentials change as the conditions
become non-standard, and this must be taken into account when discussing the
feasibility of the corrosion process. The iron half reaction coupled to a half reaction
such as described above produces what is known as a corrosion cell, with the
corrosion process being the cell reaction. Seeing corrosion in electrolytic terms
allows the understanding of the inhibition of corrosion by galvanizing. This process
involves coating the iron object with zinc. Since the standard electrode potential for
the process is favoured, and the iron is protected even if the zinc coating is breached.
Tin, however, with a potential of –0.14 V drives the iron to oxidation if the tin coating
is breached.
(a) Choosing any two suitable half reactions, describe a specific ‘conventional
electrochemical cell’. How is the e.m.f. of the cell related to the electrode
potentials of the half cells from which it is made?
(b)
What is the cell reaction in the hydrogen/oxygen fuel cell? What is the cell
reaction in a hydrocarbon/air fuel cell, if the hydrocarbon is methane?
(c)
For the typical conditions under which a hydrocarbon/air cell is operated,
suggest reasons for
(i) the use of platinum electrodes,
(ii)
the use of a strong acid as electrolyte,
(iii) a temperature of 100oC.
(d)
Explain what is meant by the phrase ‘indication of the likelihood of
corrosion’. Why can it not be predicted with certainty that corrosion will
occur?
CHMS704E1P1
P.7
(e) Derive balance equations for the corrosion reactions implied by Part II (1)-(2)
in the passage.
(f)
In the last paragraph in the passage the effect of non-standard conditions is
noted. What effect will changing the pH from acid to neutral have on the
feasibility of the hydrogen ion  hydrogen gas half reaction? What ironcontaining reaction product is likely at pH 14? Explain your answer.
(g)
Describe the electrochemical cell that results if a tin-coated iron can
containing acidic fruit juice has the tin coating breached.
(17 marks)
6.
(a)
State two compounds of chlorine, other than sodium chloride, with largescale uses and describe one such use for each.
(b)
Explain the chemistry that is occurring during the following series of
reactions that all take place under aqueous conditions. When sodium
chloride is added to silver nitrate, a white precipitate forms which
dissolves in an excess of dilute ammonia. The subsequent addition of
sodium bromide to this latter solution causes the precipitation of a creamcolored solid.
Suggest why this latter precipitate dissolves when sodium cyanide is added.
CHMS704E1P1
P.8
(7 marks)
Section B
Answer ALL questions in this Section. Write your answers in the spaces provided.
7
(a)
The following two chromatograms were obtained by analysis of one
sample of a multi-component mixture.
(i) Describe and explain how these chromatograms were obtained.
(ii) How many components can be identified from these chromatograms?
(4 marks)
(b) Iron can is a common container of the drink. If the can start to rust, Iron(III)
ions will produce.
(i) Briefly describe an experiment, using chemicals and apparatus
commonly available in a school laboratory to determine the
concentration of Iron(III) ions in drink.
CHMS704E1P1
P.9
(ii) Give one precaution of the experiment.
(6 marks)
The ferrate ion, FeO42-, which contains iron in its highest known oxidation
state, can be prepared by reacting solid iron(III) oxide at 60oC, with
concentrated aqueous sodium hydroxide through which chlorine is passing.
The reaction mixture gradually turns deep purple and is filtered hot
through a sintered glass filter to give a solution containing sodium ferrate.
This solution is then treated with saturated aqueous barium chloride to
precipitate small crystals of deep-red barium ferrate. The precipitate is
removed by suction filtration and dried in air. It is stable only in strongly
alkaline solution, oxidizing water to oxygen if the pH is decreased.
The purity of a sample of barium ferrate prepared as above was determined
by adding 0.267 g to excess acidified potassium iodide solution. The
liberated iodine was titrated with sodium thiosulphate solution of
concentration 0.100 mol dm-3, 30.0 cm3 being required.
8.
(a)
(i)
Draw a labelled diagram of a suitable apparatus in which the reaction
might be carried out in the laboratory.
(ii)
Why is a filter paper not used for the first filtration?
(iii) Draw a diagram showing the apparatus required for suction filtration
and give an advantage of this method.
CHMS704E1P1
P.10
(b)
Suggest and justify one safety precaution which you would take.
(c)
(i)
What is the highest oxidation state which iron might reasonably be
expected to form? Explain your answer.
(ii)
Suggest a reason why compounds containing iron in this oxidation
state have not been prepared.
(10 marks)
CHMS704E1P1
P.11
Section C
Answer ONE question only and write your answers in the single-lined paper.
Marks will be allocated approximately as follow:
Chemical knowledge
Organization
Presentation (including use of language)
50%
30%
20%
Equations, suitable diagrams and examples are expected where appropriate.
The examiners are looking for the ability to analyse, to evaluate and to express ideas
clearly.
9.
Write an essay on phase equilibrium of two miscible liquids.
(20 marks)
10. Give an account of primary aliphatic alcohols. Your answer should include
laboratory methods of preparation, and an account of the chemical behaviour of
the alcohols, where appropriate, give mechanisms for the reactions you have
mentioned and indicate how they would differ for tertiary alcohols.
Explain why phenol is more acidic than ethanol and ethanol has a higher boiling
point than ethane.
End of Paper
CHMS704E1P1
P.12
Useful Constants
Gas constant, R = 8.31 J K-1 mol-1
Faraday constant, F = 9.65 x 104 C mol-1
Avogadro constant, L = 6.02 x 1023 mol-1
Plank constant, h = 6.63 x 10-34 Js
Speed of light in vacuum, c = 3.00 x 108 ms-1
Ionic product of water at 298 K, Kw = 1.00 x 10-14 mol2 dm-6
Specific heat capacity of water = 4.18 J g-1 K-1
Bond
C=C
C=O
C C
C N
O-H
C-H
O-H
N-H
Characteristic Infra-red Absorption Wavenumber Ranges
Compound type
Wavenumber range /cm-1
Alkenes
1610 to 1680
Aldehydes, ketones, carboxylic acids, esters
1680 to 1750
Alkynes
2070 to 2250
Nitriles
2200 to 2280
Acids (hydrogen-bonded)
2500 to 3300
Alkanes, alkenes, arenas
2840 to 3095
Alcohols, phenols (hydrogen-bonded)
3230 to 3670
Amines
3350 to 3500
CHMS704E1P1
P.13
S 7C
Carmel Secondary School
Final Examination (04-05)
Chemistry (Paper 1)
Name:_________________ ( )
1.
(a) (i) 2 peaks
m/e
Ion
Rel. abundance
35
Cl+
3 (most intense)
35
Date: 23 Feb., 2005
Time allowed: 3 hrs.
Total no. of pages:13
Total marks: 100
37
Cl+
1
37
(2)
(ii)
3 peaks
m/e
Ion
Rel. abundance
70
(35Cl2)+
9
(most intense)
72
(35Cl37Cl)+
6
74
(37Cl2)+
1
(3)
(b)
2O3(g)  3O2(g)
(i)
Rate = k[O3]2
(1)
Order of reaction gives exact relationship between variation of rate with
concentration of reactant, k is the proportionality constant in rate equation. (2)
(ii)
(1)
Units of rate : mol dm–3 s–1
(1)
(2)
Rate = k[O3]2
= 3.38  10–5  (2.5  10–3)2
(1)
= 2.11  10–10 mol dm–3 s–1
2
(1)
(a) Ionization energy is the energy required to remove one mole of electrons from one
mole of its gaseous atoms to form one mole of gaseous positive ions.
(1)
(b)
(i)
.
CHMS704E1P1
P.14
(ii)
(4)
(1/2)
(1/2)
3. (a) (i) Primary alcohol —CH2OH
Secondary alcohol) —CHOH
Phenol
(1/2)
(1/2)
2° amine —N—HR
(ii) Primary and secondary alcohol are oxidised to acid and ketone respectively.
(1)
(iii) Alkaline
Presence of 2° amine —NHR with lone pair of electrons-proton acceptor.
(1)
(1)
present is only weakly acidic.
3(b) (i)
2-hydroxypropanoic acid
(1)
(1)
(ii)
(1)
(iii)
Optical isomerism
(1)
3(c) (i)
(1)
CHMS704E1P1
P.15
(ii)
(1)
(iii)
(1)
4.
(a)
I — Cl2 dissolved in CCl4, absence of light
II — alcoholic KOH, reflux
(1)
(1)
(b)
Electrophilic addition
(1)
(2)
(c)
Addition polymerisation
(1)
(1)
(d)
PVC is a polymeric chloroalkane which can be hydrolysed by aq. NaOH (C—Cl
bonds cleaved by OH–). Poly(ethene) contains only C—H bonds which are not
affected by NaOH.
(1)
Similarly poly(phenylethene) contains only C—H bonds and (
) which
are not
affected by NaOH and hence is stable in NaOH.
5
(a)
(b)
CHMS704E1P1
Zn(s)Zn2+(aq, 1 mol/dm3)Fe2+(aq, 1 mol/dm3)Fe(s)
E cell = Eredn – Eoxidn
= 0.44 – (–0.76)
= +0.32 V
H2 (g) +
1
O2(g)  H2O(l)
2
P.16
(1)
(2)
(1)
(1)
(c)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
(1)
(i) Inert, resistant to corrosion by acid and O2
(ii) To provide H+ for reduction of O2 to H2O.
O2 + 4H+ + 4e–  2H2O
(iii) To speed up reactions at the electrodes.
(1)
(1)
(d)
Predictions are based on the consideration of energetic feasibility and
standard conditions. The reaction may not occur because the rate is too
slow or the actual reaction conditions are not standard.
(2)
(e)
2H+ + 2e–  H2
O2 + 4H+ + 4e–  2H2O
(f)
Increment in pH (from acid to neutral) reduces the tendency for H+ to be
reduced (lower [H+]).
(1)
(1)
(1)
Product at pH 14: Fe(OH)2
Reduction of O2 in neutral conditions produces OH–.
O2 + 2H2O + 4e–
4OH–
(g)
When Sn coating is cracked, Fe is exposed.
Sn2+/Sn
E = –0.14 V
Fe2+/Fe
E = –0.44 V
i.e. Fe is oxidised instead of Sn.
Fe  Fe2+ + 2e–
 Fe is corroded.
6.
(1)
(1)
(1)
(1)
(1)
(a)
— Bleach and germicide
— Manufacture of chlorine-compounds e.g. chloroethene for PVC
(1)
(1)
(b)
Aq. Cl– forms white ppt AgCl with aq. Ag+.
Cl– + Ag+  AgCl
(1)
AgCl dissolves in aq. NH3 to form soluble complex.
AgCl + 2NH3  [Ag(NH3)2]+ + Cl–
(1)
Addition of Br– ppts AgBr (cream colored):
[Ag(NH3)2]+ + Br–  AgBr + 2NH3
(1)
AgBr has a lower solubility product than AgCl.
Hence, AgBr ppt are formed instead of AgCl.
(1)
AgBr dissolves in aq. CN– to form a soluble complex.
AgBr + 2CN–  [Ag(CN)2]– + Br–
(1)
Section B
7
(a)
(i)
CHMS704E1P1
Run 1st chromatogram using one solvent.
(1)
Turn the developed chromatogram by 90° and run again using a second
solvent.
(2)
P.17
(ii)
(b)
8.
(a)
3 components
(1)
Add known excess amount of I- to the drink.
Dilute to 250 cm3 using 250 cm3 volumetric flask.
Pipette 25 cm3 of the mixture to the conical flask.
Titrate with sodium thiosulphate using starch as indicator. (5)
(i)
(1)
(2)
(3)
(4)
(ii)
Starch should be added only when the reaction mixture becomes pale
yellow
(1)
(i)
(2)
(ii)
Filtration is done while solution is hot, solute passes through expanded
pores of filter paper.
(1)
(iii)
(2)
(b)
Reaction should be carried out in fume cupboard. It is because Cl2 used is an
extremely reactive and choking gas.
(2)
(c)
(i)
(ii)
+8
Fe [Ar] 3d6 4s2
By losing up to a maximum of 8e
(1)
Ionisation energy is too great to be feasible.
(1)
(1)
Section C:
9.
Ideal Solutions
Raoult’s Law states that when two miscible liquids are mixed, the
CHMS704E1P1
P.18
(Max. 4)
vapour pressure of each component is equal to the product of mole
fraction of that component and the pure vapour pressure of that
component.
An ideal solution is a solution which obeys Raoult’s law.
The boiling point composition curve is as follows:
For two liquids A and B, if they form an ideal solution, the
attractive force between A and B is the same as that between A
and A and B and B.
For example: hexane and heptane.
Non-ideal solution
Positive Deviation from Raoult’s Law
For two molecules A and B, if the attraction between A and B is
weaker than that between A and A and B and B. The molecules in
the mixture have a higher escaping tendency and thus have a
higher vapour pressure.
(Max. 3)
For example: a mixture of ethanol and heptane.
P > XAPA + XBPB
Negative deviation from Raoult’s Law
For two molecules A and B, if the attraction between A and B is
CHMS704E1P1
P.19
(Max. 3)
stronger than that between A and A and B and B. The molecules
in the mixture have a lower escaping tendency and thus have a
smaller vapour pressure.
For example: a mixture of trichloromethane and ethyl ethanoate.
P < XAPA + XBPB
Others:
For ideal solution, both the two pure components can be separated
by fractional distillation.
For non-ideal solution, only one of the pure components and an
azeotropic mixture will be obtained after fractional distillation.
(Max. 2)
10.
Preparation of a primary alcohol: R-CH2OH
(max. 1)
(i) From halogenoalkanes or alkyl halides:
A primary halogenoalkane, R-CH2-X, where X is Cl, Br or I, and R is an alkyl
group, undergoes a nucleophilic substitution reaction when refluxed with an
aqueous solution of potassium hydroxide to form a primary alcohol.
For example: On refluxing iodoethane, CH2CH2I, with an aqueous solution of
potassium hydroxide ethanol is formed.
CH3CH2I(l) + KOH(aq)  CH3CH2OH(l) + KI(aq)
(ii)
From catalytic hydration of alkene
An alkene can undergo hydration to form primary alcohol.
For example:
Ethene can react with water to form ethanol with the present of Ni catalyst.
CH2=CH2 + H2O  CH3CH2OH
Ni
CHMS704E1P1
P.20
Chemical properties of alcohols:
(max 7 marks)
The chemical properties of primary alcohol can be divided into two categories:
(i)
(ii)
Reactions involving fission of O-H bond and
Reactions involving fission of C-OH bond.
The chemical properties of a primary alcohol and illustrated using ethanol as an
example.
Reactions involving fission of O-H bond:
(i) Reaction with sodium:
All primary alcohols react with sodium to give off hydrogen gas.
2C2H5OH + Na  2C2H5O-Na+ + H2
(ii)
Esterification
All primary alcohols react with carboxylic acids in the presence of a few drops
of concentrated sulphuric acid to form esters.
C2H5OH + CH3COOH  CH3COOC2H5 + H2O
The primary alcohols also react with acid chlorides, acid anhydrides to form
esters. The reaction takes place at much faster rate.
C2H5OH + CH3COCl  CH3COOC2H5 + HCl
Mechanism for esterification:
The probable mechanism for the reaction involves the following stages:
(i) Protonation of ethanoic acid
(ii) The alcohol then attacks the electron deficient carbon atom in the protonated
acid. An intermediate compound is formed.
(iii) Proton transfer takes place initially and finally ester is formed and the catalyst is
regenerated.
+
O
CH3
C
OH
+
+ H
CH3
+
OH
C
OH
C
OH
+
CH3
OC2H5
+
CH3
C
+
H3O
OH +
C2H5OH
C
OC2H5
+ OH2
OH
C
CH3
OH
OH H
CH3
OH
O
OC2H5 + H2O
CH3
C
OC2H5
+
Reaction involving fission of C-OH bonds:
(i) Reaction with phosphours(V) chloride:
All organic compounds containing an –OH group react with phosphorus(V)
chloride and give off hydrogen chloride. Primary alcohols react in the same
manner.
C2H5OH(l) + PCl5(s)  C2H5Cl(l) + POCl3
CHMS704E1P1
P.21
Primary alcohols also react with sulphur dichloride oxide (thionyl chloride)
giving off hydrogen chloride.
C2H5OH(l) + SOCl2(g)  C2H5Cl(l) + SO2(g) + HCl(g)
Primary alcohols react with phosphorus(III) chloride forming chloroalkanes but
no hydrogen chloride is given off in this case.
3C2H5OH(l) + PCl3(g)  3C2H5Cl(l) + H3PO3(aq)
(ii)
Reaction with concentrated sulphuric acid
Primary alcohols react with concentrated sulphuric acid under three different
conditions:
Equal amounts: On warming equal amounts of alcohol and concentrated
sulphuric acid alkyl hydrogen sulphate is formed.
C2H5OH + H2SO4  C2H5O-SO3H + H2O
(1)
Excess of concentrated sulphuric acid
Primary alcohols are dehydrated to form an alkene when heated to
approximately 443 K with an excess of concentrated sulphuric acid.
C2H5OH  C2H4 + H2O
(2)
Excess of alcohol
On heating an excess of a primary alcohol with concentrated sulphuric acid to
400 K and ether is formed.
(3)
Excess of concentrated sulphuric acid:
2C2H5OH  C2H5OC2H5 + H2O
Mechanism for the formation of ether
(i) The primary alcohol is protonated by sulphuric acid to form protonated alcohol.
(ii) The protonated alcohol loses a water molecule to form a carbonium ion.
(iii) Nucleophilic attack of the carbonium ion takes place by the alcohol forming an
intermediate compound.
(iv) The intermediate compound loses a proton to form ether and the catalyst is
regenerated.
H
CH3CH2
+
+ H
OH
C2H5
+
CH3CH2
O
+
CH3CH2
+ C2H5OH
H
C2H5
+
O
+
+ H2O
C2H5
H
CH3CH2
O
C2H5
+
+
H
Reaction with hydrogen halides: HCl, HBr and HI
The primary alcohols react with hydrogen halides to form halogenoalkanes. The
reaction takes place more easily with HI and HCl.
C2H5OH + HBr  C2H5Br + H2O
CHMS704E1P1
P.22
Mechanism for the reaction:
The first two stages are exactly the same as that for the formation of ether.
(i) The primary alcohol is protonated by HBr.
(ii) The protonated alcohol loses a molecule of water to form a carbonium ion.
(iii)The carbonium ion is then attacked by Br- ion to give the product.
H
CH3CH2
+
+ H
OH
C2H5
+
+
CH3CH2
O
+
CH3CH2
Br
H
C2H5
+
+ H2O
Br
Reaction involving the carbon-hydrogen skeleton
Primary alcohols on mild oxidation form aldehydes, which on further oxidation form
carboxylic acid.
Mild oxidation: On warming a primary alcohol with an acidified solution of
potassium dichromate(VI), the alcohol is oxidized to the aldehyde whilst the orange
dichromate(VI) is converted to blue Cr3+ ions. The aldehyde is distilled off as soon
as it is formed.
3 C2H5OH + Cr2O72- + 8H+  3CH3CHO + 2Cr3+ + 4H2O
The reaction can also be carried by heating the alcohol with an alkaline solution of
potassium manganate(VII) or by passing the vapour of alcohol over heated copper at
673 K.
Further oxidation: The aldehyde formed above can be further oxidized to ethanoic
acid. This can be brought about by heating the aldehyde with an acidified solution of
potassium dichromate(VI) or acidified solution of potassium manganate(VIII)
Difference between primary and tertiary alcohols
Oxidation:
Tertiary alcohols are resistant to oxidation.
Lucas Test: Reaction with hydrogen chloride in the presence of anhydrous zinc
chloride: Both primary and tertiary alcohols react with hydrogen chloride in the
presence of anhydrous zinc chloride to produce turbidity due to formation of
halogenoalkane. The rate at which they react is different.
3o alcohols produce turbidity almost immediately whilst 1o alcohols take relatively a
much longer time. (Even no observation)
Acidity of phenol:
(1 mark)
The lone pair of electrons on the oxygen atom of the –OH group are delocalised over
the ring. This weaken the oxygen-hydrogen bond and facilitates the loss of a proton.
C6H5OH
H+ + C6H5Othe above equilibrium lies more to the left.
O
CHMS704E1P1
P.23
H
Higher boiling point of ethanol
(1 mark)
Ethanol is a polar covalent compound capable of forming intermolecular hydrogen
bonds as such has a higher boiling point than ethane. Ethane is a non-polar compound
which exists as discrete molecules which are held together by weak van der Waals’
forces; there are no hydrogen bonds formed in this case.
CHMS704E1P1
P.24