Download 2010 `A` Levels Suggested Solutions

2010 ‘A’ Levels Suggested Solutions
Paper 1
1
2
3
4
5
B
B
B
D
B
6
7
8
9
10
C
A
A
D
A
11
12
13
14
15
B
D
C
B
C
16
17
18
19
20
B
C
C
B
C
21
22
23
24
25
D
B
B
D
D
26
27
28
29
30
A
B
C
C
D
31
32
33
34
35
A
B
A
B
C
36
37
38
39
40
B
C
A
A
C
Paper 2
Question 1
(a)
Overall eqn: 3C2O42- +2FeO42- + 16H+ à 6CO2 + 2Fe3+ + 8H2O
Note: Qn stated that the ions are reduced to 2Fe3+ (and NOT Fe2+) under acidic conditions.
You are tested on the concept of balancing redox half eqns, and then combining them to
come up with the overall eqn..
Half eqns:
C2O42- à 2CO2 + 2e (given)
FeO42- + 8H+ + 3eà 6CO2 + Fe3+ + 4H2O
Overall eqn: 3C2O42- +2FeO4- + 16H+ à 6CO2 + 2Fe3+ + 8H2O
(b)
If volume of FeO42- is 25.0 cm3 (measured using a pipette), burette reading would
be: By proportion, A burette reading over 50cm3 would require careful refilling of the burette, making
the titration difficult.
If volume of C2O42- is 25.0 cm3 (measured using a pipette), burette reading would
be: By proportion, A small burette reading of 8 cm3 would lead to high percentage error.
(c)
*When planning the procedure, quantities are to be provided as stated by the question,
hence some calculation must be done to find out the mass / volumes of the reagents to be
use.
The general procedure is
i.
weighing of the chosen mass of sodium ethanedioate
ii.
dissolving the solid and preparation of the standard solution of sodium ethanedioate
iii.
carrying out titration (burette, pipette, conical flask, colour change at end point)
(i)
Calculation of mass of sodium ethanedioate to be weighed
Calculation:
Let 1 drop be 1 dm3
Amount of C2O42- used = 0.100 x 32 = 3.2 mol
Amount of FeO42- required = 2/3 (3.2) = 2.13 mol
Since 10 drops was used, [FeO42-] = 2.13/10 = 0.213 mol dm-3
.
We let volume of FeO42- to be used = 25 cm3
Amount of FeO42- in 25 cm3 = 0.213 x 25/1000 = 5.33 x 10-3 mol
Amount of C2O42- required to react = 5.33 x 10-3 x 3/2 = 8.00 x 10-3 mol
For the volume of solution used to be 25 cm3,
[C2O42-] = 8.00 x 10-3 / 25 x 1000 = 0.320 mol dm-3
Amount required to be dissolved in 250 cm3 volumetric flask
= 0.320 x 250/1000
=0.0800 g
Mass required to be dissolved in 250 cm3
= 0.0800 x [23.0(2) + 12.0(2) + 16.0(4)]
= 10.72 g
Procedure:
1. Using a weighing balance, weigh accurately between 10.5-10.9 g of Na2C2O4 in a clean
and dry weighing bottle. (state a suitable range ±0.2g to weigh out)
2. Transfer the Na2C2O4 into a 100 cm3 beaker and weigh the weighing bottle with the
residue. The difference in mass will give the mass of the Na2C2O4 that was used to
prepare the solution.
3. Add 100 cm3 of H2SO4 (aq) to dissolve the solid. An excess of acid must be added to
provide an acidic medium.
(As seen in the balanced eqn in (a), acidic medium is required and we add H2SO4 to
provide the acidic medium. H2SO4 is the best choice to acidify solutions as it will not be
oxidized as easily as other acids, such as HCl)
4. Transfer the solution and the washings into a 250 cm3 volumetric flask and make up to
mark with water.
5. Stopper and shake the volumetric flask to ensure thorough mixing.
6. Pipette 25.0 cm3 of the solution prepared into a 250 cm3 conical flask.
7. Top up the burette with FA1.
8. Titrate the solution against FA1 and stop when the first trace of orange is observed.
(dark red FA1 will change to yellow (Fe3+) in this reaction. The end point is marked by
the appearance of orange, due to 1 excess drop of FA1)
9. Repeat the titration until the readings are consistent (difference within 0.10 cm3).
Note:
Considering the colour change at the end point, if we have the C2O42- solution in the
conical flask, the colour change is colourless to yellow and finally to orange at end point.
If FA1 is in the conical flask instead, the colour change is red to orange to yellow at end
point. The colour change from yellow to orange is more obvious than orange to yellow.
That is why we choose to have the C2O42 solution in the conical flask.
Of the three mineral acids (HNO3, HCl and H2SO4) available, only sulfuric acid is
suitable for use because :
• Ferrate (VI) ions would oxidise HCl to Cl2 (just as manganate (VII) ions do)
• HNO3 itself is an oxidising agent and will interfere with the oxidising action of
Ferrate (VI) ions..
Hence, the titration readings would not be accurate if HCl or HNO3 are used.
(d)
M
) mol
1000
2
M
(y x
) mol
n(FA1) =
3
1000
2
M
2
(y ×
)
( yM )
2yM
1000 × 1000 = 3
[FA1] = 3
=
mol dm-3
x
x
3x
NB:
Be very careful with your units when doing such questions.
n(FA2) = (y x
(e)
FeO42- is a very strong oxidizing agent, hence it is corrosive/ may cause burns. Use
gloves to avoid contact with the chemical.
NB:
SOME other reagents that must be handled with care:
- SOCl2, PCl5 reacts vigorously with water, giving fumes of strong acid HCl.
Hence reaction must be carried out in a fumehood.
- LiAlH4 reacts with water, making it ineffective. Hence, reaction must be
carried out in an anhydrous condition.
- Organic compounds/solvents, are usually flammable, hence, reaction is to be
carried out in a fumehood, and away from naked flame.
- Organic compounds that contain benzene rings are cancer causing, hence
reaction is to be carried out in fumehood, and use gloves.
Question 2
(a)
NaBr + H2SO4 à NaHSO4 + HBr (Group VII chemistry)
CH3(CH2)2CH2OH + HBr à CH3(CH2)2CH2Br + H2O
(b)
25 × 0.81
= 0.274 mol
74
n(NaBr) = n(HBr) = 35 / 103 = 0.340 mol (Note: use Mr data from table)
n(butan-1-ol) =
Since butan-1-ol reacts with HBr in a 1:1 mole ratio, NaBr is in excess.
(c)
(i) Concentrated H2SO4 reacts very exothermically when diluted with water due to
the highly exothermic enthalpy change of hydration by water.
(ii) Inorganic by-product : Br2(g) + SO2(g)
NaBr + H2SO4 à NaHSO4 + HBr
2HBr(g) + H2SO4(l) Br2(g) + SO2(g) + 2H2O(l)
Organic by-product: But-1-ene
CH3CH2CH2CH2OH à CH3CH2CH2C=CH2 + H2O
It was expected that candidates would consider the well-known reactions of
concentrated H2SO4 with NaBr or with HBr to produce Br2 and SO2. Many
candidates did this while fewer were able to write a correct equation for the reaction.
Correct answers for the identity of the organic by-product were less common. The
dehydration of alcohols by concentrated H2SO4 is well known when asked directly
but many candidates did not bring such knowledge to this question. A significant
number of those who did deduce what would happen did not identify the alkene as
but-1-ene, giving the ambiguous answer ‘butene’ or ‘C4H8’ which Examiners did not
accept.
(d)
Heating for 45 min is required as the reaction has a high activation energy and is
slow, because strong covalent bonds need to be broken in the reactants.
(This is the reason to why heating is required for most of the organic reactions as
organic reactions involves breaking (and forming) covalent bonds.)
(e)
Butan-1-ol and water
Their b.p. are very close to that of 1-bromobutane.
(f)
Lower
1-bromobutane (1.35 g cm-3) is more dense than water (1.00 g cm-3).
(g)(i)
Butan-1-ol
(g)(ii)
CH3(CH2)2CH2OH + H+ à CH3(CH2)2CH2OH2+
This reaction is unusual as you’ve been taught that alcohols are neutral. But note
that the use of conc HCl is to remove the organic impurity that cannot be separate
by Step 5. The ONLY organic reactant used is butan-1-ol making it the only possible
candidate as the organic impure that the question is looking for. (refer to (e))
Butan-1-ol is immiscible with water due to the 4C alkyl chain that makes it non polar.
Hence, something must be happening between the acid and the butan-1-ol to cause
it to be soluble in the water layer. The only viable option is the protonation of the
alcohol occurred in order for the resulting ion to form ion-dipole with water.
(g)(iii) It is able to form strong ion-dipole attraction with the water molecules.
(h)
Conc. HCl
2HCl + Na2CO3 à 2NaCl + H2O + CO2
(i)
Water
Traces of water that was used in step 5 and 6 could be still present in the product.
Anhydrous calcium chloride is a drying agent that can be used to remove water.
i.e. ahydrous calcium chloride is added to the reaction mixture, and it will absorb any
traces of water present. The calcium chloride is then removed by filtration.
(j)
99 – 103 oC (note the range)
Note Bpt of 1-bromobutane is 102 OC. ‘General rule’ is -3 +1 (as taught by Mr Lee
during Prelims review of P2 Q2)
Question 3
(a)
NO is a radical hence it would lead to ozone depletion
NO is readily oxidized to NO2, a brown toxic gas that causes respiratory illness.
(b)
initial rate / Nm-2s-1
(ρNO)2 / N2 m-4
graph cuts through the origin.
If ρNO is choosen to be the x-axis instead, a curve will be obtained, the relation
between rate and ρNO cannot be deduced immediately without showing some
calculation.
(c)
The order of reaction with respect to ρNO is 2.
The graph is linear, hence rate increases linearly with (ρNO)2.
(d)
First order.
When the partial pressure of oxygen is halved, the gradient of the graphs is halved,
i.e. the rate of the reaction is halved.
(e)(i)
Rate = k(ρNO)2(ρO2)
Must be expressed in terms of partial pressure, not [ ]!
(e)(ii)
N-2 m4 s-1
Question 4
(ai)
Magnitude of the LE decrease from MgSO4 to BaSO4.
(aii)
Ionic radius of Mg2+: 0.065 nm
Ionic radius of Ca2+: 0.099 nm
Ionic radius of Sr2+: 0.113 nm
Ionic radius of Ba2+: 0.135 nm
|q q |
Since | LE |∝ + − where q is the charge on the ion, and r is the ionic radius
r+ + r−
The ionic radius increases from magnesium ion to barium ion, hence, |LE|
decreases.
(b)(i)
(b)(ii)
The charge density decreases from magnesium ion to barium ion due to the
increase in size of the ions. Hence the ion-dipole
ion dipole formed with water decreases in
strength.
(c)(i)
Nucleophilic addition
(c)(ii)
Hydrolysis
OH
(d)
H
C
CH3
(e)
CH3CH2CH2COO−
CH3CH2CH2COOH
Question 5
(a)(i)
When proteins are denatured, the R–group interactions are disrupted, causing only
secondary, tertiary and quaternary structures to be destroyed permanently, hence
losing its biological activity. The primary structure remains intact.
(a)(ii)
Collagen may be denatured by addition of acid to it. This will disrupt the ionic
bonding originally present between the R groups.
(b)
6 (note that there is a total of 9, but there are repeat amino acids. The question asks
for different amino acids)
(c)(i)
DISPLAYED formula
H
O
C
O
H
H
H
N
C
OH
C
N
H
H
O
c(ii)
N
H
d(i)
d(ii)
H
*
*
C
HO
O
O
Add H2SO4 (aq), heat under reflux for several hours
(NB: This reaction requires the breaking of the numerous strong covalent peptide
bonds, heating for a prolonged period is necessary. Also, as this question asked
about “in the laboratory”, a more detailed answer about the procedure will be wise.)
O
O
C
HO
C
H2N
NH2
OH
R
d(iii)
H2N
C
COOH
e)
H
Proteins is made up of amino acid monomers with general formula
There are both amine and carboxylic acid functional groups in the amino acid
monomer.
But monomers of Kevlar consists of only carboxylic acid or amine groups.
HOCH2CO2CH(CH3)CO2H
CH3CH(OH)CO2CH2CO2H
f(i)
f(ii)
Hydrolysis of the ester linkages
The products are able to form hydrogen bonds with water.
.
Paper 3
Question 1
a(i)
Entropy is the measure of the disorder of the system. [1]
a(ii)
1 mol of Cl2 added to 1 mol of N2 at 298 K
∆S is positive/entropy increases because there is mixing of gaseous particles,
leading to a higher state of disorder in the system. [1]
1 mol of Cl2 heated to 373 K
∆S is positive/entropy increases as the system possess more kinetic energy. This
implies that there are more ways to distribute the energy among the particles giving
rise to greater disorder. [1]
It is NOT enough to state there is an increase in KE. The increase in entropy
comes about because in the broadening of the Boltzmann energy distribution
as there are more ways of arranging energy quanta in the hotter gas.
1 mol of Cl2 reacts iodine: Cl2 (g) +
1
2
I2 (s)à
à ICl3 (l)
3
3
∆S is negative/entropy decreases as there is a decrease in number of gaseous
particles, leading to a higher state of disorder in the system [1]
1 mol of Cl2(g) is photolysed: Cl2 (g) à 2Cl (g)
∆S is positive/entropy increases as there is an increase in number of gaseous
particles, leading to a higher state of disorder in the system [1]
Be very careful with you terms. Entropy (S) increases/decreases. ∆S is
positive or negative.
bi)
H2 + Cl2 à 2HCl (fast – due to higher bond energy of HCl)
H2 + Br2 à 2HBr (slow) [1]
The important comparison here is that of the RATE of their reaction, not on
the extent of the reaction. No further merit could be gained by detail
analysis of the enthalpy changes in bonds being formed or broken.
bii)
H2 + F2 à 2HF (very fast) [1]
ci)
Conditions: UV light/Heat [1]
cii)
Free Radical Substitution. [1]
Cl
Initiation:
Propagation:Cl
u.v.
Cl
H
Cl
Cl [1]
CH3CH2•
CH2CH3
CH3CH2•
Cl
CH3CH2
Cl
HCl
Cl
Cl
[1]
Termination:
CH3CH2•
Cl
Cl
Cl
CH3CH2•
ciii)
CH3CH2
Cl
CH3CH2
CH3CH2•
Cl
Cl
CH2CH3[1]
Iodoalkanes cannot be formed because the reaction is endothermic.
∆H (iodoalkane) = (+410 + 151) – (240+299) = + 22 kJ mol-1
∆H (chloroalkane) = (+410 + 244) – (340+431) = – 117 kJ mol-1
[1]
Candidates needed to point out that iodination does not occur because the first
propagation step is endothermic, or point out the relevance of the BE (H—I) bond.
d)
CH3
CH3
Cl
CH3
Cl
CH3
Cl
H3C
CH2Cl
H3C
Br
CH2Cl
B
A
Br
C
D
[1] for each structure
ei)
They are inert and stable/non-flammable/non-toxic
eii)
UV light causes the C—Cl bonds to break and releases chlorine radicals, which
attack ozone and destroys the ozone layer. [1]
eiii)
Alkanes are flammable. [1]
[1]
Question 2
ai)
HA + H2O
Ka =
H3O+ + A—
[H3O+ ][A - ]
[H O+ ][A - ]
; pK a = - lg 3
[1]
[HA]
[HA]
pKa measures the strength of a weak acid. The lower the pKa, the stronger is the
acid. [1]
+
+
NH3
NH3
aii)
pH 1 HOOC
COOH
pH 3 HOOC
COO
+
NH2
NH3
-
pH 7 O OC
COO
-
-
-
pH 11O OC
COO
-
[1] for each structure
bi)
The protein is denatured during coagulation. The R group interactions are
destroyed, disrupting the tertiary and quaternary structures. [1]
The protein disintegrates to form random coils of polypeptide chains. [1]
Note: Under conditions described, secondary structures remain intact. This
is because hydrogen bonds in secondary structures are not affected. The
hydrogen bonds are formed between peptide linkages since peptide linkages
do not undergo protonation/deprotonation.
bii)
Reaction with Ca2+ will bind to carboxylic anions [1] as found in glutamic acid
residues, hence disrupting the ionic interactions [1]
Before denaturation:
After denaturation:
------ H3N+−−
−−COO
−−
Ionic bonds
−−COO
Ca2+
−−
H3N+−−
(Ionic bonds broken)
NB: Only consider the disruption of S—S when cysteine amino acid is
present!
Only the heavy transition metal ions Cu+ and Hg+ can denature proteins by
disrupting the S—S.
biii)
H+ protonates the ionic R groups, and disrupt the ionic bonds[1] which
stabilises the tertiary or quaternary structures. [1]
Low pH:
biv)
[GDL] =
Kc =
– COO- + H+ à –COO–NH2 + H+ à –NH3+
1
= 0.02 g dm-3 = 0.112 mol dm-3
50 / 1000
[gluatamic acid]
[1]
[GDL][H2O]
= =
[0.0670]
[0.112 − 0.0670][55.5]
= 0.0269 mol-1 dm3[1]
Many candidates omitted water into their Kc expression. Some candidates
did not calculate the equilibrium [GDL].
ci)
Br
HO
O
Br
Br
Br
Br
O
OH
Br
[1] for at least 1 Br substituted into each of two phenolic rings
[1] for 2 Br added across C=C
cii)
HO
O
HO
+
H2
O
OH
OH
OH
diadzein
HO
O
F
HO
O
O
K2Cr2O7
O
OH
OH
F
G
OH
[2] for structures of F and G
[1] for all 3 chiral carbon identified.
Alkenes and ketones in Diadzein undergoes reduction to give F
F contains 3 –OH groups (2 phenol and 1 alcohol) since it reacts with 3 mole of Na
2o alcohol in F then undergoes oxidation with K2Cr2O7 to give G.
G contains a carbonyl group since it gives a positive test with 2,4-DNPH.
[1] for identifying type of reaction that occurs.
Question 3
ai)
O2 + 4H+ + 4e− à 2H2O
aii)
2CH3OH + 3 O2 à 4H2O + 2CO2 [1]
aiii)
EO2/H2O = +1.23
+1.18 = + 1.23 – Eoxi
Eoxi = +0.05 V
aiv)
[1]
[1]
[1]
Easy storage and safety – CH3OH and H2O are in liquids and easier to transport.
H2 and O2 are flammable gasses and need to be kept under pressure.
bi)
C2H5OH + 3 O2 à 3H2O + 2CO2 [1]
bii)
∆G
biii)
C2H5OH + 3 H2O à 2CO2 + 12H+ + 12e− [1]
= ∆H – T∆S
= (–1367) – (298)( –140/1000) = –1325.28 = –1330 kJ mol–1 [1]
−∆Gθ
−( −1325280)
=
Ecell =
= +1.14V
zF
(12)(96500)
ci)
cii)
∆H
[1]
=
=
=
[1]
∑ bonds broken - ∑ bonds formed
(5 C—H + 1C—C + 1 C—O + 1 O—H + 3 O=O) –(4 C=O+ 6O—H)
[1]
– 1272 kJ mol-1
Bond energies assumes compounds to be in gaseous state. At
exists as a liquid.
298
Hence the difference would be due to the ∆Hvapourisation of ethanol.
K,
ethanol
[1]
di)
CxHyOH + Na à CxHyO—Na+ + ½ H2
amount of H2, nH2
=
amount of CxHyOH = 2 x nH2
Let initial VO2 be z cm3.
= 2 x 4.54 x 10-3 = 9.08 x 10-4 mol
CxHyOH (l)
(
Initial amount/mol
9.08 x 10-4
After
combustion
and cooling/mol
After reaction with
NaOH/mol
10.9 × 10−3
= 4.54 x 10-3 mol
24.0
4x + y − 1
)O2 (g)
4
xCO2 (g)
(
0
0
z
0
0
z – 54.4
109 (formed)
z – 109 – 54.4
(leftover O2)
Volume of O2 used
= z – (z– 109 – 54.4) = 163 cm3
Volume of CO2 formed
= 109 cm3
Amount of O2 used
Amount of CO2 formed
163 x 10 −3
=
= 6.79 x 10-3 mol
24.0
109 x 10 −3
=
= 4.54 x 10-3 mol
24.0
Comparing ratio:
Mole ratio of CxHyOH : CO2 is 1:x.
1
x
4.54 ×10−3
=
∴x =
=5
Hence
9.08 ×10−4 4.54 ×10−3
9.08 ×10−4
Comparing mole ratio of CxHyOH and O2 = 1 :
4(5) + y − 1
1
4
=
Since x=5,
9.08 ×10−4 6.79 ×10−3
Hence J C5H11OH [3]
4x + y − 1
4
∴ y = 10.91 ≈ 11
y +1
) H2O (l)
2
dii)
H
H
C
C
H
OH
O
C
O
CH3
H
CH3
H
H
CH3
C
C
C
H
OH
CH3
H
J
H
H
H
CH3
C
C
C
H
OH
CH3
J
H
H
CH3
C
C
H 3C
CH3
K
[1]
[1]
J is a secondary alcohol, hence can be oxidised by K2Cr2O7
K can be obtained by the dehydration of J.
Explanation [1]
A high proportion of candidates who struggled with (i) abandoned the question at
this point, despite the fact that (ii)could be solve directly without reference to (i).
Those who tackled it were usually successful. (So don’t give up! Just move on!!)
diii)
Reagents and Conditions: concentrated H2SO4, heat at 170 oC [1]
div)
No. K cannot exhibit geometric isomerism.
There are 2 methyl groups attached to the same carbon atom in the C=C bond [1]
It is necessary to state it was one of the carbon atoms in the C=C that has 2
identical groups. It is WRONG to say these identical groups are on the same side of
the C=C bond, or just that there is a carbon having two identical groups
Question 4
a)
Nucleon number refers to the total number of protons and neutrons in the nucleus
of an atom. [1]
b)
Electrons would deflect towards the positively charged electrode.
The angle of deflection would be greater in electron than in proton.
[1] for both points
c)
(i)
x
x
Bond pairs/Lone Pairs
(ii) Bond angles
ciii)
xx
O
xx
N
x
x
xx
O
xx
x
x
xx
O
xx
O
x
x
xx
O
xx
[1]
[1]
2 BP, 1 lone electron (exerts 2 BP 1 LP
less repulsion than a lone
PAIR)
Any value between x and Any value x:
110 < x < 120[1]
170 [1]
Cl is in period 3 and can expand its octet to accommodate a maximum of 18
electrons as it has energetically accessible 3d orbitals. [1]
Cl
O
O
F is in period 2, thus it is unable to expand its octet.
Moreover, F is too electronegative to form covalent bonds with O. [1]
Most candidates pointed out that ClO2 would involve chlorine expanding its octet of
electrons, which it is able to do through 3d orbitals which is energetically feasible.
Very few candidates went further with the discussion.
If F cannot expand its octet, then the other alternative would be for F to provide 2
dative bonds, which it cannot because it is too electronegative for dative bonding.
di)
2Ca + O2 à 2CaO
Heat Ca strongly in air. [1]
A brick-red flame is seen, white solid of CaO is seen after the experiment. [1]
For Ca to react with oxygen, it is necessary to apply heat, ideally by holding Ca in
tongs inside a gas jar. A noticeable red flame should be seen and a white solid
should be seen after the experiment.
dii)
Solubility of Group II hydroxides increases down the group.
[OH-] increases, thus pH increases down the group. [1]
pH values of solutions increases from Mg to Ba when shaken with water, because
each oxide reacts with water to give its hydroxide. The solubility of Group II
hydroxides increases down the group, hence the [OH-] increases.
e)
fi)
fii)
Ba(O3)2 + H2O à Ba(OH)2 +
5
O2 [1]
2
15.0
× 0.100 = 0.0015mol
1000
Amount of iodine = ½ x 0.0015 = 0.00075 mol [1]
Amount of thiosulfate =
amount of ozone = amount of iodine = 0.00075 mol
Volume of ozone = 0.00075 x 22.4 = 0.0168 dm3[1]
16.8
Percentage of ozone present =
× 100 = 3.36% [1]
500
gi)
To form aldehydes: Warm a primary alcohol and immediately distill to prevent
[1]
further oxidation
To form carboxylic acids: Heat under reflux for some time to ensure oxidation has
been completed. [1]
gii)
CH3
CH3
H
C
CH2
C
OH
H
L
[1] for each structure
M
N
CH3
Question 5
a)
impure copper
anode
pure copper
CuSO4(aq)
[1] Correctly assigned cathode and anode
[1] Electrolyte
o Since E(Cu2+/Cu) is less positive than E(Zn2+/Zn), Zn will also be oxidized at
anode.
o They dissolve into the solution as cations and migrate to the cathode.
o They will not be reduced at the cathode since E(Cu2+/Cu) is less positive
than E(Zn2+/Zn)
o Since E(Ag+/Ag) is more positive than E(Zn2+/Zn) for (4), Ag (and any other
less reactive metals e.g. Au) will not be oxidised.
o Ag drops off the electrode as the copper around dissolves, and fall to the
bottom of the electrolytic tank to form anode sludge.
b)
When dilute aqueous NH3 is gradually added, precipitation of Cu(OH)2 (pale blue)
occurs. The hydroxide ions comes from aqueous ammonia.
NH3(aq) + H2O
NH4+ OH-
[Cu(H2O)6]2+ (aq) + 2OH– (aq)
blue soln
•
When excess aqueous NH3 is added, ligand exchange takes place. NH3 is a [1]
stronger ligand than H2O and thus displaces water from [Cu(H2O)6]2+ to form
deep blue complex[Cu(NH3)4(H2O)2]2+.
[Cu(H2O)6]2+ (aq) + 4NH3 (aq)
blue soln[1]
•
Cu(OH)2 (s) + 6H2O (l) …………………..(1)
blue ppt [1]
[Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l) ……..(2)
deep blue complex [1]
This causes [Cu(H2O)6]2+ to drop, hence eqm (1) shifts to the left, and the
precipitate Cu(OH)2 (s) dissolves.
c)
O2- + NH3 à OH- + NH2- [1]
Since it contains Cu2+, colour of the solution would be blue/ blue-green. [1]
d)
[Cu(H2O)6]2+ + 4Clblue soln
[CuCl4]2yellow
+ 6H2O [1]
When concentrated HCI is added, ligand exchange takes place. Cl- displaces water
ligand to form the yellowish [CuCl4]2- complex. [1]
Note: The solution appears yellowish green due to the yellow [CuCl4]2complex and blue [Cu(H2O)6]2+.
When water is added, the position of equilibrium shits to the left, hence reforming
the blue solution [Cu(H2O)6]2+ [1]
ei)
2Cu2+(aq) + 4I− (aq) à 2CuI(s) + I2(aq)
white ppt
brown solution [1]
eii)
ECu2+/Cu = + 0.15 V
E I2/I −= +0.54 V
Ecell = +0.15 – 0.54 = – 0.39 V [1]
Since Ecell <0, Reaction is energetically not feasible [1]
eiii)
2Cu2+(aq) + 4I− (aq)
2CuI(s) + I2(aq)
2+
−
+
Cu (aq) ----(1)
Cu (aq) + e
Because CuI is insoluble in aqueous solution, causes the [Cu+(aq)] to be very low
and results in the position of equilibrium (1) to shift right, causing more Cu2+ to be
reduced to Cu+, allowing the redox reaction to take place. [1]
ER: Only a small number of candidates realize that the fact that CuI is insoluble in
aqueous solutions would mean that the precipitate would extract the Cu+(aq) ions
from the solution, leaving only a low concentration, and disturbing the equilibrium
fi)
Brick-red ppt form [1]
fii)
Positive Test: [1] displayed formulae
H
C
O
H
H
H
H
C
C
C
C
H
H
H
H
O
C
H
H
H
H
H
C
C
C
H
H
H
C
H
Negative Test [1] displayed formula
H
H
H
O
H
H
C
C
C
C
C
H
H
H
H
H
H
H