Chemistry Paper 3 (Mon, 5 Nov 07, 8-10am)

This is a reminder for students taking GCE A Level H2 Chemistry (9746).

Chemistry Paper 3

Date: 5 Nov 07, Mon

Time: 0800 - 1000

Venue: Refer to notice boards

Format of Paper: Free Response, answer 4 out of 5 questions (80 marks)

Items to Bring:

• Entry Proof (Not confirmation slip! Duh!)
• EZ Link card
• Graphic / Scientific calculator (optional: extra batteries)
• Stationery (including curve rule)

Data Booklet will be provided by examiners. DO NOT scribble anything on the booklet.

Some Advice:

• Do invest 2 to 5 min to read through the whole paper and decide which question you want to answer first.
• Plan your answer. Save time by not writing excessively or in a haphazard manner.
• Start each question on a fresh sheet of paper. No correction tape or fluid.
• Circle the question you have answered and enter the question number on the cover page (if applicable).
• Spend about 1.5 min for every 1 mark given.

Have confidence in yourselves. All the best!

2007 NJC Prelims Paper 3 FAQs

A Note for Q1b
2 mol of OH- reacts with 1 mol of A to give a white ppt B. The thermal decomposition of B yields a white ppt C. Thus M is likely to be a Group II element.

NOTE: If M = Zn, you will also get similar observations (except that the mass of B formed is NOT 0.0215 g).

Explanation for Q3b(i)
The pH of the solution is 2.5, so [H+] = 10^(-2.5) mol/dm3.
The initial concentration of CrO42- is 1.00 x 10^(-3) mol/dm3.
2 mol of CrO42- react with 2 mol of H+ to form 1 mol of Cr2O72-.
Thus, [CrO42-] : [Cr2O72-] = (1.00 x 10^(-3) - 2y) : y = 200 : 1
[Cr2O72-] = y = 4.98 x 10^(-6) mol/dm3
[CrO42-] = 1.00 x 10^(-3) - 2[4.98 x 10^(-6)] = 9.90 x 10^(-4) mol/dm3
[H+] = 10^(-2.5) - 2[4.98 x 10^(-6)] = 3.15 x 10^(-3) mol/dm3

2007 VJC Prelims Paper 2 FAQs

Explanation for Q2c
The temperature at which Rb melts is 39.5 deg C or (273 + 39.5) K. This is the temperature at which an eqm is established between Rb(s) and Rb(l). When the two species are at eqm, delta G of fusion is zero. delta G = delta H - T x delta S. Hence, delta H = T x delta S.

Use delta H from the table and substitute T = 273 + 39.5 to find delta S.

When delta G is +ve (T < g =" 0" t =" 39.5"> 39.5 deg C), melting of Rb takes place spontaneously.

A Note for Q3
The question told you that Ni(OH)2 and Cd(OH)2 are formed during discharge (i.e. when the cell behaves as a voltaic cell). Thus, electrode Y (Cd electrode) must be corroded to form Cd(OH)2 and since Cd is oxidised, electrode Y is the anode.

When the battery is being recharged, Cd(OH)2 must change back to Cd. Hence, Cd2+ is reduced at electrode Y. The cell has become an electrolytic cell! When all Cd(OH)2 has changed back to Cd, the electric current will bring about the electrolysis of KOH(aq). In this case, electrode Y will be where H+ ions get reduced.

2007 TJC Prelims Paper 3 FAQs

A Note for Q2e
Fe(III) is stabilised by CN- ligands, so it is less likely to reduce to Fe(II) in aqueous solution.
Another example: [Ni(NH3)6]2+ vs [Ni(H2O)6]2+

2007 PJC Prelims Paper 3 FAQs

A Note for Q3c
"Do I have to memorise the colours of all transition element complexes?"
The answer is NO!

Only for Cr, Mn, Fe and Cu. Examples: Cr3+, CrO42-, Cr2O72-, Mn2+, MnO2, MnO4-, Fe2+, Fe3+, Cu2+, [Cu(NH3)4]2+.

Explanation for Q5b
NaX + H2SO4 --> NaHSO4 + HX, X = Cl, Br or I
This is NOT a redox reaction!!!

SO42- does not oxidise Cl- to Cl2.
SO42- + 4H+ + 2e- <--> SO2 + 2H2O, E = +0.17V
Cl2 + 2e- <--> 2Cl-, E = +1.36V
Ecell = +0.17 - (+1.36) = -1.19V (< 0)

SO42- oxidises HBr to Br2,
eqn: 2HBr + H2SO4 --> Br2 + SO2 + 2H2O

SO42- oxidises HI to I2,
eqn: 2HI + H2SO4 --> I2 + SO2 + 2H2O or 8HI + H2SO4 --> 4I2 + H2S +4H2O
SO2 has an irritating (or pungent) smell while H2S smells like rotten eggs.

2007 HCI Prelims Paper 3 FAQs

Explanation for Q4b(i)
CrCl3.6H2O is the general formula for a series of complexes formed.
The complexes can be [Cr(H2O)6]Cl3, [Cr(H2O)5(Cl)]Cl2 . H2O, etc.
Ag+ ions can only precipitate AgCl with the Cl- ions that are NOT bonded to Cr(III).

Explanation for Q5a(i)
Imagine that you have a "half-cell" containing Fe2+ ions (in acidic medium) initially. This is connected to a standard hydrogen electrode via a salt bridge. Now, add Cr2O72- dropwise to Fe2+ so that Fe3+ is formed. Cr2O72- added before equivalence point will change to Cr3+; there is no excess of Cr2O72-. The mixture contains Fe2+ and Fe3+. At X, [Fe2+] = [Fe3+], so we can ESTIMATE the E value to be +0.77 V using "Fe3+ + e- <--> Fe2+, E = +0.77V". After equivalence point (at 16.5 cm3), there is no more Fe2+ (all oxidised to Fe3+). When excess Cr2O72- is added into the mixture, there will be Cr3+ and Cr2O72-. At Y, [Cr3+] = [Cr2O72-], so we can ESTIMATE the E value to be +1.33 V using "Cr2O72- + 14H+ + 6e- <--> 2Cr3+ + 7H2O, E = +1.33V".

Do note that all concentrations (inclusive of H+) must be 1 mol/dm3 to measure the suggested E values at X and Y.

2007 ACJC Prelims Paper 3 FAQs

A Note for Q1d(ii)
Hot alkaline aqueous iodine (i.e. I2(aq), Na(OH)(aq), heat) will not distinguish the compounds because there is neither methyl ketone nor RCH(OH)CH3 after alkaline hydrolysis.

Explanation for Q2a(i)
M3+ + 3e- <--> M, E= x V
A more positive x means that the reduction of M3+ to M is more feasible. It also implies that M3+ oxidises other substances more readily (while M3+ itself is reduced).

Explanation for Q3b(ii)
Relative oxidising power of halogens can be compared using:
(1) E values of X2/X-, e.g. E of F2/F- is more +ve than Cl2/Cl-, so F2 is more oxidising than Cl2
(2) delta H of "X2 + 2e- --> 2X-", e.g. delta H is more negative for X = Cl than X = Br, so the reduction of X2 to X- is more energetically feasible for X = Cl, i.e. Cl2 oxidises other substances more readily because Cl2 is reduced more readily
(3) displacement, e.g. Cl2 displaces Br-(aq) from solution
(4) oxidation state of product formed, e.g. Cl2 oxidises S2O32- to SO42- but I2 oxidises S2O32- to S4O62-