A Level Chemistry Paper (Nov 2006 / Jun 2007)

For students who are using the Dyna ten-year series (2007 Edition), please take note of the following errata.

Nov 2006 Paper 1 (MCQ)
Q4 Answer is A.
Q8 Answer is A.
Q32 Answer is A.

In addition, please take note of the following.

Nov 2006 Paper 1
Q33 Option 3 reads "The enthalpy change of combustion of diamond is greater than that of graphite". The word "greater" is used to describe the magnitude of the enthalpy change.

Nov 2006 Paper 3
Q1a(ii) When NaBr(aq) is electrolysed using inert electrodes, H2 is evolved at the cathode while Br2 (not O2) is evolved at the anode. (See explanation below)

------------------------------------------------------

CASE 1: Consider the electrolysis of NaCl(aq).

Cl2 + 2e- <--> 2Cl-, E=+1.36V ---(1)
O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V ---(2)
O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ---(3)

In NaCl(aq), there is Na+(aq), Cl-(aq) and H2O(l) [or write as "H+(aq) and OH-(aq)"].

At the anode, one of these reactions occur.
From (1), 2Cl- --> Cl2 + 2e-, E=-1.36V
From (2), 2H2O --> O2 + 4H+ + 4e-, E=-1.23V ---(a)
From (3), 4OH- --> O2 + 2H2O + 4e-, E=-0.40V ---(b)

From (a) and (b), the E values tell us that H2O and OH- are more easily oxidised than Cl-. Thus, O2 (and not Cl2) is produced at the anode.

NOTE: Cl2 is produced at the anode if concentrated aqueous NaCl is used.

CASE 2a: Consider the electrolysis of NaBr(aq).

Br2 + 2e- <--> 2Br-, E=+1.07V ---(4)
O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V ---(5)
O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ---(6)

In NaBr(aq), there is Na+(aq), Br-(aq) and H2O(l).

At the anode, one of these reactions occur.
From (4), 2Br- --> Br2 + 2e-, E=-1.07V
From (5), 2H2O --> O2 + 4H+ + 4e-, E=-1.23V ---(c)
From (6), 4OH- --> O2 + 2H2O + 4e-, E=-0.40V ---(d)

From (c), the E value tells us that Br- is more likely to be oxidised than H2O. Thus, Br2 (and not O2) is produced at the anode.

IMPORTANT NOTE:
(d) will give us the wrong conclusion, i.e. O2 is produced at the anode.
Thus, we should use "O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V" whenever we deal with the electrolysis of aqueous metal halides.

-------------------------------------------------------

O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V OR O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ?

"O2 + 4H+ + 4e- <--> 2H2O" is used when we discuss the electrolysis of aqueous metal halides, e.g. NaBr(aq), KI(aq). Also applies when the electrolyte is H2SO4(aq).

"O2 + 2H2O + 4e- <--> 4OH-" is NOT used for the electrolysis of metal halides.

In general, we will consider using "O2 + 2H2O + 4e- <--> 4OH-" for the anodic reaction when the electrolyte is an aqueous solution which does NOT contain halide ions, e.g. CuSO4(aq), NaNO3(aq), KOH(aq).

-------------------------------------------------------

CASE 2b: Consider the electrolysis of NaI(aq).
The answer is: I2 is produced at the anode because I- is more easily oxidised than H2O.
2I- --> I2 + 2e-, E=-0.54V
2H2O --> O2 + 4H+ + 4e-, E=-1.23V

CASE 2c: Consider the electrolysis of NaF(aq).
The answer is: O2 is produced at the anode because H2O is more easily oxidised than F-.
2F- --> F2 + 2e-, E=-2.87V
2H2O --> O2 + 4H+ + 4e-, E=-1.23V

-------------------------------------------------------

Jun 2007 Paper 4
Q1b examines the electrolysis of molten ZnCl2, concentrated ZnCl2(aq) and dilute ZnCl2(aq).

At cathode:
2H+ + 2e- --> H2, E=0.00V --- (7)
Zn2+ + 2e- --> Zn, E=-0.76 V ---(8)

In the case of molten (or liquid) ZnCl2, there are only Zn2+ and Cl- ions. Thus, Zn is produced at the cathode and Cl2 is produced at the anode.

In the case of concentrated ZnCl2(aq), Cl2 is produced at the anode. From (7) and (8), we conclude that H+ is more easily reduced than Zn2+. Hence, H2 should be formed at the cathode.

In the case of dilute ZnCl2(aq), O2 (and not Cl2) is produced at the anode. H2 is produced at the cathode.

-------------------------------------------------------

There is no need to panic! The troublemaker seems to be metal chlorides. Hence, just be more careful when you see questions involving the electrolysis of metal chlorides.

-------------------------------------------------------

Jun 2007 Paper 2 Q3a (Erratum)
The electronic configuration of Sr2+ is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d0.
(The electronic configuration of Sr is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d0 5s2.)