This question was unpopular among the candidates as shown by the number
and performance of the candidates who attempted it.
In part (a), majority of the candidates could not correctly explain anode and
cathode in terms or oxidation and reduction. Also, majority of the candidates
could not write half cell reactions of sodium and aluminium.
In part (b), majority of the candidates were able to explain why the given reaction
is redox but could not write balanced half equations for the reactions and therefore
could not write the overall balance reaction equation.
In part (c), majority of the candidates calculated the quantity of electricity involved
but could not go beyond that. Most candidates could not write the half equation for aluminium.
The expected answers include:
(a) (i) I . Anode - Electrode at which oxidation occurs / electron loss
II. Cathode - Electrode at which reduction occurs / electron gain
(ii) I. Cathodic reaction
Na+(l) + e- → Na(s)
Anodic reaction
2Cl-(l) → Cl2(g) + 2e-
II. Cathodic reaction
Al3+(l) + 3e- → Al(s)
Anodic reaction
2O2-(l) → O2(g) + 4e-
(iii) In the extraction of aluminium oxygen is produced
whereas in the extraction of sodium, chlorine is produced.
Chlorine is an environmental pollutant whereas oxygen is not
(b) (i) The reaction is redox because both oxidation and reduction
occur simultaneously. The oxidation number of Cr changes
from +6 (in K2Cr207) to +3 (in CrCl3). Cr is therefore
reduced. The oxidation number of Cl changes from -1 (in HCl)
to 0 ( in Cl2). Cl is therefore oxidized.
OR
Removal of hydrogen from HCl is oxidation
Removal of oxygen from K2Cr2O7 is reduction
(ii) Reduction
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Oxidation
2Cl- → Cl2 + 2e
(iii) Cr2O72- + 14H+ + 6Cl- → 2Cr3+ + 7H2O + 3Cl2
Accept K2Cr2O7 + 14HCl → 2CrCl3 +7H2O + 3Cl2
(c) Q = I x t
= 6 x (30 + 60) x 60
= 32400 C
Al3+ + 3e- → Al(s)