This question was attempted by only few candidates and their performance was below average. In (a), candidates made a fair attempt to write the equation at the cathode and anode but could not satisfactorily explain why the electrolysis of molten alumina is considered environmentally friendly while that of sodium chloride is not.
In (b), a fair attempt was made in describing electrolysis or CuSO4(aq) using copper electrodes. some candidates were able to state the colour of the electrolyte before end after the electrolysis but could not give a correct reason for the answer.
In (c), majority of the candidates could neither name the impurities present in bauxite nor state how they are removed.
In (d), some candidates were able to write and balance the equation for the reaction between aluminum and excess hydrochloric acid at s.t.p. although many were not able to calculate the mass and the volume of hydrogen produced at s.t.p.
The expected answers include:
7.
(a) (i) I. Cathode: Na+ + e- → Na (mercury cathode)/
2H+ + 2e- H2 (Graphite cathode)
II. Anode: 2Cl- → Cl2 + 2e-
(ii) During the electrolysis of molten alumina oxygen gas is produced.
During the electrolysis of molten sodium chloride chlorine gas is
produced.
Oxygen gas is not a pollutant but chlorine gas is a pollutant.
(b) (i) At the cathode Cu2+ ions are discharged preferentially and deposited as
metallic copper on the cathode.
At the anode, no ions is discharged but the conversion of copper atoms to ions is favoured because it required less energy.
OR
Cathode Anode
Cu2+(aq)+ 2e- Cu(s) Cu(s) Cu2+(aq) + 2e-
Preferential discharge Copper atoms go into solution because it
requires less energy
(ii) I. blue
II. blue
(iii) No colour change in the electrolyte because the copper (II) ions discharged at the cathode are replaced by the dissolving copper anode.
(c) (i) Iron (III) oxides and trioxosilicates (IV).
(ii) The impurities in bauxite can be removed by heating the bauxite with concentrated sodium hydroxide (under high pressure) to form (soluble) sodium aluminate where the undissolved impurities are filtered off.
(d) (i) 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 (g)
(ii) I. mole (Al) = n =
=
= 0.1163mol
But =
= n(H2) =
= n (H2) = 0.174mol
= Molar mass H2 = 2 x 1 =2
=M (H2) = 2 x 0.174
= 0.3489(g) = 0.349g
II. 1 mole of a gas at s.t.p = 22.4 dm3
\ 0.174 mole of H2 gas = 0.174 x 22.4
3.8976 dm3= 3.90 dm3
ALTERNATIVE
(d) (i) 2Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3H2(g)
(ii) I. 2 x 27 g of Al requires 3 x 2 g of H2
54 g of Al produce 6 g of H2
3.14 g of Al = 6 x 3.14 g
54
= 3.49 g
II. 2 g of H2 ≡ 22.4 dm3
0.349 g of H2 ≡ 22.4 x 0.349
2
= 3. 91 dm3
