Majority of the candidates that attempted this question did poorly.
In (a), majority of the candidates could not properly define standard electrode potential. They could not also state the functions of a salt bridge. However, some candidates were able to state factors affecting the discharge of ions during electrolysis.
In (b) majority of the candidates could not give all the ions present in CuSO4(aq) there by leading to loss of some marks. Candidates’ explanation of the ionization of CuSO4 was fair.
In (c) majority of the candidates got the calculations correct.
In (d), majority of candidates correctly gave the characteristics of a catalyst. However, only a few candidates were able to give the manufacturing processes involving the use of the named catalyst.
The expected answers include:
(a) (i) Standard electrode potential is the measure of the tendency of an element to form ions in solution relative to the tendency of hydrogen atoms to form ions in solution at standard conditions / Is the potential difference that exists between an electrode and the hydrogen electrode under standard conditions of 1 moldm-3 concentration at 298K and 1 atm
(ii) - position of the ions in the electrochemical series/standard electrode potential value;
- nature or reactivity of electrode used;
- relative concentration of ions.
(iii) - allows electrical contact between the two solutions/complete the electric
circuit
- maintains electrical neutrality in each half cell/ allows ions flow
in and out of salt bridge
(b) CuSO4(aq) contains Cu2+, SO42- , H+ / H3O+ , OH -
Cu2+ and H3O+ migrate to cathode
Cu2+ discharged / Cu metal deposited /Cu cathode increases in mass
Copper anode dissolves/ionizes/decreases in size to give Cu2+ ions
Solution remains blue/concentration of Cu2+ ions is constant
(c) Cu2+ + 2e → Cu
Q = It = 0.2 x 35 x 60 = 420 C
2F 2e 1 mole Cu / 2 x 96500 C≡ 64 g
420 C = x 64 g Cu
= 0.14 g Cu
ALTERNATIVE SOLUTION
Cu2+ + 2e → Cu
n =It
ZF
I= 0.2 A
t= 35 x 60 = 2100 seconds
Z = 2
F = 96500
n = 0.2 x 2100
2 x 96500
n = 0.00218 mol
n = mass
molar mass
mass = n x molar mass
= 0.00218 x 64
= 0.13952 g
= 0.139 g
(d) (i) - remains unchanged in chemical nature and mass
- does not affect the types of product formed
- alters (increase or decrease) the rate of chemical reaction
- they are reactions specific
- cannot initiate a reaction that is not feasible
- has no effect on the equilibrium of a reversible reaction
- required in small quantity
(ii) I iron – used in Haber process
II. nickel– used in hydrogenation of vegetable oil / alkenes
III. platinum – used in preparation of hydrogen iodide from hydrogen and iodine/
contact process for H2SO4 production/manufacture of HNO3/used in
hydrogenation