This question was attempted by majority of the candidates and the performance was fair.
In (a)(i), candidates correctly defined oxidizing agent and reducing agent in terms of electron transfer as follows:
(i) Oxidizing agent is a substance which accepts electrons/is an electron acceptor.
(ii) Reducing agent is a substance which donates electrons/is an electron donor.
In (a)(ii) and (iii),most of the candidates could neither write a balanced equation to show that carbon is a reducing agent nor state the change in oxidation number of the specie that reacted with carbon. The expected answers from candidates were:
(ii) 2CuO(s) + C(s) → 2Cu(s) + CO2(g)
H2O(g) + C(s) → 2CO(g) + H2(g)
CO2 (g) + C(s) → 2CO(g)
Cu in CuO from +2 to O
H in H2O from +1 to O
C in CO2 from + 4 to +2
In (b), candidates correctly identified gas X, stated two properties exhibited by X and gave two uses of X as follows:
X is sulphur (IV) oxide/sulphur dioxide
(ii) - heavier than air
- reducing/oxidizing agent
- colourless (poisonous) gas with irritating smell
(iii) - used for bleaching
- for the manufacture of H2SO4
- as germicide and fumigant
- for preservation
- as refrigerant e.t.c.
In (c)(i) – (iv), most candidates were able to correctly give their responses as follows:
lead (II) iodide
In (d)(i), candidates correctly defined salt as a compound formed when all or parts of the hydrogen of an acid is replaced by metal or ammonium ion.
They were also able to correctly mention two types of salt with corresponding example in (d)(ii) and (iii) as follows:
(ii) - normal salt
- acid salt
- basic salt
- double salt
- complex salt
(iii) NaCl/ZnSO4/ KHSO4/ NaH2PO4/ Zn(OH)Cl/Mg(OH)NO3/NH4Fe(SO4)2. 6H2O
In (e)(i), candidates correctly wrote a balanced equation for the reaction between dilute tetraoxosulphate (VI) acid with sodium hydroxide solution as follows:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O (1)
In (e)(ii), candidates correctly determined the number of moles of sodium hydroxide that would be required for the complete neutralization of the given tetraoxosulphate (VI) acid thus:
From the reaction 1 mole of H2SO4 ≡ 2 moles of NaOH
∴ 0.50 moles of H2SO4 ≡ 2 x 0.5 of NaOH
= 1 mole of NaOH