Social Science
Literature Arabic Islamic Studies C.R.K HistoryMusicVisual Art Clothing/Textile Home Management Shorthand
Chemistry Paper 2 (Essay) ,May/June 2008  
Questions:   1 2 3 4 5 6 7 8   Main
General Comments

Question 2

(a)  (i)  Define in terms of electron transfer
I.  oxidizing agent;
II.  reducing agent.

(ii) Write a balanced equation to show that carbon is a reducing agent.
(iii) State the change in oxidation number of the specie that reacted with carbon in 2 (a)(ii). [5 marks]

(b) A gas X has a vapour density of 32.  It reacts with sodium hydroxide solution to form salt and water only.  It decolourizes acidified potassium tetraoxomanganate (VII) solution and reacts with H2S to form sulphur.  Using the information provided:

  1. identify gas X;
  2. state two properties exhibited by X;
  3. give two uses of X.    [5 marks]

(c) Consider the following substances:
(i) sodium;
(ii) lead (II) iodide;
(iii) hydrogen;
(iv) magnesium;
(v) oxygen.

Which of the substances
(i) conducts electricity?
(ii) is produced at the cathode during electrolysis of H2SO4(aq)?      
(iii) corresponds to the molecular formula AB2  ?  
(iv) is an alkaline earth metal?    [5 marks]

(d) (i) Define the term salt.
(ii) Mention two types of salt.
(iii) Give an example of each of the salts mentioned in 2(d)(ii) above.  [6 marks]

(e) In a neutralization reaction, dilute tetraoxosulphate (VI) acid completely reacted with sodium hydroxide solution.
(i) Write a balanced equation for the reaction.
(ii) How many moles of sodium hydroxide would be required for the complete neutralization of 0.50 moles of  tetraoxosulphate (VI) acid?   [4 marks]


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)

(iii) 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:
(i) X is sulphur (IV) oxide/sulphur dioxide

(ii) -   heavier than air
-   acidic
-    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:

(i) sodium/magnesium
(ii) hydrogen
(iii) lead (II) iodide
(iv) magnesium

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
Powered by Sidmach Technologies(Nigeria) Limited .
Copyright © 2012 The West African Examinations Council. All rights reserved.