waecE-LEARNING
Chemistry Paper 2,Nov/Dec 2012  
Questions:   1 2 3 4 7 8   Main
General Comments
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Question7


(aQuestion 7
(a)        (i)         Sodium hydroxide can be prepared by the electrolysis of concentrated solution of                          sodium chloride.  Write the equation at the:
                        I.  cathode;
                       II.  anode.
           (ii)         Explain briefly why the electrolysis of molten alumina is considered environmentally friendly while that of molten sodium chloride is not.
(b)     (i)        Describe briefly the electrolysis of copper (II) tretraoxosulphate (VI) solution using        copper electrodes.
            (ii)        State the colour of the electrolyte in 7(b)(i):

      1.  before the electrolysis;
      2. after the electrolysis.

           (iii)        Give a reason for your answer in 7(b)(ii).
 (c)       (i)  Name the impurities present in bauxite.
             (ii)  State how the impurities in bauxite can be removed..
(d)       An aluminium of mass 3.14g reacted with hydrochloric acid at s.t.p.
(i)    Write a balanced equation for the reaction
(ii)  Calculate the:

  1.  mass of hydrogen produced;
  2. volume of hudrogen produced at s.t.p.

            [ H = 1.00; Al = 27.0; Molar volume = 22.4 dm3 ]

             

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OBSERVATION

 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 

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