Many candidates could distinguish between a primary cell and a secondary cell with correct examples.
Good knowledge of application of preferential discharge of ions and electrochemical cell during electrolysis was adequately displayed.  The determination of the e.m.f. of the given cell was correctly calculated.
Many candidates could not give a complete correct explanation of nuclear fission as well as how electrical energy could be obtained form nuclear fission.
A good attempt was made at calculating the mass of turpentine from the given data and equation.
The expected answers were:
7. (a)    (i)         A primary cell cannot be recharged while a secondary cell can be recharged.
            (ii)        Primary cell – Leclanche/Daniel                                                                                                         Secondary cell – Lead acid accumulator
    (b)    (i)         Standard hydrogen electrode
            (ii)        Voltmeter/galvanometer
(iii)      Y is a salt bridge
            To connect the two half cells and to balance the electric charges at the electrode OR To allow the movement of ions between the two compartments
(iv)       e.m.f. = Ered - Eoxn
                        =  0.80V – 0.34V
            e.m.f.   =          Ecathode -  Eanode
                                =             0.80 – 0.34 Eanode
                        =          0.46 V

(c)   In aqueous solution both Na+(aq) and H+(aq) ions are present.  Hydrogen is lower in the activity series than sodium thus requires less energy to produce hydrogen than sodium.  Therefore hydrogen is preferentially discharged.

     (d)  (i)    Nuclear fission – it is a process which involves the breaking down of a large unstable           nucleus into small and more stable nuclei with the release of energy in the form of                                         heat.
  (ii)     The heat evolved is used to generate steam which is then used to drive turbine to produce electrical energy.

 Molar mass of chlorine = 2 x 35.5

   =  71 gmol-1
               Molar mass of C10 H16 = (12 x 10) + (1 x 16)  = 136 gmol-1
                    8 x 71g of Cl2 = 136g of C10 H16
                  
  21.3g of Cl2  = 136    x 21.3                                                                        
                         8 x 7
                        = 136   x 21.3                                                                        
                           568
                                    =  5.1g

ALTERNATIVE METHOD

Molar mass of chlorine = 2 x 35.5  = 71 gmol-1
No. of mole of U2 = 21.3    =  0.3 mol
                                  171
Mole ratio of Cl2 :  C10 H16 = 8:1
No. of mole of C10 H16 = 0.3    =  0.0375 mol
                                             8
Molar mass of C10 H16  =   10 x 12 + 1 x 16
                                      =    136 gmol-1
Mass of turpentine  =  136 x 0.375
                                =   5.1g