Answers 
Assignment – 1
Electrochemistry

Ans 1.  5F

        Ans 2. Lead sulphate

Ans 3.         I = 0.5 A
                    t = 2 hours = 2 × 60 × 60 s = 7200 s
                    Q = It    =>  0.5 A x 7200 s = 3600 C
                    96500 C is equivalent of 1 mole = 6.02x1023 electrons
                    3600 C is equivalent of 1mole = 6.02x1023 x 3600/96500
                    = 2.246x1022 electrons.

Ans 4. I). It completes the cell circuit by connecting the solutions of two half cells.
           ii). it maintains the electrical neutrality of the two solution in the two half   
           cells.

Ans 5. The cell which converts combustion energy of fuel into electricity. Methane  
           and hydrogen can be used as fuel 2H2(g) + 4OH–(aq) → 4H2O(l) + 4e–
                                                     O2 + 2H2O (l) + 4e– → 4OH–(aq).

Ans 6. Current = 2A   Time = 3h = 3(60)(60)s
w = ZIt 
           Z for Hg in compound = 200.6/2F   
           w= 200.6x2x (3)(60)(60)/2(96500)
           w= 22.45g
           number of moles = 22.45/200.6 = 0.112 mol

Ans 7. i) Zn electrode is negatively charged 
            ii) At anode Zn(s) → Zn2+ + 2e– 
                 At cathode Ag+(aq) + e– → Ag(s) 
            iii) Ions are the carrier of current within the cell.
      
Ans 8. Statement  Cu2++2e- → Cu
           Charge required for the reduction of 1 mole Cu2+ = 2F    
            =  2 (96500 C)  = 193000C

Ans 9. Zncl2 combines with NH3 to forms a complex salt otherwise due to
           pressure of NH3 cell may crack.

Ans 10. i). Do not cause any type of pollution &
              ii). They have high efficiency (60% - 70%).

Ans 11.  n=2
                   So, Î”G° = -nFE0
                    = - 2 × 96500 × 1.05
                   = - 202650 Jmol-1 = -202.65 kJ mol-1
                    log Kc =    nE0/0.0591
                   = 2 x 1.05/0.0591
                   log Kc = 35.533
                   Kc = 3.412 × 1035 

Ans 12.
           
                
     

Ans 13. As the reaction in cell precedes the concentration of ions in one half cell
             increases while in other half cell decreases hence electrode potential
             also changes. When concentration become equal the E.M.F become
             zero.

Ans 14. i). Λm = λ0m - A√c  
             ii). A straight and B curved
             iii). To determine the value of limiting molar conductivity for electrolyte   
                  B, indirect method based upon Kohlrausch law of independent        
                  migration of ions is used.


Ans 15.   λm = Æ™ x 1000/molarities
               =0.0248x1000/0.20
               = 124.5 cm2mole

Ans 16. i) Cell constant G = conductivity (resistance) = 1.29 S m–1 (100Ω) 129
              m–1 or 1.29 cm–1
             ii) conductivity of 0.02 M KCl solution (k) = cell constant /resistance =    
             1.29 cm–1/520Ω 2.48 x 10–3 S cm–1
             iii) molar conductivity = k x 1000\M = 2.48 x 10–3 x 1000/0.02 = 124 S
              cm2 mol

Ans 17. λom (H Ac) =     λom ( HCl ) +     λom (NaAc) -   λo m(NaCl)
              = 425.9 + 91 – 126.4
              = 390.55 cm2 mol-1

Ans 18. ∆G0 = -nF Eo cell (n=2)
             = -2x1.1v x 96500 C mol-1
             = -212300 J mol-1 = -21.230 KJ/ml

Ans 19. i) This is due to the alkalinity prevent the availability of H+ ions.
              ii). In GI pipes Zinc coating acts as anode & exposed iron acts as  
              cathode. If zinc undergoes corrosion protecting iron from rusting.

Ans 20. i) Molar conductivity – It is defined as the conductance of the solution  
              which contain one mole of electrolyte such that entire solution is in                                 between two electrodes kept one centimetre apart sharing unit area of
               cross section. 
                     (ii) Given conductivity K = 0.146 x (10–3) S cm–1  
                     Cell constant G = k(R)  = 0.146 x 10–3 x 1500= 0.219 cm–1