3.71 plz

Asked by lovemaan5500 | 19th Aug, 2019, 11:13: PM

Expert Answer:

Given:
 
I = 25 ampere
 
Volume = 20 L
 
Mass of Cl2 = 1 kg
 
                 = 1000 gm
 
(i) Reactions taking place at electrodes:
 
       Cathode:  2H2O  + 2e → H2(g)  + 2OH(aq) 
 
       Anode : 2Cl(aq)  →  Cl2 + 2e 
 
(ii)
      Mole of Cl2 
 
      equals fraction numerator Mass over denominator Molar space mass end fraction

equals 1000 over 71

equals 14.08 space mol
 
 
    Electrons needed to carry out electrolysis = 2 × 14.08
 
                                                            = 28.16 mol
 
   Quantity of electricity carried by these electrons =  28.16 × 96500 C
 
    Time required, 
 
   Current efficiency is 62%
 
 
 
      straight I space equals 62 over 100 cross times 25

equals 15.5 space ampere
 
    straight t equals straight Q over straight I

space equals fraction numerator 28.16 cross times 96500 over denominator 15.5 end fraction

space equals 175374.82 space straight s

equals space fraction numerator 175318.7 over denominator 3600 end fraction

space equals space 48.69 space hr
 
 
(iii) Molarity of solution:
 
       Amount of OH− ions released = 2× 14.08
 
                                                   = 28.16 mol
 
       Molarity of OH− ions 
 
 
        equals fraction numerator 28.16 over denominator 20 end fraction

equals space 1.408 space straight M

Answered by Varsha | 20th Aug, 2019, 11:06: AM