10 litres of 0.01M NaCl solution subjected to electrolysis by passing a direct current of 2 amp with a current efficiency 50% for 16 min and 5sec. What is the change is PH of the solution as a result of electrolysis.

Using Faraday’s first Law of Electrolysis  m=MctnF ( M =Atomic mass, n=Valency, ct=charge of Electricity,F=Faraday)Equivalent weight=Atomicweightvalency= 1F(96500 Coloumbs) Of electricity required to deposit  1 gm.eq.wt of substance For NaCl   Atomic weight is equal to equivalent weightNaCl  PH=7                Q=it                                                                   Q= charge of electricity                   =2×965 C                                                 i= strength of current in ampere       50% efficiency    of charge                                                 t= time in seconds∴ charge passed=965 C            96500 C−−1 eq.wt                                          ∵  0.01M=0.01N                  965    −−10−2 eq                                                                    N=10−210=10−3                                        N×V=  no of gm.eq.wts        ∴  [H+]  =10−3         so   PH=3 change in PH=7−3=4