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 $P^H$ of the solution as a result of electrolysis.

Using Faraday’s first Law of Electrolysis

$m=\frac{Mct}{nF}$ ( $M$ =Atomic mass, n=Valency, ct=charge of Electricity,F=Faraday)

Equivalent weight=$\frac{Atomicweight}{valency}=$

$1F\left(96500Coloumbs\right)$ Of electricity required to deposit

 1 gm.eq.wt of substance

For $NaCl$   Atomic weight is equal to equivalent weight

$NaCl{P}^H=7$

               $Q=it$                                                                   $Q=$ charge of electricity

                   $=2\times 965C$                                                 i= strength of current in ampere

       $50\%efficiency$   of charge                                                 t= time in seconds

$\therefore \text{charge}\text{passed}=965C$

           $96500C--1eq.wt$                                          $\because$  $0.01M=0.01N$

                 $965--{10}^{-2}eq$                                                           

         $N=\frac{{10}^{-2}}{10}={10}^{-3}$                                        $N\times V=$  no of gm.eq.wts

       $\therefore$  $\left[H^{+}\right]$  $={10}^{-3}$

        so   $P^H=3$

$changein{P}^H=7-3=4$