A current strength of 1A is passed for 96.5 sec through 200ml of a solution of 0.05M KCl. P^H of the resulting solution is

The reaction taking place are 

$$\begin{gathered} Anode :\hspace{0.17em}2C{l}^{-}\to C{l}_2+ 2e^{-} \\ Cathode :\hspace{0.17em}2H_{2}O+ 2e^{-}\to H_2+\hspace{0.17em}2O{H}^{-}\hspace{0.17em} \end{gathered}$$

$$\begin{gathered} Charge flow q = it = 1 \times 96.5 = 96.5 c \\ 1 F\hspace{0.17em}= 96500 c \\ 96.5 c = 0.001F\hspace{0.17em} \end{gathered}$$

0.00F of electricity reduce 0.001 M H⁺

Therefore we can assume that 0.001 M OH^- are left 

Hence 

$molarity of \left[O{H}^{-}\right]=\frac{moles of O{H}^{- }\times 1000}{Volume \left(ml\right)}=\frac{0.001\times 1000}{200}=0.005M$

$$\begin{gathered} {\mathrm{p}}^{\mathrm{OH}}= -\log \left[{\mathrm{OH}}^{-}\right] \\ =-\log (0.005) \\ =-\log (5 \times {10}^{-3}) \\ {\mathrm{p}}^{\mathrm{OH}}= 2.3 \\ {\mathrm{p}}^{\mathrm{H}}=14-\mathrm{pOH} \\ {\mathrm{p}}^{\mathrm{H}}=14-2.3=11.7 \end{gathered}$$