(N/A) There is no change in the $pH$ value. The $pH$ value depends on the $H^{+}$ ion concentration of the solution.
In the electrolysis of dilute $H_{2}SO_{4}$ solution,the following reactions occur:
$H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$
At the cathode,$H^{+}$ ions are reduced:
$2H^{+} + 2e^{-} \rightarrow H_{2(g)}$ $(i)$
At the anode,$H_{2}O$ is oxidized rather than $SO_{4}^{2-}$ ions:
$H_{2}O_{(l)} \rightarrow \frac{1}{2}O_{2(g)} + 2H^{+} + 2e^{-}$ $(ii)$
Overall reaction = $(i)$ + $(ii)$:
$H_{2}O_{(l)} \rightarrow H_{2(g)} + \frac{1}{2}O_{2(g)}$
Since the net reaction involves the consumption of $H_{2}O$ and the production of $H_{2}$ and $O_{2}$ gases,the concentration of $H^{+}$ ions in the solution remains constant. Therefore,the $pH$ of the solution does not change.