A current strength of 9.65 A is passed for 10 seconds through 1L of solution of 0.1 M aqueous  $CuS{O}_4$. The pH of solution is .... (Assuming that during the process there is no change in volume). 

$$\begin{gathered} C{u}^{2+}+2e^{-}\to Cu{H}_{2}O\to 2H^{+}+\frac{1}{2}{O}_2+2e^{-} \\ Q=9.65\times 10=96.5C \end{gathered}$$

Moles of  $H_{2}S{O}_4$  produced = moles of $C{u}^{2+}$ reduced  $=\frac{96.5}{2\times 96500}=5.0\times {10}^{-4}$
Conc. Of   $H_{2}S{O}_4\left(M\right)=\frac{5\times {10}^{-4}}{1}=5\times {10}^{-4}mole/L$
$$\begin{gathered} \left[H^{+}\right]=2\times 5\times {10}^{-4}={10}^{-3}mole/L \\ pH=-\log \left[H^{+}\right]=3 \end{gathered}$$