A current of 0.965 A is passed for 100 minutes through 500 mL of 0.100 M CuSO4 solution using inert platinum electrodes. Assume the cathodic reaction…

Charge passed $Q=0.965\times6000=5790\ \mathrm{C}$, corresponding to $5790/96500=0.0600\ \mathrm{mol}$ of electrons. The cell contains $0.0500\ \mathrm{mol}$ of $\mathrm{Cu^{2+}}$, which would need $0.100\ \mathrm{mol}$ of electrons for complete deposition, so only $0.0300\ \mathrm{mol}$ of $\mathrm{Cu}$ is deposited and $\mathrm{Cu^{2+}}$ is not exhausted; hence no hydrogen is evolved at the cathode. At the anode $2\mathrm{H_2O}\rightarrow\mathrm{O_2}+4\mathrm{H^+}+4e^-$, so each mole of electrons liberates one mole of $\mathrm{H^+}$: $n(\mathrm{H^+})=0.0600\ \mathrm{mol}$. Then $[\mathrm{H^+}]=0.0600/0.500=0.120\ \mathrm{M}$ and $\mathrm{pH}=-\log0.120=0.92$. The sulphate ions remain and balance the charge of the $\mathrm{Cu^{2+}}$ and $\mathrm{H^+}$ present.