Chemistry - Electrochemistry Question with Solution | TestHub

Charge passed: $Q=I\cdot t=96.5\times 10=965C$

Moles of electrons: $n_e=Q/F=965/96500=0.01mol$

Moles of Cu²⁺ reduced: $n_{Cu}=n_e/2=0.005mol$

Remaining [Cu²⁺] = $(0.1-0.005)/1=0.095M$

Approx. H⁺ from hydrolysis $\to [H^{+}]\approx 0.01M$

pH $=-\log [H^{+}]=-\log 0.01=2$

pH = 2