Chemistry - Electrochemistry Question with Solution | TestHub

Anode $\Rightarrow {\mathrm{H}}_2\left(\mathrm{g}\right)⟶2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}$

Cathode $\Rightarrow \underline{{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶\mathrm{Cu}\left(\mathrm{s}\right)}$

Overall $\Rightarrow \underset{¯}{{\mathrm{H}}_2\left(\mathrm{q}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)⟶2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)}$

${\mathrm{E}}_{\mathrm{cell}}^{°}={\mathrm{E}}_{\overline{){\mathrm{Cu}}^{2+}}\mathrm{Cu}}^{°}-{\mathrm{E}}_{\overline{){\mathrm{H}}^{+}}{\mathrm{H}}_2}^{°}=0.34 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{°}-\frac{0.06}{2}\log \frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\left[{\mathrm{Cu}}^{2+}\right]}$

$0.576=0.34+0.03\left[-\log {\left[{\mathrm{H}}^{+}\right]}^2+\log \left[{\mathrm{Cu}}^{2+}\right]\right.$

$0.576=0.34+0.03\left[2\mathrm{pH}+\log \left[{\mathrm{Cu}}^{2+}\right]\right]$

$0.236=0.03\left[2\mathrm{pH}-2\right]$

$7.866=2\mathrm{pH}-2$

$\mathrm{pH}=4.93\approx 5$