Larutan natrium sulfat \(\ce{(Na2SO4)}\) dengan elektrode platina
Reaksi Elektrolisis
Anode
Elektrode \(\ce{Pt}\) (zat inert)
Mengandung sisa asam oksi \(\ce{(SO4^{2-})}\)
\(\ce{H2O}\) akan mengalami oksidasi
\(\ce {2 H2O(l) -> 4 H+(aq) + O2(g) + 4e} \)
Katode
Kation \(\ce{Na^{+}}\) merupakan logam aktif
\(\ce{H2O}\) akan mengalami reduksi
\(\ce {2 H2O(l) + 2e -> 2 OH-(aq) + H2(g) }\)
Reaksi lengkap dan reaksi ion bersih
\begin{align*}
\begin{array} {llllllllll}
\text{Larutan} & : \ce {Na2SO4(aq) && -> 2 Na+(aq) & + SO4^{2-}(aq)} && |\times 1\\\\
\text{Anode} & : \ce {2 H2O(l) && -> 4 H+(aq) & + O2(g) & + 4e} & |\times 1\\\\
\text{Katode} & : \ce {2 H2O(l) & + 2e & -> 2 OH-(aq) & + H2(g)} && | \times 2\\
\hline\\
& : \ce {Na2SO4(aq) & + 6 H2O(l) & -> 2 Na+(aq) & + SO4^{2-}(aq) & + 4 H+(aq) & + 4 OH-(aq) & + 2H2(g) & + O2(g) } \\\\
& : \ce {Na2SO4(aq) & + 6 H2O(l) & -> Na2SO4(aq) & + 4 H2O(l) & + 2H2(g) & + O2(g)} \\\\
& : \ce {2 H2O(l) && -> 2H2(g) & + O2(g) }
\end{array}
\end{align*}