Question

Why does the electrolysis of water produce twice as much Hydrogen and Oxygen?

Answer 1

1. Electrolysis is a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution or molten state.
2. Water is weakly ionized and hence is a poor conductor of electricity.
3. So, acid is added to it for better conductivity.
   $$\begin{gathered} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \end{gathered}$$ --------------(1)
4. When an electric current is passed through the solution, ${\mathrm{H}}^{+}$ ions move towards the cathode.
5. ${\mathrm{OH}}^{-}$ and ${\mathrm{SO}}_4^{2-}$ ions move towards the anode.
6. The discharge potential of ${\mathrm{OH}}^{-}$ is much lower than ${\mathrm{SO}}_4^{2-}$ ions. So, ${\mathrm{OH}}^{-}$ ions are discharged at the anode.
   At anode (oxidation) - $$\begin{gathered} {\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to \mathrm{OH}+{\mathrm{e}}^{-} \\ 4\mathrm{OH}\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \end{gathered}$$ ----------------(2)
7. ${\mathrm{H}}^{+}$ ions are discharged at the cathode.
   At the cathode (reduction) - $2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right)$ -----------(3)
8. Adding equations (1),(2), and (3) we get
   $$\begin{gathered} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ 4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{e}}^{-} \\ 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right) \end{gathered}$$
9. Multiplying the (1) by 4 and the (3) equation by 2. We get
   $$\begin{gathered} 4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ 4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{e}}^{-} \\ 4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+4{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\left(\mathrm{g}\right) \\ 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\stackrel{\mathrm{electrolysis}}{\to }2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{overall}\mathrm{reaction} \end{gathered}$$
10. Hence, the electrolysis of water produces twice as much hydrogen and oxygen.