Question

What is the molar ratio of ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ in the electrolysis of water?

Answer 1

1. The electrolysis is the electrochemical process in which current is passed into water.
2. This splits the water molecule into their constituent units.
3. The principle of electrolysis involves the use of a DC power supply.
4. The power is given to water through electrodes, in which Hydrogen and Oxygen is released from the different electrodes.
5. The chemical reaction takes places as follows:
6. At the cathode:
   $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2{\mathrm{E}}^{°}=0.0\mathrm{V}$
7. At the anode:
   $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\mathrm{E}°=1.23\mathrm{V}$
8. The overall reaction is as follows:
   $2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{E}°=-1.23$
9. Thus, we concluded that the molar ratio of ${\mathrm{H}}_2\mathrm{and}{\mathrm{O}}_2\mathrm{is}2\mathrm{and}1$respectively.