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# Explain what is meant by electrolysis of water. Write the electrode reactions and explain them.

Last updated date: 21st Feb 2024
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Hint:Electrolysis of water is the decomposition into oxygen and hydrogen gas of water. We have to do electrolysis of water and write the equations for anode and cathode. Water $\left( {{H_2}O} \right)$ will decompose in ${O_2}$ and ${H_2}$ gases because water $\left( {{H_2}O} \right)$ use the composition of these two gases.

Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them.
Electricity is the flow of electrons or ions.
It is also called water splitting.
It requires a minimum potential difference of $1.23$ volts to split water.

At anode:
$2{H_2}{O_{(l)}}\xrightarrow{{}}{O_2}_{(g)} + 4{H^ + }_{(aq)} + 4{e^ - }$
Or ${{\text{H}}_2}{O_{(l)}}\xrightarrow{{}}2{H^ + } + \dfrac{1}{2}{O_2} + 2{e^ - }$
At cathode:-
$2{H^ + } + 2{e^ - } \to {H_2}$
Net reaction:-${{\text{H}}_2}O \to {H_2} + \dfrac{1}{2}{O_2}$
When we apply $DC$ current in the $CKT$ then at anode ${H^ + }$ ions will produce and they will attract cathode. ${O_2}$ Gas will also be there.
At cathode: ${H_2}$ gas will produce and $O{H^ - }$ ions attract to anode.
We can see in the equation that the amount of ${O_2}$ gas will help than ${H_2}$ gas.
Volume of ${H_2}$ is double than ${O_2}_{(g)}$
In the net reaction we get one mole of ${H_2}_{(g)}$ and half mole of ${O_2}_{(g)}$ from one mole of water.
The electrolyte and the electrodes used in electrolysis form an electrolyte, the positive ions of the electrolyte move toward the cathode (negative electrode), where they gain electrons to become a neutral substance.

-After electrolysis of water, ${H_2}_{(g)}$ is collected at negative cathode and ${O_2}_{(g)}$ is collected at the positive anode.
-Electrolysis water is one such most capable method for production of hydrogen because it uses renewable $\left( {{H_2}O} \right)$ and produces only pure oxygen as by product.