Question

Acidified water is taken for electrolysis of water because:
(A) It undergoes complete ionization
(B) It undergoes the oxidation
(C) It is a bad conductor of electricity
(D) It is in a pure state

Answer 1

Hint: The electrolysis is a process in which electrolytes breaks down into its ions on the passage of current. On the application of electricity, water generates the hydrogen ion and hydroxide ion. Since, the water has a very low ionization constant, the electrolysis of water is a difficult process.

Complete step by step answer:
Electrolysis is defined as the process by which the ionic substances can easily decompose into simpler forms like ions on the application of an electric current through the solution.
A solution conducts electricity, only when the ions of the solution are randomly dissolved in it.
The water $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ dissociates in hydrogen ion ${{\text{H}}^{\text{+}}}$ and hydroxyl ion $\text{O}{{\text{H}}^{-}}$ .
The electricity conduction is based on the resistance offered by the solution. The water has extremely high resistance towards the current. Thus, the electrolysis of water requires very high energy in the form of the potential to overcome the energy barrier. In absence of an excess of energy, the electrolysis of pure water is a very slow process, rather it does not take place.
The self –ionization is also a key factor behind the conductivity of pure water. Water molecules self-dissociates into an equal amount of hydronium ion $\text{ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}\text{ }$and hydroxide ion$\text{O}{{\text{H}}^{-}}$. The general reaction for the self –ionization or dissociation of water is as shown below:
$\text{ 2}{{\text{H}}_{\text{2}}}\text{O }\rightleftarrows \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}}\text{ }$
The ionization constant for the self-ionization of water is$({{10}^{-14}})$. The value is small that it can even be considered that pure water does not undergo electrolysis and cannot conduct electricity. Ideally, pure water has a zero conductance.
Therefore, the electrolysis of water is studied in presence of strong acidic electrolytes like $\text{HCl}$ or${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$. The $\text{HCl}$ or the ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ tendency to undergo $100\text{ }{\scriptstyle{}^{0}/{}_{0}}$ dissociation in their ion in solution. They are good conductors of electricity.
\[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}_{\text{(aq)}}\text{ }\to \text{ 2}{{\text{H}}^{\text{+}}}_{\text{(aq)}}\text{ + SO}{{_{4}^{2-}}_{\text{(aq)}}}\]
Consider an electrolytic cell and the dilute sulfuric acid (acidified water) as the solution under the study. During the passage of the current the ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ splits into its corresponding ions. This is one of the methods to show that water is a compound of oxygen and hydrogen by splitting the water molecules into hydrogen and oxygen gas.
Water has a very low ionization constant $({{10}^{-14}})$ , so in presence of a high concentration of hydrogen ions and sulphate ions $(\text{SO}_{4}^{2-})$ from acid, the water becomes a better conductor of electricity.
The half-reactions which are associated with the acidified water as follows:
1) Reaction at the negative cathode (reduction reaction):
The hydrogen ion is attracted to the cathode and discharged as the hydrogen gas. Hydrogen ions are reduced as the hydrogen gas molecules.
$2{{\text{H}}^{\text{+}}}_{\text{(aq)}}+2{{\text{e}}^{-}}\to \text{ }{{\text{H}}_{\text{2}}}_{\text{(g)}}$
2) Reaction at the positive anode (oxidation reaction):
The negatively charged sulphate ions or the traces of the hydroxide are attracted to the positively charged anode. As sulphate ion is stable it does not undergo the oxidation, instead that hydroxide is discharged or bubbled as the oxygen gas.
$4\text{O}{{\text{H}}^{-}}_{\text{(aq)}}\text{ }\to \text{ 4}{{\text{H}}^{^{_{\text{+}}}}}_{\text{(aq)}}\text{ + 4}{{\text{e}}^{-}}\text{ + }{{\text{O}}_{\text{2}}}_{\text{(g) }}$
Here, we conclude that the dilute ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ or the acidified water is preferred over the pure water for electrolysis of water because it undergoes complete ionization.
Hence, (A) is the correct option.

Note: The alkaline water (dil.$\text{NaOH}$ ) is also an alternate method to study the electrolysis of water. Here, the hydrogen and oxygen gases are bubbled at the electrodes. The sodium ions $\text{N}{{\text{a}}^{\text{+}}}$are stable, thus it is not reduced at the cathode. The hydrogen ion from the self-ionization of water is reduced as the hydrogen gas.
Remember that, the hydrogen gas produced has a wide application as the hydrogen fuel in a space mission. It is obtained by the electrolysis of a brine solution. Where water is reduced as the hydrogen gas is used for the production of chemicals.