
Electrolysis of water is a decomposition reaction. The mole ratio of hydrogen and oxygen gases liberated during electrolysis of water is :
(A) 1:1
(B) 2:1
(C) 4:1
(D) 1:2
Answer
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Hint :Water electrolysis is the process of breaking down water into oxygen and hydrogen gas using electricity. This hydrogen gas can be used as hydrogen fuel, or it can be combined with oxygen to make oxyhydrogen gas, which is used in welding and other uses. Electrolysis, often known as water splitting, requires a minimum potential difference of 1.23 volts.
Complete Step By Step Answer:
A reduction process occurs in pure water at the negatively charged cathode, with electrons (e) from the cathode being supplied to hydrogen cations to create hydrogen gas. The acid-balanced half reaction is as follows:
$ 2{\text{ }}{H^ + }(aq){\text{ }} + {\text{ }}2{e^ - }\; \to {\text{ }}{H_2}(g) $
An oxidation process happens at the positively charged anode, producing oxygen gas and supplying electrons to the anode to complete the circuit:
$ 2{\text{ }}{H_2}O(l){\text{ }} \to {\text{ }}{O_2}(g){\text{ }} + {\text{ }}4{\text{ }}{H^ + }(aq){\text{ }} + {\text{ }}4{e^ - } $
The total breakdown of water into oxygen and hydrogen is the same when either half reaction pair is combined:
$ 2{\text{ }}{H_2}O(l){\text{ }} \to {\text{ }}2{\text{ }}{H_2}(g){\text{ }} + {\text{ }}{O_2}(g) $
As a result, the number of hydrogen molecules generated is double that of oxygen molecules. Assuming that both gases have the same temperature and pressure, the produced hydrogen gas has double the volume of the produced oxygen gas. The number of electrons pushed through the water is twice that of hydrogen molecules created and four times that of oxygen molecules generated.
Hence option b is correct.
Note :
To overcome different activation obstacles, electrolysis of pure water needs extra energy in the form of overpotential. The electrolysis of pure water proceeds very slowly or not at all without the extra energy. This is due in part to water's limited self-ionization. The electrical conductivity of pure water is roughly one-millionth that of seawater. Many electrolytic cells may be missing electrocatalysts as well. The employment of electrocatalysts and the inclusion of an electrolyte (such as a salt, an acid, or a base) improves the efficiency of electrolysis.
Complete Step By Step Answer:
A reduction process occurs in pure water at the negatively charged cathode, with electrons (e) from the cathode being supplied to hydrogen cations to create hydrogen gas. The acid-balanced half reaction is as follows:
$ 2{\text{ }}{H^ + }(aq){\text{ }} + {\text{ }}2{e^ - }\; \to {\text{ }}{H_2}(g) $
An oxidation process happens at the positively charged anode, producing oxygen gas and supplying electrons to the anode to complete the circuit:
$ 2{\text{ }}{H_2}O(l){\text{ }} \to {\text{ }}{O_2}(g){\text{ }} + {\text{ }}4{\text{ }}{H^ + }(aq){\text{ }} + {\text{ }}4{e^ - } $
The total breakdown of water into oxygen and hydrogen is the same when either half reaction pair is combined:
$ 2{\text{ }}{H_2}O(l){\text{ }} \to {\text{ }}2{\text{ }}{H_2}(g){\text{ }} + {\text{ }}{O_2}(g) $
As a result, the number of hydrogen molecules generated is double that of oxygen molecules. Assuming that both gases have the same temperature and pressure, the produced hydrogen gas has double the volume of the produced oxygen gas. The number of electrons pushed through the water is twice that of hydrogen molecules created and four times that of oxygen molecules generated.
Hence option b is correct.
Note :
To overcome different activation obstacles, electrolysis of pure water needs extra energy in the form of overpotential. The electrolysis of pure water proceeds very slowly or not at all without the extra energy. This is due in part to water's limited self-ionization. The electrical conductivity of pure water is roughly one-millionth that of seawater. Many electrolytic cells may be missing electrocatalysts as well. The employment of electrocatalysts and the inclusion of an electrolyte (such as a salt, an acid, or a base) improves the efficiency of electrolysis.
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