
Electrolysis of dil. ${H_2}S{O_4}$ liberates gases at anode and cathode respectively
A.${O_2},S{O_3}$
B.${O_2},S{O_2}$
C.${O_2},{H_2}$
D.${H_2}S{O_2}$
Answer
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Hint: Electrolysis alludes to the decay of a substance by an electric flow. Electrolysis is the cycle by which electric flow is gone through a substance to impact a compound change. The compound change is one in which the substance loses or increases an electron (oxidation or decrease). The cycle is completed in an electrolytic cell, an apparatus consisting of positive and negative terminals held separated and dunked into an answer containing emphatically and adversely charged particles.
Complete step by step answer:
On electrolysis of dil. ${H_2}S{O_4}$ gives us;
$2{H^ + } + 2{e^ - } \to {H_2}$ at the cathode and
$4O{H^ - } \to 2{H_2}O + {O_2} + 4{e^ - }$ at the anode. This electrolysis takes place because of the less reduction power of $O{H^ - }$ when compared to the $S{O_4}^{2 - }$. This is the reason the oxidation happens to the $O{H^ - }$ ion.
Thus, the correct answer is option C.
Additional Information:
What is the significance of electrolysis?
Electrolysis straightforwardly or by implication assumes a significant part in our day by day life. Some of them are as per the following:
Electroplating helps in avoidance of rusting of metals. This makes all the metal segments we utilize strong. Cleaning of metals is finished by filtration. This aids in reusing the metal for sometime later. A significant number of the metals, just as non-metals, are extricated by the strategy for electrolysis.
Note:
The substance to be changed may shape the anode, may comprise the arrangement, or might be broken down in the arrangement. Electric flow (i.e. electrons) enters through the contrary charged anode (cathode); emphatically charged segments of the arrangement travel to this terminal, join with the electrons, and are changed to unbiased components or atoms. The adversely charged parts of the arrangement travel to the next cathode (anode), surrender their electrons, and are changed into nonpartisan components or particles. In the event that the substance to be changed is the anode, the response is commonly one in which the cathode breaks down by surrendering electrons.
Complete step by step answer:
On electrolysis of dil. ${H_2}S{O_4}$ gives us;
$2{H^ + } + 2{e^ - } \to {H_2}$ at the cathode and
$4O{H^ - } \to 2{H_2}O + {O_2} + 4{e^ - }$ at the anode. This electrolysis takes place because of the less reduction power of $O{H^ - }$ when compared to the $S{O_4}^{2 - }$. This is the reason the oxidation happens to the $O{H^ - }$ ion.
Thus, the correct answer is option C.
Additional Information:
What is the significance of electrolysis?
Electrolysis straightforwardly or by implication assumes a significant part in our day by day life. Some of them are as per the following:
Electroplating helps in avoidance of rusting of metals. This makes all the metal segments we utilize strong. Cleaning of metals is finished by filtration. This aids in reusing the metal for sometime later. A significant number of the metals, just as non-metals, are extricated by the strategy for electrolysis.
Note:
The substance to be changed may shape the anode, may comprise the arrangement, or might be broken down in the arrangement. Electric flow (i.e. electrons) enters through the contrary charged anode (cathode); emphatically charged segments of the arrangement travel to this terminal, join with the electrons, and are changed to unbiased components or atoms. The adversely charged parts of the arrangement travel to the next cathode (anode), surrender their electrons, and are changed into nonpartisan components or particles. In the event that the substance to be changed is the anode, the response is commonly one in which the cathode breaks down by surrendering electrons.
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