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Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2020

ChemistryEasysingle choice

During electrolysis of water the volume of $O_{2}$ liberated is $2.24 {dm}^{3} .$ The volume of hydrogen liberated, under same conditions will be

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Answer:

Solution:

$H_{2} O \to H_{2} + \frac{1}{2} O_{2}$
During electrolysis, volumes of $O_{2}$ and $H_{2}$ liberated are in the ratio of 1 : 2 Hence, volume of $H_{2}$ liberated $= 2 \times 2.24 d m^{3}$
$= 4.48 d m^{3}$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2020

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