Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsEasy2 minPYQ_2024

ChemistryEasystatement

The amount of electricity in Coulomb required for the oxidation of $1 mol$ of $H_{2} O$ to $O_{2}$ is ______ $\times 10^{5} C .$

Answer:

Solution:

$H_{2} O \to \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-}$

Electricity required to oxidise $1$ mol $H_{2} O to O_{2} = 2 F$

$\frac{W}{E} = \frac{Q}{96500}$

mole $\times n$ -factor $= \frac{Q}{96500}$

$1 \times 2 = \frac{Q}{96500}$

$Q = 2 \times 96500 C$

$= 1.93 \times 10^{5} C$

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Faraday's Law of Electrolysis/Electrolytic Cells

⏱ 2mℹ️ Source: PYQ_2024

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