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ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsEasy2 minPYQ_2022

ChemistryEasynumerical

The quantity of electricity in Faraday needed to reduce $1 mol$ of ${Cr}_{2} O_{7}^{2 -}$ to ${Cr}^{3 +}$ is

Answer:

6.00

Solution:

1 faraday of charge is the magnitude of charge of one mole of electrons.

According to the given reaction:
${C r}_{2} O_{7}^{2 -} + {6 e}^{-} ⟶ {2 C r}^{3 +}$

Reduction of $1 mol$ of ${C r}_{2} O_{7}^{2 -}$ to ${C r}^{3 +}$ requires $6$ moles of electrons. Hence, charge required $= 6 Faraday$

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

⏱ 2mℹ️ Source: PYQ_2022

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