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ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsMedium2 minQB

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1 Faraday of electricity is passed through molten $Al_{2} O_{3}$ , molten $CuSO_{4}$ and molten NaCl taken in three different electrolytic cells. The amount of $Al, Cu$ and Na deposited at the cathodes will be in the ratio of

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Answer:

Solution:

$\frac{^{1}}{6} (A l_{2})^{+ 6} + e^{-} \to \frac{1}{3} A l$

$\frac{1}{2} C u^{2 +} + e^{-} \to \frac{1}{2} C u$

$N a^{+} + e^{-} \to N a$

ratio $\frac{1}{3} : \frac{1}{2} : 1 = 2 : 3 : 6 = 1 : 1.5 : 3$ .

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

⏱ 2mℹ️ Source: QB

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