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ChemistryElectrochemistryFaraday's Law of Electrolysis/Electrolytic CellsMedium2 minPYQ_2024
ChemistryMediumstatement

A constant current was passed through a solution ofAuCl4-ion between gold electrodes. After a period of10.0minutes, the increase in mass of cathode was1.314 g. The total charge passed through the solution is ___×10-2 F. (Given atomic mass ofAu=197)

Answer:
2
Solution:

The half-cell reaction is,

Au3+ + 3e  Au

Given W = 1.314 g

Atomic mass of Au=197

When one Faraday of charge is passed through a circuit, then one equivalent of charge is deposited at the respective electrode.

WE= charge1 F

1.3141973=Q1F

Q=2×10-2 F

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Faraday's Law of Electrolysis/Electrolytic Cells
2mℹ️ Source: PYQ_2024

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