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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020
ChemistryEasynumerical

An alloy of Pb-Ag weighing 1.08 g was dissolved in diluteHNO3and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set-up, Pt(s),H2(g)|H+(1M)||Ag+(aq)Ag(s)was 0.62 V. IfEcellois 0.80 V, what is the percentage of Ag in the alloy? (At25°C, RT/F = 0.06)

Answer:
50.00
Solution:

Overall cell reaction is
H2(g)+2Ag+2Ag(s)+2H+(aq)
E=E°-0.06×2.3032log[H+]2[Ag+]2pH2
0.62=0.80+2×0.06×2.3032log[Ag+]
[Ag+]=0.05M
Number of moles ofAg+in 100 mL=MV1000=0.05×1001000=0.005
Mass of silver=0.005×108g
Percentage of Ag in 1.08 g of alloy=0.005×108×1001.08=50%

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed
2mℹ️ Source: PYQ_2020

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