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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minPYQ_2021
ChemistryMediumstatement

Emf of the following cell at 298 K in V is x×10-2

ZnZn2+0.1MAg+0.01MAg

The value of x is ________________ (Rounded off to the nearest integer)

Given: EZn2+/Znθ=-0.76 V; EAg+/Agθ=+0.80 V;  2.303RTF=0.059

Question diagram: Emf of the following cell at 298 K in V is x × 10 - 2 Zn Zn
Answer:
147
Solution:

Ecell0=EAg+/Ag0-EZn2+/Zn0

=0.80--0.76

=1.56 V

Ecell=1.56-0.0592logZn2+Ag+2

=1.56-0.0592log0.10.012

=1.56-0.0592×3

=1.56-0.0885

=1.4715

=147.15×10-2

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells
2mℹ️ Source: PYQ_2021

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