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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2021
ChemistryHardstatement

Consider the cell at 25°C

ZnZn2+aq,1 MFe3+(aq),Fe2+aqPts

The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x×10-2. The value of x is ______. (Nearest integer)

Given:EFe3+|Fe2+=0.77V, EZn2+|Zn=-0.76V

Question diagram: Consider the cell at 25 ° C Zn Zn 2 + aq , 1 M ‖ Fe 3 + ( aq
Answer:
24
Solution:

Ecell0=0.77-0.76

=1.53 V

1.50=1.53-0.062logFe2+Fe3+2

logFe2+Fe3+=0.030.06=12

Fe2+Fe3+=101/2=10

Fe3+Fe2+=110

Fe3+Fe2++Fe3+=11+10=14.16

=0.2402

=24×10-2

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

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