Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryCommercial Cells/Fuel cellHard2 minPYQ_2023
ChemistryHardstatement

Consider the cell Pt(s)H2( s)(latm)H+aq,H+=1||Fe3+(aq),Fe2+(aq)Pt(s)

Given: EFe3+/Fe2+°=0.771 V and EH+/12H2°=0 V, T=298 K

If the potential of the cell is 0.712 V the ratio of concentration of Fe2+ to Fe3+ is
(Nearest integer)

Answer:
10
Solution:

Cell reaction which occurs:
12H2( g)+Fe3+ (aq.) H+(aq)+Fe2+ (aq.) 
Using Nernst' equation:

E=Eo-0.0591logFe2+Fe3+

0.712=(0.771-0)-0.0591logFe2+Fe3+

logFe2+Fe3+=(0.771-0712)0.059=1

Fe2+Fe3+=10

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Commercial Cells/Fuel cell
2mℹ️ Source: PYQ_2023

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