The cell in which the following reaction occurs `2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+2I_(2) " has "E_(cell)^(@)=0.236 V " at "298 K`. Calculate st

Question

The cell in which the following reaction occurs `2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+2I_(2) " has "E_(cell)^(@)=0.236 V " at "298 K`. Calculate st

The cell in which the following reaction occurs
`2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+2I_(2) " has "E_(cell)^(@)=0.236 V " at "298 K`.
Calculate standard Gibbs energy and equilibrium constant for the reaction.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The two half reactions are :
`2Fe^(3+)+2e^(-)to2Fe^(2+)" and " 2I^(-)toI_(2)+2e^(-)`
For the above reaction, n=2
`DeltaG^(@)=-nFE_(cell)^(@)=(-2" mol")xx(96500" C mol"^(-1))xx(0.236" V")`
`=-455480" CV" =-455480" J"=-45.55" kJ"`
Now, `" " logK_(c )=-(DeltaG^(@))/(2.303" RT")=-((-45.55" kJ"))/(2.303xx(8.314xx10^(-3)" kJ "K^(-1))xx(298 K))=7.983`
`K_(c)="Antilog" (7.983)=9.616xx10^(7)`