The cell in which the following cell reaction occurs,
`2Fe _((aq))^(3+)+2I_((aq))^(-)to 2Fe _((aq))^(2+)+I_(2(s))`
has `E_(cell)^(0)=0.236V` at 298 K. Calculate the standard Gibbs energy and the equilibrium costant of the cell reaction.

Explanation

Two half reactions for the given redox reaction may be written as :
`2Fe_((aq))^(3+)+2e^(-)to2Fe_((aq))^(2+),2I^(-)to I_(2)+2e^(-)`
2 moles of electrons are involved in the reactin, so `n=2`
`DeltaG^(0)=-nFE_(cell)^(0)=(-2mol)xx(96500mol ^(-1))(0.236V)`
`=- 45548CV=-454548J`
`Delta_(r)G^(0) =-45. 55kj`
`log K_(c)=(DeltaG^(0))/(2.303RT)`
`=((-45.55kj))/(2.303xx(8.314xx10^(-3)kJK^(-1))xx(298K))=7.983`
`K_(c)="Antilig"(7.983)=9.616xx10^(7)`
`K_(c)=9.616xx10^(7)`