Calculate `DeltaG^(@)` and equilibrium constant for the cell reaction,
`Cl_(2)+2l^(-) hArr 2Cl^(-)+I_(2)`
Given that: `E^(@)(Cl_(2),Cl^(-))=1*36V,E^(@)(l_(2),l^(-))=0*536V`

Explanation

`Cl_(2)+2l^(-) hArr 2Cl^(-)+l_(2)`
Half cell reactioni are,
`Cl_(2)+2e^(-) to 2Cl^(-)`
`2l^(-) to I_(2)+2e^(-)`
`2I^(-)+Cl_(2) to I_(2)+2Cl^(-)`
Hence the cell can be represented as follows:
`Pt,I_(2)//I^(-)||Cl|Cl_(2)Pt`
Nernst equator at 298K
`EMF^(@)=(E_(Cl_(2)//Cl^(-))^(@)-E_(I_(2)//I^(-))^(@))`
`=(1*36-0*536)V`
`=0*824V`
Here `n=2` mol
`DeltaG^(@)=nE^(@)F`
`=(-2xx0*824V)xx(96500C" "mol^(-1))`
`=-2xx824xx96500Vc" "mol^(-1)`
`=-1590932J" "mol^(-1)`
`=-159*032kJ" "mol^(-1)`
At equilibrium, EMF=0
`=EMF^(@)-(0.0591)/(2)logK_(c)`
`therefore` by `K_(C)=(2xxEMF^(@))/(0.0591)`
`=(2xx0*824)/(0.0591)=278849`
`K_(C)`=antilog `27*8849`
`=8*555xx10^(27)`