The cell in which the following reaction occurs
`2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(aq)+I_(2)(s)` has `E_(cell)^(0)=0.236V` at 298 K.
Calculate the stadard gibbs energy and the equilibrium constant of the cell reaction.

Answer

`9.616xx10^(7)`

Explanation

`2Fe^(3+)+2e^(-)rarr 2Fe^(2+)or 2I^(c-)rarr I_(2)+2e^(-)`
Hence, for the given cell reaction, `n=2`.
`Delta_(r)G^(c-)=-nFE_(cell)^(c-)-2xx96500xx0.236J`
`=-45.55kJ mol^(-1)`
`Delta_(r)G^(c-)=-2.303RTlog K_(c)`
`logK_(c)=(Delta_(r)G^(c-))/(2.303RT)`
`=(-45.55kJ mol^(-1))/(2.303xx8.314xx10^(-3)k J K^(-1)mol^(-1)xx298K)`
`=7.983`
`:. K_(c)=Antilog(7.983)=9.616xx10^(7)`