The cell in which the following reaction takes place : `2Fe^(3+)(aq)+2I^(-)(aq)rarr 2Fe^(2+)(aq)+I_(2)(s)" has the cell potential, "E_("cell")^(@)=0.236V` at 298K. Calculate the standard Gibbs energy and equilibrium constant.

Explanation

The half - cells are represented as :
`2Fe^(3+)(aq)+2e rarr 2Fe^(2+)(aq), 2I^(-)(aq)rarr I_(2)(s)+2e`
Applying the relation `Delta_(r)G^(@)=-nFE^(@)` gives
`Delta_(r)G^(@)=-nFE_("cell")^(@)=-2xx96500xx0.236" "[because n = 2]`
`=-"45.55 kJ.mol"^(-1)`
Also, `Delta_(r)G^(@)=-"2.303RT logK_(c)`
`"or, "logK_(c)=-(Delta_(r)G^(@))/(2.303RT)`
`"or, "log K_(c)=((-45.44))/(2.303xx8.314xx10^(-3)xx298)=7.983`
Therefore, `K_(c)=9.616 xx 10^(7)`