A solution of `Fe^(2+)` is titrated potentiaometrically using `Ce^(4+)` solution . Calculate the `EMF` of the redox electrode thus formed when
`a. 50%` of `Fe^(2+)` is titrated
`b. 90%` of `Fe^(2+)` is titrated
`c. 110%` titration is done
Given `:E^(c-)._(Fe^(2+)|Fe^(3+))=-0.77V` and `Fe^(2+)Ce^(4+)rarrFe^(3+)+Ce^(3+),K=10^(14)`

Explanation

`a.` During titration, the redox electrode is `Pt|Fe^(2+),Fe^(3+)` electrode
`E=E^(c-)._(cell)-(0.059)/(n)log.([Fe^(3+)])/([Fe^(2+)])`
`Fe^(3+) +e^(-) rarr e^(2+)`
At `50%` titration, `[Fe^(3+)]=[Fe^(2+)]`
`:. E=E^(c-)=-0.77V`
`b.` At `90%` titration, `([Fe^(3+)])/([Fe^(2+)])=(90)/(10)`
`{:(,Fe^(2+),rarr,Fe^(3+),+e^(-)),(Initial,90,,10,),(Fi nal,10,,90,):}`
`:. E=-0.77-0.059log 9 =-0.826 V`
`c.`
At `100%` titration, the electrode becomes,
`Pt|Ce^(3+),Ce^(4+)` electrode,where`([Ce^(4+)])/([Ce^(3+)])=(10)/(100)`
Cell reaction `:`

At equilibrium, `E_(cell)=0`
`E=E^(c-)._(cell)-(0.059)/(1) log K `
`E^(c-)._(cell)=(0.059)/(1) log 10^(14)`
`=0.826`
`:. x-0.77=0.822impliesx=1.596`
`:. E^(c-)._(Ce^(4+)//Ce^(3+))=1.596V`
`:.E^(c-)._(Ce^(3+)//Ce^(4+))=-1.596V`
So at `110%` titration,
`E_(Ce^(3+)//Ce^(4+))=E^(c-)._(Ce^(3+)//Ce^(4+))-0.059log .([Ce^(4+)])/([Ce^(3+)])`
`=-1.596-0.0596 log ((10)/(100))`
`=-1.596+0.059=-1.537`