An excess of liquid mercury is added to an acidicfied solution of `1.0xx10^(-3) M Fe^(3+)` . It is found that `5%` of `Fe^(3+)` remains at equilibrium at `25^(@)C`. Calculate `E^(c-)._((Hg_(2)^(2+)|Hg))` assuming that the only reaction that occurs is
`2Hg+2Fe^(3+) rarr Hg_(2)^(2+)+2Fe^(2+)`
Given `: E^(c-)._((Fe^(3+)|Fe^(2+)))=0.77V`

Explanation

`2Hg+2Fe^(3+) rarr Hg_(2)^(2+) + 2Fe^(2+)`
At equilibrium, excess `(10^(-3)xx5)/100" "(10^(-3)xx95)/(2xx100)" "(10^(-3)xx95)/100`
At equilibrium, `E_(cell)=0`
`0=E_(cell)^(@)-0.0591/2"log" ([Hg_(2)^(2+)][Fe^(2+)]^(2))/([Fe^(3+)]^(2))`
`=(E_(Hg//Hg_(2)^(2+))^(@)+E_(Fe^(3+)//Fe^(2+))^(@))`
`-0.0591/2"log"(((10^(-3)xx5)/(2xx100))((10^(-3)xx95)/100))/(((10^(-3)xx5)/100)^(2))`
`E_(Hg//Hg_(2)^(2+))^(@)=-0.77+0.0591/2"log"((95)^(3)xx10^(-5))/(25xx2)`
`=-(0.77+0.0226)`
`=-0.7926` volt
`E_(Hg_(2)^(2+)//Hg)^(@)=+0.7926` volt