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

For
`{:("For",2Hg+,2Fe^(3+) hArr,Hg_(2)^(2+)+,2Fe^(2+),),("Before reaction","Excess",10^(-3),0,0,),("After reaction","Excess",10^(-3)xx(5)/(100),(95)/(2xx100)xx10^(-3),(95)/(100)xx10^(-3),):}`
For cell at equilibrium
`E_(cell) = 0 = E_(OP_(Hg//Hg_(2)^(2+))) + E_(RP_(Fe^(3+)//Fe^(2+)))`
`0 = E_(OP_(Hg//Hg_(2)^(2+)))^(@) - (0.059)/(2)log_(10)[Hg_(2)^(2+)] + E_(RP_(Fe^(3+)//Fe^(2+)))^(@)`
`+ (0.059)/(2)log_(10).([Fe^(3+)]^(2))/([Fe^(2+)]^(2))`
`0 = E_(OP_(Hg//Hg_(2)^(2+)))^(@)+0.77+(0.059)/(2)log_(10).([Fe^(3+)]^(2))/([Fe^(2+)]^(2)[Hg_(2)^(2+)])`
`(because = E_(OP_(Fe^(2+)//Fe^(3+)))^(@) = -0.77V :. E_(RP_(Fe^(3+)//Fe^(2+)))^(@) = +0.77V)`
` E_(OP_(Hg//Hg_(2)^(2+)))^(@) = -0.771 - (0.059)/(2)`
`log.([5/100 xx 10^(-3)]^(2))/([(95 xx 10^(-3))/(100)]^(2)[(95 xx 10^(-3))/(2 xx 100)])`
`= -0.793V`