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

Explanation

Given : `2Hg(l)+2Fe^(3+)(aq)rarr Hg_(2)^(2+)(aq)+2Fe^(2+)(aq)`
`[Fe^(3+)]=1.0xx10^(-3)(M)`
`therefore" "[Fe^(3+)]" at equilibrium "=5%" of "1.0xx10^(-3)(M)`
`=(5)/(10)xx1.0xx10^(-3)(M)=5xx10^(-5)(M)`
`therefore" "[Fe^(2+)]=(1.0xx10^(-3))-(5xx10^(-5))=0.95xx10^(-3)M`
`therefore" "[Hg_(2)^(2+)]" at equilibrium "=(0.95xx10^(-3))/(2)M`
`therefore" "E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([Hg_(2)^(2+)][Fe^(2+)]^(2))/([Fe^(3+)]^(2))`
At equilibrium, `E_("cell")=0`
`therefore" "E_("cell")^(@)=(0.059)/(2)log.([(0.95xx10^(-3))/(2)][0.95xx10^(-3)]^(2))/([5xx10^(-5)]^(2))`
`"or, "E_("cell")^(2)=(0.0591)/(2)log.([95]^(3)xx10^(-5))/(50)=-0.0276`
`"Also, "E_("cell")^(@)=E_(Fe^(3+)|Fe^(2+))^(@)-E_(Hg_(2)^(2+)|Hg)^(@)`
`"or, "-0.0276 =0.77-E_(Hg_(2)^(2+)|Hg)^(@)`
`"or, "E_(Hg_(2)^(2+)|Hg)^(@)=0.77+0.0276=0.7976V`