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An excess of liquid mercury is added to ...

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`

Text Solution

Verified by Experts

The correct Answer is:
`0.7976V`
`2Hg+2Fe^(3+) rarr Hg_(2)^(2+)+2Fe^(2+)`
`Fe^(3+)` ion concentration originally `=1.0xx10^(-3)M`
`Fe^(3+)` ion at `Eq=5%` of `1.0xx10^(-3)M`
`=(5)/(100)xx10^(-3)=5xx10^(-5)M`
`Fe^(3+)` converted into `Fe^(2+)` ion
`=(1.0xx10^(-3))-(5xx10^(-5))=0.95xx10^(-3)`
`Hg_(2)^(2+)` ions at `Eq=(0.95xx10^(-3))/(2)`
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.([H_(2)^(2+)][Fe^(2+)]^(2))/([Fe^(3+)]^(2))`
At equivalent, `E_(cell)=0`
`:. " "0=E^(c-)-(0.0591)/(2)log.([(0.95xx10^(-3))/(2)][0.95xx10^(-3)]^(2))/([5xx10^(-5)]^(2))`
`E^(c-)._(cell)=(0.0591)/(2)log.([95]^(3)xx10^(-5))/(50)=-0.0276`
`E^(c-)._(cell)=E^(c-)._(Fe^(3+)|Fe^(2+))-E^(c-)._(Hg_(2)^(2+)|Hg)`
`=-0.0276=0.77-E^(c-)._(Hg_(2)^(2+)|Hg)`
`E^(c-)._(Hg_(2)^(2+)|Hg)=0.77+0.0276`
`=0.7976V`
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