Calculate the standard potential for the reaction,
`Hg_(2)Cl_(2)+Cl_(2) rarr 2Hg^(2+)+4Cl^(-)`
Given :
`{:(Hg_(2)Cl_(2)+2e^(-) rarr 2Hg+2Cl^(-),,E^(@)=0.270" volt"),(Hg_(2)^(2+) rarr 2Hg^(2+)+2e^(-),,E^(@)=-0.92" volt"),(2Hg rarr Hg_(2)^(2+)+2e^(-),,E^(@)=-0.79"volt"),(Cl_(2)+2e^(-) rarr 2Cl^(-),,E^(@)=1.36" volt"):}`

Answer

`-0.08` volt

Explanation

First determine `E^(@)` for
`Hg_(2)Cl_(2) rarr 2Hg^(2+) +2Cl^(-) +2e^(-)` electrode.
It comes to -1.44 volt.
This electrode is now coupled with `Cl_(2)+2e^(-) rarr 2Cl^(-)` electrode.