Calculate the emf of the cell having the cell reaction `2Ag^(+)+Zn iff 2Ag+Zn^(2+) " and " E^(@)""_(cell)=1.56V " at " 25^(@)C` when concentration of `Zn^(2+)=0.1 M " and " Ag^(+)=10M` in the solution.
`" "["Hint :" E_(cell)=E^(@)""_(cell)-(RT)/(nF)In([Zn^(2+)])/[Ag]^(2)]`

Explanation

Given :
`E^(@)""_(cell)=1.56V" "[Zn^(2+)]`
`" "=0.1 M [Ag^(+)]=10M`
Formula :
`E_("Cell")=E^(@)""_(cell)-(RT)/(nF)In.([Zn^(2+)])/[Ag^(+)]^(2)`
Solution : `" "=1.56-0.02955 log 0.001,`
`" "=1.56-(-0.08865)`
`" "=1.56+0.08865=1.6486V`
`E_("Cell")=1.6486V`.