1.

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)]

Answer»

Solution :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`.


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