1.

For the cell reaction Ni (s)|NI^(2+)(aq)||Ag^(+)(aq)|Ag(s), calculate the equilibrium constant at 25^(@)C. How much maximum work would be obtained for the operation of this cell ? (Given E_(Ni^(2+)//Ni)^(@)=-0.25" V " and E_(Ag^(+)//Ag)^(@)=0.80" V ")

Answer»


SOLUTION :`Ni(s)+2Ag^(+)(aq) hArr Ni^(2+)(aq)+2Ag(s)`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.80-(-0.25)=1.05" V "`
`E_(cell)^(@)=(0.0591)/(n)log K_(c )"or""log "K_(c )=(nE^(@)cell)/(0.0591)`
`log K_(c )=(1.05xx2)/(0.0951)=35.6 "or"K_(c )="Antilog "35.6=3.981xx10^(35)`
`(-DeltaG^(@))=Max.work =nFE^(@) cell`
`=2" MOL"xx(96500" C mol"^(-1))xx(1.05" V")=202650" J "=202.65 KJ`


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