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For the cell reaction Ni (s)|NI^(2+)(aq)...

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 "`)

Text Solution

Verified by Experts

The correct Answer is:
`K_(c )=3.981xx10^(35);Max.work=202.65 kJ`
`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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