Question

The `EMF` of the cell `:`
`Pt|Ce^(4+)(90%),Ce^(3+)(10%)|` Normal calomel electrode is `1.464 V` at `25^(@)C`. Find the value of equilibrium constant of the reaction `:`
`2Ce^(3+)+2H^(o+) rarr 2Ce^(4+)+H_(2)`
The electrode potential of the normal calomel electrode is `+0.28V`.

Answer 1

`E_(cell)=E_(c)-E_(a_)=1.464V`
`E_(Ce^(4+)|Ce^(3+))=E^(c-)._(Ce^(4+)|Ce^(3+))-(0.059)/(1) log .([Ce^(3+)])/([Cr^(4+)])`
`=E^(c-)._(Ce^(4+)|Ce^(3+))-(0.059)/(1) log .(10)/(90)`
`E_(cell)=E_(cal)-E_(Ce^(4+)|Ce^(3+))`

`1.464=E_(cal)-[(E^(c-)._(Ce^(4+)|Ce^(3+))-0.059log .(1)/(8))]`
`1.464=0.28-(E^(c-)._(Ce^(4+)|Ce^(3+))-0.059log.(1)/(8))`
`E^(c-)._(Ce^(4+)|Ce^(3+))=-1.24V`
Now ,
At `Eq, E_(cell)=0`
`:.E^(c-)=(0.059)/(2) log K`
At anode `:2Ce^(3+)rarr2Ce^(4+)+2e^(-)(` Oxidation`)`
At cathode `: 2H^(o+)+2e^(-)rarrH_(2)(` Reduction `)`
`ulbar(Cell reaction : 2Ce^(3+)+2H^(o+)rarr 2Ce^(4+)+H_(2))`
`E^(c-)._(cell)=E_(c-)._(red(cathode))-E^(c-)._(red(anode))`
`=E^(c-)._(red(H_(2))-E^(c-)._(red(Ce^(4+)|Ce^(3+))`
`=0-(-1.24)=1.24`
`:. E^(c-)=(0.059)/(2) log K`
`1.24=(0.059)/(2) log K`
`log K=42.03implies K =1.08 xx 10^(42)`