1.

Given the data at 25^(@)C, Ag+I^(-)toAgI+e^(-),E^(@)=0.152V AgtoAg^(+)+e^(-),E^(@)=-0.800V What is the value of log K_(sp) for AgI? (2.303RT/F=0.059V)

Answer»

`-37.83`
`-16.13`
`-8.13`
`+8.612`

Solution :`Ag+I^(-)toAgI+e^(-),""E^(@)=0.152V`
`underset(Ag^(+)+e^(-)toAg,""E^(@)=0.800V)`
`Ag^(+)+E^(-)toAgI,""E_(CELL)^(@)=0.952V`
At equilibrium: `E^(@)=(2303RT)/(F)logK_(c)`
but `K_(c)=([AGI])/([Ag^(+)][I^(-)])=(1)/(K_(sp))`
`therefore0.952=-(2.303RT)/(F)"LOG "K_(sp)=-0.059" log "K_(sp)`
or `lgoK_(sp)=-16.13`


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