Question

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)
A. `-37.83`
B. `-16.13`
C. `-8.13`
D. `+8.612`

Answer 1

Correct Answer - B
`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`