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Given the data at 25^(@)C, Ag + I^(-) t...

Given the data at `25^(@)C,`
`Ag + I^(-) to AgI + e^(-), E^(@) = 0.152V`, `Agto Ag^(+)+ e^(-), E^(@) = -0.800V` What is the value of log K-sp For AgI ? `((2.303 (RT)/(F) = 0.059V))`

A

`-37.83`

B

`-16.13`

C

`-8.13`

D

`+8.612`

Text Solution

Verified by Experts

The correct Answer is:
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`
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