The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
Given the following half cell `: YI+e^(-) rarr Y-I^(c-): " "E^(c-)=-0.27 V`
Solubility product of the iodide salt `YI` is

Options

A. `2xx10^(-3)`
B. `2xx10^(-12)`
C. `2xx10^(-14)`
D. `6.8xx10^(-16)`

Answer

c

Explanation

Also, create a cell such that the cell reaction is `:`

Cathod`:YI+e^(-)rarr Y+I^(c-)`
Anode `: Y rarr Y^(o+)+e^(-)`
`ulbar(YI rarr y^(O+)+I^(c-))`
`E^(c-)._(cell)=-0.27-0.53=-0.8V`
At equilibrium,
`log _(10)K_(sp)=(E^(c-)._(cell).n_(cell))/(0.059)=(-0.80xx1)/(0.059)~~-13.56`
, `implies K_(sp)~~2xx10^(-14)`