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. `2 xx 10^(-3)`
B. `2 xx 10^(-12)`
C. `10^(-15)`
D. `6.8 xx 10^(-16)`

Answer

C

Explanation

When connected, electron flow from `X` to `Y` means,
`E_(Y^(2+)//Y)^(@) gt E_(X^(2+)//X)^(@)`
and when `X` is connected to `SHF`, e flow from `X` to `SHE` means.
`E_(X^(2+)//X)^(@)`
`{:(YI +e^(-) rarr Y +I^(-)E^(@) =- 0.37 V),(Y rarr Y^(+) +e^(-) " "E^(@) = - 0.53 V):}/(Yi hArr Y^(+) +I^(-) " "E^(@) =- 0.90 V)`
So, `E^(@) = (0.06)/(n) log K_(p)`
`- 0.90 = (0.06)/(1) log K_(sp)`
`log K_(sp) =- 15`
`K_(sp) = 10^(-15)`