The electrochemical cell shown below is a concentration cell. `M|M^(2+)(` saturated solution of sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M`
The `emf` of the cell depends on the difference in the concentration of `M^(2+)` ions at the two electrodes. The `emf` of the cell at `298` is `0.059V`.
The solubility product `(K_(sp),mol^(3) dm^(-9))` of `MX_(2)` at 298 based on the information available the given concentration cell is `(` Take `2.303xxRxx298//F=0.059V)`

Options

A. `1xx10^(-15)`
B. `4xx10^(-15)`
C. `1xx10^(-15)`
D. `4xx10^(12)`

Answer

B

Explanation

For concentration cell,
`E_(cell)=(0.0591)/(n)log.(C_(2(RHS)))/(C_(1(LHS)))`
`E_(cell)=0.059V,C_(2(RHS))=0.001`
`0.059=(0.0591)/(2)log.(0.001)/(C_(1))`
`or (2xx0.059)/(0.0591)=log.(0.001)/(C_(1))" or anti log2"=(0.001)/(C_(1))`
`thereforeC_(1)=(0.001)/(100)=10^(-5)`
`C_(1)"=concentration or solubility of M"^(2+)=10^(-5)`
`MX_(2)hArrM_(s)^(2+)+2underset(2S)X^(-)`
`K_(sp)=S(2S)^(2)=4S^(3)`
`K_(sp)=4xx(10^(-5))^(3)=4xx10^(-15)mol^(3)dm^(-9)`