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The electrochemical cell shown below is ...

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 concentrations of `M^(2+)` ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.
The solubility product `(K_(sp), mol^(3) dm^(-9))` of `MX_(2)` at 298 K based on the information available for the given concentration cell is `("take "2.303xxRxx298//F=0.059 V)` :

A

`4xx10^(-15)`

B

`5.7`

C

`1xx10^(-12)`

D

`4xx10^(-12)`

Text Solution

Verified by Experts

The correct Answer is:
A
`M|M^(2+)underset(0.001M)((aq)||M^(2+))(aq)|M`
Anode: `MtoM^(2+)(aq)+2e^(-)`
Cathode: `underline(M^(2+)(aq)+2e^(-)toM" ")`
`M^(2+)(aq)hArrM^(2+)(aq)`
`E_(cell)=0-(0.059)/(2)"log"((M^(2+)(aq))/(10^(-3)))`
`0.059=-(0.059)/(2)"log"((M^(2+)(aq))/(10^(-3)))implies-2="log"((M^(2+)(aq))/(10^(-3)))`
`10^(-2)xx10^(-3)=M^(2+)(aq)="solubility"=s`
`K_(sp)=4S^(3)=4xx(10^(-5))^(3)=4xx10^(-15)`.
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