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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 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)`

A

`1 xx 10^(-15)`

B

`4 xx 10^(-15)`

C

`1 xx 10^(-12)`

D

`4 xx 10^(-12)`

Text Solution

Verified by Experts

The correct Answer is:
B
`M|M^(2+) (aq)` (saturated solution ) `||M^(2+) (aq) (0.001 M) |1 M`

`E = E^(@) - (0.059)/(2) "log" ([M^(2+) (aq)]_("anode"))/([M^(2+) (aq)]_("cathode"))`
`0.059 = 0 - (0.059)/(2) "log" ([M^(2+) (aq)]_("anode"))/(10^(-3))`
`-2 = 3 log [M^(2+) (aq)]_("anode")`
or `10^(-2) = 10^(3) [M^(2+) (aq)]_("anode")`
`therefore [M^(2+) (aq)]_("anode") = (10^(-2))/(10^(3)) = 10^(-5)`
For `MX_(2) , MX_(2) hArr M^(2+) + 2 X^(-)`
If s is the solubility , then
`[M^2+] = s , [X^-] = 2 s`
`K_(sp) = [M^(2+)] [X^(-)]^(2)`
`= s xx (2 s)^(2) = 4 s^(3)`
`= 4 xx (10^(-5))^(3) = 4 xx 10^(-15) dm^(-9) mol^(3)`
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