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JEE Advance 2012
paper 2
Question
The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt, $\left.\mathrm{MX}_2\right)\left|\left|\mathrm{M}^{2+}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right)\right| \mathrm{M}\right.$ The emf of the cell depends on the difference in concetration of $\mathrm{M}^{2+}$ ions at the two electrodes. The emf of the cell at 298 is 0.059 V
The solubility product $\left(\mathrm{K}_{\mathrm{sp}} ; \mathrm{mol}^3 \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_2$ at 298 based on the information available the given concentration cell is (take $2.303 \times \mathrm{R} \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )
Select the correct option:
A
$1 \times 10^{-15}$
B
$4 \times 10^{-15}$
C
$1 \times 10^{-12}$
D
$4 \times 10^{-12}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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