IIT-JEE 2012 · Paper 2 · Chemistry Q31

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ChemistrySingle correct+3 / -1Question 31 of 60

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Paragraph for Questions 31 and 32

The electrochemical cell shown below is a concentration cell.

M | $\ce{M^{2+}}$ (saturated solution of a sparingly soluble salt, $\ce{MX2}$ ) || $\ce{M^{2+}}$ (0.001 mol dm⁻³) | M

The emf of the cell depends on the difference in concentrations of $\ce{M^{2+}}$ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

The solubility product ( $K_{sp}$ ; mol³ dm⁻⁹) of $\ce{MX2}$ at 298 K based on the information available for the given concentration cell is (take $2.303\times R\times298/F=0.059$ V)

1\times10^{-15}

4\times10^{-15}

1\times10^{-12}

4\times10^{-12}

Answer and worked solution

Official answer: B

Concentration cell. The right electrode (0.001 M) is the cathode:
$E=\frac{0.059}2\log\frac{0.001}{[\ce{M^{2+}}]_\text{sat}}=0.059\implies\frac{0.001}{[\ce{M^{2+}}]}=100\implies[\ce{M^{2+}}]=10^{-5}\text{ M}.$

Solubility product. With solubility $s=10^{-5}$ , $[\ce{X-}]=2s$ :
$K_{sp}=s(2s)^2=4s^3=4\times10^{-15}.$

(B)

Topics:Nernst equation Solubility product

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IIT-JEE 2012 · Paper 2, Chemistry question 31