The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt, $\mathrm{MX}_{2}$ ) \| $\mathrm{M}^{2+}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid \mathrm{M}$ The emf of the cell depends on the difference in concentrations of $\mathrm{M}^{2+}$ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. The solubility product $\left(K_{s p} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_{2}$ at 298 K based on the information available for the given concentration cell is (take $2.303 \times \mathrm{R} \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Redox Reactions and Electrochemistry

The electrochemical cell shown below is a concentration cell.

M ∣ M^{2 +}

(saturated solution of a 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_{s p}; mol^{3} dm^{- 9})

MX_{2}

at 298 K based on the information available for the given concentration cell is (take

2.303 \times R \times 298/ F = 0.059 V

)

(A) $1 \times 1 0^{- 15}$
(B) $4 \times 1 0^{- 15}$ ✓ Correct
(C) $1 \times 1 0^{- 12}$
(D) $4 \times 1 0^{- 12}$

Answer: (B)

Solution

For the concentration cell: E = (0.059/n) log([M²⁺]cathode/[M²⁺]anode). With n=2 and E=0.059 V: 0.059 = (0.059/2) log(0.001/[M²⁺]anode). Solving: [M²⁺]anode = 10⁻⁵ M = S (solubility). For MX₂: Ksp = [M²⁺][X⁻]² = S(2S)² = 4S³ = 4×(10⁻⁵)³ = 4×10⁻¹⁵. The answer is (B).

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Redox Reactions and Electrochemistry

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