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JEE Advanced · Chemistry · 17. Electrochemistry

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

  1. A \(1 \times 10^{-15}\)
  2. B \(4 \times 10^{-15}\)
  3. C \(1 \times 10^{-12}\)
  4. D \(4 \times 10^{-12}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4 \times 10^{-15}\)

Step-by-step Solution

Detailed explanation

\(M \mid M^{2+}\) (aq) \(\| M^{2+}\) (aq) \(\mid M\)

\(0.001 \mathrm{M}\)

Anode : \(M \longrightarrow M^{2+}(\mathrm{aq})+2 e^{-}\)

Cathode : \(M^{2+}(\mathrm{aq})+2 e^{-} \longrightarrow M\)

\(M^{2+}(\mathrm{aq})_{c} \rightleftharpoons M^{2+}(\mathrm{aq})_{a}\)

\(E_{\text {cell }}=0-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \)
\( \Rightarrow 0.059=-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \)
\( -2=\log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\}\)

\(\Rightarrow 10^{-2} \times 10^{-3}=\mathrm{M}^{2+}(\mathrm{aq})_{a}=\) solubility \(=s\)

\(M X_{2} \rightleftharpoons M^{2+}+2 X^{-}\)

\(K_{s p}=S .(2 S)^{2} 2 S\)

\(\Rightarrow K_{s p}=4 s^{3}=4 \times\left(10^{-5}\right)^{3}=4 \times 10^{-15}\)
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