JEE Challenger
More from Electrochemistry
ChemMedAdv 2012 • Paper 2

Solubility Product of Sparingly Soluble Salt in Concentration Cell

Comprehension Passage

Paragraph for Questions 31 and 32

The electrochemical cell shown below is a concentration cell.

M∣M2+(saturated solution of a sparingly soluble salt, MX2)∥M2+(0.001 mol dm−3)∣M\text{M} \mid \text{M}^{2+} (\text{saturated solution of a sparingly soluble salt, MX}_2) \parallel \text{M}^{2+} (0.001 \text{ mol dm}^{-3}) \mid \text{M}

The emf of the cell depends on the difference in concentrations of M2+\text{M}^{2+} ions at the two electrodes. The emf of the cell at 298 K298 \text{ K} is 0.059 V0.059 \text{ V}.

The solubility product (KspK_{sp}; mol3 dm−9\text{mol}^3 \text{ dm}^{-9}) of MX2\text{MX}_2 at 298 K298 \text{ K} based on the information available for the given concentration cell is (take 2.303×R×298/F=0.059 V2.303 \times \text{R} \times 298/\text{F} = 0.059 \text{ V})

Options

A

1×10−151\times 10^{-15}

B

4×10−154\times 10^{-15}

Correct
C

1×10−121\times 10^{-12}

D

4×10−124\times 10^{-12}

Topics & Concepts

ElectrochemistryElectrochemical CellSolubility ProductNernst Equation

Step-by-Step Solution

To find the solubility product (KspK_{sp}) of the sparingly soluble salt MX2\text{MX}_2, we analyze the given concentration cell:

M∣M2+(saturated solution of MX2)∥M2+(0.001 mol dm−3)∣M\text{M} \mid \text{M}^{2+} (\text{saturated solution of } \text{MX}_2) \parallel \text{M}^{2+} (0.001 \text{ mol dm}^{-3}) \mid \text{M}

Step 1: Cell Reactions and Nernst Equation

The half-cell reactions occurring at the electrodes are:

  • Anode (oxidation): M(s)→Manode2++2e−\text{M}(s) \rightarrow \text{M}^{2+}_{\text{anode}} + 2e^-
  • Cathode (reduction): Mcathode2++2e−→M(s)\text{M}^{2+}_{\text{cathode}} + 2e^- \rightarrow \text{M}(s)

Overall Cell Reaction: Mcathode2+⇌Manode2+\text{M}^{2+}_{\text{cathode}} \rightleftharpoons \text{M}^{2+}_{\text{anode}}

Here, the number of electrons transferred is n=2n = 2. Since this is a concentration cell, the standard cell potential Ecell∘=0 VE^\circ_{\text{cell}} = 0 \text{ V}.

Applying the Nernst equation at T=298 KT = 298 \text{ K}: Ecell=Ecell∘−2.303RTnFlog⁡([M2+]anode[M2+]cathode)E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 R T}{n F} \log \left( \frac{[\text{M}^{2+}]_{\text{anode}}}{[\text{M}^{2+}]_{\text{cathode}}} \right)

Given:

  • Ecell=0.059 VE_{\text{cell}} = 0.059 \text{ V}
  • 2.303RTF=0.059 V\frac{2.303 R T}{F} = 0.059 \text{ V}
  • [M2+]cathode=0.001 mol dm−3=10−3 M[\text{M}^{2+}]_{\text{cathode}} = 0.001 \text{ mol dm}^{-3} = 10^{-3} \text{ M}

Substituting these values into the equation: 0.059=0−0.0592log⁡([M2+]anode10−3)0.059 = 0 - \frac{0.059}{2} \log \left( \frac{[\text{M}^{2+}]_{\text{anode}}}{10^{-3}} \right)

Dividing both sides by 0.0590.059: 1=−12log⁡([M2+]anode10−3)1 = -\frac{1}{2} \log \left( \frac{[\text{M}^{2+}]_{\text{anode}}}{10^{-3}} \right)

−2=log⁡([M2+]anode10−3)-2 = \log \left( \frac{[\text{M}^{2+}]_{\text{anode}}}{10^{-3}} \right)

Taking the antilog on both sides: [M2+]anode10−3=10−2\frac{[\text{M}^{2+}]_{\text{anode}}}{10^{-3}} = 10^{-2}

[M2+]anode=10−5 mol dm−3[\text{M}^{2+}]_{\text{anode}} = 10^{-5} \text{ mol dm}^{-3}

Step 2: Calculation of Solubility Product (KspK_{sp})

The dissolution equilibrium of the sparingly soluble salt MX2\text{MX}_2 in water is: MX2(s)⇌M2+(aq)+2X−(aq)\text{MX}_2(s) \rightleftharpoons \text{M}^{2+}(aq) + 2\text{X}^-(aq)

Let SS be the molar solubility of MX2\text{MX}_2.

  • [M2+]=S=10−5 mol dm−3[\text{M}^{2+}] = S = 10^{-5} \text{ mol dm}^{-3}
  • [X−]=2S=2×10−5 mol dm−3[\text{X}^-] = 2S = 2 \times 10^{-5} \text{ mol dm}^{-3}

The expression for the solubility product KspK_{sp} is: Ksp=[M2+][X−]2K_{sp} = [\text{M}^{2+}][\text{X}^-]^2

Substituting the equilibrium concentrations: Ksp=(S)(2S)2=4S3K_{sp} = (S)(2S)^2 = 4S^3

Ksp=4×(10−5)3=4×10−15 mol3 dm−9K_{sp} = 4 \times (10^{-5})^3 = 4 \times 10^{-15} \text{ mol}^3 \text{ dm}^{-9}

Thus, the correct option is B.

PreviousNext