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JEE Advanced 2024 Paper 1 · Chemistry · Electrochemistry · medium

A galvanic cell is built as Ag(s) | AgX(s) | X^-(0.010 M) \| Ag^+(0.10 M) | Ag(s), where AgX is a sparingly soluble 1:1 salt. At 298 K the measured…

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A galvanic cell is built as Ag(s) ∣ AgX(s) ∣ X−(0.010 M) ∥ Ag+(0.10 M) ∣ Ag(s)\mathrm{Ag(s)\,|\,AgX(s)\,|\,X^-(0.010\ M)\,\|\,Ag^+(0.10\ M)\,|\,Ag(s)}, where AgX\mathrm{AgX} is a sparingly soluble 1:1 salt. At 298 K298\ \mathrm{K} the measured cell emf is 0.354 V0.354\ \mathrm{V}. Take 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm{V}. The solubility product KspK_{sp} of AgX\mathrm{AgX} is
  1. A. 1.0×10−101.0\times10^{-10}
  2. B. 1.0×10−81.0\times10^{-8}
  3. C. 1.0×10−91.0\times10^{-9}
  4. D. 1.0×10−121.0\times10^{-12}
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Answer: (C) 1.0×10−91.0\times10^{-9}

Both electrodes are silver, so this is a concentration cell with E∘=0E^\circ=0. The left electrode (anode) has [Ag+][\mathrm{Ag^+}] fixed by the solubility equilibrium: [Ag+]a=Ksp/[X−]=Ksp/0.010[\mathrm{Ag^+}]_a=K_{sp}/[\mathrm{X^-}]=K_{sp}/0.010. The right electrode (cathode) has [Ag+]c=0.10[\mathrm{Ag^+}]_c=0.10 M. For one electron, E=0.059log⁡[Ag+]c[Ag+]aE=0.059\log\dfrac{[\mathrm{Ag^+}]_c}{[\mathrm{Ag^+}]_a}. Hence log⁡0.10×0.010Ksp=0.3540.059=6\log\dfrac{0.10\times0.010}{K_{sp}}=\dfrac{0.354}{0.059}=6, so Ksp=10−3/106=1.0×10−9K_{sp}=10^{-3}/10^{6}=1.0\times10^{-9}. Using 0.10/Ksp0.10/K_{sp} without the factor 0.0100.010 would give 10−710^{-7}-type errors, and forgetting that the anode concentration is Ksp/[X−]K_{sp}/[\mathrm{X^-}] gives the other distractors.
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Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake