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JEE Main 2024 · Chemistry · Electrochemistry · hard

The cell Ag | AgX(s), X^-(aq) \| Ag^+(0.1 M) | Ag is built using a saturated solution of the sparingly soluble salt AgX on the left, with [Ag^+]=s…

Single correct option

The cell Ag ∣ AgX(s), X−(aq) ∥ Ag+(0.1 M) ∣ Ag\mathrm{Ag\,|\,AgX(s),\,X^-(aq)\ \|\ Ag^+(0.1\ M)\,|\,Ag} is built using a saturated solution of the sparingly soluble salt AgX on the left, with [Ag+]=s[\mathrm{Ag^+}]=s there. Its emf at 298 K298\ \mathrm{K} is 0.236 V0.236\ \mathrm{V}. Taking 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm{V}, the solubility product of AgX is
  1. A. 1×10−101\times10^{-10}
  2. B. 1×10−51\times10^{-5}
  3. C. 1×10−91\times10^{-9}
  4. D. 1×10−121\times10^{-12}
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Answer: (A) 1×10−101\times10^{-10}

This is a concentration cell, so E=0.059log⁡[Ag+]right[Ag+]left=0.059log⁡0.1sE=0.059\log\dfrac{[\mathrm{Ag^+}]_\text{right}}{[\mathrm{Ag^+}]_\text{left}}=0.059\log\dfrac{0.1}{s}. Then log⁡(0.1/s)=0.236/0.059=4\log(0.1/s)=0.236/0.059=4, so s=0.1×10−4=10−5 Ms=0.1\times10^{-4}=10^{-5}\ \mathrm{M}. AgX is a 1:1 salt, hence Ksp=s2=10−10K_{sp}=s^2=10^{-10}. Choosing 10−510^{-5} confuses solubility with KspK_{sp}.
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