PrepRung
Sign inSign up free

JEE Main 2021 · Chemistry · Electrochemistry · medium

A concentration cell is set up as Ag | Ag^+(saturated AgBr solution) \| Ag^+(0.1 M) | Ag. Its emf at 298 K is 0.354 V. Taking 2.303RT/F=0.059 V, the…

Single correct option

A concentration cell is set up as Ag ∣ Ag+(saturated AgBr solution) ∥ Ag+(0.1 M) ∣ Ag\mathrm{Ag}\,|\,\mathrm{Ag^+}(\text{saturated AgBr solution})\,\|\,\mathrm{Ag^+}(0.1\ \mathrm{M})\,|\,\mathrm{Ag}. Its emf at 298 K298\ \mathrm{K} is 0.354 V0.354\ \mathrm{V}. Taking 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm{V}, the solubility product of AgBr\mathrm{AgBr} is
  1. A. 1×10−71\times10^{-7}
  2. B. 1×10−141\times10^{-14}
  3. C. 1×10−121\times10^{-12}
  4. D. 1×10−101\times10^{-10}
Show answer and solution

Answer: (B) 1×10−141\times10^{-14}

For a concentration cell, E=0.059log⁡[Ag+]cathode[Ag+]anodeE=0.059\log\dfrac{[\mathrm{Ag^+}]_{\text{cathode}}}{[\mathrm{Ag^+}]_{\text{anode}}}. So log⁡0.1s=0.3540.059=6\log\dfrac{0.1}{s}=\dfrac{0.354}{0.059}=6, giving s=0.1106=1×10−7 Ms=\dfrac{0.1}{10^6}=1\times10^{-7}\ \mathrm{M} as the [Ag+][\mathrm{Ag^+}] in the saturated solution. Because AgBr dissociates 1:1, [Br−]=[Ag+][\mathrm{Br^-}]=[\mathrm{Ag^+}] and Ksp=s2=1×10−14K_{sp}=s^2=1\times10^{-14}. The value 10−710^{-7} is the solubility, not KspK_{sp}.
Report a mistake

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake