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JEE Advanced 2019 Paper 1 · Chemistry · Electrochemistry · medium

At 298 K the standard reduction potentials are E^°(Ag^+/Ag)=+0.80 V and E^°(AgCl/Ag,Cl^-)=+0.22 V. Take 2.303RT/F=0.059 V. The molar solubility of…

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At 298 K298\ \mathrm{K} the standard reduction potentials are E∘(Ag+/Ag)=+0.80 VE^\circ(\mathrm{Ag^+/Ag})=+0.80\ \mathrm{V} and E∘(AgCl/Ag,Cl−)=+0.22 VE^\circ(\mathrm{AgCl/Ag,Cl^-})=+0.22\ \mathrm{V}. Take 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm{V}. The molar solubility of AgCl\mathrm{AgCl} in 0.01 M0.01\ \mathrm{M} aqueous NaCl\mathrm{NaCl} is closest to
  1. A. 1.2×10−5 M1.2\times10^{-5}\ \mathrm{M}
  2. B. 1.5×10−6 M1.5\times10^{-6}\ \mathrm{M}
  3. C. 1.5×10−8 M1.5\times10^{-8}\ \mathrm{M}
  4. D. 1.5×10−10 M1.5\times10^{-10}\ \mathrm{M}
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Answer: (C) 1.5×10−8 M1.5\times10^{-8}\ \mathrm{M}

Combine the two half cells so that the net process is the dissolution AgCl(s)⇌Ag++Cl−\mathrm{AgCl(s)\rightleftharpoons Ag^+ + Cl^-}: cathode AgCl+e−→Ag+Cl−\mathrm{AgCl+e^-\to Ag+Cl^-} (0.22 V) and anode Ag→Ag++e−\mathrm{Ag\to Ag^++e^-} (0.80 V). So Ecell∘=0.22−0.80=−0.58 VE^\circ_{cell}=0.22-0.80=-0.58\ \mathrm{V} and log⁡Ksp=nE∘/0.059=−0.58/0.059=−9.83\log K_{sp}=nE^\circ/0.059=-0.58/0.059=-9.83, giving Ksp≈1.48×10−10K_{sp}\approx1.48\times10^{-10}. In pure water the solubility would be Ksp=1.2×10−5\sqrt{K_{sp}}=1.2\times10^{-5} M, which is the trap (option A). With 0.01 M Cl−0.01\ \mathrm{M}\ \mathrm{Cl^-} already present, s(0.01+s)=Ksps(0.01+s)=K_{sp}; since s≪0.01s\ll0.01, s≈Ksp/0.01=1.48×10−8s\approx K_{sp}/0.01=1.48\times10^{-8} M. Option D is just KspK_{sp} itself and option B comes from dividing by 10−410^{-4}. The answer is 1.5×10−81.5\times10^{-8} M.
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