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JEE Main 2022 · Chemistry · Electrochemistry · medium

At 298 K the cell Ag | Ag^+(saturated AgBr) \| Ag^+(0.1 M) | Ag is set up. Given Ksp(AgBr)=1.0×10^-13 and 2.303RT/F=0.059 V, the emf of the cell is

Single correct option

At 298 K298\ \mathrm K the cell Ag ∣ Ag+(saturated AgBr) ∥ Ag+(0.1 M) ∣ Ag\mathrm{Ag\,|\,Ag^+(\text{saturated AgBr})\,\|\,Ag^+(0.1\ M)\,|\,Ag} is set up. Given Ksp(AgBr)=1.0×10−13K_{sp}(\mathrm{AgBr})=1.0\times10^{-13} and 2.303RT/F=0.059 V2.303RT/F=0.059\ \mathrm V, the emf of the cell is
  1. A. 0.32 V0.32\ \mathrm V
  2. B. 0.71 V0.71\ \mathrm V
  3. C. 0.16 V0.16\ \mathrm V
  4. D. 0.65 V0.65\ \mathrm V
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Answer: (A) 0.32 V0.32\ \mathrm V

This is a concentration cell with n=1n=1. In the left half-cell [Ag+]=Ksp=3.16×10−7 M[\mathrm{Ag^+}]=\sqrt{K_{sp}}=3.16\times10^{-7}\ \mathrm M (a common mistake is to use KspK_{sp} itself). Right electrode (higher concentration) is the cathode: E=0.059log⁡0.13.16×10−7=0.059log⁡(3.16×105)=0.059×5.5=0.32 VE=0.059\log\dfrac{0.1}{3.16\times10^{-7}}=0.059\log(3.16\times10^{5})=0.059\times5.5=0.32\ \mathrm V. Using KspK_{sp} directly gives 0.710.71 V and halving for n=2n=2 gives 0.160.16 V, both wrong.
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