PrepRung
Sign inSign up free

JEE Main 2018 · Chemistry · Electrochemistry · medium

At 298 K the cell Ag | AgCl(sat.) \| Ag^+(0.1 M) | Ag is set up, where the left half-cell contains silver metal dipped in a saturated solution of…

Single correct option

At 298 K298\ \mathrm{K} the cell Ag ∣ AgCl(sat.) ∥ Ag+(0.1 M) ∣ Ag\mathrm{Ag\,|\,AgCl(sat.)\,\|\,Ag^+(0.1\ M)\,|\,Ag} is set up, where the left half-cell contains silver metal dipped in a saturated solution of AgCl\mathrm{AgCl} (Ksp=1.8×10−10K_{sp}=1.8\times10^{-10}). Using 2.303RTF=0.0591 V\dfrac{2.303RT}{F}=0.0591\ \mathrm{V}, the emf of the cell is closest to
  1. A. 0.11 V0.11\ \mathrm{V}
  2. B. 0.23 V0.23\ \mathrm{V}
  3. C. 0.46 V0.46\ \mathrm{V}
  4. D. 0.51 V0.51\ \mathrm{V}
Show answer and solution

Answer: (B) 0.23 V0.23\ \mathrm{V}

This is a silver concentration cell, E∘=0E^\circ=0. In saturated AgCl\mathrm{AgCl}, [Ag+]=Ksp=1.34×10−5 M[\mathrm{Ag^+}]=\sqrt{K_{sp}}=1.34\times10^{-5}\ M. E=0.0591log⁡[Ag+]right[Ag+]left=0.0591log⁡0.11.34×10−5=0.0591×3.873=0.229 VE=0.0591\log\dfrac{[\mathrm{Ag^+}]_{\text{right}}}{[\mathrm{Ag^+}]_{\text{left}}}=0.0591\log\dfrac{0.1}{1.34\times10^{-5}}=0.0591\times3.873=0.229\ \mathrm{V}. The common errors are using KspK_{sp} itself as the concentration (0.510.51 V) or doubling for two electrons (0.460.46 V).
Report a mistake

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake