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JEE Main 2019 · Chemistry · Electrochemistry · medium

A silver electrode (E^°Ag^+/Ag = +0.80 V) is dipped in a saturated solution of AgBr, Ksp = 5.0×10^-13 at 298 K (no other source of Ag^+ or Br^-).…

Single correct option

A silver electrode (EAg+/Ag∘=+0.80 VE^\circ_{\mathrm{Ag^+/Ag}} = +0.80\ \mathrm{V}) is dipped in a saturated solution of AgBr, Ksp=5.0×10−13K_{sp} = 5.0\times10^{-13} at 298 K298\ \mathrm{K} (no other source of Ag+\mathrm{Ag^+} or Br−\mathrm{Br^-}). Taking 2.303RTF=0.059 V\dfrac{2.303RT}{F} = 0.059\ \mathrm{V} and log⁡5=0.70\log 5 = 0.70, the electrode potential of this half cell is nearest to
  1. A. 0.80 V0.80\ \mathrm{V}
  2. B. 0.44 V0.44\ \mathrm{V}
  3. C. 1.16 V1.16\ \mathrm{V}
  4. D. 0.62 V0.62\ \mathrm{V}
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Answer: (B) 0.44 V0.44\ \mathrm{V}

In a saturated AgBr solution [Ag+]=Ksp=5×10−13=7.07×10−7 M[\mathrm{Ag^+}] = \sqrt{K_{sp}} = \sqrt{5\times10^{-13}} = 7.07\times10^{-7}\ \mathrm{M}, log⁡[Ag+]=12(log⁡5−13)=12(0.70−13)=−6.15\log[\mathrm{Ag^+}] = \tfrac12(\log5-13) = \tfrac12(0.70-13) = -6.15. Nernst: E=E∘+0.059log⁡[Ag+]=0.80−0.059×6.15=0.80−0.363=0.437 VE = E^\circ + 0.059\log[\mathrm{Ag^+}] = 0.80 - 0.059\times6.15 = 0.80 - 0.363 = 0.437\ \mathrm{V}. Using [Ag+]=Ksp[\mathrm{Ag^+}] = K_{sp} would give 0.620.62 (wrong).
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