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JEE Main 2023 · Chemistry · Electrochemistry · medium

The emf of the cell Zn|Zn^2+(0.1 M) \| Cu^2+(0.001 M)|Cu at 298 K is (given E^°cell=1.10 V, (2.303RT)/(F)=0.059 V)

Single correct option

The emf of the cell Zn∣Zn2+(0.1 M) ∥ Cu2+(0.001 M)∣Cu\mathrm{Zn|Zn^{2+}(0.1\ M)\,\|\,Cu^{2+}(0.001\ M)|Cu} at 298 K298\ \mathrm{K} is (given Ecell∘=1.10 VE^\circ_{\mathrm{cell}}=1.10\ \mathrm V, 2.303RTF=0.059 V\frac{2.303RT}{F}=0.059\ \mathrm V)
  1. A. 1.041 V1.041\ \mathrm V
  2. B. 1.159 V1.159\ \mathrm V
  3. C. 1.10 V1.10\ \mathrm V
  4. D. 0.982 V0.982\ \mathrm V
Show answer and solution

Answer: (A) 1.041 V1.041\ \mathrm V

The cell reaction is Zn+Cu2+→Zn2++Cu\mathrm{Zn+Cu^{2+}\to Zn^{2+}+Cu} with n=2n=2 and Q=[Zn2+][Cu2+]=0.10.001=100Q=\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=\frac{0.1}{0.001}=100. Nernst: E=E∘−0.0592log⁡Q=1.10−0.0295×2=1.041 VE=E^\circ-\frac{0.059}{2}\log Q=1.10-0.0295\times2=1.041\ \mathrm V. Inverting QQ gives the wrong 1.1591.159 V; the cell emf drops because product concentration is larger than reactant concentration.
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Original question written in the JEE Main 2023 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake