JEE Main · 2025 · Shift-IhardEC-135

For a Mg | Mg2+(aq) || Ag+(aq) | Ag cell, the correct Nernst Equation is:

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

For a \ceMg∣Mg2+(aq)∣∣Ag+(aq)∣Ag\ce{Mg | Mg^{2+}(aq) || Ag^+(aq) | Ag} cell, the correct Nernst Equation is:

Options
  1. a

    Ecell=Ecell∘−\dRT2Fln⁡\d[\ceAg+][\ceMg2+]E_{\text{cell}} = E^\circ_{\text{cell}} - \d\frac{RT}{2F} \ln \d\frac{[\ce{Ag^+}]}{[\ce{Mg^{2+}}]}

  2. b

    Ecell=Ecell∘+\dRT2Fln⁡\d[\ceAg+]2[\ceMg2+]E_{\text{cell}} = E^\circ_{\text{cell}} + \d\frac{RT}{2F} \ln \d\frac{[\ce{Ag^+}]^2}{[\ce{Mg^{2+}}]}

    ✓
  3. c

    Ecell=Ecell∘−\dRT2Fln⁡\d[\ceMg2+][\ceAg+]E_{\text{cell}} = E^\circ_{\text{cell}} - \d\frac{RT}{2F} \ln \d\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^+}]}

  4. d

    Ecell=Ecell∘−\dRT2Fln⁡\d[\ceAg+]2[\ceMg2+]E_{\text{cell}} = E^\circ_{\text{cell}} - \d\frac{RT}{2F} \ln \d\frac{[\ce{Ag^+}]^2}{[\ce{Mg^{2+}}]}

Correct Answerb

Ecell=Ecell∘+\dRT2Fln⁡\d[\ceAg+]2[\ceMg2+]E_{\text{cell}} = E^\circ_{\text{cell}} + \d\frac{RT}{2F} \ln \d\frac{[\ce{Ag^+}]^2}{[\ce{Mg^{2+}}]}

Detailed Solution

Strategy: Resistan\ce (RR) is inversely proportional to conductivity (κ\kappa). First find the cell constant (G∗G^*) from the known \ceKCl\ce{KCl} data, then find the unknown κ\kappa and its related molar conductivity.

Step 1: Find Cell Constant R1=200ΩR_1 = 200 \Omega, κ1=0.12Sm−1\kappa_1 = 0.12 \mathrm{S m^{-1}}. G∗=κ1⋅R1=0.12×200=24 m−1G^* = \kappa_1 \cdot R_1 = 0.12 \times 200 = 24 \text{ m}^{-1}

Step 2: Find Unknown κ\kappa R2=1000ΩR_2 = 1000 \Omega. κ2=G∗/R2=24/1000=0.024Sm−1\kappa_2 = G^* / R_2 = 24 / 1000 = 0.024 \mathrm{S m^{-1}}

Step 3: Calculate Λm\Lambda_m C=0.2 M=200 mol m−3C = 0.2 \text{ M} = 200 \text{ mol m}^{-3}. Λm=0.024200=1.2×10−4Sm2mol−1\Lambda_m = \frac{0.024}{200} = 1.2 \times 10^{-4} \mathrm{S m^2 mol^{-1}} Λm=1.2Scm2mol−1=120×10−2\Lambda_m = 1.2 \mathrm{S cm^2 mol^{-1}} = 120 \times 10^{-2}

Value =120= 120.

Answer: 120\boxed{\text{Answer: 120}}

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