JEE Main · 2019 · Shift-IImediumEC-123

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)\mathrm{Zn(s)+Cu^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Cu(s)} at 300 K is approximately: (R=8 J K−1 mol−1R=8\ \mathrm{J\ K^{-1}\ mol^{-1}}, F=96000 C mol−1F=96000\ \mathrm{C\ mol^{-1}})

Options
  1. a

    e−160e^{-160}

  2. b

    e−80e^{-80}

  3. c

    e160e^{160}

    ✓
  4. d

    e320e^{320}

Correct Answerc

e160e^{160}

Detailed Solution

Strategy: Use the relationship lnK=(nFE∘)/(RT)ln K = (nFE^\circ) / (RT). Ensure consistency between the units of RR, TT, and FF.

Calculation: n=2n = 2 (for \ceZn/Cu\ce{Zn/Cu} system) E∘=2 VE^\circ = 2 \text{ V}, T=300 KT = 300 \text{ K}, R=8 J/K/molR = 8 \text{ J/K/mol}, F=96000 C/molF = 96000 \text{ C/mol}. lnK=2×96000×28×300=3840002400=160ln K = \frac{2 \times 96000 \times 2}{8 \times 300} = \frac{384000}{2400} = 160 K=e160K = e^{160}

Answer: (c) e160\boxed{\text{Answer: (c) } e^{160}}

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