JEE Main · 2019 · Shift-IhardEC-088

The standard electrode potential E° and its temperature coefficient (dEdT) for a cell are 2 V and -5 10-4\ V\ K-1 at…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

The standard electrode potential E°E° and its temperature coefficient (dEdT)\left(\frac{dE}{dT}\right) for a cell are 2 V and −5×10−4 V K−1-5 \times 10^{-4}\ \mathrm{V\ K^{-1}} at 300 K, respectively. The reaction is Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}. The standard reaction enthalpy (ΔrH°\Delta_r H°) at 300 K in kJ mol−1\mathrm{kJ\ mol^{-1}} is ____\_\_\_\_.

[Use R=8 J K−1 mol−1R = 8\ \mathrm{J\ K^{-1}\ mol^{-1}} and F=96500 C mol−1F = 96500\ \mathrm{C\ mol^{-1}}]

Options
  1. a

    -412.8

    ✓
  2. b

    206.4

  3. c

    -384.0

  4. d

    192.0

Correct Answera

-412.8

Detailed Solution

Strategy: Use the thermodynamic relation between cell potential, its temperature coefficient, and enthalpy: ΔH=nF[T(dE/dT)−E]\Delta H = nF [T(dE/dT) - E].

Step 1: Set up the equation n=2n = 2 (for \ceZn/Cu\ce{Zn/Cu} cell). E=2.0 VE = 2.0 \text{ V}, T=300 KT = 300 \text{ K}, dE/dT=−5×10−4 V/KdE/dT = -5 \times 10^{-4} \text{ V/K}. F=96500 C/molF = 96500 \text{ C/mol}.

Step 2: Calculate ΔH\Delta H ΔH=2×96500×[300⋅(−5×10−4)−2.0]\Delta H = 2 \times 96500 \times [300 \cdot (-5 \times 10^{-4}) - 2.0] ΔH=193000×[−0.15−2.0]=193000×[−2.15]\Delta H = 193000 \times [-0.15 - 2.0] = 193000 \times [-2.15] ΔH=−414,950 J/mol=−414.95 kJ/mol\Delta H = -414,950 \text{ J/mol} = -414.95 \text{ kJ/mol}

The closest option provided is −412.8-412.8.

Answer: (a) -412.8\boxed{\text{Answer: (a) -412.8}}

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