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Question #44229

Difficulty: 8/10 | 🟢 Free

The Question

An electrochemical cell is constructed as follows:

\ceM(s)∣M2+(aq,1M)∣∣X−(aq,1M)∣X2(g,V,T)∣Pt(s)\ce{M(s) | M^{2+}(aq, 1 M) || X^{-}(aq, 1 M) | X_2(g, V, T) | Pt(s)}

The cathodic compartment initially contains 10 moles10 \text{ moles} of a real gas \ceX2\ce{X_2} in a volume of 2.0 L2.0 \text{ L} at 400 K400 \text{ K}. The gas undergoes adiabatic free expansion against a vacuum until its volume becomes 10.0 L10.0 \text{ L}.

Given for \ceX2(g)\ce{X_2(g)}: van der Waals constants: a=6.0 L2 atm mol−2a = 6.0 \text{ L}^2 \text{ atm mol}^{-2}, b=0b = 0. Molar heat capacity at constant volume, Cv,m=3RC_{v,m} = 3 R.

Standard electrode potentials (assumed independent of temperature): E\ceM2+/M∘=−0.76 VE^\circ_{\ce{M^{2+}/M}} = -0.76 \text{ V} E\ceX2/X−∘=+1.36 VE^\circ_{\ce{X_2/X^-}} = +1.36 \text{ V}

Assume the activity of \ceX2(g)\ce{X_2(g)} in the Nernst equation is numerically equal to its pressure in atm. Use R=0.08 L atm K−1 mol−1R = 0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}, 2.303R×300F=0.06 V\frac{2.303 R \times 300}{F} = 0.06 \text{ V}. log⁡102=0.30\log_{10} 2 = 0.30, log⁡103=0.48\log_{10} 3 = 0.48.

Calculate the EMF of the cell after the expansion is complete.

Options

  • A.

    2.158 V2.158 \text{ V}

  • B.

    2.180 V2.180 \text{ V}

  • C.

    2.161 V2.161 \text{ V}

  • D.

    2.082 V2.082 \text{ V}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Adiabatic Free Expansion|Electrode Potential Dependencies

Why This Question Is Tricky

Common Mistakes

The problem tests the intersection of real gas thermodynamics and electrochemistry.

Students often rote-memorize that adiabatic free expansion is isothermal (ΔT=0\Delta T = 0).

However, this is only true for ideal gases.

For a real gas, the internal pressure (∂U/∂V)T=a/V2(\partial U/\partial V)_T = a/V^2 is non-zero, leading to a temperature drop (Joule-Thomson cooling effect) during expansion.

Failing to account for this ΔT\Delta T leads to an incorrect temperature for the Nernst equation and an incorrect final pressure.

Furthermore, students must use the van der Waals equation to find the true final pressure, not the ideal gas law.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptAdiabatic Free Expansion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree