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ChemistryNumerical+4Question 33 of 54

JEE Advanced 2020 · Paper 1, Chemistry question 15

Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is

HX2(g)+12 OX2(g)→HX2O(l).\ce{H2(g) + 1/2 O2(g) -> H2O(l)}.

The work derived from the cell on the consumption of 1.0×10−31.0\times10^{-3} mol of HX2(g)\ce{H2(g)} is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?

The standard reduction potentials for the two half-cells are given below.

OX2(g)+4 HX+(aq)+4 eX−→2 HX2O(l),E0=1.23 V,\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)},\quad E^0 = 1.23\ \text{V},

2 HX+(aq)+2 eX−→HX2(g),E0=0.00 V.\ce{2H+(aq) + 2e- -> H2(g)},\quad E^0 = 0.00\ \text{V}.

Use F=96500 C mol−1F = 96500\ \mathrm{C\,mol^{-1}}, R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}.

Answer and worked solution

Official answer: 13

Cell. E∘=1.23−0.00=1.23E^\circ=1.23-0.00=1.23 V, with n=2n=2 per mole of HX2\ce{H2}.

Useful work.
∣ΔG∣=nFE∘=2×96500×1.23=237 390 J mol−1,|\Delta G|=nFE^\circ=2\times96500\times1.23=237\,390\ \mathrm{J\,mol^{-1}},
W=0.70×237 390×1.0×10−3=166.2 J.W=0.70\times237\,390\times1.0\times10^{-3}=166.2\text{ J}.

Compressing the gas. In an insulated container q=0q=0, so the work goes into internal energy:
W=nCvΔT=1.00×32×8.314 ΔT  ⟹  ΔT=166.212.47=13.32 K.W=nC_v\Delta T=1.00\times\tfrac32\times8.314\,\Delta T\implies\Delta T=\frac{166.2}{12.47}=13.32\text{ K}.
Answer: 13.32 (key: 13.00 to 13.60)

Topics:Gibbs energy and EMFFirst law and enthalpy
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