Q.
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is H2(g)+12O2(g)→H2O(ℓ)The work derived from the cell on the consumption of 1.0 × 10–3mol of H2(g) is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulted 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. O2(g)+4H+(aq.)+4e−→2H2O(ℓ),Eo=1.23V2H+(aq.)+2e−→H2(g),Eo=0.00VUse F = 96500 C mol–1 , R = 8.314 J mol–1K–1
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answer is 13.32.
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Detailed Solution
Ecell 0=1.23−0.00=1.23VΔGcell 0=−nFEcell 0=−2×96500×1.23J∴ Work derived from this fuel cell=70100×−ΔGcell0×10−3=xJSince insulated vessel, hence q = 0From equation, for monoatomic gas, w=ΔU ⇒ x=nCV,mΔTCv,m=3R2 or, 70100×(2×96500×1.23)×10−3=1×32×8.314×ΔT∴ ΔT=13.32
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