Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ)$
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.
$O_{2} (g) + 4 H^{+} (aq .) + 4 e^{-} \to 2 H_{2} O (ℓ), E^{o} = 1.23 V$
$2 H^{+} (aq .) + 2 e^{-} \to H_{2} (g), E^{o} = 0.00 V$
Use F = 96500 C mol^–1 , R = 8.314 J mol^–1K^–1

Question

Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ)$
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.
$O_{2} (g) + 4 H^{+} (aq .) + 4 e^{-} \to 2 H_{2} O (ℓ), E^{o} = 1.23 V$
$2 H^{+} (aq .) + 2 e^{-} \to H_{2} (g), E^{o} = 0.00 V$
Use F = 96500 C mol^–1 , R = 8.314 J mol^–1K^–1

Infinity Learn · Accepted Answer

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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