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

Very difficult

Solution

$\begin{array}{l} E_{cell}^{0} = 1.23 - 0.00 = 1.23 V \\ {ΔG}_{cell}^{0} = - {nFE}_{cell}^{0} = - 2 \times 96500 \times 1.23 J \end{array}$
$∴$ Work derived from this fuel cell
$= \frac{70}{100} \times (- {ΔG}_{cell}^{0}) \times 10^{- 3} = xJ$
Since insulated vessel, hence q = 0
From equation, for monoatomic gas,
$\begin{array}{l} w = ΔU \Rightarrow x = {nC}_{V, m} ΔT \{C_{v, m} = \frac{3 R}{2}\} \\ or, \frac{70}{100} \times (2 \times 96500 \times 1.23) \times 10^{- 3} = 1 \times \frac{3}{2} \times 8.314 \times ΔT \\ ∴ ΔT = 13.32 \end{array}$

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