Question

Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$.

The work derived from the cell on the consumption of $1.0 \times 10^{-3}$ mol of $H_2(g)$ is used to compress 1.00 mol of a monoatomic ideal gas in 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 : $O_2(g) + 4H^+(aq) + 4e^- \longrightarrow 2H_2O(l), E^\circ = 1.23 V$, $2H^+(aq) + 2e^- \longrightarrow H_2(g), E^\circ = 0.00 V$.

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

Answer 1

13.32 K

Answer 2

1. Electrical Energy from the Fuel Cell:

   • Reaction: $H_2 + \frac{1}{2}O_2 \to H_2O$.
   • Number of electrons per $H_2$ = 2.
   • For $1.0 \times 10^{-3}$ mol $H_2$: $n_e = 2 \times 1.0 \times 10^{-3} = 2.0 \times 10^{-3}$ mol electrons.

2. Calculate Ideal Electrical Energy (ΔG):

   • $E^\circ_{\text{cell}} = 1.23 \,V$
   • Energy = $n_e \, F \, E^\circ = 2.0 \times 10^{-3} \times 96500 \times 1.23$.
   • $\approx 2.0 \times 10^{-3} \times 96500 \approx 193\,J$ and then $193\,J \times 1.23 \approx 237.39\,J$.

3. Taking 70% Efficiency:

   • Work done, $W = 0.70 \times 237.39 \,J \approx 166.17 \,J$.

4. Relating Work to Temperature Change in an Adiabatic Process:

   • Since compression is adiabatic (no heat exchange), the work done increases the internal energy.
   • For 1 mole of a monoatomic ideal gas, $\Delta U = \frac{3}{2}R\,\Delta T$ and $\Delta U = W$.
   • Therefore, $\Delta T = \frac{2W}{3R} = \frac{2 \times 166.17}{3 \times 8.314}$.
   • $\Delta T \approx \frac{332.34}{24.942} \approx 13.33\,K$.

Final Answer: Change in temperature = 13.32 K (approximately).