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

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 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.
$$O_{2}(g) + 4H^{+} (aq) + 4e^{-} \rightarrow 2 H_{2}O (l), E^{0} = 1.23 V$$,

$$2H^{+} (aq) + 2e^{-} \rightarrow H_{2} (g), E^{0} = 0.00 V$$.

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

The overall fuel-cell reaction is $$H_2(g)+\frac12 O_2(g)\rightarrow H_2O(l)$$.
For one mole of $$H_2$$ oxidised, 2 electrons are transferred, so $$n_e = 2$$.

Standard cell potential:
$$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}} = 1.23\,\text{V}-0.00\,\text{V}=1.23\,\text{V}$$

The maximum (reversible) electrical work equals $$-\Delta G^{\circ}$$, where
$$\Delta G^{\circ} = -n_e F E^{\circ}_{\text{cell}}$$

Thus for 1.00 mol of $$H_2$$:
$$\Delta G^{\circ} = -(2)(96500\;\text{C mol}^{-1})(1.23\;\text{V}) = -2.3739\times 10^{5}\;\text{J mol}^{-1}$$

Only $$1.0\times10^{-3}$$ mol of $$H_2$$ is consumed, so the maximum work becomes
$$W_{\text{max}} = -\Delta G^{\circ}\times 1.0\times10^{-3} = 237.39\;\text{J}$$

The fuel cell is 70 % efficient, therefore the useful work actually obtained is
$$W = 0.70 \times 237.39\;\text{J}=166.17\;\text{J}$$

This work is used to compress 1.00 mol of a monatomic ideal gas in a thermally insulated container. Because the container is adiabatic, $$q = 0$$ and the first law gives $$\Delta U = W$$ (work done on the gas increases its internal energy).

For an ideal monatomic gas $$C_V = \frac32 R$$, hence
$$\Delta U = n C_V \Delta T = (1.00\;\text{mol})\left(\frac32 R\right)\Delta T$$

Equating to the work:
$$n C_V \Delta T = W$$
$$\Delta T = \frac{W}{\frac32 R} = \frac{166.17\;\text{J}} {\tfrac32 (8.314\;\text{J mol}^{-1}\text{K}^{-1})} = \frac{166.17}{12.471} = 13.32\;\text{K}$$

Hence the temperature of the ideal gas rises by 13.32 K.