Consider the following two half-cell reactions along with the...

The half-cell reactions for the fuel cell are:
Anode (Oxidation): $CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^- \quad E^{\circ}_{ox} = -0.02$ V
Cathode (Reduction): $\dfrac{3}{2}O_2 + 6H^+ + 6e^- \rightarrow 3H_2O \quad E^{\circ}_{red} = 1.23$ V

The overall cell reaction is:
$CH_3OH + \dfrac{3}{2}O_2 \rightarrow CO_2 + 2H_2O$

The standard cell potential is:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.23 - 0.02 = 1.21$ V

The number of electrons transferred per mole of $CH_3OH$ is $n = 6$.

The maximum electrical work that can be obtained from the cell is given by the decrease in Gibbs free energy:
W_\max = nFE^{\circ}_{cell} = 6 \times 96500 \times 1.21 = 700590 J

Since the fuel cell operates at $80\%$ efficiency, the actual work derived is:
$W_{actual} = 0.80 \times 700590 = 560472$ J

This work is used to compress an ideal gas isothermally against a constant external pressure of $1$ kPa ($1000$ Pa). The work done on the gas during compression is positive:
$W = -P_{ext} \Delta V$

Substituting the values:
$560472 = -1000 \times \Delta V$
$\Delta V = -560.472$ m$^3$

Rounding to the nearest integer, the change in volume is $-560$ m$^3$.

Answer: $-560$