Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
$C{H}_{3}OH\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
At 298 K standard Gibb’s energies of formation for $C{H}_{3}OH\left(l\right),H_{2}O\left(l\right)$ and $C{O}_2\left(g\right)$ are -166.2, -237.2 and -394.4 $kJmo{l}^{-1}$ respectively. If standard enthalphy of combustion of methanol is -726 $kJmo{l}^{-1}$ , efficiency of the fuel cell will be

• A. 87%
• B. 90%
• C. 97%
• D. 80%

Answer 1

Answer: (C)

$C{H}_{3}OH\left(l\right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
$\Delta H=-726kJmo{l}^{-1}$
Also $\Delta G_f^{o}C{H}_{3}OH\left(I\right)=-166.2kJmo{l}^{-1}$
$\Delta G_f^o{H}_{2}O\left(I\right)=-166.2kJmo{l}^{-1}$
$\Delta G_f^{o}C{O}_2\left(I\right)=-394.4kJmo{l}^{-1}$
$\Delta G_f^{}=\sum \Delta G_f^{o}products-\sum \Delta G_f^{o}reactants$
$=-394.4-2\left(237.2\right)+166.2=702.6kJmo{l}^{-1}$
now efficiency of fuel cell
$=\frac{\Delta G}{\Delta H}\times 100=\frac{702.6}{726}\times 100=97\%$