Question

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

• A. 87%
• B. 90%
• C. 97%
• D. 80%

Answer 1

Answer: (C)

$CH_{3}OH\left(l\right)+\frac{3}{2}O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right)+2H_{2}O\left(l\right)$
$\Delta H=-726kJmol^{- 1}$
Also $\Delta G_{f}^{o}CH_{3}OH\left(I\right)=-166.2kJmol^{- 1}$
$\Delta G_{f}^{o}H_{2}O\left(I\right)=-166.2kJmol^{- 1}$
$\Delta G_{f}^{o}CO_{2}\left(I\right)=-394.4kJmol^{- 1}$
$\Delta G_{f}^{}=\displaystyle \sum \Delta G_{f}^{o}products-\displaystyle \sum \Delta G_{f}^{o}reactants$
$=-394.4-2\left(237.2\right)+166.2=702.6kJmol^{- 1}$
now efficiency of fuel cell
$=\frac{\Delta G}{\Delta H}\times 100=\frac{702.6}{726}\times 100=97\%$