Question

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

• A. $80\%$
• B. $87\%$
• C. $90\%$
• D. $97\%$

Answer 1

Answer: (D)

$C{H}_{3}OH\left(\ell \right)+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(\ell \right)\Delta H=-726kJmo{l}^{-1}$
Also $\Delta G_f^{o}C{H}_{3}OH\left(\ell \right)=-166.2kJmo{l}^{-1}$
$\Delta G_f^{o}C{O}_2\left(\ell \right)=-394.4kJmo{l}^{-1}$
$\because \Delta G=\sum \Delta G_f^o$ 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 $=\frac{\Delta G}{\Delta H}\times 100$
$=\frac{702.6}{726}\times 100$
$=97\%$