Verified Solution • 2025–26 Session moderate

QJEEM09Ch9a32: JEE Main 2009 Chemistry - Class 12 JEEM09Ch9a32 Solution: In a fuel cell methanol is used as fuel and oxygen gas is used…

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OH(l) + 3/2O2 →\to CO2 (g) + 2H2O (l)
At 298K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726 kJ mol−1, efficiency of the fuel cell will be

Multiple Choice Options

  • A 87%
  • B 90%
  • C 97%
  • D 80%

Step-by-Step Verified Solution

CH3OH(l)+32O2(g)C{H_3}OH\left( l \right) + {3 \over 2}{O_2}\left( g \right)

                   →CO2(g)+2H2O(l)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right)

ΔGr=ΔGf(CO2,g)+2ΔGf(H2O,l)−\Delta {G_r} = \Delta {G_f}\left( {C{O_2},g} \right) + 2\Delta {G_f}\left( {{H_2}O,l} \right) -

                    ΔGf(CH3OH,l)−32ΔGf(O2,g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G_f}\left( {C{H_3}OH,l} \right) - {3 \over 2}\Delta {G_f}\left( {{O_2},g} \right)

=−394.4+2(−237.2)−(−166.2)−0= - 394.4 + 2\left( { - 237.2} \right) - \left( { - 166.2} \right) - 0

=−394.4−474.4+166.2= - 394.4 - 474.4 + 166.2

=−702.6 kJ= - 702.6\,kJ

%  \% \,\, efficiency   =702.6726×100\,\, = {{702.6} \over {726}} \times 100 =97%= 97\%

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