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Question: Answered & Verified by Expert
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OHl+32O2gCO2g+2H2Ol
At 298 K standard Gibb’s energies of formation for CH3OHl, H2Ol and CO2g are -166.2, -237.2 and -394.4 kJ mol-1 respectively. If standard enthalphy of combustion of methanol is -726 kJmol 1 , efficiency of the fuel cell will be
ChemistryElectrochemistryNEET
Options:
  • A 87%
  • B 90%
  • C 97%
  • D 80%
Solution:
2739 Upvotes Verified Answer
The correct answer is: 97%
CH3OHl+32O2gCO2g+2H2Ol
ΔH=-726kJmol-1
Also ΔGfoCH3OHI=-166.2kJmol-1
ΔGfoH2OI=-166.2kJmol-1
ΔGfoCO2I=-394.4kJmol-1
ΔGf=ΔGfoproducts-ΔGforeactants
=-394.4-2237.2+166.2=702.6kJmol-1
now efficiency of fuel cell
=ΔGΔH×100=702.6726×100=97%

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