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Question

# Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for H2O(l),CO2(g) and pentane (g), respectively. The value of E0cell for the pentane-oxygen fuel cell is:

A
1.0968V
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B
0.0968V
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C
1.968V
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D
2.0968V
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Solution

## The correct option is A 1.0968VC5H12+8O2⟶5CO2+6H2OΔG=G(products)−G(reactants)=5×(−394.4)+6×(−237.2)−(−8.2)=−3387Kjmol−1To get Ecell use the equation:ΔG=−nFEcellAs O is moving from 0 to -2 oxidation state, n = 32 for this reaction−3387×1000=−32×96500×EcellEcell=1.0968V

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