Standard free energies of formation (in KJ/mol) at 298K are -237.2, -394.4, and -8.2 for H2O(l), CO2(g) and pentane (g) respectively. The value of E(standard)cell for pentane oxygen fuel cell is?
1- 0.0968V 2- 1.968V
3- 2.0968V 4- 0.059V
Answers
Answered by
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Hey dear,
● Answer -
E° = 1.096 V
● Solution -
Combustion of pentane in fuel cell is shown by -
C5H12 + 8O2 --> 5CO2 + 6H2O
Here , n = 32 e-
Standard free energy of reaction is calculated by -
∆G° = ∆G(products) ∆G(reactants)
∆G° = 5∆Gf(CO2) + 6∆Gf(H2O) - ∆Gf(C5H12) - 8∆Gf(O2)
∆G° = 5×(-394) + 6(-237) - (-8) - 8(0)
∆G° = -3384 kJ
EMF of a cell is calculated by -
E° = -∆G° / nF
E° = -(-3384×10^3) / (32×96500)
E° = 1.096 V
Therefore, E° for pentane-oxygen cell is 1.096 V.
Hope this helped...
● Answer -
E° = 1.096 V
● Solution -
Combustion of pentane in fuel cell is shown by -
C5H12 + 8O2 --> 5CO2 + 6H2O
Here , n = 32 e-
Standard free energy of reaction is calculated by -
∆G° = ∆G(products) ∆G(reactants)
∆G° = 5∆Gf(CO2) + 6∆Gf(H2O) - ∆Gf(C5H12) - 8∆Gf(O2)
∆G° = 5×(-394) + 6(-237) - (-8) - 8(0)
∆G° = -3384 kJ
EMF of a cell is calculated by -
E° = -∆G° / nF
E° = -(-3384×10^3) / (32×96500)
E° = 1.096 V
Therefore, E° for pentane-oxygen cell is 1.096 V.
Hope this helped...
ashubh:
how n=32 electrons please explain
Same doubt, pls explain
cauz 1 O2 consumes 4e- in combustion
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