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Electrochemistry question

2008 · Q99
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Electrochemistry question

2008 · Q99

NEETChemistryElectrochemistryMCQ+4 / −1
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 Eocell for the pentane-oxygen fuel cell is
  1. A
    1.0968 V
  2. B
    0.0968 V
  3. C
    1.968 V
  4. D
    2.0968 V
View written solutionFree

Correct answer: A

At Anode:

C5H12 + 10H2O →\to→ 5CO2 + 32H+ + 32e-

At Cathode:

8O2 + 32H+ + 32e- →\to→ 16H2O
-------------------------------------------------
C5H12(g)

  • 8O2(g) →\to→ 5CO2(g)
  • 6H2O(l)

    Δ\Delta ΔG = 5×Δ\Delta ΔGCO2
  • 6 Δ\Delta ΔG(H2O) – [Δ\Delta ΔG(C5H12) +8 × Δ\Delta ΔGO2 ]

    = 5 × (– 394.4) + 6 × (–237.2) – (– 8.2 + 0)

    = – 3387 kJ mol–1

    = – 3387 × 103 J mol–1

    Δ\Delta ΔG = - nFEocell

    From the overall equation we find n = 32

    – 3387 × 103 = -32 ×\times× 96500 ×\times× Eocell

    Eocell = −3387×103−32×96500{{ - 3387 \times {{10}^3}} \over { - 32 \times 96500}}−32×96500−3387×103​ = 1.0968 V
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