JEE Challenger
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ChemHardAdv 2025 • Paper 2

Cell Potential Calculation for Butane Fuel Cell using Gibbs Energy

An electrochemical cell is fueled by the combustion of butane at 1 bar1\text{ bar} and 298 K298\text{ K}. Its cell potential is XF×103 volts\frac{X}{F} \times 10^3\text{ volts}, where FF is the Faraday constant. The value of XX is ______.

Use: Standard Gibbs energies of formation at 298 K298\text{ K} are: ΔfGCO2o=−394 kJ mol−1\Delta_f G_{\text{CO}_2}^o = -394\text{ kJ mol}^{-1}; ΔfGwatero=−237 kJ mol−1\Delta_f G_{\text{water}}^o = -237\text{ kJ mol}^{-1}; ΔfGbutaneo=−18 kJ mol−1\Delta_f G_{\text{butane}}^o = -18\text{ kJ mol}^{-1}

Official Numerical Answer105.4 to 105.6

Topics & Concepts

ElectrochemistryElectrochemical CellGibbs Free Energy

Step-by-Step Solution

To find the value of XX, we analyze the combustion reaction of butane (C4H10\text{C}_4\text{H}_{10}) in the electrochemical cell:

C4H10(g)+132O2(g)→4CO2(g)+5H2O(l)\text{C}_4\text{H}_{10}(g) + \frac{13}{2} \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 5 \text{H}_2\text{O}(l)

Step 1: Calculate the Standard Gibbs Energy Change of Combustion (ΔrG∘\Delta_r G^\circ)

The standard Gibbs energy of reaction per mole of butane is given by: ΔrG∘=[4⋅ΔfG∘(CO2)+5⋅ΔfG∘(H2O)]−[ΔfG∘(C4H10)+132⋅ΔfG∘(O2)]\Delta_r G^\circ = \left[ 4 \cdot \Delta_f G^\circ(\text{CO}_2) + 5 \cdot \Delta_f G^\circ(\text{H}_2\text{O}) \right] - \left[ \Delta_f G^\circ(\text{C}_4\text{H}_{10}) + \frac{13}{2} \cdot \Delta_f G^\circ(\text{O}_2) \right]

Given standard Gibbs energies of formation (ΔfG∘\Delta_f G^\circ):

  • ΔfG∘(CO2)=−394 kJ mol−1\Delta_f G^\circ(\text{CO}_2) = -394 \text{ kJ mol}^{-1}
  • ΔfG∘(H2O)=−237 kJ mol−1\Delta_f G^\circ(\text{H}_2\text{O}) = -237 \text{ kJ mol}^{-1}
  • ΔfG∘(C4H10)=−18 kJ mol−1\Delta_f G^\circ(\text{C}_4\text{H}_{10}) = -18 \text{ kJ mol}^{-1}
  • ΔfG∘(O2)=0 kJ mol−1\Delta_f G^\circ(\text{O}_2) = 0 \text{ kJ mol}^{-1} (elementary state)

Substituting these values into the reaction Gibbs energy equation: ΔrG∘=[4(−394)+5(−237)]−[−18+0] kJ mol−1\Delta_r G^\circ = \left[ 4(-394) + 5(-237) \right] - \left[ -18 + 0 \right] \text{ kJ mol}^{-1} ΔrG∘=[−1576−1185]+18 kJ mol−1\Delta_r G^\circ = \left[ -1576 - 1185 \right] + 18 \text{ kJ mol}^{-1} ΔrG∘=−2761+18=−2743 kJ mol−1=−2743×103 J mol−1\Delta_r G^\circ = -2761 + 18 = -2743 \text{ kJ mol}^{-1} = -2743 \times 10^3 \text{ J mol}^{-1}

Step 2: Determine the Number of Electrons Transferred (nn)

In the combustion process, oxygen is reduced from an oxidation state of 00 in O2\text{O}_2 to −2-2 in CO2\text{CO}_2 and H2O\text{H}_2\text{O}. The number of oxygen atoms involved per mole of butane is 1313. n=13 atoms of O×2 electrons/atom=26 moles of e−n = 13 \text{ atoms of O} \times 2 \text{ electrons/atom} = 26 \text{ moles of } e^-

Step 3: Relate ΔrG∘\Delta_r G^\circ to Cell Potential (E∘E^\circ)

The relation between the cell potential and standard Gibbs free energy change is: ΔrG∘=−nFE∘\Delta_r G^\circ = -n F E^\circ

Given that E∘=XF×103 VE^\circ = \frac{X}{F} \times 10^3 \text{ V}: −2743×103=−26×F×(XF×103)-2743 \times 10^3 = -26 \times F \times \left(\frac{X}{F} \times 10^3\right)

Canceling common terms (10310^3 and FF) on both sides: 2743=26X2743 = 26 X X=274326=105.5X = \frac{2743}{26} = 105.5

Final Answer: The value of XX is 105.5.

Cell Potential Calculation for Butane Fuel Cell using Gibbs Energy | Chemistry PYQ Solution - JEE Challenger