An electrochemical cell is fuelled by combustion of butane at 1 bar and 298 K. Its cell potential is

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Question

An electrochemical cell is fuelled by combustion of butane at 1 bar and 298 K. Its cell potential is (X × 10³/F) volts, where F is Faraday constant. The value of X is (Standard Gibbs energies of formation at 298 K: ΔfG°(CO₂) = −394 kJ/mol, ΔfG°(H₂O,l) = −237 kJ/mol, ΔfG°(C₄H₁₀,g) = −18 kJ/mol)

Accepted Answer

Correct Answer: X = 105.50Step-by-step Solution:Step 1: Write the balanced combustion reaction for butane.C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂OOr: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂OStep 2: Calculate ΔrG° for combustion.ΔrG° = Σ ΔfG°(products) − Σ ΔfG°(reactants)= [4×(−394) + 5×(−237)] − [1×(−18) + 0] (O₂ has ΔfG° = 0)= [−1576 + (−1185)] − [−18]= −2761 + 18 = −2743 kJ/molStep 3: Find number of electrons transferred.Butane C₄H₁₀: oxidation state of C changes from (on average) −5/2 in C₄H₁₀ to +4 in CO₂.Carbon atoms in C₄H₁₀: average oxidation state = (−2×10 + 0)/4 = ... using the formula:For C₄H₁₀: 4x + 10(+1) = 0 → x = −10/4 = −2.5. In CO₂: C = +4. Change per C = +4−(−2.5) = +6.5. Total electrons per mol butane = 4 × 6.5 = 26 electrons. For H: changes from +1 to +1 (in H₂O), no change. So n = 26 electrons per mole of butane.Step 4: Calculate cell potential.ΔrG° = −nFE_cell−2743×10³ J/mol = −26 × F × E_cellE_cell = 2743×10³/(26F) = 105.5×10³/F = (105.5×10³)/F VSo X×10³/F = 105.5×10³/F → X = 105.50Key Concepts:Relation between ΔrG° and E_cell: ΔrG° = −nFE_cell where n = number of electrons transferred per mole of reaction and F = Faraday constant (96485 C/mol ≈ 96500 C/mol). More negative ΔrG° means more positive (spontaneous) E_cell.Calculating n for organic combustion: For CₓHᵧ + O₂ → CO₂ + H₂O: n = 2x + y/2 (or equivalently, count electron loss: each C goes from oxidation state in the hydrocarbon to +4 in CO₂; each H stays at +1). For C₄H₁₀: n = 2(4) + 10/2 = 8 + 5 = 13... wait: this formula gives n = 2(4) + 10/2 = 13 for C₄H₁₀/2. For 1 mol C₄H₁₀: n = 2×4 + 1×10/2 = 8 + 5 = 13. But from the oxidation state method: 4 × (change per C) = 4 × 6.5 = 26. There's a discrepancy. The correct way: n = 4(4 − (−2.5)) = 4 × 6.5 = 26 (from oxidation states). The formula n = 4(4a + b)/2 for CₐH_b: n = 4(4×4+10)/2 = 4×26/2 = 52? No. Using the standard formula: for CₙH₂ₙ₊₂: n_electrons = 6n+2. For C₄H₁₀: 6(4)+2 = 26. ✓Gibbs energy of formation: ΔfG° is the standard Gibbs free energy change for formation of 1 mol of compound from its elements in standard states. For elements (O₂, N₂, etc.) in standard state: ΔfG° = 0.Electrochemical fuel cell efficiency: The maximum electrical work obtainable from a fuel cell equals −ΔrG°. This is always greater than the heat obtainable from simple combustion (−ΔrH°) when ΔrS° > 0, making fuel cells potentially more efficient than heat engines.

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