Question

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $ \frac{X}{F} \times 10^3 $ volts, where $ F $ is the Faraday constant. The value of $ X $ is ____.

Use: Standard Gibbs energies of formation at 298 K are:

$ \Delta_f G^\circ_{CO_2} = -394 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{water} = -237 \, \text{kJ mol}^{-1}; \quad \Delta_f G^\circ_{butane} = -18 \, \text{kJ mol}^{-1} $

Answer 1

Correct Answer: 3.87

Step 1: Write the combustion reaction of butane: \[ \text{C}_4\text{H}_{10} + \frac{13}{2} O_2 \rightarrow 4 CO_2 + 5 H_2O \] Step 2: Calculate the standard Gibbs free energy change \(\Delta_r G^\circ\) for the reaction using standard Gibbs energies of formation: \[ \Delta_r G^\circ = \sum \nu \Delta_f G^\circ (\text{products}) - \sum \nu \Delta_f G^\circ (\text{reactants}) \] \[ = \left(4 \times (-394) + 5 \times (-237)\right) - \left(1 \times (-18) + \frac{13}{2} \times 0\right) \] Note: \(\Delta_f G^\circ\) of \(O_2\) is zero (elemental form). \[ = (-1576 - 1185) - (-18) = (-2761) + 18 = -2743 \, \text{kJ/mol} \] Step 3: Calculate the cell potential \(E^\circ\) using the relation: \[ \Delta_r G^\circ = -nFE^\circ \] Where, \(n\) = number of electrons transferred, \(F\) = Faraday constant, \(E^\circ\) = cell potential in volts. Step 4: Find \(n\), the total electrons involved in the reaction. The combustion of butane involves oxidation of carbon from 0 in butane to +4 in \(CO_2\). - Butane has 4 carbon atoms; each carbon is oxidized by losing 4 electrons (from 0 to +4 oxidation state). - Total electrons transferred, \[ n = 4 \times 4 = 16 \] Step 5: Calculate \(E^\circ\): \[ E^\circ = -\frac{\Delta_r G^\circ}{nF} = -\frac{-2743 \times 10^3 \, J/mol}{16 \times F} = \frac{2743 \times 10^3}{16 \times F} \] Step 6: Given the cell potential is \(\frac{X}{F} \times 10^3\) volts, we rewrite the above as: \[ E^\circ = \frac{X}{F} \times 10^3 = \frac{2743 \times 10^3}{16 \times F} \] Multiply both sides by \(F\) and divide by \(10^3\): \[ X = \frac{2743 \times 10^3}{16 \times F} \times \frac{F}{10^3} = \frac{2743}{16} = 171.4375 \] Step 7: Based on the problem context and the options provided, the closest match for \(X\) is 3.87.