Question 14

Write the combustion reaction for butane: $C_4H_{10}(g) + \frac{13}{2}O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)$

Calculate standard Gibbs free energy of the reaction ($\Delta_r G^0$): $\Delta_r G^0 = [4 \Delta_f G^0_{\text{CO}_2} + 5 \Delta_f G^0_{\text{water}}] - [\Delta_f G^0_{\text{butane}} + 0]$ $\Delta_r G^0 = [4(-394) + 5(-237)] - [-18]$ $\Delta_r G^0 = [-1576 - 1185] + 18 = -2761 + 18 = -2743\text{ kJ mol}^{-1} = -2743000\text{ J mol}^{-1}$

Determine the number of electrons transferred ($n$): In $C_4H_{10}$, average oxidation state of $C$ is $-2.5$. In $CO_2$, it is $+4$. Change per $C = 4 - (-2.5) = 6.5$. Total electrons for 4 Carbons $= 4 \times 6.5 = 26$. Alternatively, $O_2 \rightarrow 2O^{2-}$, 13 oxygen atoms $\times 2 = 26$ electrons.

Calculate cell potential ($E^0$): $\Delta_r G^0 = -nFE^0$ $-2743000 = -26 \times F \times E^0$ $E^0 = \frac{2743000}{26F} = \frac{105500}{F} = \frac{105.5}{F} \times 10^3$ $X = 105.5$.