Animated Solution for Chemistry - Electrochemistry: 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 : ${\Delta }_f{G}_{{\text{CO}}_2}^{\circ }=-394{\text{ kJ mol}}^{-1}$; ${\Delta }_f{G}_{\text{water}}^{\circ }=-237{\text{ kJ mol}}^{-1}$; ${\Delta }_f{G}_{\text{butane}}^{\circ }=-18{\text{ kJ mol}}^{-1}$

The intersection of thermodynamics and electrochemistry is one of the most fascinating areas in physical chemistry. In this problem, we are tasked with finding the standard cell potential of a fuel cell powered by the combustion of butane. This requires us to bridge the gap between the Gibbs free energy of a chemical reaction and the electrical work it can perform.

We are given a fuel cell operating at standard conditions (1 bar and 298 K) using butane ($C_4{H}_{10}$) as the fuel. The first step in any such problem is to write down the balanced chemical equation for the combustion process. Butane reacts with oxygen gas to produce carbon dioxide and liquid water:

This balanced equation is our roadmap. It tells us exactly how many moles of each product are formed per mole of butane consumed, which is crucial for calculating the overall energy change.

To find the cell potential, we first need to determine the standard Gibbs free energy change (${\Delta }_r{G}^{\circ }$) for the reaction. We can calculate this using the standard Gibbs free energies of formation (${\Delta }_f{G}^{\circ }$) provided in the problem:

Substituting the given values, and remembering that the formation energy of an element in its standard state (like $O_2$) is zero, we get:

Calculating the terms inside the brackets:

The negative sign indicates that the combustion of butane is a highly spontaneous process, which is exactly what we want for a fuel cell!

Now comes the crucial step: linking this thermodynamic quantity to the cell potential ($E^{\circ }$). The relationship is given by the famous equation:

Here, $n$ is the number of moles of electrons transferred in the balanced reaction, and $F$ is the Faraday constant. To find $n$, we look at the change in oxidation states. The average oxidation state of carbon in butane ($C_4{H}_{10}$) is $-2.5$. In carbon dioxide ($C{O}_2$), it is $+4$.

The change in oxidation state per carbon atom is $4-(-2.5)=6.5$. Since there are 4 carbon atoms in a molecule of butane, the total number of electrons transferred is:

With ${\Delta }_r{G}^{\circ }$ and $n$ in hand, we can solve for $E^{\circ }$. A critical trap here is unit consistency. The Gibbs free energy is in kilojoules, but the Faraday constant is typically used with joules. We must convert ${\Delta }_r{G}^{\circ }$ to J/mol by multiplying by $10^3$:

Solving for $E^{\circ }$:

The problem states that the cell potential is $\frac{X}{F}\times 10^3$ volts. By directly comparing our result with this expression, we find that $X=105.50$.

This problem beautifully demonstrates how the chemical energy stored in the bonds of a hydrocarbon can be quantified and translated into an electrical potential, the very principle that powers modern fuel cell technology!