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: $\Delta_f G_{CO_2}^o = -394$ kJ mol$^{-1}$; $\Delta_f G_{water}^o = -237$ kJ mol$^{-1}$; $\Delta_f G_{butane}^o = -18$ kJ mol$^{-1}$

Answer 1

105.5

Answer 2

The balanced chemical equation for the combustion of butane is:

$\text{C}_4\text{H}_{10}\text{(g)} + 6.5\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 5\text{H}_2\text{O(l)}$

We are given the standard Gibbs energies of formation at 298 K:

$\Delta_f G_{CO_2}^o = -394$ kJ mol$^{-1}$

$\Delta_f G_{water(l)}^o = -237$ kJ mol$^{-1}$

$\Delta_f G_{butane(g)}^o = -18$ kJ mol$^{-1}$

The standard Gibbs energy of formation of $\text{O}_2\text{(g)}$ in its standard state is 0 kJ mol$^{-1}$.

The standard Gibbs free energy change for the reaction ($\Delta_r G^o$) is calculated as:

$\Delta_r G^o = \sum (\text{stoichiometric coefficient} \times \Delta_f G^o_{\text{products}}) - \sum (\text{stoichiometric coefficient} \times \Delta_f G^o_{\text{reactants}})$

$\Delta_r G^o = [4 \times \Delta_f G_{CO_2}^o + 5 \times \Delta_f G_{H_2O(l)}^o] - [1 \times \Delta_f G_{C_4H_{10}(g)}^o + 6.5 \times \Delta_f G_{O_2(g)}^o]$

$\Delta_r G^o = [4 \times (-394 \text{ kJ/mol}) + 5 \times (-237 \text{ kJ/mol})] - [1 \times (-18 \text{ kJ/mol}) + 6.5 \times (0 \text{ kJ/mol})]$

$\Delta_r G^o = [-1576 \text{ kJ/mol} - 1185 \text{ kJ/mol}] - [-18 \text{ kJ/mol}]$

$\Delta_r G^o = -2761 \text{ kJ/mol} + 18 \text{ kJ/mol}$

$\Delta_r G^o = -2743 \text{ kJ/mol}$

Convert $\Delta_r G^o$ to Joules per mole:

$\Delta_r G^o = -2743 \times 10^3$ J/mol

The relation between standard Gibbs free energy change and standard cell potential ($E^o$) is:

$\Delta_r G^o = -nFE^o$

where $n$ is the number of moles of electrons transferred in the reaction, and $F$ is the Faraday constant.

To find $n$, we examine the changes in oxidation states.

In $\text{C}_4\text{H}_{10}$, the oxidation state of C is -2.5. The total oxidation state for 4 carbons is $4 \times (-2.5) = -10$.

In $\text{CO}_2$, the oxidation state of C is +4. The total oxidation state for 4 carbons is $4 \times (+4) = +16$.

The change in total oxidation state of carbon is $+16 - (-10) = +26$. This corresponds to the loss of 26 electrons.

In $\text{O}_2$, the oxidation state of O is 0.

In $\text{CO}_2$ and $\text{H}_2\text{O}$, the oxidation state of O is -2.

There are $6.5$ moles of $\text{O}_2$, which contain $6.5 \times 2 = 13$ moles of oxygen atoms.

Each oxygen atom changes its oxidation state from 0 to -2, which is a gain of 2 electrons.

The total number of electrons gained by oxygen is $13 \times 2 = 26$.

So, the number of electrons transferred is $n = 26$.

Now, we can calculate $E^o$:

$E^o = \frac{-\Delta_r G^o}{nF}$

$E^o = \frac{-(-2743 \times 10^3 \text{ J/mol})}{26 \times F \text{ C/mol}}$

$E^o = \frac{2743 \times 10^3}{26F}$ V

The cell potential is given in the form $\frac{X}{F} \times 10^3$ volts.

Comparing this with our calculated $E^o$:

$\frac{X}{F} \times 10^3 = \frac{2743}{26} \times \frac{10^3}{F}$

$X = \frac{2743}{26}$

Calculate the value of $X$:

$X = \frac{2743}{26} = 105.5$

The value of $X$ is 105.5.