A fuel cell involves combustion of the butane at 1 atm and 298 K `C_(4)H_(10)(g)+(13)/(2)O_(2)(g)rarr 4CO_(2)(g)+5H_(2)O(l)`
`DeltaG^(@)=-"2744 kJ/mole"`
The value of `E_("cell")^(@)` Report your answer by rounding it upto nearest whole number.

Answer

To find the value of \( E_{\text{cell}}^\circ \) for the combustion of butane, we can use the relationship between Gibbs free energy change (\( \Delta G^\circ \)), the number of moles of electrons transferred (n), and the Faraday constant (F). The formula we will use is:

\[ \Delta G^\circ = -n E_{\text{cell}}^\circ F \]

### Step 1: Write the balanced equation for the combustion of butane
The balanced equation for the combustion of butane is:

\[ C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4 CO_2(g) + 5 H_2O(l) \]

### Step 2: Determine the change in oxidation state
To find the number of electrons transferred in the reaction, we need to determine the oxidation states of carbon in butane and carbon dioxide.

- In butane (\( C_4H_{10} \)), the oxidation state of carbon can be calculated as follows:
- Let the oxidation state of carbon be \( x \).
- The equation for the total charge is:
\[ 4x + 10(-1) = 0 \implies 4x - 10 = 0 \implies x = +2.5 \]
(This indicates an average oxidation state; however, for the purpose of electron transfer, we consider the extremes.)

- In carbon dioxide (\( CO_2 \)), the oxidation state of carbon is:
\[ x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4 \]

### Step 3: Calculate the total change in oxidation state
Each carbon atom in butane goes from an average oxidation state of approximately +2.5 to +4. Therefore, the change in oxidation state for each carbon atom is:

\[ \Delta \text{oxidation state} = 4 - 2.5 = 1.5 \]

Since there are 4 carbon atoms, the total change in oxidation state for all carbon atoms is:

\[ \text{Total change} = 4 \times 1.5 = 6 \text{ (for carbon)} \]

### Step 4: Calculate the total number of electrons transferred
In the combustion reaction, each carbon atom loses 4 electrons (from +2.5 to +4), thus:

\[ \text{Total electrons} = 4 \text{ (carbons)} \times 4 \text{ (electrons per carbon)} = 16 \text{ electrons} \]

### Step 5: Use the Gibbs free energy to find \( E_{\text{cell}}^\circ \)
Given that \( \Delta G^\circ = -2744 \, \text{kJ/mol} \), we convert this to joules:

\[ \Delta G^\circ = -2744 \times 1000 \, \text{J/mol} = -2744000 \, \text{J/mol} \]

Now, substituting into the equation:

\[ -2744000 = -n E_{\text{cell}}^\circ F \]

Where \( n = 26 \) (from the previous calculation) and \( F = 96500 \, \text{C/mol} \):

\[ -2744000 = -26 E_{\text{cell}}^\circ (96500) \]

### Step 6: Solve for \( E_{\text{cell}}^\circ \)
Rearranging gives:

\[ E_{\text{cell}}^\circ = \frac{2744000}{26 \times 96500} \]

Calculating the denominator:

\[ 26 \times 96500 = 2509000 \]

Now substituting back:

\[ E_{\text{cell}}^\circ = \frac{2744000}{2509000} \approx 1.09 \, \text{V} \]

### Step 7: Round to the nearest whole number
Rounding \( 1.09 \) to the nearest whole number gives us:

\[ \text{Final Answer: } 1 \, \text{V} \]