A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K
`C_4H_(10)(g)+6.5O_2(g)to4CO_2(g)+5H_2O(l), /_\_rG^(@)=-2746kJ//mol`
what is `E^(@)` of a cell?
(a)4.74V
(b)0.547V
(c)4.37V
(d)1.09V

Options

A. 4.74V
B. 0.547V
C. 4.37V
D. 1.09V

Answer

To find the standard cell potential \( E^\circ \) of the fuel cell from the given reaction, we can follow these steps:

### Step 1: Write down the reaction and identify the change in oxidation states.
The reaction provided is:
\[ C_4H_{10}(g) + 6.5O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l) \]

In this reaction:
- The oxidation state of carbon in butane (\( C_4H_{10} \)) is -3.
- The oxidation state of carbon in carbon dioxide (\( CO_2 \)) is +4.
- The oxidation state of oxygen in \( O_2 \) is 0 and in \( H_2O \) is -2.

### Step 2: Calculate the number of electrons transferred (n).
Each carbon atom undergoes an increase of 7 in oxidation state (from -3 to +4). Since there are 4 carbon atoms, the total change for carbon is:
\[ 4 \times 7 = 28 \text{ electrons} \]

However, we also need to consider the oxygen. The \( O_2 \) is reduced to \( H_2O \). Each \( O_2 \) molecule (6.5 moles) will gain 4 electrons (2 electrons per oxygen atom). Thus, for 6.5 moles:
\[ 6.5 \times 4 = 26 \text{ electrons} \]

### Step 3: Use the Gibbs free energy change to find the cell potential.
The relationship between Gibbs free energy change and cell potential is given by:
\[ \Delta_r G^\circ = -nFE^\circ \]
Where:
- \( \Delta_r G^\circ = -2746 \, \text{kJ/mol} = -2746000 \, \text{J/mol} \)
- \( n = 26 \) (from our previous calculation)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)

### Step 4: Rearranging the equation to solve for \( E^\circ \).
Rearranging the equation gives:
\[ E^\circ = -\frac{\Delta_r G^\circ}{nF} \]

Substituting the values:
\[ E^\circ = -\frac{-2746000 \, \text{J/mol}}{26 \times 96500 \, \text{C/mol}} \]

### Step 5: Calculate \( E^\circ \).
Calculating the denominator:
\[ 26 \times 96500 = 2509000 \, \text{C/mol} \]

Now substituting back:
\[ E^\circ = \frac{2746000}{2509000} \approx 1.09 \, \text{V} \]

### Final Answer:
The standard cell potential \( E^\circ \) of the cell is:
**(d) 1.09 V**

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