The efficiency of a fuel cell si `80%` and the standard heat of reaction is `-300kJ`. The reaction involves two electrons in redox change. `E^(c-)` for the cell is
(a)`1.24V`
(b)`2.48V`
(c)`0V`
(d)`0.62V`

Options

A. `1.24V`
B. `2.48V`
C. `0V`
D. `0.62V`

Explanation

To solve the problem, we need to find the standard cell potential \( E^\circ \) for a fuel cell given its efficiency, standard heat of reaction, and the number of electrons involved in the redox change. Here’s how to approach the problem step by step:

### Step 1: Understand the relationship between efficiency, Gibbs free energy, and enthalpy.
The efficiency of the fuel cell can be expressed as:
\[ \text{Efficiency} = \frac{\Delta G^\circ}{\Delta H^\circ} \]
Given that the efficiency is 80%, we can write:
\[ 0.80 = \frac{\Delta G^\circ}{\Delta H^\circ} \]

### Step 2: Convert the standard heat of reaction to joules.
The standard heat of reaction is given as \(-300 \text{ kJ}\). We need to convert this to joules:
\[ \Delta H^\circ = -300 \times 10^3 \text{ J} = -300000 \text{ J} \]

### Step 3: Calculate \(\Delta G^\circ\).
Using the efficiency equation:
\[ \Delta G^\circ = 0.80 \times \Delta H^\circ \]
Substituting the value of \(\Delta H^\circ\):
\[ \Delta G^\circ = 0.80 \times (-300000) = -240000 \text{ J} \]

### Step 4: Relate Gibbs free energy to cell potential.
The relationship between Gibbs free energy, cell potential, and the number of electrons is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( n \) = number of moles of electrons (given as 2)
- \( F \) = Faraday's constant (\( 96500 \text{ C/mol} \))
- \( E^\circ \) = standard cell potential

### Step 5: Rearrange the equation to find \( E^\circ \).
Rearranging the equation gives:
\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]
Substituting the values:
\[ E^\circ = -\frac{-240000}{2 \times 96500} \]

### Step 6: Calculate \( E^\circ \).
Calculating the above expression:
\[ E^\circ = \frac{240000}{193000} \approx 1.24 \text{ V} \]

### Conclusion:
The standard cell potential \( E^\circ \) for the fuel cell is approximately \( 1.24 \text{ V} \). Therefore, the correct answer is:
**(a) 1.24 V**