For a `H_(2) - O_(2)` fuel cell , the theoretical voltage has been found to be 1.23 V and `Delta H` to be `-285 kJ mol^(-1)` . The efficiency of the fuel cell is

Options

A. `76%`
B. `83 %`
C. `89%`
D. `72 %`

Explanation

To find the efficiency of the `H2 - O2` fuel cell, we will follow these steps:

### Step 1: Understand the relationship between ΔG and ΔH
The efficiency of a fuel cell can be calculated using the formula:
\[ \text{Efficiency} = \left( \frac{|\Delta G|}{|\Delta H|} \right) \times 100 \]
where:
- \( \Delta G \) is the change in Gibbs free energy.
- \( \Delta H \) is the change in enthalpy.

### Step 2: Calculate ΔG
We know that:
\[ \Delta G = -nFE \]
where:
- \( n \) is the number of moles of electrons transferred.
- \( F \) is Faraday's constant (approximately \( 96500 \, C/mol \)).
- \( E \) is the theoretical voltage of the cell.

From the problem, we have:
- The theoretical voltage \( E = 1.23 \, V \).
- For the reaction \( 2H_2 + O_2 \rightarrow 2H_2O \), 2 moles of electrons are transferred for every 2 moles of \( H_2 \) consumed, so \( n = 2 \).

Now we can calculate \( \Delta G \):
\[ \Delta G = -nFE = -2 \times 96500 \, C/mol \times 1.23 \, V \]
Calculating this gives:
\[ \Delta G = -237390 \, J/mol = -237.39 \, kJ/mol \]

### Step 3: Substitute values into the efficiency formula
Now we have:
- \( |\Delta G| = 237.39 \, kJ/mol \)
- \( |\Delta H| = 285 \, kJ/mol \)

Substituting these values into the efficiency formula:
\[ \text{Efficiency} = \left( \frac{237.39}{285} \right) \times 100 \]

### Step 4: Calculate the efficiency
Calculating the above expression:
\[ \text{Efficiency} \approx 0.832 \times 100 \approx 83.2\% \]

### Final Answer
Thus, the efficiency of the `H2 - O2` fuel cell is approximately **83%**.

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