In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is `:`
`CH_(3)OH_((l))+(3)/(2)O_(2(g))rarr CO_((g))+2H_(2)O_((l))`
At `298K` standard Gibb's energies of formation for `CH_(3)OH(l), H_(2)O(l)` and `CO_(2)(g)` are `-166.2,-237.2` and `-394.4kJ mol^(-1)` respectively. If standard enthalpy of combustion of methanol is `-726kJ mol^(-1)`, efficiency of the fuel cell will be `:`

Options

A. `80%`
B. `87%`
C. `90%`
D. `97%`

Explanation

To calculate the efficiency of the fuel cell using methanol and oxygen, we will follow these steps:

### Step 1: Write the Reaction
The reaction for the fuel cell is given as:
\[ \text{CH}_3\text{OH}_{(l)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{(g)} + 2 \text{H}_2\text{O}_{(l)} \]

### Step 2: Identify Given Data
- Standard Gibbs energies of formation:
- \( \Delta G_f^\circ (\text{CH}_3\text{OH}_{(l)}) = -166.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{H}_2\text{O}_{(l)}) = -237.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{CO}_{(g)}) = -394.4 \, \text{kJ/mol} \)
- Standard enthalpy of combustion of methanol:
- \( \Delta H_{comb} = -726 \, \text{kJ/mol} \)

### Step 3: Calculate Gibbs Free Energy of the Reaction
Using the formula for Gibbs free energy change:
\[ \Delta G = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} \]
Substituting the values:
\[ \Delta G = \left[ \Delta G_f^\circ (\text{CO}_{(g)}) + 2 \cdot \Delta G_f^\circ (\text{H}_2\text{O}_{(l)}) \right] - \left[ \Delta G_f^\circ (\text{CH}_3\text{OH}_{(l)}) + \frac{3}{2} \cdot \Delta G_f^\circ (\text{O}_{2(g)}) \right] \]
Since the Gibbs energy of formation for \( O_2 \) in its standard state is zero:
\[ \Delta G = \left[ -394.4 + 2 \cdot (-237.2) \right] - \left[ -166.2 + 0 \right] \]
Calculating the values:
\[ \Delta G = \left[ -394.4 - 474.4 \right] - \left[ -166.2 \right] \]
\[ \Delta G = -868.8 + 166.2 = -702.6 \, \text{kJ/mol} \]

### Step 4: Calculate Efficiency of the Fuel Cell
Efficiency (\( \eta \)) is calculated using the formula:
\[ \eta = \left( \frac{\Delta G}{|\Delta H_{comb}|} \right) \times 100 \]
Substituting the values:
\[ \eta = \left( \frac{-702.6}{726} \right) \times 100 \]
Calculating the efficiency:
\[ \eta = \left( -0.966 \right) \times 100 \approx 96.6\% \]

### Final Answer
The efficiency of the fuel cell is approximately **96.6%**.

---