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_2((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. `96.7%`
B. `66.2%`
C. `41.3%`
D. `85.1%`

Answer

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

### Step 1: Calculate the Gibbs Free Energy Change (ΔG°) for the reaction
The Gibbs free energy change for the reaction can be calculated using the formula:

\[ \Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_{\text{products}} - \sum \Delta G^\circ_{\text{reactants}} \]

For the reaction:
\[ \text{CH}_3\text{OH}_{(l)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2 \text{H}_2\text{O}_{(l)} \]

The standard Gibbs energies of formation are given as:
- \(\Delta G^\circ_f \text{(CH}_3\text{OH}_{(l)}) = -166.2 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f \text{(H}_2\text{O}_{(l)}) = -237.2 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f \text{(CO}_2_{(g)}) = -394.4 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f \text{(O}_2_{(g)}) = 0 \, \text{kJ/mol}\) (as it is in its elemental form)

Now, substituting these values into the equation:

\[ \Delta G^\circ_{\text{reaction}} = \left[-394.4 + 2 \times (-237.2)\right] - \left[-166.2 + 0\right] \]

Calculating the products:

\[ \Delta G^\circ_{\text{reaction}} = \left[-394.4 - 474.4\right] - \left[-166.2\right] \]
\[ = -868.8 + 166.2 = -702.6 \, \text{kJ/mol} \]

### Step 2: Use the standard enthalpy of combustion (ΔH°)
The standard enthalpy of combustion of methanol is given as:

\[ \Delta H^\circ = -726 \, \text{kJ/mol} \]

### Step 3: Calculate the efficiency of the fuel cell
The efficiency (η) of the fuel cell can be calculated using the formula:

\[ \eta = \left(\frac{\Delta G^\circ}{\Delta H^\circ}\right) \times 100 \]

Substituting the values we calculated:

\[ \eta = \left(\frac{-702.6}{-726}\right) \times 100 \]

Calculating this gives:

\[ \eta = \left(0.9667\right) \times 100 = 96.67\% \]

### Final Answer
The efficiency of the fuel cell is approximately **96.7%**.

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