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In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is:
${CH}_{3} OH (l) + \frac{3}{2} O_{2} (g) ⟶ {CO}_{2} (g) + 2 H_{2} O (l)$
At 298 K standard Gibbs energies of formation for ${CH}_{2} OH (l), H_{2} O (l)$ and ${CO}_{2} (g)$ are $- 166.2, - 237 . 2$ and $- 394.4 kJ {mol}^{- 1}$ respectively. If standard enthalpy of combustion of methanol is $- 726 kJ {mol}^{- 1}$ , efficiency of the fuel cell will be :

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87%

90%

97%

80%

answer is C.

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Detailed Solution

$\begin{array}{l} Δ G_{R x^{n}}^{\circ} = - 394.4 - 237.2 \times 2 + 166.2 \\ = - 702.6 kJ / mol \\ η = \frac{| Δ G |}{| Δ H |} \times 100 = \frac{702.6}{726} \times 100 = 97 % \end{array}$

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