In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is C H 3 O H l + 3 2 O 2 g → C O 2 g + 2 H 2 O l At 298 K standard Gibb’s energies of formation for C H 3 O

Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
$C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
At 298 K standard Gibb’s energies of formation for $C H_{3} O H (l), H_{2} O (l)$ and $C O_{2} (g)$ are -166.2, -237.2 and -394.4 $k J {m o l}^{- 1}$ respectively. If standard enthalphy of combustion of methanol is -726 ${kJmol}^{- 1}$ , efficiency of the fuel cell will be

TestHub · Accepted Answer

C H 3 O H l + 3 2 O 2 g → C O 2 g + 2 H 2 O l Δ H = - 726 k J m o l - 1 Also Δ G f o C H 3 O H I = - 166.2 k J m o l - 1 Δ G f o H 2 O I = - 166.2 k J m o l - 1 Δ G f o C O 2 I = - 394.4 k J m o l - 1 Δ G f = ∑ Δ G f o p r o d u c t s - ∑ Δ G f o r e a c t a n t s = - 394.4 - 2 237.2 + 166.2 = 702.6 k J m o l - 1 now efficiency of fuel cell = Δ G Δ H × 100 = 702.6 726 × 100 = 97 %

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