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 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

Step-by-step solution

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 %