A fuel cell involves combustion of the butane at at 1 atm and 298 K C 4 H 10 ( g ) + 13 2 O 2 ( g ) → 4 C O 2 ( g ) + 5 H 2 O ( l ) Δ G ° = - 2746 kJ/mole The value of E c e l l o is nearly?

Question

A fuel cell involves combustion of the butane at at 1 atm and 298 K C 4 H 10 ( g ) + 13 2 O 2 ( g ) → 4 C O 2 ( g ) + 5 H 2 O ( l ) Δ G ° = - 2746 kJ/mole The value of E c e l l o is nearly?, 0.8 V, 1 V, 1.2 V, 1.4 V

MARKS App · Accepted Answer

C 4 - 10 H 10 ( g ) + 13 2 O 2 ( g ) → 4 C O 2 + 4 ( g ) + 5 H 2 O ( l ) Here total change in oxidation number of C-atoms = + 16 - ( - 10 ) = + 26 It means total number of e - involved = 26 As E c e l l o - - Δ G ° n F = - ( - 2746 ) × 1000 26 × 96500 = + 1.09 V ≈ 1.0 V

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