A fuel cell involves combustion of the butane at 1 atm and 298 K,
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$
$Δ G ° = - 2746 k J / m o l$
What is $E °$ of a cell?

Question

A fuel cell involves combustion of the butane 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 k J / m o l What is E ° of a cell?

TestHub · Accepted Answer

In the reaction C 4 H 10 g - 10 + 10 + 13 2 O 2 g → 4 C O 2 g + 4 - 4 + 5 H 2 O l Change in oxidation number of carbon = + 16 - ( - 10 ) = + 26 So number of electrons involved in cell process will be 26. E ° = - Δ G ° n F = ( - 2746 ) × 1000 26 × 96500 = + 1.09 V

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