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_4H_10(g)+13/2O_2(g)→4CO_2(g)+5H_2O(l) ΔG°=-2746 kJ/mol What is E° of a cell?

MARKS Content Team · Accepted Answer

Correct answer: (C) +1.09 V. In the reaction C_4H_10g-10+10+13/2O_2g→4CO_2g+4-4+5H_2Ol Change in oxidation number of carbon =+16-(-10)=+26 So number of electrons involved in cell process will be 26. E°=(-ΔG°)/nF=((-2746)×1000)/(26×96500) =+1.09 V

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?

Solution

More Electrochemistry Questions

A fuel cell involves combustion of the butane at 1 atm and 298 K, C4H10(g)+132O2(g)→4CO2(g)+5H2O(l) ΔG°=-2746 kJ/mol What is E° of a cell?

Solution

More Electrochemistry Questions

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?