A fuel cell involves combustion of the butane at 1 atm and 298 K, C4H10(.g.)+(13/2)O2(.g.) arrow 4CO2(.g.)+5H2O(.l.) Δ G° =-2746 kJ/mol What is E° of a cell?

Question

A fuel cell involves combustion of the butane at 1 atm and 298 K, C4H10(.g.)+(13/2)O2(.g.) arrow 4CO2(.g.)+5H2O(.l.) Δ G° =-2746 kJ/mol What is E° of a cell? (A) +4.74 V (B) +0.547 V (C) +1.09 V (D) +4.37 V

Tardigrade Admin · Accepted Answer

In the reaction overset- 10 + 10C4 H10 (g)+(13/2)O2(g) arrow overset+ 4 - 44 C O2 (g)+5H2O(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

Q. A fuel cell involves combustion of the butane at 1 atm and 298 K, C4​H10​(g)+213​O2​(g)→4CO2​(g)+5H2​O(l) ΔG∘=−2746kJ/mol What is E∘ of a cell?

+4.74 V

+0.547 V

+1.09 V

+4.37 V

In the reaction C4​H10​(g)−10+10​+213​O2​(g)→4CO2​(g)+4−4​+5H2​O(l) Change in oxidation number of carbon =+16−(−10)=+26 So number of electrons involved in cell process will be 26. E∘=nF−ΔG∘​=26×96500(−2746)×1000​ =+1.09V

Q. A fuel cell involves combustion of the butane at 1 atm and 298 K,

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$

$Δ G^{\circ} = - 2746 k J / m o l$

What is $E^{\circ}$ of a cell?