A fuel cell involves combustion of the butane at at 1 atm and 298 K mathrmC4 mathrmH10( mathrm~g)+(13/2) mathrmO2( mathrm~g) arrow 4 mathrmCO2( mathrm~g)+5 mathrmH2 mathrmO(ℓ) Δ G° =-2744 kJ/mole The value of Ec e l lo Report your answer by rounding it upto nearest whole number.

Question

A fuel cell involves combustion of the butane at at 1 atm and 298 K mathrmC4 mathrmH10( mathrm~g)+(13/2) mathrmO2( mathrm~g) arrow 4 mathrmCO2( mathrm~g)+5 mathrmH2 mathrmO(ℓ) Δ G° =-2744 kJ/mole The value of Ec e l lo Report your answer by rounding it upto nearest whole number. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C4H10(g)+(13/2)O2(g) arrow 4(CO)2(g)+5H2O(l) Here total change in oxidation number of C-atoms =+16-(.-10.)=+26 It means total number of e- involved = 26 As Ec e l lo=(- Δ G ° /n F)=(- (. - 2744 .) × 1000/26 × 96500) =+1.09V ≈ 1.0V

Q. A fuel cell involves combustion of the butane at at 1 atm and 298 K C4​H10​( g)+213​O2​( g)→4CO2​( g)+5H2​O(ℓ) ΔG∘=−2744 kJ/mole The value of Ecello​ Report your answer by rounding it upto nearest whole number.

C4​H10​(g)+213​O2​(g)→4(CO)2​(g)+5H2​O(l) Here total change in oxidation number of C-atoms =+16−(−10)=+26 It means total number of e− involved = 26 As Ecello​=nF−ΔG∘​=26×96500−(−2744)×1000​ =+1.09V ≈1.0V

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

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

$Δ G^{\circ} = - 2744$ kJ/mole

The value of $E_{ce ll}^{o}$

Report your answer by rounding it upto nearest whole number.