A fuel cell develops an electrical potential form the combustion of butane at 1 bar and 298 KC4H10g-6-5O2g- 4CO2g-5H2Ol- - rG-2746 kJ-mol What is Eo in volts of cell- Give your answer upto two decimal only

To find n we break the cell reaction into two half cell reduction. Anode: 13H_2O(l)→ 6.5O_2(g)+26H^+(aq.)+26e^ Cathode: 4CO_2(g)+26H^+(aq.)+26e^ → C_4H_10(g)+8 ...

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Standard XII

Chemistry

Introduction

A fuel cell d...

Question

A fuel cell develops an electrical potential form the combustion of butane at 1 bar and 298 K
$C_{4} H_{10} (g) + 6.5 O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) : Δ_{r} G^{\circ} = - 2746 kJ/mol$
What is $E^{o}$ (in volts) of cell? ( Give your answer upto two decimal only)

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Solution

To find n we break the cell reaction into two half cell reduction.
Anode:
$13 H_{2} O (l) \to 6.5 O_{2} (g) + 26 H^{+} (a q .) + 26 e^{-}$
Cathode:
$4 C O_{2} (g) + 26 H^{+} (a q .) + 26 e^{-} \to C_{4} H_{10} (g) + 8 H_{2} O (l)$
$∴ Δ G = - n F E^{\circ}$
$E^{\circ} = - \frac{Δ G^{\circ}}{n F} = - \frac{(- 2746) \times 1000}{26 \times 96500} = 1.09 V$

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A fuel cell develops an electrical potential form the combustion of butane at 1 bar and 298 K C4H10(g)+6.5O2(g)→4CO2(g)+5H2O(l): ΔrG∘=−2746 kJ/mol What is Eo (in volts) of cell? ( Give your answer upto two decimal only)

To find n we break the cell reaction into two half cell reduction. Anode: 13H2O(l)→6.5O2(g)+26H+(aq.)+26e− Cathode: 4CO2(g)+26H+(aq.)+26e−→C4H10(g)+8H2O(l) ∴ΔG=−nFE∘ E∘=−ΔG∘nF=−(−2746)×100026×96500=1.09V

A fuel cell develops an electrical potential form the combustion of butane at 1 bar and 298 K
$C_{4} H_{10} (g) + 6.5 O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) : Δ_{r} G^{\circ} = - 2746 kJ/mol$
What is $E^{o}$ (in volts) of cell? ( Give your answer upto two decimal only)