A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K - C 4 H 10 g -6-502 g - 4 CO 2 g - 5 H 2 O - - r G -2746 KJ - mole The E - of the cell is -A- 4-74 VB- 0-547 VC- 4-37 VD- 1-09 V

The correct option is D 1.09 VE^∘ = Δ G^∘/nF = ( 2746 × 10^3)/26 × 96500 = 1.09 V ...

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Gibb's Energy and Nernst Equation

A fuel cell d...

Question

A fuel cell develops an electrical potential from the combustion of butane at 1 bar and $298 K . C_{4} H_{10} (g) + {6.50}_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (ℓ) Δ_{r} G^{\circ} = - 2746$ KJ/mole The $E^{\circ}$ of the cell is ?

4.74 V

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0.547 V

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4.37 V

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1.09 V

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1 b a r

298 K

C_{4} H_{10} (g) + 6.5 O_{2} (g) ⟶ 4 C O_{2} (g) + 5 H_{2} O (l); △_{r} G^{\circ} = - 2746 k J / m o l

E^{\circ}

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 kJ/mol

E^{\circ}

1 a t m

298 K

C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) △ G^{\circ} = - 2746 k J / m o l

E^{\circ}

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

E^{o}

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