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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 $1 b a r$ and $298 K$ . $C_{4} H_{10} (g) + 6.5 O_{2} (g) ⟶ 4 C O_{2} (g) + 5 H_{2} O (l); △_{r} G^{\circ} = - 2746 k J / m o l$
What is $E^{\circ}$ of a cell?

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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Solution

The correct option is D $1.09 V$
$Δ G = - n F E_{c e l l}$
where $n = 6.5 \times 4$
$2746000 = 6.5 \times 4 \times 96500 E_{c e l l}$
$E_{c e l l} = 1.09 V$

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Similar questions

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

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}

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}

C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 C O_{2} (g) + 5 H_{2} O (l); Δ H = - 2878 k J

Δ H

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