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A fuel Cell develops an electrical potential from the combustion of butane at 1 bar and 298 k
$C_{4} H_{10 (g)} + 6.5 O_{2 (g)} \to 4 C O_{2 (g)} + 5 H_{2} O_{(l)}; Δ G^{0} = - 2746 k J / m o l$

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$1.09 V$

$4.74 V$

$4.37 V$

$0.547 V$

answer is D.

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Detailed Solution

$\begin{array}{l} Δ G^{o} = - n F {E_{c e l l}}^{o} \\ - 274600 = - 26 \times 96, 500 \times {E_{c e l l}}^{o} \end{array}$

$E_{cell}^{0}$ = 1.09V

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