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At 298 \mathrm{~K} and \mathrm{1} bar pressure, what is the \mathrm{E_{\text {cell }}^0} of a fuel well which develops an electric potential from combustion of propane if \Delta_\gamma G^{\circ}=-1382 \mathrm{~kJ} / \mathrm{mol}
\mathrm{C}_3 \mathrm{H}_8(g)+5 \mathrm{O}_2(g) \rightarrow 3 \mathrm{CO}_2(g)+4 \mathrm{H}_2 \mathrm{O}(l)

Option: 1

\mathrm{0.647\ V}


Option: 2

\mathrm{4.74\ V}


Option: 3

\mathrm{0.71\ V}


Option: 4

\mathrm{3.26\ V}


Answers (1)

best_answer

To find \mathrm{n}, we break the cell reaction into two half cell reactions
\mathrm{Anode: 10 \mathrm{H}_2 \mathrm{O} \rightarrow 5 \mathrm{O}_2+2 \mathrm{OH}^{+}+2 \mathrm{OC}^{-}}\\ \mathrm{ Cathode: \quad 3 \mathrm{CO}_2+2 \mathrm{OH}^{+}+2 \mathrm{Oe}^{-} \rightarrow \mathrm{C}_3 \mathrm{H}_8+6 \mathrm{H}_2 \mathrm{O}}
\mathrm{\begin{aligned} E^{\circ}=\frac{-\Delta G^0}{n F} & =\frac{-1382 \times 1000}{20 \times 96500} \\ & =0.71 \mathrm{~V} \end{aligned}}

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