An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is F X × 1 0 3 volts, where F is the Faraday constant. The value of X is . Use : Standard Gibbs

Question

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{X}{F} \times 1 0^{3}$ volts, where $F$ is the Faraday constant. The value of $X$ is .
Use : Standard Gibbs energies of formation at 298 K are : $Δ_{f} G_{CO_{2}}^{\circ} = - 394 kJ mol^{- 1}$ ; $Δ_{f} G_{water}^{\circ} = - 237 kJ mol^{- 1}; Δ_{f} G_{butane}^{\circ} = - 18 kJ mol^{- 1}$

TestHub · Accepted Answer

​ C 4 ​ H 10 ​ ( g ) + 2 13 ​ O 2 ​ ( g ) → 4 CO 2 ​ ( g ) + 5 H 2 ​ O ( l ) Δ r ​ G ∘ = 4 Δ f ​ G CO 2 ​ ∘ ​ + 5 Δ f ​ G H 2 ​ O ∘ ​ − Δ f ​ G C 4 ​ H 10 ​ ∘ ​ = 4 × ( − 394 ) + 5 ( − 237 ) + 18 = − 2743 kJ / mol Δ r ​ G ∘ = − nFE ∘ − 2743 × 1000 = − 26 × FE ∘ E ∘ = F 105.5 ​ × 1 0 3 = 105.50 ​

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