The efficiency of a fuel cell is given by:

\frac{Δ S}{Δ G}

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\frac{Δ H}{Δ G}

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\frac{Δ G}{Δ S}

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\frac{Δ G}{Δ H}

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Solution

The correct option is B $\frac{Δ G}{Δ H}$
The fuel cell thermodynamic efficiency is given by the ratio of the Gibbs function change to the Enthalpy change in the overall cell reaction. The Gibbs function change measures the electrical work and the enthalpy change is a measure of the heating value of the fuel.
Efficiency = $\frac{d G}{d H}$ $= \frac{Δ G}{Δ H}$

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d (Δ G) = - (Δ S_{reaction}) . d T

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Δ S

Δ G = Δ H - T Δ S

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For $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ .The correct set of thermodynamic parameter is

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