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The efficiency of a fuel cell is given by
A
$\dfrac{\Delta G}{\Delta S}$
B
$\dfrac{\Delta G}{\Delta H}$
C
$\dfrac{\Delta S}{\Delta G}$
D
$\dfrac{\Delta H}{\Delta G}$
Detailed Solution
A fuel cell converts the chemical energy of a fuel directly into electrical energy.
The maximum useful (electrical) work obtainable from the cell reaction is equal to the decrease in Gibbs energy: $W_{max} = \Delta G = -nFE_{cell}$
The total heat energy that the same reaction would release on direct combustion is the enthalpy change $\Delta H$.
Efficiency = useful work obtained / total heat energy of the reaction
Efficiency of a fuel cell: $\eta = \dfrac{\Delta G}{\Delta H}$
For example, for the $H_2$-$O_2$ fuel cell the theoretical efficiency is $\dfrac{237}{286} \times 100$, about 83 percent.
The maximum useful (electrical) work obtainable from the cell reaction is equal to the decrease in Gibbs energy: $W_{max} = \Delta G = -nFE_{cell}$
The total heat energy that the same reaction would release on direct combustion is the enthalpy change $\Delta H$.
Efficiency = useful work obtained / total heat energy of the reaction
Efficiency of a fuel cell: $\eta = \dfrac{\Delta G}{\Delta H}$
For example, for the $H_2$-$O_2$ fuel cell the theoretical efficiency is $\dfrac{237}{286} \times 100$, about 83 percent.