Question

The thermodynamic efficiency of cell is given by:
(A) ${}^{\mathrm{\Delta }H}/{}_{\mathrm{\Delta }G}$
(B) ${}^{nFE}/{}_{\mathrm{\Delta }G}$
(C) ${}^{nFE}/{}_{\mathrm{\Delta }H}$
(D) nFE

Answer 1

Hint: The fuel cell thermodynamic efficiency in the overall cell reaction, is generally given as the ratio of the Gibb’s function change to the Enthalpy change. It is given as,
     $$Efficiency={\displaystyle \frac{dG}{dH}}$$

Complete answer:
Let us see the fuel cell efficiency in detail;
The study of thermodynamics of the Carnot cycle for a perfectly reversible heat engine will show that there is a limiting efficiency which would restrict the usefulness of any heat engine and this fact can be compared with the possible efficiency of a fuel cell.
Thermodynamic efficiency of a fuel cell is given as the ratio of the Gibb’s function change to the Enthalpy change. Gibb's function measures the electrical work whereas, the enthalpy change is a measure of the heating value of the fuel. This is given as;
     $$Efficiency={\displaystyle \frac{dG}{dH}}$$
Now, by Nernst equation;
The cell potential E which is associated with the electrochemical reaction taking place is defined as the decrease in the Gibbs free energy per coulomb of charge transferred during the process. This can be stated as;
     $$\mathrm{\Delta }G=-nFE$$
where,
F is Faraday's constant.
n is the number of electrons being transferred during the reaction.
Thus, for the given illustration;
The thermodynamic efficiency of a cell is given as ${}^{nFE}/{}_{\mathrm{\Delta }H}$ (negative sign may be ignored while stating equation).

Therefore, option (C) is correct.

Note:
Here, we have given the thermodynamic efficiency otherwise we have another measure too for fuel cell efficiency i.e. voltage efficiency. Voltage efficiency is the ratio of the actual voltage under operating conditions to the theoretical cell voltage.