Question

The thermodynamic efficiency of a cell is given by:
A. $\vartriangle \text{H/}\vartriangle \text{G}$
B. $\text{nFE/}\vartriangle \text{G}$
C. $\text{nFE/}\vartriangle \text{H}$
D. $\text{nFE}{}^\circ $

Answer 1

Hint: For this problem, we have to study the complete concept of thermodynamics and its efficiency in a cell. Then we can easily write the relationship between enthalpy and Gibbs free energy according to the thermodynamic efficiency.

Complete Step-by-step answer:
- In the given question, we have to determine the correct relationship between enthalpy and Gibbs free energy according to the thermodynamic efficiency of a cell.
- So, firstly we should know that Gibbs free energy is related to the chemical reaction and it is used to determine the work which is done by the thermodynamic constant.
- The Gibbs free energy is denoted by the $\vartriangle \text{G}$.
- Whereas enthalpy is also a property of thermodynamics which is used to measure the energy or heat in the system.
- The main difference between the two is that Gibbs free energy is used to measure the electrical work whereas enthalpy measures the heat.
- Now, the thermodynamic efficiency of the cell is defined as the ratio of the change in Gibbs free energy to the change in the enthalpy.
- The formula is given by:
$\vartriangle \text{G/}\vartriangle \text{H}$ ….. (1)
- Gibbs Free energy is also given by the relation nFE where n is the number of the electron transferred in the half cell, F is the Faraday's constant and E is the cell potential.
So, after putting the value of Gibbs free energy equation (1) will become written as:
$\dfrac{\text{nFE}}{\vartriangle \text{H}}$

Therefore, option C is the correct answer.

Note: Another method to measure fuel cell efficiency is known as voltage efficiency. Voltage efficiency is the ratio of the actual voltage to the theoretical voltage under certain conditions.