Question

The efficiency of the fuel cell is given by –
A.${\displaystyle \frac{\mathrm{\Delta }S}{\mathrm{\Delta }G}}$
B.${\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }S}}$
C.${\displaystyle \frac{\mathrm{\Delta }H}{\mathrm{\Delta }G}}$
D.${\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }H}}$

Answer 1

Hint: The energy efficiency of the system or device that converts energy is measured by the ratio of amount of useful energy put out by the system to the total amount of energy that is put in or by the useful output energy as percentage of the total input energy.

Complete step by step solution:
A fuel cell is an electrochemical cell that converts the chemical energy of the fuel and an oxidizing agent into electricity through the pair of redox reactions. Fuel cells are different from most batteries requiring a continuous source of fuel and oxygen to sustain the chemical reaction whereas in the battery the chemical energy usually comes from the metals and their ions or oxides that are already present in the battery, except the flow of batteries. There are many types of fuel cells, but they all consist of an anode, a cathode, and an electrolyte that allows ions, often positively charged hydrogen ions (protons), to move between the two sides of the fuel cell. At the anode a catalyst causes the fuel to undergo oxidation reactions that generate ions (often positively charged hydrogen ions) and electrons.
Here, $\mathrm{\Delta }G$ is the Gibbs free energy which measures the electrical work and $\mathrm{\Delta }H$ is the enthalpy change which is the measure of heating value of the fuel.
Efficiency of fuel cell $={\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }H}}$
For the hydrogen oxygen reaction –
$\Rightarrow$ $dH=-68,317cal/g$ mol of $H_2$ and
$\Rightarrow$ $dG=-56,690cal/g$ mol of $H_2$
The efficiency of the ideal fuel cell is therefore –
$\Rightarrow$ Efficiency: $\left({\displaystyle \frac{56,690}{68,317}}\right)=83\mathrm{\%}$

Hence, the correct option is (D).

Note: Another measure of fuel cell efficiency is known as voltage cell efficiency and is the ratio of actual voltage under operating conditions to the theoretical cell voltage.
Voltage efficiency $={\displaystyle \frac{V_A}{1.23}}$