Fuel Cell: Definition. Application, Examples, Solved Problems

The cell potential ( ${\mathrm{E}}_{\text{cell}}$  ) of a fuel cell depends on the redox reaction. For the ${\mathrm{H}}_2$ ${\mathrm{O}}_2$   fuel cell under standard conditions $\left({25}^{\circ }\mathrm{C},1\mathrm{M},1\mathrm{a}\mathrm{t}\mathrm{m}\right),{\mathrm{E}}_{\text{cell}}^{\circ }$  is calculated using standard reduction potentials: $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$

For ${\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}\left({\mathrm{E}}^{\circ }=+1.23\mathrm{V}\right)$  and ${\mathrm{H}}_2/{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}_{\text{cell}}^{\circ }\approx 1.23\mathrm{V}$ . Under non-standard conditions, the Nernst equation applies: $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\mathrm{l}\mathrm{o}\mathrm{g}Q$

where electrons are transferred, and involves gas pressures and ion concentrations.

Fuel cells are versatile:

• Tranportation:
  • Used in vehicles
  • Produces zero emission
• Stationary Power Generation
  • Used for continous power supply at hospitals, factories, offices, and home.
• Portable Power Devices:
  • Fuel cell can be used to power the laptops, military equipments, and mobile phones.
• Space Application:
  • NASA has already used the alkaline fuel cell in Apollo mission for electricity and drinking water.
• Industrial Use:
  • In industries, fuel cell are used in forklifts, warehouse vehicles, and material handling equipments for long operation.

Problem: Calculate the ${\mathrm{E}}_{\text{cell}}$  for a ${\mathrm{H}}_2-{\mathrm{O}}_2$ fuel cell at ${25}^{\circ }\mathrm{C}$  with ${\mathrm{P}}_{{\mathrm{H}}_2}=2\mathrm{a}\mathrm{t}\mathrm{m},{\mathrm{P}}_{{\mathrm{O}}_2}=1.5$  atm, and $\left[{\mathrm{O}\mathrm{H}}^{-}\right]=0.1\mathrm{M}$ . Given ${\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^{\circ }=+1.23\mathrm{V},{\mathrm{E}}_{{\mathrm{H}}_2/{\mathrm{H}}_2\mathrm{O}}^{\circ }=0\mathrm{V}$ .

Solution: $E_{\text{cell}}^{\circ }=1.23-0=1.23\mathrm{V}$

Reaction: $2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right),n=4$ .

$\begin{array}{c}Q=\frac{1}{P_{{\mathrm{H}}_2}^2\cdot P_{{\mathrm{O}}_2}\cdot {\left[{\mathrm{O}\mathrm{H}}^{-}\right]}^4}=\frac{1}{(2{)}^2\cdot (1.5)\cdot (0.1{)}^4}=\frac{1}{0.06}=16.67\\ E_{\text{cell}}=1.23-\frac{0.0591}{4}\mathrm{l}\mathrm{o}\mathrm{g}16.67=1.23-0.019=1.211\mathrm{V}\end{array}$

This tests Nernst equation application, a JEE Main staple.