Final answer:
The half reaction at the cathode in the H₂/O₂ fuel cell is O₂+4H+ + 4e- → 2H₂O. However, the fuel cell achieves a voltage of only about 0.9 V instead of the expected 1.23 V under standard conditions due to the slow reduction of O₂(g) at the cathode. Research is focused on improving the catalyst for the reduction of O₂(g) in fuel cells. The correct answer in option 3.
Step-by-step explanation:
The half reaction at the cathode in the H₂/O₂ fuel cell is:
The overall electricity-producing reaction in the fuel cell is 2H₂ + O₂ → 2H₂O.
However, it's important to note that in practice, the fuel cell achieves a voltage of only about 0.9 V, instead of the expected 1.23 V under standard conditions. This is mainly due to the slow four-electron reduction of O₂(g) at the cathode, limiting the current that can be achieved. Major research programs are focused on improving the catalyst for the reduction of O₂(g) in fuel cells.