In a fuel cell based on the oxidation of methane, the equation for the anode half reaction is CH₄(g) + 2H₂O(l) → CO₂(g) + 8H⁺(aq) + 8e⁻ The corresponding equation for the half reaction at the cathode is: A - VCE - SSCE Chemistry - Question 5 - 2004 - Paper 1

Let's analyze each option:

• A. 2H₂O(l) + 4e⁻ → 4H⁺(aq) + O₂(g): This does not correctly represent the reduction process, as it suggests that water is broken down instead of forming it.
• B. 4H⁺(aq) + O₂(g) → 2H₂O(l): This option correctly shows the reduction process where protons and oxygen combine to form water, which matches the requirement at the cathode.
• C. 2H₂O(l) + 4H⁺(aq) + O₂(g) + 4e⁻: This does not represent a complete reaction and does not demonstrate the formation of water.
• D. 4H⁺(aq) + O₂(g) → 2H₂O(l): Similar to option B, this correctly shows the reduction of protons and oxygen to form water.

Considering the proper stoichiometry and reaction direction, both B and D appear to be reductions, but option D gives a more concise representation.

After careful evaluation, the correct equation for the half reaction at the cathode is: D. 4H⁺(aq) + O₂(g) → 2H₂O(l). This corresponds perfectly to the reduction scenario in the context of the fuel cell process.