Which one of the following is the cathodic reaction in a fuel cell?

\(O_2 (g) + 2H_2O (l) + 4e^- \longrightarrow 4OH^- (aq)\)

The correct answer is option 2. \(O_2 (g) + 2H_2O (l) + 4e^- \longrightarrow 4OH^- (aq)\).

In a fuel cell, the cathodic reaction involves the reduction of a substance. Let's break down the cathodic reaction represented by option 2:

\[ O_2 (g) + 2H_2O (l) + 4e^- \longrightarrow 4OH^- (aq) \]

This equation describes the reduction of oxygen gas (\( O_2 \)) in the presence of water (\( H_2O \)) and electrons (\( e^- \)) to form hydroxide ions (\( OH^- \)).

Here's a detailed explanation of each component of the reaction:

1. \( O_2 (g) \): This represents oxygen gas, which is the oxidant (or oxidizing agent) involved in the reduction reaction. Oxygen gas is being reduced by gaining electrons.

2. \( 2H_2O (l) \): This represents water molecules, which serve as both the solvent and the source of protons (\( H^+ \)). In this reaction, water molecules are being consumed.

3. \( 4e^- \): These are four electrons that are being transferred to oxygen gas. In a fuel cell, electrons flow through an external circuit from the anode to the cathode, and this flow of electrons is what generates electrical energy.

4. \( 4OH^- (aq) \): These are four hydroxide ions that are formed as a result of the reduction of oxygen gas. The hydroxide ions are produced in the aqueous (liquid) phase of the fuel cell.

Overall, the cathodic reaction described by option 2 involves the reduction of oxygen gas to hydroxide ions in the presence of water and electrons. This reduction reaction occurs at the cathode of the fuel cell, where oxygen gas from the surrounding environment is consumed, and electrical energy is generated as a result of the flow of electrons through an external circuit.