Final answer:
The balanced equations for the electrode reactions are:
Anode (oxidation): Fe(s) → Fe²⁺(aq) + 2e⁻
Cathode (reduction): 2H₂O(l) + O₂(g) + 4e⁻ → 4OH⁻(aq)
The overall cell reaction is: 4Fe(s) + O₂(g) + 4H₂O(l) → 4Fe²⁺(aq) + 4OH⁻(aq)
Step-by-step explanation:
In the galvanic cell setup described, iron (Fe) acts as the anode and undergoes oxidation, releasing electrons to form Fe²⁺ ions in the solution according to the equation: Fe(s) → Fe²⁺(aq) + 2e⁻. Meanwhile, at the cathode, oxygen (O₂) from the gas phase, along with water (H₂O) and electrons, forms hydroxide ions (OH⁻) in an overall reduction reaction: 2H₂O(l) + O₂(g) + 4e⁻ → 4OH⁻(aq).
The overall cell reaction is obtained by adding the anode and cathode half-reactions. The balanced overall cell reaction is: 4Fe(s) + O₂(g) + 4H₂O(l) → 4Fe²⁺(aq) + 4OH⁻(aq). This equation represents the combination of the oxidation of iron and the reduction of oxygen and water to form Fe²⁺ ions and hydroxide ions, respectively.
In the cell diagram, the anode (Fe(s)) and cathode (Pt(s)) should be clearly labeled. Electrons flow from the anode to the cathode through the external circuit, while Fe²⁺ ions move from the anode into the solution. Simultaneously, oxygen gas reacts at the cathode, contributing to the formation of OH⁻ ions. This cell setup and the associated reactions illustrate the electron and ion flow in a galvanic cell configuration, showcasing the redox reactions occurring at the electrodes.