Final answer:
The electrochemical potential and free-energy change for the given conditions can be determined using the reduction half-reaction for MnO₂ and the Nernst equation. The equilibrium concentration of Mn²⁺ can be calculated based on the equilibrium condition of zero cell potential.
Step-by-step explanation:
To determine the electrochemical potential and the free-energy change for a system where a 0.1 M Mn²⁺ solution at pH 1 is shaken in air at 0.2 bar pressure of O₂, resulting in the production of MnO₂ (s), we must first look at the half-reactions for each species involved.
We consider the reduction half-reaction for manganese dioxide as given:
- MnO₂(s) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(l)
The cell potential can then be estimated using the Nernst equation to account for the non-standard conditions of Mn²⁺ concentration, pH, and O₂ pressure. The equilibrium concentration of Mn⁻²⁺ in the presence of excess MnO₂ and under the specified pH and O₂ pressure can be determined based on the requirement that the electrochemical potential of this reaction under equilibrium is zero.
To find the free-energy change, ΔG°, we use the equation:
ΔG° = -nFE°, where
- n is the number of electrons transferred in the reaction,
- F is the Faraday constant,
- E° is the standard cell potential.
Considering the electrochemical cell potential at equilibrium is zero, we can then calculate the concentration at which the system will be at equilibrium.