Use tabulated electrode potentials to calculate ∆G°_rxn for each reaction at 25 °C. b. Br₂(l) + 2 Cl^–(aq) → 2 Br^–(aq) + Cl₂(g) c. MnO₂(s) + 4 H⁺(aq) + Cu(s) → Mn²⁺(aq) + 2 H₂O(l) + Cu²⁺(aq)

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Identify the half-reactions involved in the overall reaction. For the reduction half-reaction: MnO2(s) + 4 H+(aq) + 2 e- → Mn2+(aq) + 2 H2O(l). For the oxidation half-reaction: Cu(s) → Cu2+(aq) + 2 e-.

Look up the standard electrode potentials (E°) for each half-reaction from the standard electrode potential table. You will find E° for MnO2/Mn2+ and Cu/Cu2+.

Calculate the standard cell potential (E°cell) for the overall reaction using the formula: E°cell = E°cathode - E°anode, where MnO2/Mn2+ is the cathode (reduction) and Cu/Cu2+ is the anode (oxidation).

Use the relationship between the Gibbs free energy change (∆Gr°) and the cell potential to calculate ∆Gr°xn. The formula to use is: ∆Gr° = -nFE°cell, where n is the number of moles of electrons transferred in the balanced equation (which is 2 in this case), and F is the Faraday constant (approximately 96485 C/mol).

Ensure all units are consistent, particularly that temperature is in Kelvin if needed, and calculate ∆Gr°xn for the reaction at 25 °C (298 K).

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주요 개념

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Electrode Potentials

Electrode potentials, measured in volts, indicate the tendency of a chemical species to be reduced. Standard electrode potentials (E°) are measured under standard conditions (1 M concentration, 1 atm pressure, and 25 °C). These values are crucial for predicting the direction of redox reactions and calculating the Gibbs free energy change (∆Gr°) for reactions.

Gibbs Free Energy (∆Gr°)

Gibbs free energy change (∆Gr°) is a thermodynamic quantity that indicates the spontaneity of a reaction at standard conditions. It can be calculated using the equation ∆Gr° = -nFE°, where n is the number of moles of electrons transferred, F is Faraday's constant, and E° is the cell potential. A negative ∆Gr° indicates a spontaneous reaction.

Redox Reactions

Redox reactions involve the transfer of electrons between species, where one species is oxidized (loses electrons) and another is reduced (gains electrons). In the given reaction, manganese dioxide (MnO2) is reduced to manganese ions (Mn2+), while copper (Cu) is oxidized to copper ions (Cu2+). Understanding the oxidation states and half-reactions is essential for calculating overall cell potentials.

Use tabulated electrode potentials to calculate ∆G°rxn for each reaction at 25 °C. b. Br2(l) + 2 Cl–(aq) → 2 Br–(aq) + Cl2(g) c. MnO2(s) + 4 H+(aq) + Cu(s) → Mn2+(aq) + 2 H2O(l) + Cu2+(aq)