The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential (V) Li + (aq) + e-→Li (s) - 3.04Cl2 (g) + 2e-→2Cl - (aq) + 1.36In a galvanic cell, the two half-reactions combine to 2Li (s) + Cl2 (g) →2Li + (aq) + 2Cl - (aq) A) Calculate the cell potential of this reaction under standard reaction conditions. B) Calculate the free energy ΔG∘ of the reaction.

A) E° = 4.40 V

B) ΔG° = - 8.49 * 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li (s) + Cl₂ (g) → 2 Li⁺ (aq) + 2 Cl⁻ (aq)

We can write the corresponding half-reactions.

Cathode (reduction) : Cl₂ (g) + 2 e⁻ → 2 Cl⁻ (aq) E°red = 1.36 V

Anode (oxidation) : 2 Li (s) → 2 Li⁺ (aq) + 2 e⁻ E°red = - 3.04

A) Calculate the cell potential of this reaction under standard reaction conditions.

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

B) Calculate the free energy ΔG° of the reaction.

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = - n. F. E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol * (96468 J/V. mol) * 4.40 V = - 8.49 * 10⁵ J