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

b. acid is the correct answer @jayxboof all social media

acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.

c. is the answer

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