Question

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.

Answer 1

Answer:

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

Answer 2

The electrochemical cell standard cell potential has been 4.4 V. The Gibbs free energy of the reaction is $-8.49\;\times\;10^5$ J.

What is an electrochemical cell?

The electrochemical cell allows the interconversion of the electrical and the chemical energy.

The cell potential for the reaction has been given  as:

$\rm E^\circ=E_{reduction}-E_{oxidation}$

Substituting the values of reduction and oxidation half cell:

$\rm E^\circ=1.36 \;V - (-3.04 \;V) \\E^\circ=4.4 \;V$

The cell potential of the given galvanic cell has been 4.4 V.

The Gibbs free energy of the reaction has been given as:

$\Delta \rm G^\circ=-nFE^\circ$

Substituting the value of electrons transfer per mole (n), Faraday's constant (F), and the cell potential:

$\rm \Delta G=-2\;\times\;96500\;\times\;4.4\\\Delta G=-8.49\;\times\;10^{5}\;J$

The value of Gibbs free energy for the given galvanic cell is $-8.49\;\times\;10^5$ J.

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