Cell Potential and Free Energy of a Lithium–Chlorine Cell G = n F E cell Part A

Question

Cell Potential and Free Energy of a Lithium–Chlorine Cell

G=nFEcell

Part A

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

Express your answer with the appropriate units.

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Part B

Calculate the free energy G of the reaction.

Express your answer with the appropriate units.

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Part C

What can be said about the spontaneity of this reaction?

The reaction is

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Cell Potential and Free Energy of a Lithium–Chlorine Cell

In thermodynamics, we determine the spontaneity of a reaction by the sign of G. In electrochemistry, spontaneity is determined by the sign of Ecell. The values of G and Ecell are related by the following formula:

G=nFEcell

where n is the number of moles of electrons transferred and F=96,500J/Vmol e is the Faraday constant.

Part A

The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction Reduction potential
(V) Li+(aq)+eLi(s) 3.04 Cl2(g)+2e2Cl(aq) +1.36 In a galvanic cell, the two half-reactions combine to

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

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

Express your answer with the appropriate units.

E =

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Part B

Calculate the free energy G of the reaction.

Express your answer with the appropriate units.

G =

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Part C

What can be said about the spontaneity of this reaction?

The reaction is

What can be said about the spontaneity of this reaction? spontaneous as written. spontaneous in the reverse direction. at equilibrium. nonspontaneous in either direction.

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Explanation / Answer

Part A

Eo = +3.04 + 1.36 = 4.4 V   (note : Li is oxidised so it will be +3.04 V)

Part B

dGo = -nFEo

dGo = -2*96485*4.4

dGo = -849 kJ

Part C

Reaction is spontaneous as written

because it has -ve dG value, if dG is +ve reaction would ne reverse

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