Question

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined by the sign of E∘cell. The values of ΔG and E∘cell are related by the following formula: ΔG∘=−nFE∘cell where n is the number of moles of electrons transferred and F=96,500J/V⋅mol 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)+e−→Li(s) −3.04
Cl2(g)+2e−→2Cl−(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.

Part B: Calculate the free energy ΔG∘ of the reaction. Express your answer with the appropriate units.

Part C:

What can be said about the spontaneity of this reaction?

The reaction is:

a. spontaneous as written.
b. spontaneous in the reverse direction.
c. at equilibrium.
d. nonspontaneous in either direction.

Answer 1

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

• For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction:  $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction:  $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

• For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

• For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.