Question

In: Chemistry

Where n is the number of moles of electrons and F=96,500J/V⋅mol e− is the Faraday constant.

Part A

Cell Potential and Free Energy of a Lithium–Chlorine Cell 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 and F=96,500J/V⋅mol e− is the Faraday constant.

Part B

Calculate the free energy ΔG∘ of the reaction.

Express your answer in kilojoules.

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.

 

Solutions

Expert Solution

Part A:

The overall cell reaction of the Lithium - Chlorine cell is as follows:

\(2 \mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \operatorname{LiCl}(a q)\)

The oxidation and reduction half reactions that are taking place at anode

and cathode respectively are as foll ows:

Anode:

\(\mathrm{Li}(s) \longrightarrow \mathrm{Li}^{+}(a q)+e^{-} \quad E_{\mathrm{Li}^{+} / \mathrm{L}}=-3.04 \mathrm{~V}\)

Cathode:

\(\mathrm{Cl}_{2}(\mathrm{~g})+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) E_{\mathrm{cucr}}=1.36 \mathrm{~V}\)

Add the above 2 half reactions as foll ows:

\(2 \mathrm{Li}(s) \longrightarrow 2 \mathrm{Li}^{+}(a q)+2 e^{-}\)

\(\mathrm{Cl}_{2}(\mathrm{~g})+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q)\)

\(2 \mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{LiCl}(a q)\)

Therefore, the number of electrons transferred are 2 .

Part B:

Calculate the \(E_{\text {cen }}^{\circ}\) as follows:

\(\begin{aligned} E_{\text {cell }}^{o} &=E_{\text {cathode }}-E_{\text {modt }}^{o} \\ &=[1.36-(-3.04)] \mathrm{V} \\ &=4.4 \mathrm{~V} \end{aligned}\)

Calculate the \(\Delta G^{\circ}\) as foll ows:

\(\Delta G^{0}=-n F E_{\mathrm{cell}}^{0}\)

\(\begin{array}{l} =-2 \mathrm{~mol} \mathrm{e}^{-} \times 96500 \frac{\mathrm{J}}{\mathrm{V} \mathrm{mol} \mathrm{e}^{-}} \times 4.4 \mathrm{~V} \\ =-849200 \mathrm{~J} \\ =-849.2 \mathrm{~kJ} \end{array}\)

Part C:

The condition for spontainety of a reaction is \(\Delta G^{\circ}<0\)

Therefore, the calulated value of \(\Delta G^{\circ}\) for the above reaction is negative.

Therefore, the reaction is spontaneous as written.

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