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.

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.

## Related Solutions

##### If f(x) = exg(x), where g(0) = 3 and g'(0) = 4, find f '(0).
If $$f(x)=e^{x} g(x),$$ where $$g(0)=3$$ and $$g(0)=4,$$ find $$f(0)$$
##### Consider the circuit shown in (Figure 1) . Suppose that E = 15 V
Consider the circuit shown in (Figure 1). Suppose that E = 15 V . include units with answers. Part A: Find the current through the resistor a. Part B: Find the potential difference across the resistor a. answer: 7.5 V Part C: Find the current through the resistor b. Part D: Find the potential difference across the resistor b. Part E: Find the current through the resistor c. Part F: Find the potential difference across the resistor c. Part G: Find...
##### Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown.
Let P(x) = F(x)G(x) and Q(x) = F(x)/G(x), where F and G are the functions whose graphs are shown.(a) Find P ' (2)(b) Find Q ' (7)
##### if f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x).
if f and g are the functions whose graphs are shown, let u(x) = f(x)g(x) and v(x) = f(x)/g(x) (a) Find u'(1) (b) Find v'(5).
##### How many electrons in an atom could have these sets of quantum numbers? n=3
How many electrons in an atom could have these sets of quantum numbers? (a) n=3 (b) n=4, l=2 (c) n=7, l=3, ml=-1
##### You buy a 75-W lightbulb in Europe, where electricity is delivered to home at 240 V
You buy a 75-W lightbulb in Europe, where electricity is delivered to home at 240 V. If you use the lightbulb in the United States at120 V (assume its resistance does not change). how bright will it be relative to 75-W 120-V Bulbs? [ Hint: assume roughly that brightness is proportional to power consumed]