Question

Given: $$\begin{array}{ll} \mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~F}^{-}(\mathrm{aq}) & E^{\circ}=+2.87\mathrm{~V} \\ \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) & E^{\circ}=+1.36\mathrm{~V} \end{array}$$ determine whether this reaction is product-favored: $$\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{~F}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{F}_{2}(\mathrm{~g})$$

Short answer

The reaction is product-favored with a positive cell potential of 1.51 V.

Step-by-step solution

Identify Redox Reactions

Review the given half-reactions:1. \( \mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~F}^{-}(\mathrm{aq}) \) with \( E^{\circ}=+2.87\mathrm{~V} \) 2. \( \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq}) \) with \( E^{\circ}=+1.36\mathrm{~V} \)

Reverse Oxidation Reaction

For the given reaction \( \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{~F}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{F}_{2}(\mathrm{~g}) \), identify which half-reaction needs to be reversed. This is the chlorine reduction reaction, as it will be oxidized to \( \mathrm{Cl}_{2} \). Reverse the chlorine reaction:\( 2 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g}) + 2 \mathrm{e}^{-} \) with \( E^{\circ}=-1.36\mathrm{~V} \)

Calculate Overall Cell Potential

Add the reversed chlorine potential to the fluorine reduction potential: \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{F}_{2}} + E^{\circ}_{\mathrm{Cl}_{2}} \] \[ E^{\circ}_{\text{cell}} = 2.87\mathrm{~V} + (-1.36\mathrm{~V}) \] \[ E^{\circ}_{\text{cell}} = 1.51\mathrm{~V} \]

Determine Reaction Favorability

Since the overall cell potential \( E^{\circ}_{\text{cell}} \) is positive \( (+1.51\mathrm{~V}) \), the reaction is product-favored under standard conditions.

Key concepts

Redox Reactions

Redox reactions, also known as oxidation-reduction reactions, involve the transfer of electrons between two species. These reactions are essential in chemistry as they are responsible for many processes, such as energy production and combustion. In a redox reaction, one substance loses electrons (oxidation), and another gains electrons (reduction).
To better understand, consider the reaction components separately as half-reactions. Each half-reaction either donates or accepts electrons.

• In the given example, the half-reaction for fluorine (\( \mathrm{F}_{2} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{F}^{-} \)) represents a reduction process, as fluorine molecules gain electrons to become fluoride ions.
• On the other hand, chlorine in the half-reaction (\( \mathrm{Cl}_{2} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \)) also illustrates reduction when viewed initially, but when reversed for our reaction scenario (\( 2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2 \mathrm{e}^{-} \)), it undergoes oxidation.

Understanding the concept of electron flow is crucial in identifying which elements undergo oxidation and which undergo reduction in redox reactions.

Standard Cell Potential

The standard cell potential, denoted as \( E^{\circ} \), is a measure of the driving force behind a redox reaction. It's determined by the difference in potential energy between the reduction and oxidation half-reactions. Importantly, the standard cell potential helps chemists understand whether a reaction can do work under standard conditions (1 M concentration, 25°C, and 1 atm pressure for gases).

• To calculate the standard cell potential, you combine the potentials of two half-reactions. In our example, you need to add the reduction potential of the fluorine reaction and the oxidation potential of the chlorine reaction (converted from its reduction potential).
• For this example, the calculations are: \( E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{F}_{2}} + (-E^{\circ}_{\mathrm{Cl}_{2}}) \).
• The resulting \( E^{\circ}_{\text{cell}} \) is 1.51 V, indicating an energetic and favorable reaction.

A positive standard cell potential such as this indicates that the reaction is spontaneous under standard conditions.

Reaction Favorability

Reaction favorability refers to the tendency of a chemical reaction to proceed in a given direction under specified conditions. When assessing whether a reaction is product-favored, a key indicator is the sign and magnitude of the standard cell potential.

• A positive \( E^{\circ}_{\text{cell}} \) value is a hallmark of a product-favored reaction. This implies that the reaction is thermodynamically spontaneous.
• In the case of our example reaction (\( \mathrm{Cl}_{2} + 2 \mathrm{F}^{-} \rightarrow 2 \mathrm{Cl}^{-} + \mathrm{F}_{2} \), a \( E^{\circ}_{\text{cell}} \) of 1.51 V confirms that the forward reaction is naturally inclined to occur.
• Product-favored reactions are important in applications such as battery design and industrial chemical production, where efficient energy transformation is critical.

Understanding the standard cell potential can inform predictions about reaction progress and efficiency in both theoretical studies and practical applications.