Sure, let's go through the solution step-by-step!
To determine the standard cell potential ([tex]\(E^{\circ}_{\text{cell}}\)[/tex]) for the given reaction:
[tex]\[ 2 F^-_{\text{(aq)}} + Cl_{2(g)} \rightarrow F_{2(g)} + 2 Cl^-_{\text{(aq)}} \][/tex]
we need to look at the standard reduction potentials for the half-reactions involved and then combine them appropriately.
First, let's write the relevant half-reactions and their standard reduction potentials:
1. Reduction of [tex]\( Cl_2 \)[/tex] to [tex]\( Cl^- \)[/tex]:
[tex]\[ Cl_2(g) + 2e^- \rightarrow 2Cl^-_{\text{(aq)}} \][/tex]
The standard reduction potential for this half-reaction is [tex]\( E^\circ_{Cl_2/Cl^-} = +1.36 \, \text{V} \)[/tex].
2. Reduction of [tex]\( F_2 \)[/tex] to [tex]\( F^- \)[/tex]:
[tex]\[ F_2(g) + 2e^- \rightarrow 2F^-_{\text{(aq)}} \][/tex]
The standard reduction potential for this half-reaction is [tex]\( E^\circ_{F_2/F^-} = +2.87 \, \text{V} \)[/tex].
Since [tex]\( F^- \)[/tex] is being oxidized to [tex]\( F_2 \)[/tex] in the given reaction, we need to reverse the reduction half-reaction for [tex]\( F_2 \)[/tex] to obtain the oxidation potential:
3. Oxidation of [tex]\( F^- \)[/tex] to [tex]\( F_2 \)[/tex]:
[tex]\[ 2F^-_{\text{(aq)}} \rightarrow F_2(g) + 2e^- \][/tex]
The standard oxidation potential for this reaction is the negative of the reduction potential:
[tex]\[ E^\circ_{F^-/F_2} = -2.87 \, \text{V} \][/tex]
Now, combine the two half-reactions:
- [tex]\( Cl_2(g) \)[/tex] is being reduced, leveraging [tex]\( E^\circ_{Cl_2/Cl^-} = +1.36 \, \text{V} \)[/tex]
- [tex]\( F^- \)[/tex] is being oxidized, leveraging [tex]\( E^\circ_{F^-/F_2} = -2.87 \, \text{V} \)[/tex]
The overall cell potential [tex]\( E^\circ_{\text{cell}} \)[/tex] is the sum of the individual electrode potentials:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{(\text{reduction of } Cl_2)} + E^\circ_{(\text{oxidation of } F^-)} \][/tex]
[tex]\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} + (-2.87 \, \text{V}) \][/tex]
[tex]\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} - 2.87 \, \text{V} \][/tex]
[tex]\[ E^\circ_{\text{cell}} = -1.51 \, \text{V} \][/tex]
Thus, the standard cell potential for the reaction is:
[tex]\[ \boxed{-1.51 \, \text{V}} \][/tex]
Therefore, the correct answer is:
a) -1.51 V
