Find overall reaction for a galvanic cell that will have the highest cell potential. What is the overall potential of the cell? Which two equations do I use?

Well, you can see that all your given reactions are reductions; they reduce the oxidation state of the atom(s) to a more negative or less positive value.

(1) #"Li"^(+) + e^(-) -> "Li"#

#E_"red"^@ = -"3.05 V"#

(2) #"Cl"_2 + 2e^(-) -> 2"Cl"^(-)#

#E_"red"^@ = +"1.36 V"#

(3) #"F"_2 + 2e^(-) -> 2"F"^(-)#

#E_"red"^@ = +"2.87 V"# (#2.78# is a typo)

Since one of these must be reversed to obtain an oxidation half-reaction (thereby reversing the sign of this #E^@#), we are looking for two half-reaction potentials that are:

• opposite in sign
• largest in magnitude overall

Thus, we choose reactions (1) and (3). To find #E_"cell"^@#, I prefer to use the equation

#\mathbf(E_"cell"^@ = E_"red"^@ + E_"ox"^@)#,

because it does not require me to think about cathodes/anodes. Instead, I just need to look at how the oxidation states changed.

We should recognize that #E_"red"^@ = -E_"ox"^@# for the same half-reaction that we've reversed (the lithium one). The full reaction becomes:

#"F"_2(g) + cancel(2e^(-)) -> 2"F"^(-)(aq)#
#2("Li"(s) -> "Li"^(+)(aq) + cancel(e^(-)))#
#"---------------------------------------------"#
#\mathbf("F"_2(g) + 2"Li"(s) -> 2"Li"^(+)(aq) + 2"F"^(-)(aq))#

Note that #E^@# (an intensive property) does NOT depend on the stoichiometric coefficients of the half-reaction, so you do NOT scale #E_"ox"^@#.

So, you just have:

#color(blue)(E_"cell"^@) = E_"red"^@ + E_"ox"^@#

#= "2.87 V" + (-(-"3.05 V"))#

#=# #color(blue)("5.92 V")#

You can see how someone else solved this problem here.