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?
Potential Half cell reaction
-3.05 V Li+ + e- -> Li
+1.36V Cl2 + 2e- -> 2Cl-
+2.78V F2 + 2e- -> 2F-
Potential Half cell reaction
-3.05 V Li+ + e- -> Li
+1.36V Cl2 + 2e- -> 2Cl-
+2.78V F2 + 2e- -> 2F-
1 Answer
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
- opposite in sign
- largest in magnitude overall
Thus, we choose reactions (1) and (3). To find
#\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
#"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
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.