Question

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-

Answer 1

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.