# 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-

ReportAnswer 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) $\text{Li"^(+) + e^(-) -> "Li}$

${E}_{\text{red"^@ = -"3.05 V}}$

(2) ${\text{Cl"_2 + 2e^(-) -> 2"Cl}}^{-}$

${E}_{\text{red"^@ = +"1.36 V}}$

(3) ${\text{F"_2 + 2e^(-) -> 2"F}}^{-}$

${E}_{\text{red"^@ = +"2.87 V}}$ ($2.78$ is a typo)

Since one of these must be reversed to obtain an oxidation half-reaction (thereby reversing the sign of this ${E}^{\circ}$), 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 ${E}_{\text{cell}}^{\circ}$, I prefer to use the equation

$\setminus m a t h b f \left({E}_{\text{cell"^@ = E_"red"^@ + E_"ox}}^{\circ}\right)$,

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 ${E}_{\text{red"^@ = -E_"ox}}^{\circ}$ for the same half-reaction that we've reversed (the lithium one). The full reaction becomes:

${\text{F"_2(g) + cancel(2e^(-)) -> 2"F}}^{-} \left(a q\right)$
$2 \left({\text{Li"(s) -> "Li}}^{+} \left(a q\right) + \cancel{{e}^{-}}\right)$
$\text{---------------------------------------------}$
$\setminus m a t h b f \left({\text{F"_2(g) + 2"Li"(s) -> 2"Li"^(+)(aq) + 2"F}}^{-} \left(a q\right)\right)$

Note that ${E}^{\circ}$ (an intensive property) does NOT depend on the stoichiometric coefficients of the half-reaction, so you do NOT scale ${E}_{\text{ox}}^{\circ}$.

So, you just have:

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

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

$=$ $\textcolor{b l u e}{\text{5.92 V}}$

You can see how someone else solved this problem here.

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