# Calculate the potential E for the Fe3+/Fe2+ electrode when the concentration of Fe2+ is exactly five times that of Fe3+?

Aug 11, 2018

${E}_{\text{cell" ~~0.73 color(white)(l) "V}}$

#### Explanation:

Start with the standard reduction potential for a ${\text{Fe"^(3+)//"Fe}}^{2 +}$ cell where $\left[{\text{Fe"^(3+)] = ["Fe}}^{2 +}\right]$ hence $Q = 1$:

From a data sheet of standard electrode potentials,

${E}_{\text{cell"^"o"(["Fe"^(3+)]//["Fe"^(2+)]) = +0.77 color(white)(l) "V}}$ ""^([1])

Consider the "equilibrium" for this cell reaction:

${\text{Fe"^(3+) + "e"^(-) rightleftharpoons "Fe}}^{2 +}$

... for which

• there's one mole of electron transfer per mole reaction, $n = 1$;
• $Q = \left[{\text{Fe"^(2+)]//["Fe}}^{3 +}\right] = 5$ according to the question;

Apply the Nernst equation""^([2]):

E_"cell" = E_"cell"^(@) - (0.0592 color(white)(l) "V")/(n) * log Q
color(white)(E_"cell") ~~0.73 color(white)(l) "V"

[1] "Standard electrode potential (data page)", Wikipedia, https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
[2] "Cell Potential Under Nonstandard Conditions", Libretexts, https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Map:_Chemistry_-_The_Central_Science_(Brown_et_al.)/20:_Electrochemistry/20.6:_Cell_Potential_Under_Nonstandard_Conditions