Question

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

Answer 1

`E_"cell" ~~0.73 color(white)(l) "V"`

Explanation:

Start with the standard reduction potential for a `"Fe"^(3+)//"Fe"^(2+)` cell where `["Fe"^(3+)] = ["Fe"^(2+)]` hence `Q = 1`:

From a data sheet of standard electrode potentials,

`E_"cell"^"o"(["Fe"^(3+)]//["Fe"^(2+)]) = +0.77 color(white)(l) "V"` `""^([1])`

Consider the "equilibrium" for this cell reaction:

`"Fe"^(3+) + "e"^(-) rightleftharpoons "Fe"^(2+)`

... for which

• there's one mole of electron transfer per mole reaction, `n = 1`;
• `Q = ["Fe"^(2+)]//["Fe"^(3+)] = 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