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

1 Answer
Jacob T. · Truong-Son N.
Aug 11, 2018

#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

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