Question

a cell is based on the following reaction
Fe(s)+2Fe+3(aq) -----> 3Fe+2 (aq) E cell= 1.18 V.

Calculate the concentration of iron (II), in M, if the cell emf is 1.28 V. when [Fe^(3+)]= 0.5 M

the answer is 0.047 M

Answer 1

Ecell = Eocell - (0.05916/n)*log Q
You have Ecell and Eocell. n is 2 and
Q = (Fe^2+)^3/(Fe^3+)^2
Substitute and solve for (Fe^2+).
0.047M is the correct answer.

Answer 2

thank you