The ferricyanide method of determining reducing sugars is simple but I need clarity on time of addition of potassium iodide solution. I am unable to observe change in color from yellow to blue. I even tried carrying the titration out under low heat.


1 Answer 1


A ferricyanide solution contains the ion $\ce{Fe(CN)6^{3-}}$. This ion may react with iodide containing solutions producing ferrocyanide ions $\ce{Fe(CN)6^{4-}}$ ions and elemental iodine $\ce{I2}$ (which becomes triiodide $\ce{I3–}$ ions in presence of an excess of $\ce{I-}$ ions). The corresponding equations are : $$\ce{2 [Fe(CN)6]^{3-} + 2 I- -> 2 [Fe(CN)6]^{4-} + I2}$$ $$\ce{I2 + I- <=> I3-}$$ Now the iodine $\ce{I2}$ has a strange property in the presence of starch. Starch is an organic polymer formed of long helixes. These helixes look like long tubes ; and the inner diameter of these inner tubes is equal to the outer diameter of the $\ce{I2}$ molecule. So if both starch and iodine $\ce{I2}$ are present in solution, iodine molecules or triiodde ions may easily enter the inner tube of starch, and will not easily get out of this complex $$\ce{I2 + Starch <=> Starch-I2 complex}$$ $$\ce{I3- + Starch <=> Starch-triiodide complex}$$ This reaction forming the complex is an equilibrium- And for some reason, the complex Starch-Iodine or starch Starch - triiodide is blue. So any solution containing both starch and iodine is blue, provided the solution is a little acidic. If it is a bit basic, the iodine $\ce{I2}$ or the trioxide ion is destroyed and transformed according to $$\ce{6 I2 + 12 OH- -> 2 IO3- + 10 I- + 6 H2O}$$ Furthermore, if iodine molecules are destroyed by any other chemical reaction, for example by adding thiosulfate ions in solution, the blue color due to the complex $\ce{I2 - Starch}$ (or triiodide-Starch) will also disappear, as the following reaction occurs $$\ce{I2 + 2 S2O3^{2-} -> S4O6^{2-} + 2 I-}$$ All ions present in this equation are finally colorless. So the blue solution becomes colorless.

  • $\begingroup$ According to wikipedia starch forms a very dark blue-black complex with triiodide. However, the complex is not formed if only iodine or only iodide (I−) is present. $\endgroup$
    – Poutnik
    Commented May 23 at 9:48

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