It's given in my book that $\ce{Fe^3+}$ ion reacts with potassium hexacyanidoferrate(III) to give brown colour. $$\ce{Fe^3+ + [Fe(CN)6]^3- -> Fe[Fe(CN)6]}$$ Both the reactants have the same oxidation states.

My question is, can I use the same test for $\ce{Fe^2+}$?

First, the hexacyanidoferrate(III) will oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ like this: $$\ce{Fe^2+ + [Fe(CN)6]^3- -> Fe^3+ + [Fe(CN)6]^4-}$$ Then the reaction will continue to give brown colour.

  • $\begingroup$ You kinda can, but the product would be different: the Prussian blue, famous in its own right. $\endgroup$ – Ivan Neretin Feb 4 '16 at 6:24
  • $\begingroup$ @IvanNeretin I thought prussian blue is obtained when $\ce{Fe^3+} $ reacts with hexacyanidoferrate(II) or $\ce{[Fe(CN) 6]^4-} $. I am talking about $\ce{[Fe(CN)6]^3-}$. $\endgroup$ – Aditya Dev Feb 4 '16 at 7:08
  • $\begingroup$ for prussian blue, $\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$ $\endgroup$ – Aditya Dev Feb 4 '16 at 7:11
  • $\begingroup$ $\ce{Fe^3+ + [Fe(CN)6]^4-}$ is exactly what you have on the right side of your own equation. $\endgroup$ – Ivan Neretin Feb 4 '16 at 7:24
  • $\begingroup$ @IvanNeretin But can't the reactant $\ce{Fe^2+}$ combine with the product ferricyanide to give Turnbull's blue? $\endgroup$ – Aditya Dev Feb 4 '16 at 9:26

$\ce{Fe^2+}$ is oxidised to $\ce{Fe^3+}$ while $\ce{[Fe(CN)_6]^{3-}}$ is reduced to $\ce{[Fe(CN)_6]^{4-}}$. These then form the prussian blue precipitate.

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