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Chapter 23: Problem 42

The reaction to form Turnbull's blue (page 1053 ) appears to occur in two stages. First, \(\mathrm{Fe}^{2+}(\mathrm{aq})\) is oxidized to \(\mathrm{Fe}^{3+}(\mathrm{aq})\) and ferricyanide ion is reduced to ferrocyanide ion. Then, the \(\mathrm{Fe}^{3+}(\text { aq })\) and ferrocyanide ion combine. Write equations for these reactions.

Short Answer

Expert verified
1) \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-\), 2) \(\mathrm{Fe(CN)}_6^{3-} + e^- \rightarrow \mathrm{Fe(CN)}_6^{4-}\), 3) \(\mathrm{Fe}^{3+} + \mathrm{Fe(CN)}_6^{4-} \rightarrow \mathrm{Fe}\mathrm{Fe(CN)}_6^{3-}\). The formation of Turnbull's blue includes the oxidation of \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\) ions, the reduction of ferricyanide ion to ferrocyanide ion and the combination of \(\mathrm{Fe}^{3+}\) ions and ferrocyanide ion.

Step by step solution

01

Oxidation of Iron (II) ions

The oxidation of Iron (II) to Iron (III) involves the loss of one electron by the Iron (II) ion. This can be written by the following half-reaction: \[\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-\]
02

Reduction of ferricyanide ion

In the reduction of the ferricyanide ion to the ferrocyanide ion, the ferricyanide ion gains an electron. Because there are six iron atoms per ferricyanide ion the half-reaction needs to be balanced taking this into account: \[ \mathrm{Fe(CN)}_6^{3-} + e^- \rightarrow \mathrm{Fe(CN)}_6^{4-}\]
03

Combination of Iron (III) and ferrocyanide ion

In the final step, the Iron (III) and the ferrocyanide ions combine to form Turnbull's blue. This can be represented by the following equation: \[ \mathrm{Fe}^{3+} + \mathrm{Fe(CN)}_6^{4-} \rightarrow \mathrm{Fe}\mathrm{Fe(CN)}_6^{3-}\]

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Most popular questions from this chapter

Three properties expected for transition elements are (a) low melting points; (b) high ionization energies; (c) colored ions in solution; (d) positive standard electrode (reduction) potentials; (e) diamagnetism; (f) complex ion formation; (g) catalytic activity.

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