Problem 42 - General Chemistry: Principles and Modern Applications

Question

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.

Step-by-Step Solution

Answer

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 1: 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: \[\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-\]

Step 2: 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: \[ \mathrm{Fe(CN)}_6^{3-} + e^- \rightarrow \mathrm{Fe(CN)}_6^{4-}\]

Step 3: 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: \[ \mathrm{Fe}^{3+} + \mathrm{Fe(CN)}_6^{4-} \rightarrow \mathrm{Fe}\mathrm{Fe(CN)}_6^{3-}\]

Key Concepts

Oxidation-Reduction

In the world of chemistry, oxidation-reduction (or redox) reactions describe any process in which electrons are transferred between chemical species. This may sound complex, but let's break it down.

• Oxidation refers to the loss of electrons during a chemical reaction. When an atom or ion undergoes oxidation, it increases its oxidation state.
• Reduction is the opposite—it is the gain of electrons, and it decreases an oxidation state.

These two processes happen simultaneously in a redox reaction. When one species loses electrons (is oxidized), another gains those electrons (is reduced). In the exercise provided above, the oxidation step involves the transformation of Iron (II) ions (\( \mathrm{Fe}^{2+} \)) to Iron (III) ions (\( \mathrm{Fe}^{3+} \)), where the Iron (II) ions lose an electron. Meanwhile, reduction occurs when ferricyanide ions (\( \mathrm{Fe(CN)}_6^{3-} \)) become ferrocyanide ions (\( \mathrm{Fe(CN)}_6^{4-} \)) by gaining an electron.

Half-Reaction

The concept of half-reactions is crucial when discussing redox reactions. Each redox process can be split into two parts, known as half-reactions. These individually show the oxidation process and the reduction process.
In our exercise:

• **Oxidation Half-Reaction:** This equation focuses on the species losing electrons. For instance, the half-reaction for the oxidation of Iron (II) to Iron (III) is represented as:\[\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\]
• **Reduction Half-Reaction:** Here, we track the species gaining electrons. The ferricyanide ion is reduced to the ferrocyanide ion, shown by:\[\text{Fe(CN)}_6^{3-} + e^- \rightarrow \text{Fe(CN)}_6^{4-}\]

These half-reactions make it easier to understand and balance the overall redox reaction. Notice how one half-reaction's oxidized element corresponds to the other half-reaction's reduced element, creating a complete, balanced redox equation.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the same number of each type of atom appears on both sides of the equation. This reflects the law of conservation of mass, where matter cannot be created or destroyed.
In redox reactions, balancing involves ensuring that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. Let's see how it applies to the provided exercise.

• First, identify the oxidation and reduction half-reactions, as discussed earlier.
• Ensure each half-reaction is balanced regarding atoms and overall charge.
• If the electrons transferred do not equal out, multiply the half-reactions by suitable coefficients so the electrons cancel each other out when added together.

Following these steps leads to a balanced equation that respects both the material and charge conservation laws. For the creation of Turnbull's blue, ensuring both \( \text{Fe}^{3+} \) and \( \text{Fe(CN)}_6^{4-} \) species come together properly, keeping charges balanced, results in the final product you see in this reaction:\[ \text{Fe}^{3+} + \text{Fe(CN)}_6^{4-} \rightarrow \text{FeFe(CN)}_6^{3-}\] This allows the reaction to proceed and maintain balance at every step.