Problem 31 - Chemistry Matter and Change

Question

Determine The oxidation half-reaction of a redox reaction is \(5 n^{2+} \rightarrow 5 n^{4+}+2 e^{-},\) and the reduction half-reaction is \(A u^{3+}+3 e^{-} \rightarrow\) Au. What minimum numbers of tin(II) ions and gold(III) ions would have to react in order to have zero electrons left over?

Step-by-Step Solution

Answer

The minimum numbers of tin(II) ions (\(n^{2+}\)) and gold(III) ions (\(Au^{3+}\)) that would have to react in order to have zero electrons left over are 15 tin(II) ions and 2 gold(III) ions.

Step 1: Balance the half-reactions for the number of electrons exchanged

The given oxidation half-reaction is: \(5 n^{2+} \rightarrow 5 n^{4+} + 2 e^-\) The given reduction half-reaction is: \(Au^{3+} + 3 e^- \rightarrow Au\)

Step 2: Find the least common multiple (LCM) of the numbers of electrons exchanged

The oxidation half-reaction involves 2 electrons, and the reduction half-reaction involves 3 electrons. We need to find the LCM of these numbers to determine the minimum number of electrons exchanged so that there are no electrons left over. The LCM of 2 and 3 is 6.

Step 3: Adjust the half-reactions to have an equal number of electrons

To have an equal number of 6 electrons exchanged in both half-reactions, we have to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2: Oxidation half-reaction: \(3 (5 n^{2+} \rightarrow 5 n^{4+} + 2 e^-)\) → \(15 n^{2+} \rightarrow 15 n^{4+} + 6 e^-\) Reduction half-reaction: \(2 (Au^{3+} + 3 e^- \rightarrow Au)\) → \(2 Au^{3+} + 6 e^- \rightarrow 2 Au\)

Step 4: Determine the minimum number of ions that have to react

Now that both half-reactions have an equal number of electrons exchanged, we can determine the minimum number of ions that need to react. In the balanced half-reactions, we see: 15 \(n^{2+}\) ions must react to produce 15 \(n^{4+}\) ions in the oxidation half-reaction. 2 \(Au^{3+}\) ions must react to produce 2 Au atoms in the reduction half-reaction.

Step 5: Answer

The minimum numbers of tin(II) ions (\(n^{2+}\)) and gold(III) ions (\(Au^{3+}\)) that would have to react in order to have zero electrons left over are 15 tin(II) ions and 2 gold(III) ions.

Key Concepts

Oxidation Half-Reaction

An oxidation half-reaction is a part of a redox reaction where the oxidation process occurs. In this process, a species loses electrons. This means the oxidation state of the molecule, atom, or ion involved in the reaction increases. In our exercise, tin(II) ions, represented as \(5n^{2+}\), are oxidized to \(5n^{4+}\) by losing electrons. Here, the equation shows that each tin(II) ion loses two electrons, thus becoming tin(IV) in the end. Understanding this half of the redox reaction helps us see how electrons are released in the reaction.

To make it balanced and have the same number of electrons lost and gained, finding the least common multiple (LCM) of electrons is crucial.

Reduction Half-Reaction

A reduction half-reaction involves the gain of electrons by a substance. This is the opposite of oxidation and results in a decrease in the oxidation state of the substance. In the exercise example, the equation \(Au^{3+} + 3 e^- \rightarrow Au\) illustrates how gold(III) ions gain electrons to become pure gold atoms. The gold ion accepts 3 electrons, indicating it undergoes reduction. This half-reaction takes the electrons released by the oxidation half-reaction and uses them, ensuring the entire process is balanced.

For the redox reaction to work, it is crucial to have these two processes matched for the number of electrons transferred, requiring electron balancing.

Electron Balancing

Electron balancing is an essential step in ensuring that a redox reaction is balanced. It involves equalizing the number of electrons exchanged in both the oxidation and reduction half-reactions. In our example, different numbers of electrons are transferred in the oxidation (2 electrons) and reduction (3 electrons) processes. To balance this, we need to consider the Least Common Multiple (LCM) of these numbers.

• The oxidation reaction is multiplied by 3
• The reduction reaction is multiplied by 2

This process adjusts both half-reactions to involve 6 electrons, ensuring that electrons are balanced and correctly conserved throughout the reaction. This way, no electrons are left unaccounted for.

Oxidation States

Oxidation states, also known as oxidation numbers, represent the hypothetical charges an atom would have if all bonds to atoms of different elements were completely ionic. They are critical for identifying which elements are oxidized and which are reduced in a redox reaction. In the given exercise:

• Tin(II) ions, \(n^{2+}\), have an initial oxidation state of +2, which increases to +4 when they turn into \(n^{4+}\) ions during oxidation.
• Gold(III) ions, \(Au^{3+}\), have an initial oxidation state of +3, which decreases to 0 when they become elemental gold during reduction.

Tracking changes in oxidation state helps determine how electrons are being transferred within a redox reaction, which is the hallmark of these types of chemical processes.