Problem 31 - Chemistry Matter and Change

Question

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

Step-by-Step Solution

Answer

3 tin(II) ions and 2 gold(III) ions are needed.

Step 1: Understanding the Oxidation Half-Reaction

The oxidation half-reaction involves the loss of electrons. Here, tin is oxidized from a +2 oxidation state to a +4 oxidation state according to the equation \(\mathrm{Sn}^{2+} \rightarrow \mathrm{Sn}^{4+} + 2 \mathrm{e}^{-}\). Two electrons are released in this process.

Step 2: Understanding the Reduction Half-Reaction

The reduction half-reaction involves the gain of electrons. Here, gold is reduced from a +3 oxidation state to 0 according to the equation \(\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au}\). Three electrons are consumed in this process.

Step 3: Balancing Electrons Between Oxidation and Reduction

To ensure that the number of electrons lost in oxidation is equal to the number of electrons gained in reduction, we need the total electrons to be balanced. Since the oxidation releases 2 electrons and the reduction consumes 3 electrons, we need to find the smallest common multiple of 2 and 3, which is 6.

Step 4: Determining the Minimum Number of Ions Needed

In order to achieve 6 electrons in both redox processes, we multiply the oxidation equation by 3: \[3(\mathrm{Sn}^{2+} \rightarrow \mathrm{Sn}^{4+} + 2 \mathrm{e}^{-}) = 3\mathrm{Sn}^{2+} \rightarrow 3\mathrm{Sn}^{4+} + 6\mathrm{e}^{-} \]And the reduction equation by 2:\[2(\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au}) = 2\mathrm{Au}^{3+} + 6\mathrm{e}^{-} \rightarrow 2 \mathrm{Au}\]

Step 5: Conclusion

From these equations, it can be concluded that the reaction requires a minimum of 3 tin(II) ions (\(\mathrm{Sn}^{2+}\)) and 2 gold(III) ions (\(\mathrm{Au}^{3+}\)) to ensure that 6 electrons are both released and consumed.

Key Concepts

Understanding Oxidation Half-Reaction

In a redox reaction, one of the vital parts is the oxidation half-reaction. This process involves the loss of electrons from a molecule, ion, or atom. Specifically, in our example, the tin ion (\(\mathrm{Sn}^{2+}\)) is oxidized as electrons are lost. This means that the tin changes from the +2 oxidation state to the +4 oxidation state.
If we look at the equation \(\mathrm{Sn}^{2+} \rightarrow \mathrm{Sn}^{4+} + 2 \mathrm{e}^{-}\), it clearly translates that 2 electrons are released during this change. Essentially, oxidation can be remembered for having an increase in the oxidation number because of electron loss.

Understanding Reduction Half-Reaction

On the opposite end of a redox reaction sits the reduction half-reaction, which involves the gain of electrons. Reduction effectively decreases the oxidation state of a molecule, atom, or ion, triggering a more stable form.
Looking at our reduction reaction with gold, \(\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au}\), the gold ion (\(\mathrm{Au}^{3+}\)) is being reduced from a +3 to 0 by gaining 3 electrons. In simple terms, reduction is an electron-gaining process. Always remember: Reduction = Release (of electrons) = Decrease in oxidation number.

Electron Transfer in Redox Reactions

Electron transfer is the essence of redox reactions. It dictates the movement of electrons from one species undergoing oxidation to another species undergoing reduction.
The number of electrons lost during oxidation must be equal to those gained in reduction for the reaction to be balanced. In our scenario, tin releases 2 electrons per ion while gold receives 3 electrons per ion. Balancing these two requires figuring out the least common multiple of electrons lost and gained, often essential in redox reactions to maintain electron neutrality.

Balancing Chemical Equations

Balancing chemical equations is crucial for accurately describing a chemical reaction, ensuring that all mass and charge are conserved. When dealing with redox reactions, it involves more than just simple balancing of atoms.
Our main task includes ensuring the electrons taken by the reduction process equals those given by the oxidation. Here, finding the least common multiple is crucial: for tin and gold—releasing and absorbing 2 and 3 electrons, respectively—6 becomes this common ground.

• Multiply the oxidation reaction by 3: \[3(\mathrm{Sn}^{2+} \rightarrow \mathrm{Sn}^{4+} + 2 \mathrm{e}^{-}) = 3\mathrm{Sn}^{2+} \rightarrow 3\mathrm{Sn}^{4+} + 6\mathrm{e}^{-}\]
• Multiply the reduction reaction by 2: \[2(\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au}) = 2\mathrm{Au}^{3+} + 6\mathrm{e}^{-} \rightarrow 2 \mathrm{Au}\]

Having them balanced this way confirms the reaction uses up 6 electrons in both oxidation and reduction processes, achieving equilibrium and correctly describing the chemical change.