Problem 77 - Chemistry Matter and Change

Question

Write a balanced ionic redox equation using the following pairs of redox half- reactions. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}} \\ {\underline{\phantom{xxx}} \quad \mathrm{Te}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Te}} \\ {\text { b. } \mathrm{IO}_{4}^{-}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}} \\ {\quad \text { Al } \rightarrow \mathrm{Al}^{3+}+3 \mathrm{e}^{-}(\text { in acid solution })} \\\ {\text { c. } \mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}} \\\\\ \quad{\mathrm{N}_{2} \mathrm{O} \rightarrow \mathrm{NO}_{3}^{-}+4 \mathrm{e}^{-}(\text { in acid solution })}\end{array} \end{equation}

Step-by-Step Solution

Answer

(a) Fe + Te²⁺ → Fe²⁺ + Te; (b) 2Al + 3IO₄⁻ → 3IO₃⁻ + 2Al³⁺; (c) N₂O + 2I₂ → 4I⁻ + NO₃⁻.

Step 1: Identify the Oxidation and Reduction Half-Reactions

The provided half-reactions involve oxidation and reduction processes. For each pair, identify which is the reduction half-reaction and which one is oxidation. Redox reactions involve electron transfer from the oxidizing agent to the reducing agent. For pair (a), - Oxidation: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{e}^{-} \) - Reduction: \( \mathrm{Te}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Te} \). Similarly, identify the corresponding reactions in (b) and (c).

Step 2: Equalize Electron Transfer

Ensure that the electrons lost in the oxidation reaction equal the electrons gained in the reduction reaction for each pair. This often involves multiplying the half-reactions by appropriate coefficients. - For pair (a), the electrons are already balanced since both involve 2 electrons. - For (b), IO₄⁻ gains 2 electrons, and Al loses 3, so multiply the IO₄⁻ reaction by 3 and the Al reaction by 2 to equalize at 6 electrons. - For (c), I₂ gains 2 electrons, while N₂O loses 4 electrons. Multiply the I₂ half-reaction by 2 to ensure both transfer 4 electrons.

Step 3: Add and Balance the Redox Equation

Combine the balanced half-reactions for each pair, ensuring mass balance and charge neutrality. - (a): Add the two reactions: \( \mathrm{Fe} + \mathrm{Te}^{2+} \rightarrow \mathrm{Fe}^{2+} + \mathrm{Te} \).- (b): Combine \( 3\cdot(\mathrm{IO}_4^- + 2\mathrm{e}^- \rightarrow \mathrm{IO}_3^-) + 2\cdot(\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{e}^-) \) to yield \( 2\mathrm{Al} + 3\mathrm{IO}_4^- \rightarrow 3\mathrm{IO}_3^- + 2\mathrm{Al}^{3+} \).- (c): Combine \( 2 \cdot(\mathrm{I}_2 + 2\mathrm{e}^- \rightarrow 2\mathrm{I}^-) + \mathrm{N}_2\mathrm{O} \rightarrow \mathrm{NO}_3^- + 4e^- \) to form \( \mathrm{N}_2\mathrm{O} + 2\mathrm{I}_2 \rightarrow 4\mathrm{I}^- + \mathrm{NO}_3^- \).

Key Concepts

Half-Reactions

Half-reactions are a vital component of balancing ionic redox equations. They showcase the separate events of oxidation and reduction, allowing us to thoroughly understand how electrons are transferred between substances. In a half-reaction:

• The oxidation half-reaction illustrates the loss of electrons.
• The reduction half-reaction reveals the gain of electrons.

For example, consider the half-reaction: \( ext{Fe} ightarrow ext{Fe}^{2+} + 2 ext{e}^- \). This clearly shows iron losing electrons, demonstrating oxidation. Conversely, the half-reaction \( ext{Te}^{2+} + 2 ext{e}^- ightarrow ext{Te} \) demonstrates reduction, as electrons are gained.
By looking at each half-reaction separately, we can ensure the overall redox equation's mass and charge are balanced, a critical step in comprehending ionic equations in chemistry.

Electron Transfer

In redox reactions, electron transfer is the heart of the process, driving the changes in oxidation states between reactants and products. It involves electrons moving from the substance that gets oxidized to the one that gets reduced.
In any given pair of redox half-reactions, equalizing the number of electrons lost and gained is crucial. For example:

• In the pair involving iron and tellurium, both gain and lose 2 electrons naturally, hence no adjustment is necessary.
• When pairing \( ext{IO}_4^- \) and aluminum, adjustments help achieve a balance of 6 electrons each, leading to the balanced equation \( 2 ext{Al} + 3 ext{IO}_4^- ightarrow 3 ext{IO}_3^- + 2 ext{Al}^{3+} \).

By ensuring total charges match, the resulting redox equation accurately reflects the true nature of these interactions.

Oxidation

Oxidation occurs when a substance loses electrons, consequently increasing its oxidation state. This process is a cornerstone of redox reactions, helping substances undergo transformation.
Considering the half-reaction: \( ext{Al} ightarrow ext{Al}^{3+} + 3 ext{e}^- \), it is evident that aluminum oxidizes by losing three electrons. This change translates into an increase in its oxidation state from 0 to +3.

• Another example is \( ext{Fe} ightarrow ext{Fe}^{2+} + 2 ext{e}^- \), where iron loses electrons and gets oxidized.

Recognizing substances undergoing oxidation is vital for understanding redox reactions as a whole, since these transformations are continually paired with reductions.

Reduction

Reduction is the opposite of oxidation, defined by the gain of electrons and a decrease in oxidation state. It's an essential part of redox reactions, simultaneously occurring alongside oxidation.
Consider the half-reaction \( ext{Te}^{2+} + 2 ext{e}^- ightarrow ext{Te} \). Here, tellurium gains electrons, becoming reduced. The reduction process results in a decrease in its oxidation state from +2 to 0.

• In the redox pair involving iodine, \( ext{I}_2 + 2 ext{e}^- ightarrow 2 ext{I}^- \) depicts iodine gaining electrons and reducing.

Understanding where reduction occurs in chemical reactions highlights the balance and interdependence between oxidation and reduction, facilitating the comprehension of more complex chemical interactions.