Question

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

Short answer

\(a. Fe + Te^{2+} \rightarrow Fe^{2+} + Te\) \(b. 2Al + 3IO_{4}^{-} \rightarrow 2Al^{3+} + 3IO_{3}^{-}\) (in acid solution) \(c. 2I_{2} + N_{2}O \rightarrow 4I^{-} + NO_{3}^{-}\) (in acid solution)

Step-by-step solution

a. Solution for Pair 1: Fe and Te half-reactions

First, let's identify the oxidation and reduction half-reactions: Fe → Fe²⁺ + 2e⁻ (Oxidation half-reaction) Te²⁺ + 2e⁻ → Te (Reduction half-reaction) Since the number of electrons lost in the oxidation half-reaction (2e⁻) is already equal to the number of electrons gained in the reduction half-reaction (2e⁻), we can directly add the two half-reactions to get the balanced ionic redox equation: \(Fe + Te^{2+} \rightarrow Fe^{2+} + Te\)

b. Solution for Pair 2: IO₄⁻ and Al half-reactions

First, let's identify the oxidation and reduction half-reactions: IO₄⁻ + 2e⁻ → IO₃⁻ (Reduction half-reaction) Al → Al³⁺ + 3e⁻ (Oxidation half-reaction) Here, the number of electrons lost in the oxidation half-reaction (3e⁻) is not equal to the number of electrons gained in the reduction half-reaction (2e⁻). We need to equalize the number of electrons transferred in both half-reactions. Multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2: 3(IO₄⁻ + 2e⁻ → IO₃⁻) (Reduction half-reaction multiplied by 3) 2(Al → Al³⁺ + 3e⁻) (Oxidation half-reaction multiplied by 2) Now the total number of electrons transferred in both half-reactions is equal (6e⁻). We can now combine the half-reactions to form the balanced ionic redox equation: \(2Al + 3IO_{4}^{-} \rightarrow 2Al^{3+} + 3IO_{3}^{-}\) (in acid solution)

c. Solution for Pair 3: I₂ and N₂O half-reactions

First, let's identify the oxidation and reduction half-reactions: I₂ + 2e⁻ → 2I⁻ (Reduction half-reaction) N₂O → NO₃⁻ + 4e⁻ (Oxidation half-reaction) Here, the number of electrons lost in the oxidation half-reaction (4e⁻) is not equal to the number of electrons gained in the reduction half-reaction (2e⁻). We need to equalize the number of electrons transferred in both half-reactions. Multiply the reduction half-reaction by 2: 2(I₂ + 2e⁻ → 2I⁻) (Reduction half-reaction multiplied by 2) Now the total number of electrons transferred in both half-reactions is equal (4e⁻). We can now combine the half-reactions to form the balanced ionic redox equation: \(2I_{2} + N_{2}O \rightarrow 4I^{-} + NO_{3}^{-}\) (in acid solution)

Key concepts

Redox Half-Reactions

At the heart of understanding redox chemistry is grasping what happens in redox half-reactions. These are the individual processes that occur during a redox reaction, with one substance gaining electrons (reduction) and another losing electrons (oxidation).

Let's look at the example provided: One half-reaction is \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-}\), representing the loss of electrons, thus we're talking about oxidation. The other \(\mathrm{Te}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Te}\) shows the gain of electrons, and this is reduction.

Understanding these processes is crucial because every redox reaction can be broken down into these two fundamental steps. By analyzing half-reactions, students can more easily track electron transfer, identify which species are oxidized and which are reduced, and learn the roles they play in the broader context of redox chemistry.

Balancing Redox Reactions

Balancing redox reactions can sometimes be tricky, but it is a necessary skill in chemistry. The key is ensuring that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

For example, in \(\mathrm{Al} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{e}^{-}\) and \(\mathrm{IO}_{4}^{-}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}\), we have a mismatch in electron count. By multiplying the reactions, we can balance the number of electrons to achieve a compatible total, which allows us to combine the half-reactions into a complete, balanced redox equation.

Step-by-Step Balancing

First, identify the electrons exchanged in each half-reaction and then use multipliers to equate those numbers. Once both half-reactions involve the same amount of electron transfer, add them together, conserving both mass and charge, to obtain the balanced redox equation.

Oxidation and Reduction

Oxidation and reduction are two sides of the same chemical coin. Oxidation involves an atom or molecule losing electrons, and as a result, often increases in oxidation state. In contrast, reduction is the gain of electrons, accompanied by a decrease in oxidation state.

The classic mnemonic 'OIL RIG' helps remember this - Oxidation Is Loss, Reduction Is Gain. For instance, when \(\mathrm{Fe}\) becomes \(\mathrm{Fe}^{2+}\), it loses two electrons, it's oxidized. Conversely, when \(\mathrm{Te}^{2+}\) gains two electrons to become \(\mathrm{Te}\), it's reduced.

Understanding these concepts aids in interpreting chemical reactions and predicting the behavior of substances in a redox process. Both processes occur simultaneously during a redox reaction, meaning there will always be something being oxidized while something else is reduced - this is the essence of redox chemistry.