Question

Balance each of the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method. a. \(\mathrm{Al}(s)+\mathrm{H}^{+}(a q) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{S}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{S}(s)+\mathrm{NO}(g)\) c. \(\mathrm{I}_{2}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{HCl}(g)\) d. \(\mathrm{AsO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{AsO}_{3}^{-}(a q)+\mathrm{S}(s)\)

Short answer

The short version of the balanced oxidation-reduction reaction for part a is: \(2\mathrm{Al}(s)+6\mathrm{H}^{+}(a q) \rightarrow 2\mathrm{Al}^{3+}(a q)+3\mathrm{H}_{2}(g)\). Follow the same procedure for the remaining parts to find their balanced reactions.

Step-by-step solution

Identify the oxidation and reduction half-reactions

For this reaction, we have: Oxidation half-reaction: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)\) (Al loses 3 electrons) Reduction half-reaction: \(\mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2}(g)\) (H gains 2 electrons)

Balance the atoms and charges in each half-reaction

For the oxidation half-reaction: 1. Balance the Al atoms: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)\) (already balanced) 2. Balance the charges by adding electrons: \(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+3e^{-}\) For the reduction half-reaction: 1. Balance the H atoms: \(2\mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2}(g)\) 2. Balance the charges by adding electrons: \(2\mathrm{H}^{+}(a q)+2e^{-} \rightarrow \mathrm{H}_{2}(g)\)

Combine the half-reactions to achieve the balanced overall reaction

To do this, ensure the number of electrons transferred in both half-reactions are equal. Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3: Oxidation: \(2(\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+3e^{-})\) Reduction: \(3(2\mathrm{H}^{+}(a q)+2e^{-}\rightarrow \mathrm{H}_{2}(g))\) Oxidation: \(2\mathrm{Al}(s) \rightarrow 2\mathrm{Al}^{3+}(a q)+6e^{-}\) Reduction: \(6\mathrm{H}^{+}(a q)+6e^{-}\rightarrow 3\mathrm{H}_{2}(g)\) Now, add the two half-reactions and cancel out the electrons: Balanced reaction: \(2\mathrm{Al}(s)+6\mathrm{H}^{+}(a q) \rightarrow 2\mathrm{Al}^{3+}(a q)+3\mathrm{H}_{2}(g)\) Please follow similar steps for the remaining reactions.

Key concepts

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions, especially in an acidic setting. This method involves splitting the overall reaction into two separate equations that represent either oxidation (loss of electrons) or reduction (gain of electrons).

To illustrate, let's focus on balancing the reaction between solid aluminum and hydrogen ions to produce aluminum ions and hydrogen gas. The first step is to break them down into half-reactions: one for the oxidation of aluminum, and the other for the reduction of hydrogen ions. After that, each half-reaction needs to be balanced for both atoms and charges. This is done by first equalizing the number of atoms of the element being oxidized or reduced, followed by adding electrons to balance the charges.

Next, to ensure charge and mass are conserved, the half-reactions are then adjusted so that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction. A useful tip for more complex reactions involves balancing elements other than oxygen and hydrogen first, and leaving these till last, while keeping in mind the acidic environment can provide H⁺ ions and H₂O to assist in the balance.

Finally, the half-reactions are recombined, and simplifications are made where possible, typically canceling the electrons that appear on both sides of the combined equation. The end result is a correctly balanced equation that takes into account both mass and charge conservation.

Oxidation and Reduction

Oxidation and reduction are two key concepts in redox reactions. Oxidation refers to the loss of electrons by a substance, whereas reduction refers to the gain of electrons. These processes can be remembered by the mnemonic 'OIL RIG' - oxidation is loss, reduction is gain.

In our example, aluminum solid (Al) is oxidized as it loses three electrons to become aluminum ions (Al³⁺), and the hydrogen ions (H⁺) are reduced as they gain electrons to form hydrogen gas (H₂).

Oxidation and reduction always occur together; it's impossible to have one without the other because electrons must come from somewhere and go somewhere. When writing half-reactions, it is crucial to indicate clearly whether each reaction represents oxidation or reduction. This clear distinction helps in properly balancing the full redox reaction. It's important to always check that the number of electrons lost in the oxidation process equals the number of electrons gained in the reduction process.

Acidic Solution

In an acidic solution, we deal with an excess of H⁺ ions, which must be accommodated in the balancing process. When using the half-reaction method in an acidic solution, after balancing the atoms of elements being oxidized or reduced, balance the oxygen atoms next by adding H₂O as needed. Following this, hydrogen atoms are balanced by adding H⁺ ions, with the final step being the balancing of charges with electrons.

It's essential to remember that the presence of H⁺ ions in the environment of the reaction affects how we balance the half-reactions. For example, oxygen can often be balanced using water molecules, and then any additional hydrogen provided by water is balanced by adding H⁺. The H⁺ ions are integral to the balancing process in acidic solutions and cannot be ignored.

This approach helps adjust the redox equation to ensure that it gives an accurate representation of the reaction's chemistry in an acid-based context. In a basic solution, the process would be different, using OH^- ions instead of H⁺ to balance excess hydrogens.

Electrons Transfer

Electron transfer is the fundamental process that drives all redox reactions. It's also the central aspect of the half-reaction method of balancing redox equations. Each half-reaction explicitly shows the movement of electrons — either their loss in the case of oxidation or their gain in the case of reduction.

When balancing, it's crucial to ensure the same number of electrons is transferred in both half-reactions before recombining them. This is typically achieved by finding the lowest common multiple of electrons in both half-reactions and multiplying the reactions accordingly to equalize electron exchange. For instance, in our original problem, the oxidation half-reaction shows the loss of 3 electrons, while the reduction half-reaction shows a gain of 2 electrons. To balance them, we multiply the oxidation half-reaction by 2 and the reduction by 3, which gives us 6 electrons in both cases.

Paying attention to electron transfer not only helps with balancing the overall equation but also aids in understanding the chemical changes the reactants undergo during the reaction. Correctly representing this electron exchange is essential for presenting a balanced and accurate depiction of the redox reaction.