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. \(\mathrm{S}^{2-}(a q)+\mathrm{N O}_{3}^{-}(a q) \rightarrow \mathrm{S}(s)+\mathrm{N O}(g)\) c. \(\mathrm{I}_{2}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{HCl}(g)\) d. \(\mathrm{A s O}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{A s O}_{3}^{-}(a q)+\mathrm{S}(s)\)

Short answer

a: 2Al + 6H⁺ -> 2Al³⁺ + 3H₂ b: S²⁻ + 2NO₃⁻ + 4H⁺ -> S + 2NO + 2H₂O c: 2I₂ + 10H₂O + 5Cl₂ -> 2IO₃⁻ + 12H⁺ + 10Cl⁻ d: S²⁻ + 2AsO₄⁻ + 4H⁺ -> S + 2AsO₃⁻ + 2H₂O

Step-by-step solution

Identify Oxidation and Reduction half-reactions

Our oxidation half-reaction is Al -> Al³⁺, where Al loses 3 electrons. And our reduction half-reaction is H⁺ -> H₂, where H⁺ gains 2 electrons.

Balance Elements and Charges

For the oxidation half-reaction: Al -> Al³⁺ + 3e⁻ For the reduction half-reaction: 2H⁺ + 2e⁻ -> H₂

Combine half-reactions

Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3, and then add them together to obtain the balanced equation: 2Al + 6H⁺ -> 2Al³⁺ + 3H₂ b. \(\mathrm{S}^{2-}(a q)+\mathrm{N O}_{3}^{-}(a q) \rightarrow \mathrm{S}(s)+\mathrm{N O}(g)\)

Identify Oxidation and Reduction half-reactions

Our oxidation half-reaction is S²⁻ -> S, where S loses 2 electrons. And our reduction half-reaction is NO₃⁻ -> NO, where N gains 2 electrons.

Balance Elements and Charges

For the oxidation half-reaction: S²⁻ -> S + 2e⁻ For the reduction half-reaction, we first balance N and O: NO₃⁻ + 3e⁻ -> NO + 2H₂O Then balance H: NO₃⁻ + 3e⁻ + 4H⁺ -> NO + 2H₂O

Combine half-reactions

Both half-reactions have the same number of electrons, so just add them together to obtain the balanced equation: S²⁻ + 2NO₃⁻ + 4H⁺ -> S + 2NO + 2H₂O c. \(\mathrm{I}_{2}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{HCl}(g)\)

Identify Oxidation and Reduction half-reactions

Our oxidation half-reaction is Cl₂ -> 2Cl⁻, where Cl gains 2 electrons. And our reduction half-reaction is I₂ -> 2IO₃⁻, where I loses 10 electrons.

Balance Elements and Charges

For the oxidation half-reaction: Cl₂ + 2e⁻ -> 2Cl⁻ For the reduction half-reaction, we first balance I and O: 2I₂ + 10e⁻ -> 2IO₃⁻ Then balance H: 2I₂ + 10e⁻ + 6H₂O -> 2IO₃⁻ + 12H⁺

Combine half-reactions

Multiply the oxidation half-reaction by 5 and the reduction half-reaction remains the same. Then add them together to obtain the balanced equation: 2I₂ + 10H₂O + 5Cl₂ -> 2IO₃⁻ + 12H⁺ + 10Cl⁻ d. \(\mathrm{A s O}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{A s O}_{3}^{-}(a q)+\mathrm{S}(s)\)

Identify Oxidation and Reduction half-reactions

Our oxidation half-reaction is S²⁻ -> S, where S loses 2 electrons. And our reduction half-reaction is AsO₄⁻ -> AsO₃⁻, where As gains 1 electron.

Balance Elements and Charges

For the oxidation half-reaction: S²⁻ -> S + 2e⁻ For the reduction half-reaction, we first balance As and O: AsO₄⁻ + e⁻ -> AsO₃⁻ Then balance H: AsO₄⁻ + e⁻ + 2H⁺ -> AsO₃⁻ + H₂O

Combine half-reactions

Multiply the oxidation half-reaction by 1 and the reduction half-reaction by 2. Then add them together to obtain the balanced equation: S²⁻ + 2AsO₄⁻ + 4H⁺ -> S + 2AsO₃⁻ + 2H₂O

Key concepts

Half-Reaction Method

Understanding the half-reaction method is critical when it comes to balancing redox reactions in chemistry. This method simplifies complex equations by breaking them down into two separate processes – the oxidation half-reaction and the reduction half-reaction.

The first step in this technique involves separating the original redox reaction into two halves. Each half represents either the loss of electrons (oxidation) or the gain of electrons (reduction). It's similar to dissecting a convoluted story into two simpler narratives that are easier to follow. Once you've identified these halves, you'll need to balance the atoms and charges for each half-reaction individually. This often involves adding water, hydrogen ions, or electrons to equalize the number of atoms on both sides. The final step is combining these balanced half-reactions, ensuring that the electrons lost in the oxidation half-reaction are equal to those gained in the reduction half-reaction.

This step-by-step process aids in making the overall redox reaction balanced not only in terms of atoms but also electrical charge, paving the way to a smooth resolution where everything adds up perfectly.

Oxidation Half-Reaction

When diving into the depths of an oxidation half-reaction, you're exploring the part of the redox process where a substance loses electrons. In this type of reaction, we witness an increase in oxidation state, akin to someone giving away slices of their favorite pie.

An easy way to remember oxidation is the mnemonic 'LEO says GER', where LEO stands for 'Loss of Electrons is Oxidation'. The half-reaction method helps us balance these tricky chemical equations by allowing us to focus solely on the substance that's handing over electrons. For instance, in the example with aluminum, \(\mathrm{Al}\rightarrow \mathrm{Al}^{3+} + 3e^-\), aluminum sheds three electrons to form \(\mathrm{Al}^{3+}\). This standalone perspective simplifies balancing, as we can address each part without the distraction of the reaction's other elements.

Reduction Half-Reaction

Reduction half-reactions are the complementary tales to the oxidation narrative; they focus on substances gaining electrons. A reduction is essentially a chemical embrace, where a species welcomes extra electrons, reducing its oxidation state.

Remembering the other half of 'LEO says GER', GER stands for 'Gain of Electrons is Reduction'. The reduction half-reaction holds hands with the need to balance charges and elements separately from the oxidation process. As shown in our example, hydrogen ions acquire electrons to form hydrogen gas, \(2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2\). By isolating this section of the reaction, the complexity of balancing the overall equation is reduced significantly, enabling us to adjust any disparities in atom counts and charges with greater precision.

Stoichiometry

Stoichiometry serves as the mathematical backbone of chemical reactions, providing the quantitative relationship between reactants and products. Picture it as the scales of chemistry, ensuring each side weighs the same. When applying stoichiometry to the half-reaction method, it's akin to baking with the precise amount of each ingredient to get a perfect result.

In the context of balancing redox reactions, stoichiometry ensures that for every electron lost in one half-reaction, there is an electron gained in the other. As seen in our examples, we might multiply one half-reaction by a certain factor to achieve this balance, much like adjusting a recipe to get the right proportions. This numerical dance is key to ensuring both mass and charge are conserved, resulting in a completely balanced chemical equation.

Electrochemistry

Electrochemistry is the study of the movement of electrons and their role in creating electricity through chemical reactions. This field is where oxidation and reduction reactions become more than just a method for balancing chemical equations; they're the heartbeats of batteries and the breath of electronic devices.

Within this intriguing study, redox reactions are the driving force behind the conversion of chemical energy into electrical energy and vice versa. Understanding how to balance these reactions through the half-reaction method is not just academic exercise; it's foundational knowledge for developing new energy solutions and diving into the intricate world where chemistry meets electricity. Hence, the carefully balanced equations you achieve on paper mirror the precise and harmonious flow of electrons necessary for the functioning of electrochemical cells. It's a perfect blend of theory and application, chemistry and innovation.