Problem 35 - Chemistry and Chemical Reactivity

Question

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{BrO}_{3}^{-}\) (d) \(\mathrm{CaH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) (e) \(\mathrm{H}_{1} \mathrm{SiO}_{4}\) (c) \(\mathrm{F}^{-}\) (f) \(\mathrm{HSO}_{4}^{-}\)

Step-by-Step Solution

Answer

(a) Br: +5; (b) C: +3; (c) F: -1; (d) Ca: +2, H: -1; (e) Si: +7; (f) S: +6

Step 1: Oxidation Number Basics

Understand that oxidation numbers are a way to keep track of electrons in compounds and ions. The oxidation number for an element in its elemental state is 0, while in a compound or ion, it can vary.

Step 2: Determine Oxidation Number for Each Atom in BrO3^-

For the bromate ion, \(\mathrm{BrO}_3^-\), assign oxidation numbers:- Assume oxygen has an oxidation number of -2. - Let the oxidation number of bromine be \(+x\).- Write the equation: \(x + 3(-2) = -1\) from the ion charge.- Solve for \(x\): \(x - 6 = -1\), thus \(x = +5\).- So, Br: +5 and O: -2

Step 3: Determine Oxidation Number for Each Atom in CaH2

For calcium hydride, \(\mathrm{CaH}_2\):- Calcium is in Group 2, typically having an oxidation number of +2.- Let the oxidation number of hydrogen here (unusually, as hydride) be -1.- For the compound \(\mathrm{CaH}_2\), equation: \(+2 + 2(-1) = 0\) - Ca: +2 and H: -1

Step 4: Determine Oxidation Number for Each Atom in C2O4^2-

For oxalate ion, \(\mathrm{C}_2\mathrm{O}_4^{2-}\):- Assume oxygen is -2.- Let the oxidation number of carbon be \(+x\).- Equation: \(2x + 4(-2) = -2\).- Solve: \(2x - 8 = -2\), hence \(2x = 6\), \(x = +3\).- C: +3 and O: -2

Step 5: Determine Oxidation Number for Each Atom in HSiO4

For hydrogen silicate, \(\mathrm{HSiO}_4\):- Assume hydrogen is +1, oxygen is -2.- Let silicon's oxidation number be \(+x\).- Equation: \(+1 + x + 4(-2) = 0\).- Simplify: \(+1 + x - 8 = 0\), \(x = +7\)._ H: +1, Si: +7, O: -2

Step 6: Determine Oxidation Number for Atom in F^-

For fluoride ion, \(\mathrm{F}^-\):- Fluorine typically has an oxidation number of -1.- For \(\mathrm{F}^-\), no other elements are present, maintain: F: -1

Step 7: Determine Oxidation Number for Each Atom in HSO4^-

For bisulfate ion, \(\mathrm{HSO}_4^-\):- Assume hydrogen is +1, oxygen is -2.- Let sulfur's oxidation number be \(+x\).- Equation: \(+1 + x + 4(-2) = -1\).- Solve: \(+1 + x - 8 = -1\), \(x = +6\).- H: +1, S: +6, O: -2

Key Concepts

Electrochemistry

Electrochemistry is the fascinating study of the relationship between chemical reactions and electricity. In simple terms, it's about how chemical energy transforms into electric energy and vice versa. This field is crucial in your everyday life because it's involved in batteries, fuel cells, and even biological processes.

At its core, electrochemistry focuses on redox reactions – these are reactions where electrons are transferred between different substances. The oxidation numbers help us to keep track of these electrons. Understanding the flow of electrons allows us to harness energy effectively. We see this in devices like batteries, where chemical reactions produce electrical energy that powers our gadgets. In essence, by oxidizing and reducing specific compounds, we either release or store energy.

A great practical application of electrochemistry you encounter daily is in galvanic cells or batteries. Here, the redox reactions create a flow of electrons, and this current is what powers your devices. Electrochemistry also plays a role in corrosion prevention, where understanding the movement of electrons can lead to better protective measures for metals.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are essential chemical processes where there's a transfer of electrons between molecules. These reactions break into two parts: oxidation and reduction. During oxidation, a molecule loses electrons, while in reduction, a molecule gains electrons. These two processes always happen together – you can't have one without the other.

To describe how electrons move in redox reactions, we use oxidation numbers. An oxidation number is simply a tool for keeping track of how many electrons an atom has gained or lost. For example, when finding the oxidation number in \( \mathrm{BrO}_3^- \), we see that bromine is at +5 because it "gives away" five electrons to keep the compound electrically stable.

Understanding redox reactions is crucial for many chemical processes, including metabolic pathways and combustion. In biology, redox reactions take part in cellular respiration – the process your body uses to generate energy from glucose. In industry, they help create essential chemicals and materials. Knowing how to balance these reactions is also a vital skill in chemistry, ensuring that the number of electrons lost equals the number gained.

Chemical Bonds

Chemical bonds are attractions that bind atoms together, forming molecules and compounds necessary for life as we know it. The type of chemical bond determines the properties of the substance, like melting point, conductivity, and hardness. The main types include covalent bonds, ionic bonds, and metallic bonds.

Covalent bonds form when atoms share electrons, usually between non-metal atoms. In these bonds, the shared electrons allow atoms to attain stability. For instance, the compound \( \mathrm{C}_2\mathrm{O}_4^{2-} \) features covalent bonding between carbon and oxygen. Here, electrons shared allow for effective molecular stability.

Ionic bonds occur when electrons are transferred from one atom to another, leading to attraction between oppositely charged ions. This kind of bond is common in compounds like \( \mathrm{CaH}_2 \) where calcium and hydrogen form a stable lattice structure. The calcium gives up electrons while the hydrogen gains them, creating strong electrostatic forces between the ions.

Each bond type plays a role in the compound's behavior and functionality, which is why learning them can be so rewarding and fascinating. By understanding chemical bonds, you gain insight into how substances interact, why they behave the way they do, and how to manipulate these interactions for practical purposes.