Problem 89 - Chemistry: An Atoms-Focused Approach

Question

The following chemical reactions have helped to shape Earth's crust. Determine the oxidation numbers of all the elements in the reactants and products, and identify which elements are oxidized and which are reduced: a. \(3 \mathrm{SiO}_{2}(s)+2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) \rightarrow 3 \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)+\mathrm{O}_{2}(g)\) b. \(\operatorname{si} \mathrm{O}_{2}(s)+2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)\) c. \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 4 \mathrm{Fe}(\mathrm{OH})_{3}(s)\)

Step-by-Step Solution

Answer

Question: Identify the elements that got oxidized and reduced in the given reactions: a. \(3 \mathrm{SiO}_{2}(s)+2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) \rightarrow 3 \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)+\mathrm{O}_{2}(g)\) b. \(\operatorname{si} \mathrm{O}_{2}(s)+2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)\) c. \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 4 \mathrm{Fe}(\mathrm{OH})_{3}(s)\) Answer: a. Fe got reduced. b. Fe got oxidized, and O got reduced. c. Fe got oxidized, and O got reduced.

Step 1: a. Determine oxidation numbers and identify oxidation and reduction

: First, write out the given reaction, and then we'll find the oxidation numbers. \(3 \mathrm{SiO}_{2}(s)+2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s) \rightarrow 3 \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)+\mathrm{O}_{2}(g)\) Oxidation numbers for each element are: - O in \(\mathrm{SiO}_2\): -2 (oxygen usually has oxidation number -2) - Si in \(\mathrm{SiO}_2\): +4 (since there are two oxygen atoms, and the compound is neutral) - O in \(\mathrm{Fe}_3\mathrm{O}_4\): -2 - Fe in \(\mathrm{Fe}_3\mathrm{O}_4\): +8/3 (\(8/3 \cdot 3 = 8\) for three Fe atoms and four O atoms with -2 each, in a neutral compound) - O in \(\mathrm{Fe}_2\mathrm{SiO}_4\): -2 - Si in \(\mathrm{Fe}_2\mathrm{SiO}_4\): +4 - Fe in \(\mathrm{Fe}_2\mathrm{SiO}_4\): +2 (as 2 Fe atoms are there and two O atoms with -4 charges, in a neutral compound) - O in \(\mathrm{O}_2\): 0 (uncombined state) Now we can identify the oxidized and reduced elements: - Si: no change in oxidation number - O: no change in oxidation number - Fe: Oxidation number changed from +8/3 to +2, it got reduced (oxidation number decreased)

Step 2: b. Determine oxidation numbers and identify oxidation and reduction

: First, write out the given reaction, and then we'll find the oxidation numbers. \(\operatorname{si} \mathrm{O}_{2}(s)+2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{SiO}_{4}(s)\) Oxidation numbers for each element are: - O in \(\mathrm{SiO}_2\): -2 - Si in \(\mathrm{SiO}_2\): +4 - O in \(\mathrm{O}_2\): 0 (uncombined state) - Si in \(\mathrm{Fe}_2\mathrm{SiO}_4\): +4 - Fe in \(\mathrm{Fe}_2\mathrm{SiO}_4\): +2 - Fe: 0 (uncombined state) Now we can identify the oxidized and reduced elements: - Si: no change in oxidation number - Fe: Oxidation number changed from 0 to +2, it got oxidized (oxidation number increased) - O: Oxidation number changed from 0 to -2, it got reduced (oxidation number decreased)

Step 3: c. Determine oxidation numbers and identify oxidation and reduction

: First, write out the given reaction, and then we'll find the oxidation numbers. \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 4 \mathrm{Fe}(\mathrm{OH})_{3}(s)\) Oxidation numbers for each element are: - O in \(\mathrm{FeO}\): -2 - Fe in \(\mathrm{FeO}\): +2 (since there is one oxygen atom with -2 charge, and the compound is neutral) - O in \(\mathrm{O}_2\): 0 (uncombined state) - H in \(\mathrm{H}_2\mathrm{O}\): +1 (hydrogen usually has oxidation number +1) - O in \(\mathrm{H}_2\mathrm{O}\): -2 - O in \(\mathrm{Fe}(\mathrm{OH})_3\): -2 - H in \(\mathrm{Fe}(\mathrm{OH})_3\): +1 - Fe in \(\mathrm{Fe}(\mathrm{OH})_3\): +3 (since there are three oxygen and three hydrogen atoms with a total charge of -6, and the compound is neutral) Now we can identify the oxidized and reduced elements: - Fe: Oxidation number changed from +2 to +3, it got oxidized (oxidation number increased) - O: Oxidation number changed from 0 to -2, it got reduced (oxidation number decreased) - H: no change in oxidation number

Key Concepts

Oxidation Numbers

In oxidation-reduction (redox) reactions, oxidation numbers are crucial for keeping track of electron transfer. They help us identify which elements are oxidized and which are reduced. An oxidation number represents the charge an element would have if all bonds were ionic. Here are some rules to help you find oxidation numbers:

• Elements in their natural state have an oxidation number of zero. For example, the oxidation number of O in O₂ is 0.
• The oxidation number for oxygen in most compounds is -2. However, there are exceptions such as in peroxides where it's -1.
• Hydrogen typically has an oxidation number of +1 when bonded to nonmetals and -1 when bonded to metals.
• For ions, the oxidation number equals the charge of the ion.
• The sum of the oxidation numbers in a neutral compound is zero, while in a polyatomic ion, it is equal to the ion's charge.

In the provided reactions, tracking the changes in oxidation numbers helps us see which elements have lost or gained electrons. The changes in these numbers from reactants to products indicate oxidation or reduction. For example, when iron's oxidation number changes from +8/3 in Fe₃O₄ to +2 in Fe₂SiO₄, it's reduced.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and this often involves changes in oxidation states. In redox reactions, oxidation and reduction occur simultaneously.

• Oxidation is the loss of electrons, usually resulting in an increase in oxidation number.
• Reduction is the gain of electrons, leading to a decrease in oxidation number.

These reactions include

• exchange of elements, such as SiO₂ with Fe,
• the combination of new elements, like forming Fe(OH)₃ from FeO and H₂O,
• rearrangements, which alter bonds to form different structures like Fe₂SiO₄.

Observing oxidation and reduction helps us understand how elements interact. They show us that elements can become part of new compounds, contributing to the development of diverse chemical characteristics in the earth's environment.

Earth's Crust Chemistry

The chemistry of Earth's crust is shaped by redox reactions, playing a pivotal role in the formation and transformation of minerals. The crust is primarily composed of oxygen, silicon, aluminum, iron, calcium, sodium, potassium, and magnesium, which exist mostly in oxidized forms. Silicon dioxide (SiO₂) is a primary component of many minerals such as quartz.

• Redox reactions help in the formation of complex minerals like fayalite (Fe₂SiO₄), where iron and silicon combine with oxygen.
• Iron, existing in different oxidation states, can switch between ferric (Fe³⁺) and ferrous (Fe²⁺) forms, influencing mineral colors and characteristics.
• The dynamic nature of these reactions allows for the formation of diverse geological features, including mountains and valleys, through processes such as weathering and erosion.

These processes not only help shape physical landscapes but also contribute to vital cycles like the carbon and water cycles. By understanding the redox processes in these chemical reactions, we gain insights into the geological and environmental changes on Earth.