Problem 5 - Chemistry and Chemical Reactivity

Question

The following equations represent various ways of obtaining transition metals from their compounds. Balance each equation. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\) (b) \(\operatorname{TiCl}_{4}(\ell)+\operatorname{Mg}(\mathrm{s}) \longrightarrow \operatorname{Ti}(\mathrm{s})+\operatorname{Mg} \mathrm{Cl}_{2}(\mathrm{s})\) (c) \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow\) \(\mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})\) (d) \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Cr}_{2} \mathrm{O}_{3} + 2\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3} + 2\mathrm{Cr}\), (b) \(\mathrm{TiCl}_{4} + 2\mathrm{Mg} \rightarrow \mathrm{Ti} + 2\mathrm{MgCl}_{2}\), (c) \(2[\mathrm{Ag(CN)}_{2}]^{-} + \mathrm{Zn} \rightarrow 2\mathrm{Ag} + [\mathrm{Zn(CN)}_{4}]^{2-}\), (d) \(3\mathrm{Mn}_{3} \mathrm{O}_{4} + 8\mathrm{Al} \rightarrow 9\mathrm{Mn} + 4\mathrm{Al}_{2} \mathrm{O}_{3}\).

Step 1: Identify the components

First, let's identify the compounds and elements in each of the given equations that need to be balanced. By doing so, we can understand the types of atoms we’re dealing with in each reaction. For instance, reaction (a) has chromium (Cr), aluminum (Al), and oxygen (O).

Step 2: Balancing equation (a)

In equation (a), the reaction is \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s})\longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\). First, balance the chromium atoms by putting 2 \(\mathrm{Cr}\) on the product side. Next, balance aluminum by placing 2 \(\mathrm{Al}\) on the reactant side. Finally, balance the oxygen; the equation is already balanced for oxygen.

Step 3: Write balanced equation (a)

The balanced equation is: \( \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) + 2 \mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) + 2 \mathrm{Cr}(\mathrm{s}) \).

Step 4: Balancing equation (b)

In equation (b), the reaction is \( \operatorname{TiCl}_{4}(\ell)+\operatorname{Mg}(\mathrm{s})\longrightarrow \operatorname{Ti}(\mathrm{s})+\operatorname{MgCl}_{2}(\mathrm{s})\). To balance it, place 2 \(\mathrm{Mg}\) on the reactant side to balance chlorine atoms, as there are four chlorine atoms to account for.

Step 5: Write balanced equation (b)

The balanced equation is: \( \operatorname{TiCl}_{4}(\ell) + 2 \operatorname{Mg}(\mathrm{s}) \rightarrow \operatorname{Ti}(\mathrm{s}) + 2 \mathrm{MgCl}_{2}(\mathrm{s}) \).

Step 6: Balancing equation (c)

For equation (c), which is \( \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq}) \), balance the silver atoms first by putting a coefficient of 2 before \(\mathrm{Ag}\) and \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\). There are 4 CN ions now, balanced with 1 Zn.

Step 7: Write balanced equation (c)

The balanced equation is: \( 2 \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq}) + \mathrm{Zn}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}(\mathrm{s}) + \left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq}) \).

Step 8: Balancing equation (d)

Equation (d) is \( \mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \). First, place 4 \(\mathrm{Al}\) atoms on the reactant side to balance aluminum. Next, balance manganese by placing a coefficient of 3 before \(\mathrm{Mn}\).

Step 9: Write balanced equation (d)

The final balanced equation is: \( 3 \mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s}) + 8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Mn}(\mathrm{s}) + 4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \).

Key Concepts

Transition Metals

Transition metals are a unique group of elements found in the center of the periodic table. They are located in groups 3-12 and are known for their ability to form a variety of oxidation states. This means they can lose different numbers of electrons, making them very versatile in chemical reactions. Transition metals include elements such as iron (Fe), copper (Cu), and chromium (Cr). Unlike other elements, they often form colored compounds and act as catalysts in chemical reactions.

These elements are essential in industrial processes and biological systems due to their special properties.

• Variable oxidation states: Transition metals can form compounds in multiple oxidation states, which is central to their chemistry.
• Complex ions: They can form stable complexes with various ligands, which makes them important in areas like biochemistry and synthetic chemistry.
• Catalytic activity: Many transition metals serve as catalysts in reactions, speeding up processes without being consumed.

Understanding transition metals is crucial for mastering the art of balancing equations as they showcase the diversity of chemistry in practice.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It is grounded in the conservation of mass, where the mass of the reactants equals the mass of the products. This concept helps us balance chemical equations by ensuring that atoms are neither created nor destroyed, allowing us to predict the amounts of substances consumed and produced in reactions.

To balance a chemical equation, follow these steps:

• List the number of atoms: Count the number of each type of atom on both sides of the equation.
• Use coefficients judiciously: Adjust coefficients to make the number of atoms of each element equal on both sides.
• Check your work: Reassess the equation to ensure all atoms and charges are balanced.

Mastering stoichiometry is key to understanding how different chemicals interact and predicting the outcomes of various reactions.

Redox Reactions

Redox reactions are chemical reactions involving the transfer of electrons between two substances. The term "redox" is derived from "reduction" (gaining electrons) and "oxidation" (losing electrons). These reactions are essential for many processes, including the extraction of metals from their ores, like in the provided exercise with chromium and manganese. In every redox reaction, there is always one element that gets oxidized and another that gets reduced.

Here’s how they work:

• Oxidation: The substance that loses electrons is said to be oxidized, and its oxidation state increases.
• Reduction: The substance that gains electrons is said to be reduced, and its oxidation state decreases.

Redox reactions play a critical role in energy transfer processes, such as respiration and photosynthesis in living organisms.

Oxidation-Reduction Processes

Oxidation-reduction processes are central to redox reactions. These processes dictate how the transfer of electrons is carried out in a chemical reaction. In these processes, it is important to identify the oxidizing and reducing agents.

An oxidizing agent is a substance that accepts electrons, making it the entity that gets reduced itself. On the other hand, a reducing agent donates electrons, causing it to be oxidized during the process.
To identify these agents, consider the following:

• Change in oxidation states: Identify which elements have changed oxidation states.
• Oxidizing agents: A substance gains electrons, reducing its oxidation state.
• Reducing agents: A substance loses electrons, increasing its oxidation state.

Understanding these principles is vital for solving complex chemical reactions, particularly involving transition metals, as illustrated in the original exercise.