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) \(\operatorname{Ti} \mathrm{Cl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s})\) (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) 1 Cr₂O₃ + 2 Al → 1 Al₂O₃ + 2 Cr, (b) 1 TiCl₄ + 2 Mg → 1 Ti + 2 MgCl₂, (c) 2 [Ag(CN)₂]⁻ + 1 Zn → 2 Ag + 1 [Zn(CN)₄]²⁻, (d) 1 Mn₃O₄ + 4 Al → 3 Mn + 2 Al₂O₃.

Step 1: Analyzing Reaction (a)

In the reaction \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\), we need to balance the chromium (Cr), oxygen (O), and aluminum (Al) atoms. Notice that there are 2 Cr, 3 O, and 2 Al atoms in \(\mathrm{Cr}_2\mathrm{O}_3\) and \(\mathrm{Al}_2\mathrm{O}_3\) respectively. Start by balancing chromium.

Step 2: Balancing Chromium and Oxygen in Reaction (a)

Balance Chromium by writing 2Cr on the products side: \[ \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) + \mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) + 2\mathrm{Cr}(\mathrm{s}) \] Oxygen is already balanced with 3 atoms on both sides.

Step 3: Balancing Aluminum in Reaction (a)

Now, balance the aluminum atoms by placing 2Al on the reactants side: \[ \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) + 2\mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) + 2\mathrm{Cr}(\mathrm{s}) \] This gives us a balanced equation.

Step 4: Analyzing Reaction (b)

In the reaction \(\operatorname{Ti} \mathrm{Cl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{MgCl}_{2}(\mathrm{s})\), balance the titanium (Ti), chlorine (Cl), and magnesium (Mg) atoms.

Step 5: Balancing Reaction (b)

Balance by writing 2 MgCl2 to account for the 4 Cl atoms introduced by TiCl4: \[ \operatorname{Ti} \mathrm{Cl}_{4}(\ell) + 2\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s}) + 2\mathrm{MgCl}_{2}(\mathrm{s}) \] Now we have 1 Ti, 4 Cl, and 2 Mg on both sides.

Step 6: Analyzing Reaction (c)

For the reaction \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})\), balance silver (Ag), cyanide (CN), and zinc (Zn).

Step 7: Balancing Reaction (c)

Balance by placing 2 in front of \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) and 2Ag:\[ 2 \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq}) + \mathrm{Zn}(\mathrm{s}) \rightarrow 2\mathrm{Ag}(\mathrm{s}) + \left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq}) \] This equation is now balanced with the metals and cyanide ions.

Step 8: Analyzing Reaction (d)

In the reaction \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\), balance manganese (Mn), oxygen (O), and aluminum (Al) atoms.

Step 9: Balancing Manganese and Oxygen in Reaction (d)

Balance manganese by writing 3Mn on the products side:\[ \mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s}) + \mathrm{Al}(\mathrm{s}) \rightarrow 3\mathrm{Mn}(\mathrm{s}) + \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \] Oxygen atoms are balanced allowing the placement of suitable aluminum.

Step 10: Balancing Aluminum in Reaction (d)

To balance aluminum, place 4Al on the reactants side:\[ \mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s}) + 4\mathrm{Al}(\mathrm{s}) \rightarrow 3\mathrm{Mn}(\mathrm{s}) + 2\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \] The equation is balanced with respective Mn, Al, and O atoms.

Key Concepts

Transition Metals in Chemical Reactions

Transition metals are elements that can be found in the d-block of the periodic table. They are known for their ability to form various oxidation states, which is a key feature that makes them useful in industrial and chemical processes. Transition metals typically have high melting points and are capable of forming complex ions with other elements.
Compounds containing transition metals include chromium(III) oxide (\(\mathrm{Cr}_2\mathrm{O}_3\)) and titanium(IV) chloride (\(\mathrm{TiCl}_4\)), each displaying different metal + oxidation states. These differences in oxidation states allow for diverse chemical equations to occur.
Balancing chemical equations involving transition metals requires careful attention to the number of each type of atom on both the reactant and product sides, ensuring the atoms are equally distributed across the equation.

Understanding Redox Reactions

Redox reactions stand for reduction-oxidation reactions, and they involve the transfer of electrons between different species. Essentially, one element gets oxidized (loses electrons) while another gets reduced (gains electrons).
In these exercises, the transition metals are typically involved in redox processes. For example, in the reaction between chromium(III) oxide and aluminum, chromium gets reduced from an oxidation state of +3 to 0, while aluminum is oxidized from 0 to +3.

• Oxidation: Increase in oxidation state (loss of electrons)
• Reduction: Decrease in oxidation state (gain of electrons)

Recognizing and correctly balancing these changes in electron states is crucial for understanding and writing balanced redox reactions.

The Role of Stoichiometry

Stoichiometry is a term used in chemistry to describe the calculation of reactants and products within chemical reactions. It is based on the principle that mass is conserved during a chemical reaction, so the mass of reactants must equal the mass of products.
To balance a chemical equation accurately, the stoichiometric coefficients—numbers placed before compounds in a chemical equation—must be adjusted so that the number of each type of atom is the same on both sides.
For example, in reaction (b), with the equation \[ \operatorname{Ti} \mathrm{Cl}_{4}(\ell) + 2\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s}) + 2\mathrm{MgCl}_{2}(\mathrm{s}) \], the coefficient 2 before \(\mathrm{Mg}\) and \(\mathrm{MgCl}_2\) ensures the conservation of chlorine atoms and allows the magnesium to properly reduce the titanium.

Metallurgical Processes and Their Chemical Foundations

Metallurgical processes involve the extraction and refining of metals from ores. These processes often rely heavily on principles we've discussed, such as redox chemistry and stoichiometry, to effectively obtain pure metals.
A classic method involved is the thermite reaction, which is exemplified in reaction (a) and (d) from the exercise. These reactions involve the reduction of metal oxides with aluminium, a process that releases a significant amount of heat, thereby melting the metal and allowing it to be extracted.

• Aluminium acts as a reducing agent, helping achieve the necessary reduction in metallurgical extraction.
• Control over the chemical equations ensures the desired metal is obtained in pure form.

Understanding these chemical foundations allows us to efficiently manipulate and conduct metallurgical processes for industrial applications.