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Chapter 22: Problem 5

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) \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\)

Short Answer

Expert verified
(a) 1 Cr鈧侽鈧� + 2 Al 鈫� 1 Al鈧侽鈧� + 2 Cr, (b) 1 TiCl鈧� + 2 Mg 鈫� 1 Ti + 2 MgCl鈧�, (c) 2 [Ag(CN)鈧俔鈦� + 1 Zn 鈫� 2 Ag + 1 [Zn(CN)鈧刔虏鈦�, (d) 1 Mn鈧僌鈧� + 4 Al 鈫� 3 Mn + 2 Al鈧侽鈧�.

Step by step solution

01

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.
02

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.
03

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.
04

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.
05

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.
06

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).
07

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.
08

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.
09

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.
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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攏umbers placed before compounds in a chemical equation鈥攎ust 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.

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Most popular questions from this chapter

For a tetrahedral complex of a metal in the first transition series, which of the following statements concerning energies of the \(3 d\) orbitals is correct? (a) The five \(d\) orbitals have the same energy. (b) The \(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\) orbitals are higher in energy than the \(d_{x z}, d_{y z},\) and \(d_{x y}\) orbitals. (c) The \(d_{x z}, d_{y z},\) and \(d_{x y}\) orbitals are higher in energy than the \(d_{x^{2} y^{2}}\) and \(d_{z^{2}}\) orbitals. (d) The \(d\) orbitals all have different energies.

Three geometric isomers are possible for \(\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{3+} .\) One of the three is chiral; that is, it has a nonsuperimposable mirror image. Draw the structures of the three isomers. Which one is chiral?

Experiments show that \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) is paramagnetic and has two unpaired electrons. The related complex \(\mathrm{K}_{4}\left[\mathrm{Cr}(\mathrm{SCN})_{6}\right]\) is paramagnetic and has four unpaired electrons. Account for the magnetism of each compound using the ligand field model. Predict where the SCN - ion occurs in the spectrochemical series relative to CN'.

A The complex ion \(\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-},\) an octahedral complex with bidentate carbonate ions as ligands, has one absorption in the visible region of the spectrum at \(640 \mathrm{nm} .\) From this information, (a) Predict the color of this complex, and explain your reasoning. (b) Is the carbonate ion a weak- or strong-field ligand? (c) Predict whether \(\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]^{3-}\) will be paramagnetic or diamagnetic.

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} &\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\\\ &K_{f}=10^{8} \end{aligned}$$ $$\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) \quad \quad K_{\mathrm{f}}=10^{18}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{r} H^{\circ}\) for the NH seaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn},\) and \(\Delta_{r} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\) -rxn. Is the difference in \(\Delta_{r} H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.
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